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Physics 2 Electromagnetism Circuit Theory
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Circuit Theory -EM II
Dr David OttawayBraggs 412
Ph 8313 5165david.ottaway@adelaide.edu.au
Circuit Theory Outline DC Circuits Resistors and Voltage
Sources AC Circuits- Resistors, Capacitors and
Inductors Frequency Response Transfer functions Time Domain Response
Active Circuits
DC Circuits and Networks Ohms Law and Resistors Power and Resistors Ideal voltage and current sources Kirchoffs Laws (Apply to AC circuits as well) Resistor and Voltage Source Networks Thevenins Equivalent Circuits Nortons Equivalent Circuits Voltage Dividers (Apply to AC circuits as well) Parallel and Series resistors
Resistors and Ohms Law
( ) ( )resistorV t RI t=
Symbol
Ohm's Law:
V is the voltageR is the Resistance (SI Unit Ohm ())I is the current
2 possibilities
Power Dissipation in an electrical component
2
2
( ) ( ) ( )( )( ) /
Power t I t V tI t RV t R
=
=
=
Apply Ohms Law
Can be derived from the fundamental definition of Voltage ie Voltage = Energy/Charge
Ideal Voltage Sources
An ideal voltage source provides a fixed voltage regardless of the resistance placed across it
A useful model that does not exist in reality
A large battery is an excellent approximation
Battery Ideal Voltage Source
The sum of the changes in potential around any closed path of a circuit must be zero
From Giancolli
Kirchoffs Loop Rule
Based on the law of conservation of energy
Example: Resistor Combinations -Series
Req = R1 + R2 + R3
Kirchhoff s Junction RuleAt any junction point, the sum of all currents
entering the junction must equal the sum of all current leaving the junctionThis is based on the law of conservation of charge
B
At Junction A
I = I1 + I2 Also at B
Note Some currents will need to be negative
Resistor Combinations - Parallel
eq 1 2 3
1 1 1 1R R R R
= + +
Resistor Network Analysis
R1
R3
R2R3
Vs1
Vs2
Use Kirchoffs Laws to derive 5 Equations in 5 unknowns
Thevenins Equivalent Circuits Any terminal networks of voltage
sources and resistors is equivalent to a single resistor in series with a single voltage source
Figure from Wikipedia
Calculating Vth and Rth The value of Vth is value of the voltage when no
current is drawn from the circuit The value of Rth is determined by:
Where Ishort is the current drawn from the terminals when the source is short circuited
Rth is also the resistance between the terminals when all voltage sources are replaced by zero resistance
thth
short
VRI
=
Thevenins Equivalent CircuitExample: Wheatstone Bridge
R1VR2
R3 R4
Vout
Nortons Theorem
Any network of batteries and resistors is equivalent to a perfect current source in parallel with a resistor
INO is the short circuit current, Ishort (See Thevenins) and RNO = RTh (See Thevenins)
Figure from Wikipedia
AC Circuits Capacitors and Inductors required Will consider continuous periodic signals and
transient response Periodic Signals
Best way to deal with is complex circuit analysis Consider the transfer functions of
RC Circuits RL Circuits LCR Circuits
Circuit Components - Inductor
( ) ( )inductor dI tV t L dt=
Symbol
V is the voltageL is the Inductance (SI Unit Henry (H))I is the current (SI Unit Ampere (A))
Circuit Components - Capacitor
( ) ( ) ( )0t
capacitor
I t dtQ tV t
C C= =
Symbol
V is the voltageC is the capacitance (SI Unit Farad (F))I is the current
Q is the charge stored on the plates
Current vs Voltage (Capacitor)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-1
-0.5
0
0.5
1Current and Voltage for a 1uF Cap
time(s)
V
o
l
t
a
g
e
(
V
o
l
t
s
)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-1
-0.5
0
0.5
1x 10-4
C
u
r
r
e
n
t
(
A
m
p
s
)
Voltage lags the currentVoltage
Current vs Voltage (Inductor)
Voltage leads the current
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-1
-0.5
0
0.5
1Current and Voltage for a 1uH Inductor
time(s)
V
o
l
t
a
g
e
(
V
o
l
t
s
)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-1
-0.5
0
0.5
1x 104
C
u
r
r
e
n
t
(
A
m
p
s
)
Voltage
Complex Analysis
( )2
( )2
1 1( )2 2
1 1( )2 2
1 1
i t i t
i t i t
i ti t i tddt
i ti t i t
Cos t e e
Sin t e e
e i e e
e dt e ei
pi
pi
+
= +
=
= =
= =
Electronic Systems as Linear Systems
if = costhen = cos +
Vin(t) TF() Vout(t)
Voltage Amplitude Voltage Phase
Phase and amplitude are real numbers
Amplitude and Phase
0 2 4 6 8 10 12 14 16 18 20-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time (s)
V
o
l
t
a
g
e
(
v
)
VinVout
What is the transfer function, amplitude and relative phase for the above ???
Amplitude and Phase
0 2 4 6 8 10 12 14 16 18 20-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time (s)
V
o
l
t
a
g
e
(
v
)
VinVout
AinAout
T T
12
2.22 2 2.36
out
in
ATFA
Trad
T pi pi
=
= = =
Linear SystemsRemember: cos =
+
So: = cos =
+
and = cos +
= 2
+
= 2
+
=
2 +
2
Linear Systems is a complex number
=
=
, = cos + !"( )
Summary of Complex Impedance
1CZ i C
=
Inductor
Capacitor
Resistor
LZ i L=
RZ R=
Transfer Function of RC Circuit
10-2 10-1 100 101 10210-2
10-1
100Transfer Function of RC circuit
Angular frequency w/w0
A
m
p
l
i
t
u
d
e
10-2 10-1 100 101 102-100
-80
-60
-40
-20
0
Angular frequency w/w0
P
h
a
s
e
(
d
e
g
r
e
e
s
)
Transfer Function of RL Circuit
10-2 10-1 100 101 10210-2
10-1
100Transfer Function of RL circuit
Angular frequency w/w0
A
m
p
l
i
t
u
d
e
10-2 10-1 100 101 1020
50
100
Angular frequency w/w0
P
h
a
s
e
(
d
e
g
r
e
e
s
)
Transfer Function of LCR Circuit
10-2 10-1 100 101 10210-3
10-2
10-1
100Transfer Function of LCR circuit
Angular frequency w/w0
A
m
p
l
i
t
u
d
e
10-2 10-1 100 101 102-100
-50
0
50
100
Angular frequency w/w0
P
h
a
s
e
(
d
e
g
r
e
e
s
)
B=0.1B = 1B=10
Transfer Function of LCR Notch
10-2 10-1 100 101 10210-4
10-2
100Transfer Function of LCR circuit
Angular frequency w/w0
A
m
p
l
i
t
u
d
e
10-2 10-1 100 101 102-100
-50
0
50
100
Angular frequency w/w0
P
h
a
s
e
(
d
e
g
r
e
e
s
)
B=0.1B = 1B=10
Summary of Circuits Considered
( )( )
out
in
VRV i L
=
Vin() Vout()
Vin() Vout()
Vin() Vout()
RC Circuit
RL Circuit
RLC Circuit
( )( )
11
out
in
VV i RC
=
+
( )( ) ( )21
out
in
LiV RLV LC i R
=
+
Transient Response Will consider two circuits with step function
inputs RC Filter RLC
The time domain response is trivial if Laplace Transforms are used Not taught anymore in the Science part of maths(Following slides summarize this, not examinable and unfortunately cannot be used in exam)
Transient Approach1. Use Kirchoffs Loop Law to develop initial
equation, use V& = ' (, V) =*+,()
+ and V- = .
,()
-/
0
2. Differential to covert to first or second order differential equation
3. Solve
Summary of Transient Solutions
= 0 1 2 exp 2 (67
Vin(t) Vout(t)RC Circuit
Vin(t)RLC CircuitVout(t)
= 08
= 08
= 0 1 2(*0
exp 2( 2*7 9:; 0
Transient Response Using Laplace Transforms
(Not examinable and techniques cannot be used in exam/test)
Transient Response of R, L and C Circuits
Start by looking at the Differential Equations
Easier way is to use the S plane version of complex analysis
Utilize the power of Laplace Transforms Deriving equation in s space and transforming
them back to the time domain
Summary of S plane impedance
1CZ i C
=
RZ R=
LZ sL=Inductor
Capacitor
Resistor
LZ i L=
1CZ
sC=
RZ R=
S plane
Laplace Transformations
0
22
2
11
1
0
( ) { ( )} ( )
{ ( ) ( )} { ( )} { ( )}( ){ } { ( )} (0)
( ) (0){ } { ( )} (0)
( ) (0){ } { ( )} (0) .....
1{ ( ) } { ( )}
st
n nn n
n n
t
F s L f t e f t dt
L af t bg t aL f t bL g tdf tL sL f t f
dtdf t dfL s L f t sf
dt dtdf t dfL s L f t s f
dt dt
L f t dt L f ts
= =
+ = +
=
=
=
=
Definition of Laplace Transform
Linearity
Relationship for 1st derivative
Relationship for 2ndderivative
Relationship for nthderivative
Relationship for integral
Common Laplace Transforms Pairs
( )
( )
2 2
2 2
2 21
2 2
12 2
2 2
( ) { ( )} ( ) ( ) { ( )} ( )1 sin( )
1/ ( ) cos( )
1/ / ( 1)! sin( )1 ( tan( / ))1 1
sin( )( 1)! ( )( )1 c( ) ( )
n n
at
n at atn
at at
F s L f t f t F s L f t f tt t
s
ss u t t
s
ss t n t
s
e as a
t e e tn s as a
a s ae e
s s a s a
= =
+
+
+ + +
+
=
+
+ ++
+
+ + +
( )( ) ( )( )
os( )
1 1 1( ) ( )( ) ( )at bt at bt
t
se e ae be
s a s b a b s a s b b a
+ + + +
End of Transient Response Using Laplace Transforms
Operational Amplifiers Electronic Circuits made up
of transistors that electrically amplify signals
Op Amps are amongst the most utilized and versatile of all electronic components
Cost between a few cents and a few $100 dollars
Operational Amps Both Voltage and Power Gain
Transformers can increase the voltage of oscillating signals but cannot increase power Not DC and low frequency signals
Op Amps are active hence can amplify both voltage and power
Crucial for many applications such as stereo systems
CD playerLight signal->Voltage signal ->Amplified -> Sound Signal
Operational Amplifiers
V+ Non inverting inputV
-
Inverting inputVout OutputG Open loop gainVout = (V+- V-)G
V+
V-
Vout+
-
Ideal Op Amp has infinite open loop gain, infinite input impedance and zero output impedance
Operational Amplifiers -The RealityParameter Ideal Typical Values
Open Loop Gain (G) Infinite < 106-107
Input Impedance (Rin)
Infinite Typically 106-1012Ohm
Output Impedance (Rout)
Zero 40 2000 Ohms
V+
V-
Vout
+
-
+
Vs = (V+- V-)GVs
Rout
Rin
Two Golden Rules of Op AmpsRule 1 The output of an Op Amp provides
whatever voltage is necessary to make the two input voltages equal
Rule 2 The input draws no current
Apply when negative feedback is applied
Op Amp Circuit Inverting Amplifier
Vout+
-
Vin
Rb
Ra
Op Amp Circuit Non-Inverting Amplifier
Vout+
-
Vin
Rb
Ra
Op Amp Circuit Integrator
Vout+
-
VinR
C
Op Amp Circuit Differentiator
Vout+
-
Vin
R
C
Op Amp Circuit Summing Amplifier
Vout+
-
V1
Rb
RaV2
Ra
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