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Circuits and Analog ElectronicsCircuits and Analog Electronics
电路与模拟电子技术电路与模拟电子技术
Prof Li Chen School of Information Science and Technology Sun Yat-sen University
中山大学信息科学与技术学院 陈立副教授
Email chenli55mailsysueducn
10 10 级计算机科学 级计算机科学 22 ++ 22
References bull W H Hayt Jr J E Kemmerly and S M Durbin Engineering
Circuit Analysis McGraw-Hill 2005 ISBN 978-7-121-01667-7
bull R L Boylestad and L Nashelsky Electronic Devices and Circuit Theory Pearson Education 2007 ISBN 978-7-121-04396-3
bull 高玉良 电路与模拟电子技术 高教出版社 2004 ISBN 7-04-014536-7
Circuits and Analog ElectronicsCircuits and Analog Electronics
Handouts available at
sistsysueducn~chenli
References bull W H Hayt Jr J E Kemmerly and S M Durbin Engineering Circuit
Analysis McGraw-Hill 2005 ISBN 978-7-121-01667-7
bull R L Boylestad and L Nashelsky Electronic Devices and Circuit Theory Pearson Education 2007 ISBN 978-7-121-04396-3
bull 高玉良 电路与模拟电子技术 高教出版社 2004 ISBN 7-04-014536-7
Circuits and Analog ElectronicsCircuits and Analog Electronics
Weeks Chapters References
1 2 Basis concepts and laws of electronics Hayt Ch 1 2 5
3 4 Basis analysis methods to circuits Hayt Ch 3 4
5 Basis RL amp RC circuits Hayt Ch 6
6 7 8 Sinusoidal steady state analysis Hayt Ch 7
9 Midterm
10 Diodes and diodes circuits Boylestad Ch 1 2
11 12 13 Basic BJT amplifier circuits Boylestad Ch 3-6
14 15 16 Operational amplifier and Op Amp circuits Boylestad Ch 11
17 Review
Teaching ScheduleTeaching Schedule
Ch1 Ch1 Basic Concepts and Laws of Electric Basic Concepts and Laws of Electric CircuitsCircuits
11 Basic Concepts and Electric Circuits
12 Basic Quantities
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
References Hayt Ch1 2 5 Gao Ch1
Circuits and Analog ElectronicsCircuits and Analog Electronics
Signal processing and transmissiontransmission
Amplifiers
11 Basic Concepts and Electric Circuits
Electrical powerElectrical power conversion and transmissiontransmission
Power Supplies
TransmissionTransmission Loads
Circuits KinescopeAntenna
Speakertransmitter
11 Basic Concepts and Electric Circuits Electrical power conversion and transmission
11 Basic Concepts and Electric Circuits
Question What is the current through the bulb
Concept of Abstraction
Solution
In order to calculate the current we can replace the bulb with a resistor
R is the only subject of interest which serves as an abstraction of the bulb
11 Basic Concepts and Electric Circuits
Lumped circuit abstraction
bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)
bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc
bull A circuit consists of sources resistors capacitors inductors and conductors
bull Elements are lumped
bull Conductors are perfect
Resistance R = VI 1 =1VA ohm
Conductance G = 1R = 1AV siemens (S)
1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t
11 Basic Concepts and Electric Circuits
Understanding the AM radio requires knowledge of several concepts
bull Communicationssignal processing (frequency domain analysis)
bull Electromagnetics (antennas high-frequency circuits)
bull Power (batteries power supplies)
bull Solid state (miniaturization low-power electronics)
The AM Radio SystemThe AM Radio System
Transmitter Receiver
Example 1 The AM audio system
Example 2 The telephone system
11 Basic Concepts and Electric Circuits
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
References bull W H Hayt Jr J E Kemmerly and S M Durbin Engineering
Circuit Analysis McGraw-Hill 2005 ISBN 978-7-121-01667-7
bull R L Boylestad and L Nashelsky Electronic Devices and Circuit Theory Pearson Education 2007 ISBN 978-7-121-04396-3
bull 高玉良 电路与模拟电子技术 高教出版社 2004 ISBN 7-04-014536-7
Circuits and Analog ElectronicsCircuits and Analog Electronics
Handouts available at
sistsysueducn~chenli
References bull W H Hayt Jr J E Kemmerly and S M Durbin Engineering Circuit
Analysis McGraw-Hill 2005 ISBN 978-7-121-01667-7
bull R L Boylestad and L Nashelsky Electronic Devices and Circuit Theory Pearson Education 2007 ISBN 978-7-121-04396-3
bull 高玉良 电路与模拟电子技术 高教出版社 2004 ISBN 7-04-014536-7
Circuits and Analog ElectronicsCircuits and Analog Electronics
Weeks Chapters References
1 2 Basis concepts and laws of electronics Hayt Ch 1 2 5
3 4 Basis analysis methods to circuits Hayt Ch 3 4
5 Basis RL amp RC circuits Hayt Ch 6
6 7 8 Sinusoidal steady state analysis Hayt Ch 7
9 Midterm
10 Diodes and diodes circuits Boylestad Ch 1 2
11 12 13 Basic BJT amplifier circuits Boylestad Ch 3-6
14 15 16 Operational amplifier and Op Amp circuits Boylestad Ch 11
17 Review
Teaching ScheduleTeaching Schedule
Ch1 Ch1 Basic Concepts and Laws of Electric Basic Concepts and Laws of Electric CircuitsCircuits
11 Basic Concepts and Electric Circuits
12 Basic Quantities
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
References Hayt Ch1 2 5 Gao Ch1
Circuits and Analog ElectronicsCircuits and Analog Electronics
Signal processing and transmissiontransmission
Amplifiers
11 Basic Concepts and Electric Circuits
Electrical powerElectrical power conversion and transmissiontransmission
Power Supplies
TransmissionTransmission Loads
Circuits KinescopeAntenna
Speakertransmitter
11 Basic Concepts and Electric Circuits Electrical power conversion and transmission
11 Basic Concepts and Electric Circuits
Question What is the current through the bulb
Concept of Abstraction
Solution
In order to calculate the current we can replace the bulb with a resistor
R is the only subject of interest which serves as an abstraction of the bulb
11 Basic Concepts and Electric Circuits
Lumped circuit abstraction
bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)
bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc
bull A circuit consists of sources resistors capacitors inductors and conductors
bull Elements are lumped
bull Conductors are perfect
Resistance R = VI 1 =1VA ohm
Conductance G = 1R = 1AV siemens (S)
1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t
11 Basic Concepts and Electric Circuits
Understanding the AM radio requires knowledge of several concepts
bull Communicationssignal processing (frequency domain analysis)
bull Electromagnetics (antennas high-frequency circuits)
bull Power (batteries power supplies)
bull Solid state (miniaturization low-power electronics)
The AM Radio SystemThe AM Radio System
Transmitter Receiver
Example 1 The AM audio system
Example 2 The telephone system
11 Basic Concepts and Electric Circuits
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Handouts available at
sistsysueducn~chenli
References bull W H Hayt Jr J E Kemmerly and S M Durbin Engineering Circuit
Analysis McGraw-Hill 2005 ISBN 978-7-121-01667-7
bull R L Boylestad and L Nashelsky Electronic Devices and Circuit Theory Pearson Education 2007 ISBN 978-7-121-04396-3
bull 高玉良 电路与模拟电子技术 高教出版社 2004 ISBN 7-04-014536-7
Circuits and Analog ElectronicsCircuits and Analog Electronics
Weeks Chapters References
1 2 Basis concepts and laws of electronics Hayt Ch 1 2 5
3 4 Basis analysis methods to circuits Hayt Ch 3 4
5 Basis RL amp RC circuits Hayt Ch 6
6 7 8 Sinusoidal steady state analysis Hayt Ch 7
9 Midterm
10 Diodes and diodes circuits Boylestad Ch 1 2
11 12 13 Basic BJT amplifier circuits Boylestad Ch 3-6
14 15 16 Operational amplifier and Op Amp circuits Boylestad Ch 11
17 Review
Teaching ScheduleTeaching Schedule
Ch1 Ch1 Basic Concepts and Laws of Electric Basic Concepts and Laws of Electric CircuitsCircuits
11 Basic Concepts and Electric Circuits
12 Basic Quantities
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
References Hayt Ch1 2 5 Gao Ch1
Circuits and Analog ElectronicsCircuits and Analog Electronics
Signal processing and transmissiontransmission
Amplifiers
11 Basic Concepts and Electric Circuits
Electrical powerElectrical power conversion and transmissiontransmission
Power Supplies
TransmissionTransmission Loads
Circuits KinescopeAntenna
Speakertransmitter
11 Basic Concepts and Electric Circuits Electrical power conversion and transmission
11 Basic Concepts and Electric Circuits
Question What is the current through the bulb
Concept of Abstraction
Solution
In order to calculate the current we can replace the bulb with a resistor
R is the only subject of interest which serves as an abstraction of the bulb
11 Basic Concepts and Electric Circuits
Lumped circuit abstraction
bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)
bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc
bull A circuit consists of sources resistors capacitors inductors and conductors
bull Elements are lumped
bull Conductors are perfect
Resistance R = VI 1 =1VA ohm
Conductance G = 1R = 1AV siemens (S)
1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t
11 Basic Concepts and Electric Circuits
Understanding the AM radio requires knowledge of several concepts
bull Communicationssignal processing (frequency domain analysis)
bull Electromagnetics (antennas high-frequency circuits)
bull Power (batteries power supplies)
bull Solid state (miniaturization low-power electronics)
The AM Radio SystemThe AM Radio System
Transmitter Receiver
Example 1 The AM audio system
Example 2 The telephone system
11 Basic Concepts and Electric Circuits
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Weeks Chapters References
1 2 Basis concepts and laws of electronics Hayt Ch 1 2 5
3 4 Basis analysis methods to circuits Hayt Ch 3 4
5 Basis RL amp RC circuits Hayt Ch 6
6 7 8 Sinusoidal steady state analysis Hayt Ch 7
9 Midterm
10 Diodes and diodes circuits Boylestad Ch 1 2
11 12 13 Basic BJT amplifier circuits Boylestad Ch 3-6
14 15 16 Operational amplifier and Op Amp circuits Boylestad Ch 11
17 Review
Teaching ScheduleTeaching Schedule
Ch1 Ch1 Basic Concepts and Laws of Electric Basic Concepts and Laws of Electric CircuitsCircuits
11 Basic Concepts and Electric Circuits
12 Basic Quantities
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
References Hayt Ch1 2 5 Gao Ch1
Circuits and Analog ElectronicsCircuits and Analog Electronics
Signal processing and transmissiontransmission
Amplifiers
11 Basic Concepts and Electric Circuits
Electrical powerElectrical power conversion and transmissiontransmission
Power Supplies
TransmissionTransmission Loads
Circuits KinescopeAntenna
Speakertransmitter
11 Basic Concepts and Electric Circuits Electrical power conversion and transmission
11 Basic Concepts and Electric Circuits
Question What is the current through the bulb
Concept of Abstraction
Solution
In order to calculate the current we can replace the bulb with a resistor
R is the only subject of interest which serves as an abstraction of the bulb
11 Basic Concepts and Electric Circuits
Lumped circuit abstraction
bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)
bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc
bull A circuit consists of sources resistors capacitors inductors and conductors
bull Elements are lumped
bull Conductors are perfect
Resistance R = VI 1 =1VA ohm
Conductance G = 1R = 1AV siemens (S)
1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t
11 Basic Concepts and Electric Circuits
Understanding the AM radio requires knowledge of several concepts
bull Communicationssignal processing (frequency domain analysis)
bull Electromagnetics (antennas high-frequency circuits)
bull Power (batteries power supplies)
bull Solid state (miniaturization low-power electronics)
The AM Radio SystemThe AM Radio System
Transmitter Receiver
Example 1 The AM audio system
Example 2 The telephone system
11 Basic Concepts and Electric Circuits
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Ch1 Ch1 Basic Concepts and Laws of Electric Basic Concepts and Laws of Electric CircuitsCircuits
11 Basic Concepts and Electric Circuits
12 Basic Quantities
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
References Hayt Ch1 2 5 Gao Ch1
Circuits and Analog ElectronicsCircuits and Analog Electronics
Signal processing and transmissiontransmission
Amplifiers
11 Basic Concepts and Electric Circuits
Electrical powerElectrical power conversion and transmissiontransmission
Power Supplies
TransmissionTransmission Loads
Circuits KinescopeAntenna
Speakertransmitter
11 Basic Concepts and Electric Circuits Electrical power conversion and transmission
11 Basic Concepts and Electric Circuits
Question What is the current through the bulb
Concept of Abstraction
Solution
In order to calculate the current we can replace the bulb with a resistor
R is the only subject of interest which serves as an abstraction of the bulb
11 Basic Concepts and Electric Circuits
Lumped circuit abstraction
bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)
bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc
bull A circuit consists of sources resistors capacitors inductors and conductors
bull Elements are lumped
bull Conductors are perfect
Resistance R = VI 1 =1VA ohm
Conductance G = 1R = 1AV siemens (S)
1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t
11 Basic Concepts and Electric Circuits
Understanding the AM radio requires knowledge of several concepts
bull Communicationssignal processing (frequency domain analysis)
bull Electromagnetics (antennas high-frequency circuits)
bull Power (batteries power supplies)
bull Solid state (miniaturization low-power electronics)
The AM Radio SystemThe AM Radio System
Transmitter Receiver
Example 1 The AM audio system
Example 2 The telephone system
11 Basic Concepts and Electric Circuits
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Signal processing and transmissiontransmission
Amplifiers
11 Basic Concepts and Electric Circuits
Electrical powerElectrical power conversion and transmissiontransmission
Power Supplies
TransmissionTransmission Loads
Circuits KinescopeAntenna
Speakertransmitter
11 Basic Concepts and Electric Circuits Electrical power conversion and transmission
11 Basic Concepts and Electric Circuits
Question What is the current through the bulb
Concept of Abstraction
Solution
In order to calculate the current we can replace the bulb with a resistor
R is the only subject of interest which serves as an abstraction of the bulb
11 Basic Concepts and Electric Circuits
Lumped circuit abstraction
bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)
bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc
bull A circuit consists of sources resistors capacitors inductors and conductors
bull Elements are lumped
bull Conductors are perfect
Resistance R = VI 1 =1VA ohm
Conductance G = 1R = 1AV siemens (S)
1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t
11 Basic Concepts and Electric Circuits
Understanding the AM radio requires knowledge of several concepts
bull Communicationssignal processing (frequency domain analysis)
bull Electromagnetics (antennas high-frequency circuits)
bull Power (batteries power supplies)
bull Solid state (miniaturization low-power electronics)
The AM Radio SystemThe AM Radio System
Transmitter Receiver
Example 1 The AM audio system
Example 2 The telephone system
11 Basic Concepts and Electric Circuits
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits Electrical power conversion and transmission
11 Basic Concepts and Electric Circuits
Question What is the current through the bulb
Concept of Abstraction
Solution
In order to calculate the current we can replace the bulb with a resistor
R is the only subject of interest which serves as an abstraction of the bulb
11 Basic Concepts and Electric Circuits
Lumped circuit abstraction
bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)
bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc
bull A circuit consists of sources resistors capacitors inductors and conductors
bull Elements are lumped
bull Conductors are perfect
Resistance R = VI 1 =1VA ohm
Conductance G = 1R = 1AV siemens (S)
1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t
11 Basic Concepts and Electric Circuits
Understanding the AM radio requires knowledge of several concepts
bull Communicationssignal processing (frequency domain analysis)
bull Electromagnetics (antennas high-frequency circuits)
bull Power (batteries power supplies)
bull Solid state (miniaturization low-power electronics)
The AM Radio SystemThe AM Radio System
Transmitter Receiver
Example 1 The AM audio system
Example 2 The telephone system
11 Basic Concepts and Electric Circuits
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
Question What is the current through the bulb
Concept of Abstraction
Solution
In order to calculate the current we can replace the bulb with a resistor
R is the only subject of interest which serves as an abstraction of the bulb
11 Basic Concepts and Electric Circuits
Lumped circuit abstraction
bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)
bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc
bull A circuit consists of sources resistors capacitors inductors and conductors
bull Elements are lumped
bull Conductors are perfect
Resistance R = VI 1 =1VA ohm
Conductance G = 1R = 1AV siemens (S)
1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t
11 Basic Concepts and Electric Circuits
Understanding the AM radio requires knowledge of several concepts
bull Communicationssignal processing (frequency domain analysis)
bull Electromagnetics (antennas high-frequency circuits)
bull Power (batteries power supplies)
bull Solid state (miniaturization low-power electronics)
The AM Radio SystemThe AM Radio System
Transmitter Receiver
Example 1 The AM audio system
Example 2 The telephone system
11 Basic Concepts and Electric Circuits
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
Lumped circuit abstraction
bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)
bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc
bull A circuit consists of sources resistors capacitors inductors and conductors
bull Elements are lumped
bull Conductors are perfect
Resistance R = VI 1 =1VA ohm
Conductance G = 1R = 1AV siemens (S)
1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t
11 Basic Concepts and Electric Circuits
Understanding the AM radio requires knowledge of several concepts
bull Communicationssignal processing (frequency domain analysis)
bull Electromagnetics (antennas high-frequency circuits)
bull Power (batteries power supplies)
bull Solid state (miniaturization low-power electronics)
The AM Radio SystemThe AM Radio System
Transmitter Receiver
Example 1 The AM audio system
Example 2 The telephone system
11 Basic Concepts and Electric Circuits
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
Understanding the AM radio requires knowledge of several concepts
bull Communicationssignal processing (frequency domain analysis)
bull Electromagnetics (antennas high-frequency circuits)
bull Power (batteries power supplies)
bull Solid state (miniaturization low-power electronics)
The AM Radio SystemThe AM Radio System
Transmitter Receiver
Example 1 The AM audio system
Example 2 The telephone system
11 Basic Concepts and Electric Circuits
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Example 1 The AM audio system
Example 2 The telephone system
11 Basic Concepts and Electric Circuits
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time
Voltage or current in a circuit
Sound (sinusoidal wave traveling through air)
Light or radio waves (electromagnetic energy traveling through free space)
The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T
High Frequency Low Frequency
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz
FM radio and TV uses different frequencies
Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines
red green blue
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +
=
10
1 10
10
cossin
sincoscossin
)sin(
kkmkm
k kkkmkkm
kkkm
tkCtkBA
tkAtkAA
tkAA
0 0
0
0
1
2sin
2cos
T
T
km
T
km
A f t dtT
B f t k tdtT
B f t k tdtT
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
21
01)(
t
ttfExample Given function during a period
2 3 t
1
)12sin(12
14]5sin
5
13sin
3
1[sin
4)(
l
tll
ttttf
For the example 2 2
0 0 0
1 1 11 1 0
2 2 2A f t d t d t d t
2 2
0 0
00
1 1cos 1 cos 1 cos
2 2 cos sin 0
kmC f t k td t k td t k td t
k td t k tk
2 2
0 0
00 40
1 1sin 1 sin 1 sin
2 2 2 sin cos 1 cos
km
k
B f t k td t k td t k td t
k td t k t kk k
k is even
k is odd
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Example-Fourier SeriesExample-Fourier Series
基波
3次谐波
基波+3 次谐波
bull Signals can be represented in terms of their frequency components
bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals
1st series + 3rd series
1st series (k = 1)
3rd series (k = 3)
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)
freq5kHz
Frequency domain representation of input
Frequency domain representation of output
freq610kHz
ModulatorModulator
Signal
SourceModulator
Power
Amplifier
Antenna
Transmitter Block DiagramTransmitter Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Input Signal
Output Signal
Modulator Time DomainModulator Time Domain
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station
bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude
Power AmplifierPower Amplifier
The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space
AntennaAntenna
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
Audio
Amplifier
Antenna
Speaker
Receiver Block DiagramReceiver Block Diagram
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull The antenna captures electromagnetic energy and converts it to a small voltage or current
bull In the frequency domain the antenna output is
0 frequency
Undesired SignalsDesired Signal
Carrier Frequencyof desired station
AntennaAntenna
interferences interferences
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits
bull RF Amplifier also performs as a Bandpass filter for the signal
ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station
RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier
0 frequency
Undesired Signals
Desired Signal
Carrier Frequency of desired station
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
The AM Radio SystemThe AM Radio System
0 frequency
Undesired Signals
Desired Signal
455 kHz
IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals
IF AmplifierIF Amplifier
0 frequency
Desired Signal
455 kHz
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
bull Computes the envelope of its input signal
Envelope DetectorEnvelope Detector
Output Signal
Input Signal
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier
bull Amplifies signal from envelope detector
bull Provides power to drive the speaker
Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction
bull Higher levels of the model describe overall function of the system
bull Lower levels of the model describe necessary details to implement the system
bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker
bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs
SystemInputs Outputs
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model
AM ReceiverInput Signal Sound
Second Level ModelSecond Level Model
RFAmplifier
IFMixer
IFAmplifier
EnvelopeDetector
AudioAmplifier
Antenna
Speaker
Power Supply
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
11 Basic Concepts and Electric Circuits
The AM Radio SystemThe AM Radio System
Half-waveRectifier
Low-passFilter
Low Level Model Envelope DetectorLow Level Model Envelope Detector
Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector
+
-R C
+
-VoutVin
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
12 Basic Quantities
UnitsUnitsbull Standard SI Prefixes
ndash 10-12 pico (p)
ndash 10-9 nano (n)
ndash 10-6 micro ()
ndash 10-3 milli (m)
ndash 103 kilo (k)
ndash 106 mega (M)
ndash 109 giga (G)
ndash 1012 tera (T)
bull Electric charge (q)
ndash in Coulombs (C)
bull Current (I)
ndash in Amperes (A)
bull Voltage (V)
ndash in Volts (V)
bull Energy (W)
ndash in Joules (J)
bull Power (P)
ndash in Watts (W)
I t q
VI
R
IR V
W qV Pt V I t
P VI
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
CurrentCurrent
bull Time rate of change of charge t
qI Constant current tIq
dttdqti )()( Time varying current
t
dxxitq )()(
Unit mAA 3101 AmA 3101 (1 A = 1 Cs)
12 Basic Quantities
bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)
i(t) i(t)
t t
DCACTime ndash varying current Steady current
bull A mount of electric charges flowing through the surface per unit time
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
CurrentCurrent
Positive versus negative currentPositive versus negative current
2 A -2 A
P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2
Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1
mAmA
12 Basic Quantities
Current is always associated with arrows (directions)
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Negative charge of -2Cs moving
Positive charge of 2Cs moving or
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Voltage(Potential)Voltage(Potential)
baab VVV
b
a
b
aab ldE
q
ldF
q
WV
VoltageVoltage Units 1 V = 1 JC
Positive versus negative voltagePositive versus negative voltage
+
ndash
ndash
+
2 V -2 V
12 Basic Quantities
bull Energy per unit chargebull It is an electrical force drives an electric current
+- of voltage (V) tell the actual polarity of a certain point DN
Two ldquoDo Not (DN)rdquo
+- of current (I) tell the actual direction of particlersquos movement DN
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Voltage (Potential)Voltage (Potential)
a
b
VVab 5 a b which pointrsquos potential is higher
b
a
V6aV V4bV Vab =
a b +Q from point b to point a get energy Point a is
Positive or Positive or negativenegative
12 Basic Quantities
Example
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Voltage (Potential)Voltage (Potential)
ab
cacute
c d
dacute
2211
21
221121222
2
21112
1111
111
1b1bb
0
)(
)(
0
rRrR
EEI
rRrRIEEIrEVIrVV
EVV
RrRIEIRVV
rRIEIrVV
IREVEV
IRVIRVVVV
V
dda
dd
cd
cc
bc
aab
a
12 Basic Quantities
Example
I
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Voltage (Potential)Voltage (Potential)
K Open
K Close
Va=)V(521
)V(18
a
a
V
V
12 Basic Quantities
Example
I
I
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
I
11 2
a
Ev E R
R R
12 Basic Quantities
ExampleExample
I
1 21 1
1 2a
E Ev E R
R R
1 2 3 1 2 3 2 1 3 3 1 2
1 2 3 1 2 3 2 3 1 2 1 3
a a a aa
v E v E v E v E R R R E R R R E R R Rv
R R R R R R R R R R R R R R R R
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
PowerPower
bull One joules of energy is expanded per second
bull Rate of change of energy
P = Wt )()()()()( titVdt
dqtVdttdwtp abab
bull Used to determine the electrical power is being absorbed or supplied
ndash if P is positive (+) power is absorbed
ndash if P is negative (ndash) power is supplied
+
ndash
v(t)
i(t)p(t) = v(t) i(t)
v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering
12 Basic Quantities
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
PowerPower
Example
12 Basic Quantities
2A+
ndash
-5V 5 2 10WP Power is supplied delivered power to external element
+
ndash
5V
2A
5 2 10WP Power is absorbed Power delivered to
Note +
ndash
+5V
+
ndash
-5V
2A
-2A
Power absorbed
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
PowerPower
bull Power absorbed by a resistor
)()()( titvtp )(2 tiR
Rtv )(2)(2 tvG
Gti )(2
12 Basic Quantities
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
PowerPower
1
2
3 4
5
I1 I2 I3+
-
-
-
-
-
+
+
+
+-
+
+
-
+-
P15 Find the power absorbed by each element in the circuit
12 Basic Quantities
A21 I A12 IA13 I
V35 V
V41 V
V82 V V43 V
V74 V
3
16
7
4
8
535
212
734
323
111
WVIP
WVIP
WVIP
WVIP
WVIP
Supply energy element 1 3 4 Absorb energy element 2 5
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Open CircuitOpen Circuit R=
I=0 V=E P=0E
R0
Short CircuitShort Circuit R=0
E
R0
R = 0 0R
EI 00 IREV
02RIPE
12 Basic Quantities
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
RR
EI
o
0IREIRV
02RIEIVI
Loaded CircuitLoaded Circuit
E
R0 R
I
0PPP E
12 Basic Quantities
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
13 Circuit ElementsCircuit Elements
Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
bull Passive elements (cannot generate energy)
ndash eg resistors capacitors inductors etc
bull Active elements (capable of generating energy)
ndash batteries generators etc
bull Important active elements
ndash Independent voltage source
ndash Independent current source
ndash Dependent voltage source
bull voltage dependent and current dependent
ndash Dependent current source
bull voltage dependent and current dependent
13 Circuit ElementsCircuit Elements
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
ResistorsResistors
Dissipation ElementsElements
S
lR v=iR P=vi=Ri2=v2R gt0
v-i relationship
v
i
13 Circuit ElementsCircuit Elements
Resistors connected in series
ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip
R1 R2 R3
Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip
R1 R2 R3
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Capacitors
bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)
bull Charge on the two conductors creates an electric field that stores energy
bull The voltage difference between the two conductors is proportional to the charge q = C v
bull The proportionality constant C is called capacitance
bull Units of Farads (F) - CV
bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device
1F=106F 1F=106PF
13 Circuit ElementsCircuit Elements
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Capacitors
store energy in an electric field
v-i relationship
dt
dqti =)(
dt
dvC
t
dxxiC
tv )(1
)(
i(t)+
-
v(t)
Therestofthe
circuit
dt
dvcvivp 2
2
1cvcvdvpdtwEnergy stored
13 Circuit ElementsCircuit Elements
Capacitors connected in seriesndash Equivalent capacitance is found
by 1Ceq=1C1 + 1C2 + 1C3 + hellip
series
parallel
Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip
vC(t+) = vC(t-)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
v(t)5V
1s 2s(1)
00
0
1
0
2
1
1
0
1
0
1
0 0 0
11 1 0 5 1 0 5
021
2 1 5 5 2 1 5 002
0 1s
11 0 5 1 5
021s 2s
11 5 10 5 2 0
02
t
tv t i t dt v t
Ct v
v dt
v dt
t
v t dt t v
t
v t dt t v
For (1)
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Capacitors
t
i(t)1A
-1A 1s
2s
i(t)
+
-
v(t)02F
P17
13 Circuit ElementsCircuit Elements
t
w (t)
25J
1s 2s(2)
0 0
0
2 20
20
1
2
1 If 0
2Now 0 0 1 5 2 0
1 01 25 25
2 01 0 0
t t
t t
t
t
dvw t Pdt C v dt
dt
C vdv C v t v t
v t w t C v t
v v v
w
w
For (2)
For (1) (2)
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
dt
tdiLtv
)()(
t
dxxvL
ti )(1
)(
Inductors
store energy in a magnetic field that is created by electric passing through it
v-i relationship i(t) +
-
v(t)L
Inductors connected in series Leq= L1 + L2 + L3 + hellip
Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip
13 Circuit ElementsCircuit Elements
dt
diLiivP )(
2
1)( 2 tLitwL Energy stored
022
000 2)( titi
LidiLdt
dt
diiLPdttw
ti
tv
t
t
t
t
iL(t+) = iL(t-)
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Independent voltage source
+VS
RS = 0
v
i
VS
Ideal
sS
sS
IRVV
IRV
practical
13 Circuit ElementsCircuit Elements
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Independent current source
I
v
iIS
RS infin=
Ideal
SS
SS
RVII
RVI
practical
13 Circuit ElementsCircuit Elements
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
n
kSkS VV
1
Voltage source connected in series
n
kSkS RR
1
Voltage source connected in parallel
n
kSkS II
1
SnSSS
SnSSS
RRRR
RRRR
1111
21
21
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Voltage controlled (dependent) voltage source (VCVS)
+_
_
+
Sv Svv
Current controlled (dependent) voltage source (CCVS)
+_ Sriv Si
Q What are the units for and r
13 Circuit ElementsCircuit Elements
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Voltage controlled (dependent) current source (VCCS)
Current controlled (dependent) current source (CCCS)
_
+
SvSgvi
Si Sii
Q What are the units for and g
13 Circuit ElementsCircuit Elements
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Independent source
dependent source
Can provide power to the circuit
Excitation to circuit
Output is not controlled by external
Can provide power to the circuit No excitation to circuit
Output is controlled by external
13 Circuit ElementsCircuit Elements
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
bull So far we have talked about two kinds of circuit elements
ndash Sources (independent and dependent)
bull active can provide power to the circuit
ndash Resistors
bull passive can only dissipate power
Review
The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements
13 Circuit ElementsCircuit Elements
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
14 Kirchhoffs Current and Voltage Laws
Key Words Nodes Branches Loops KCL KVL
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
Node point where two or more elements are joined (eg big node 1)
Loop A closed path that never goes twice over a node (eg the blue line)
Branch Component connected between two nodes (eg component R4)
The red path is NOT a loop
Mesh A loop that does not contain any other loops in it
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
Nodes Branches Loops mesh
bull A circuit containing three nodes and five branches
bull Node 1 is redrawn to look like two nodes it is still one nodes
P18
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
KCL MathematicallyKCL Mathematicallyi1(t)
i2(t) i4(t)
i5(t)
i3(t)
n
jj ti
1
0)(
n
jjI
1
0
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
bull sum of all currents entering a node is zero
bull sum of currents entering node is equal to sum of currents leaving node
KCL
P19
DCBA iiii
14 Kirchhoffs Current and Voltage Laws
In
Out
0A B C O
I
I
i i i i
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
KCL
+
-120V
50 1W Bulbs
Is
P110
bull Find currents through each light bulb
IB = 1W120V = 83mA
bull Apply KCL to the top node
IS - 50IB = 0
bull Solve for IS IS = 50 IB = 417mA
KCL-Christmas LightsKCL-Christmas Lights
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
KCL
P111 We can make supernodes by aggregting node
0
0
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii3 amp 2 Adding
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
KCL
Current dividerCurrent divider
N VG1
G2
I+
-
I1I2
IGG
GG
G
IVGI
21
1111
IGG
GVGI
21
222
I
G
GI
n
kk
kk
1
121
21
111
11
RRR
RRI
RRI
R
VI
I
RR
RI
21
12
14 Kirchhoffs Current and Voltage Laws
In case of parallel 1 21 2
1 1 1 V=
I IG G G
R R R R G
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
sum of voltages around any loop in a circuit is zero
KVL
bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative
KVL Mathematically 0)(1
n
jj tv 0
1
n
jjV
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
KVL is a conservation of energy principle
KVL
A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage
AV
BBV)( AB VVqW
q
abV
a bq
abqVW LOSES
cdV
c dq
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
If the charge comes back to the same Initial point the net energy gain Must be zero
0)( CABCAB VVVq
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
KVL
P113 Determine the voltages Vae and Vec
14 Kirchhoffs Current and Voltage Laws
10 24 0aeV
16 12 4 6 0aeV
4 + 6 + Vec = 0
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
KVL
Voltage dividerVoltage divider
R1
R2
-
V1
+
+
-
V2
+
-
V
21
111 RR
RVIRV
21
222 RR
RVIRV
Important voltage Divider equations
NV
R
RV n
kk
kk
1
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws
KVLVoltage dividerVoltage divider
kR 151
Volume control
P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ
14 Kirchhoffs Current and Voltage Laws