Electric current Electric current ‐ 2 · Electric current • Electric current = movement of charges dq(t) Electric current tells us how much dt i(t) = ect ccu e tte s charges per

Lecture 1Lecture 1

• Charge

• Current, voltage, Power,Current, voltage, Power,

• KCL, KVL

• Voltage and current sources

• ResistorsResistors

• Book: Chapter 1– Sections: 1.2/1.3/1.4/1.5/1.6/1.7

(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS

Electric ChargesElectric Charges

• It all starts with electric charges – Symbol: qy q

– Unit: Coulomb [C]

E g an electron 1 602e 19 C– E.g. an electron = ‐1.602e‐19 C


Electric currentElectric current

• Electric current = movement of charges

tdq )( Electric current tells us how much

dttdqti )()( =

ect c cu e t te s us o uccharges per unit time flow through an area

– Symbol: i

– Unit: Ampere [A]p [ ]


Electric current 2Electric current ‐ 2

• Current = movement of charges reference direction!

baab ii −= baab


Electric current: DC & ACElectric current: DC & AC

DC AC

DC+AC


VoltageVoltage

• Voltage = energy per unit charge, needed to move charge from one node to the other nodeg– Symbol = v

Units = Volt [V] = Joule/Coulomb– Units = Volt [V] = Joule/Coulomb

• Reference polarity


Voltage 2Voltage ‐ 2

• Circuit element can generate (e.g. battery) or dissipate (e.g. resistor) energyp ( g ) gy

Element dissipates energy

Element generates energy

Note: assume vab and iab are both positive values !

energy energy


Voltage 3Voltage ‐ 3

• Voltage “of a node”– Voltage is always across a circuit elementg y

ReferenceReference node


PowerPower

• Power =– Units: Watt [W]

ivp ⋅=[ ]

– Symbol: p

Di i t d G t d?• Dissipated or Generated?– With this convention positive power = dissipated, negative power = generated


EnergyEnergy

∫=2t

dttpE )(∫1t

Is the energy delivered or generated byIs the energy delivered or generated by a circuit element between t1 and t2

Units = Joule [J]


Kirchoff’s Current Law (KCL)Kirchoff s Current Law (KCL)

• The net current entering a node = 0– Net current entering a node arriving currents g gare positive, leaving currents are negative

0iii 321 =+−


Kirchoff’s Current Law (KCL)Kirchoff s Current Law (KCL)

• Alternative formulation:– sum of entering currents = sum of leaving currentsg g

∑∑ ∑∑ = leavingenteringii

231 iii =+


Kirchoff’s Voltage Law (KVL)Kirchoff s Voltage Law (KVL)

• The algebraic sum of the voltages in a loop = 0

0vvvv 4321 =−+− 4321



• The algebraic sum of the voltages in a loop = 0

0vvv + 0vvv 542 =+−



• Alternative formulaton:– Sum of positive arrows = sum of negative arrowsp g

452 vvv =+


Voltage sourceVoltage source

• Ideal, independent voltage source– Ideal: output voltage is independent of current p g pthrough the source

– Independent: output voltage does not depend onIndependent: output voltage does not depend on other voltages or currents in the network

Voltage is known, but current depends on load network

Direction of Is chosen such that source delivers power, but this is p ,not necessarily so.


Voltage sourceVoltage source

• Ideal, independent voltage source– Regardless of Is, Vs maintains its valueg ,


Current sourceCurrent source

• Ideal, independent current source– Ideal: output current is independent of voltage p p gacross the source

– Independent: output current does not depend onIndependent: output current does not depend on other voltages or currents in the network

Current is known, but voltage depends on load network

Direction of Vs chosen such that source delivers power, but this is p ,not necessarily so.


Current sourceCurrent source

• Ideal, independent current source– Regardless of Vs, Is maintains its valueg ,


Dependent sourcesDependent sources

• Voltage‐Controlled Voltage Source (VCVS)– = ideal voltage source whose output voltage g p gdepends on another voltage

A l i12V34 vAv ⋅= AV = voltage gain


VCVS CCVS VCCS CCCSVCVS, CCVS, VCCS, CCCSVCVS VCCS

12V34 vAv ⋅= 12M34 vGi ⋅=

CCVS

AV = voltage gain [‐] GM = trans‐conductance [A/V]

CCVS CCCS

12M34 iRv ⋅= 12i34 iAi ⋅=RM = trans‐resistance [V/A] Ai = current gain [‐](c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS

Resistors and Ohm’s LawResistors and Ohm s Law

• Resistor– Linear relationship between current through the p gresistor and the voltage across the resistor.

– Resistance of a resistor ; units: Ohm [Ω]Resistance of a resistor ; units: Ohm [Ω]

– Always dissipates energy

Ohm’s Law:

RR iRv ⋅=


Resistance and ConductanceResistance and Conductance

RR vGi ⋅=

R1G = [S] SiemensRG [ ]


Resistor graphical representationResistor – graphical representation

RR iRv ⋅= RR


Resistor and physical parametersResistor and physical parameters

ρ [ ]Ω• Resistivity of a material:

l

ρ [ ]mΩ

Αl

⋅ρ=R Copper: 1.72e‐8 ΩmGold: 2.27e‐8 ΩmΑ

ASilver: 1.63e‐8 ΩmTeflon: 1e19 ΩmC b 3 5 8 ΩA Carbon: 3.5e‐8 Ωm

Exercise: what is theExercise: what is the resistance of a copper wire of 1 meter length and

lof 1 meter length and diameter of 1mm?


Resistors and powerResistors and power

iivp ⋅=

For resistors, this becomes

iRp 2⋅=

viRp

2

Rp =

Which is always positive if R is positive!(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS

HomeworkHomework

• P1.63

• P1.67P1.67

• P1.79


Lecture 2Lecture 2

• Resistors in series and parallel

• Voltage and current dividerVoltage and current divider

• Thévenin and Norton

• Superposition,

• Book chapter 2Book chapter 2– 2.1/2.2/2.3/2.4/2.5/2.6/2.7

Resistors in SeriesResistors in Series

N21t iiii ==== LKCL:

NN2211

N21t

iRiRiRvvvv+++=

+++=L

LKVL:

( )N21t RRRi +++= L

∑=N

it R

iv ∑

=1iti

Resistors in SeriesResistors in Series

∑N

t RvR ∑=

==1i

it

teq R

iR

Resistors in ParallelResistors in Parallel

N21t vvvv ==== LKVL:

+++

+++=

N21

N21t

vvviiii L

= Nt

11

ivKVL:

⎟⎞

⎜⎛

+++=N

N

2

2

1

1

111

RRRL

∑=

N

1i i

t

R1i

⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

N21t R

1R1

R1v L


1v

∑== N

t

teq 1

1ivR

∑=1i iR


NN

∑∑==

===N

1ii

N

1i ieqeq G

R1

R1GParallel circuits easier to

work with conductances!1i1i ieq

Two Resistors in ParallelTwo Resistors in Parallel

21eq RR

RRR+

=Parallel circuits smallest one dominates!

21 RR +

Resistive voltage dividerResistive voltage divider

inin RR

vi+

=R21 RR +

21

2inout RR

Rvv+

=

2it Riv = 2inout Riv

Resistive current dividerResistive current divider

21RRi21

21inin RR

iv+

=1

inoutRii =

21inout RR +

inout R

vi =2R

Resistive current dividerResistive current divider

iniv =21

in GGv

+=

2inout GG

Gii =

2inout Gvi =21

inout GG +

Node voltage analysisNode‐voltage analysis

• Self‐study

Mesh current analysisMesh‐current analysis

• Self‐study

Thévenin equivalent circuitThévenin equivalent circuitAny two‐port network consisting of …can be replaced by this resistors, current and voltage sources… Thévenin equivalent circuit

RThevR

Thevv

Equivalent Circuit (1)Equivalent Circuit (1)

Thevv

openThev vv =⇒

Equivalent Circuit (2)Equivalent Circuit (2)

ThevR

Thevv

vshort

Thev

Thev iRv

=⇒Thev

openThevThev

vvR ==⇒shortshort

Thev iiR⇒

Thévenin equivalent circuitThévenin equivalent circuit

Norton equivalent circuitNorton equivalent circuitAny two‐port network consisting of …can be replaced by this resistors, current and voltage sources… Norton equivalent circuit

NortiNortR

Nort

Norton equivalent circuitNorton equivalent circuitAny two‐port network consisting of …can be replaced by this resistors, current and voltage sources… Norton equivalent circuit

ExampleExample

R v1

21

2open v

RRRv+

=1

1short R

vi =

ExampleExample

21

21

RRRR+1

1

Rv

21

21

RRRR+

21

21 RR

Rv+ 21 RR +

Another way to determine RAnother way to determine RThev

• Zero all independent sources– Zero voltage source = short circuitg

– Zero current source = open circuit

C l l t th i t b t th t• Calculate the resistance between the two terminals– Connect a voltage source Vtest, calculate Itest, impedance between two nodes is Vtest/Itesttest test

Another way to determine RAnother way to determine RThev

Rinin

21Thevin RRRR //==

Make v1 zero volt

Example 2Example‐2 Maximum power transferMaximum power transfer

ThevRLThev

ThevL RR

vi+

=

Thv LR 2

LThev

iRp =Thevv LR

( )2L

2Thev

LLL

RRRv

iRp

=

=

( )2LThev RR +

Maximum for ThevL RR =

SuperpositionSuperposition

• For linear networks,– The response (voltage and currents) of the p ( g )network for all sources is the sum of the individual responses. p

– Individual response = response of the network for one source only when other sources are zero‐edone source only when other sources are zero ed.

• Zero voltage source = short circuit

• Zero current source = open circuit• Zero current source = open circuit

• This is the superposition principle

Superposition: exampleSuperposition: example

Superposition: exampleSuperposition: example

21

111OUT RR

RVv+

=,21 RR +

21

11

21

211OUT RR

RVRR

RRIv+

++

=

21

2112OUT RR

RRIv+

=,

HomeworkHomework

• Resistive networks:– P2.8

– P2.11

P2 40– P2.40

– P2.87

Lecture 3Lecture 3

• Capacitance

• InductanceInductance

• Mutual Inductance

• Book chapter 3– 3.1/3.2/3.3/3.4/3.5/3.6/3.73.1/3.2/3.3/3.4/3.5/3.6/3.7


Capacitors and capacitanceCapacitors and capacitance

q+

εq−q−

C CvCq ⋅=ε =permittivity of the dielectric

C = capacitance of the capacitorUnits: Coulomb/Volt = Farad [F]Typical values: fF (10‐15) mF (10‐3)Typical values: fF (10 5)…mF (10 3)


Capacitors: physicalCapacitors: physicalA

d

A

ε d

AεParallel‐plate capacitor:

dAC ε

=

0r ε⋅ε=ε Relative permittivity values

120 10858 −⋅=ε . F/m

Air: 1Mica: 7Silicon dioxide: 3.9ε = relative permittivity [‐]Water: 78.5rε = relative permittivity [‐]


Capacitors: exampleCapacitors: example

• Two metal plates of 10cm by 10cm

• Plastic foil in between (Epsr=3) with aPlastic foil in between (Epsr 3) with a thickness of 1mm

Wh i h i ?• What is the capacitance?


Capacitors: currentCapacitors: current

q+Ci+ q−

Ci+ C

Ci+Ci+

q− q−q


Capacitors: voltage and currentCapacitors: voltage and current

q+ Ci+q+ Ci+

dtdqi = dvdt

CvCq ⋅= dtdvCi C

C =


Capacitors: exampleCapacitors: exampleCv

1V

dvCi C

t1us 3us 4us 5us

dtCi C

C =

Draw iC for C = 100nF


Capacitors: power and energyCapacitors: power and energy

ddtdvvCivtp ⋅⋅=⋅=)(

∫=t

dttptE )()( Energy delivered to the ∫=t0

dttptE )()( capacitor from t0 till t

)()( tvC21dvvCdt

dtdvvCtE 2

vt

⋅⋅=⋅⋅=⋅⋅⋅= ∫∫ 2dt 0t0

∫∫


Capacitors in seriesCapacitors in series

N21t

N21t

dtdv

dtdv

dtdv

dtdv

vvvv

+++=

+++=

K

K

N21t

Ci

Ci

Ci

Ci

dtdtdtdt

+++= KN21eq

C1

C1

C1

C1

CCCC

+++= KN21eq CCCC

11

∑=

+++= Neq 1

1111

1C∑=

+++1i iN21 CCCC

K


Capacitors in parallelCapacitors in parallel

N21eq CCCC +++= Kq


Capacitor: remarksCapacitor: remarks

dvCi C

dtCi C

C =

dv=⇒= C

C 0i0dt

dvfor DC, a capacitor is an open circuit

↑↑⇒ CC i

dtdv for high frequencies, a capacitor is a short circuitdt


Inductors and inductanceInductors and inductance

• Current and magnetic flux:– Changing current creates a changing magnetic g g g g gfield

– Changing magnetic field induces a voltageChanging magnetic field induces a voltage

– Self‐inductance effect induced voltage across a coil due to its own changing currentcoil due to its own changing current

di L = (self‐)inductance

dtdiLv L

L =L = (self )inductanceUnits: Henry [H]Typical values: pH…mH


Inductors: practicalInductors: practicalVarious approximate equations depending

N22

pp q p gon the shape:

⎥⎦⎤

⎢⎣⎡ +

πμ=

lD450201l

NrL22

0

.Example: Wheeler long-coil equation (l>D)

⎥⎦⎢⎣ l

μ = permeabilityr = radius (m)N = number of turns ( )N = number of turns (-)D= diameter (m)l = coil length (m)


Inductors in seriesInductors in series

∑=N

LL ∑=

=1i

ieq LL

Assuming zero magnetic coupling ( l l )(mutual coupling) between the inductors


Inductors in parallelInductors in parallel

1

∑= Neq 1

1L Assuming zero magnetic coupling ( l l )∑

=1i iL(mutual coupling) between the inductors


Inductor: remarksInductor: remarks

diL L

dtLv L

L =

di=⇒= L

L 0v0dtdi

for DC, an inductor is a short circuit

↑↑⇒ LL v

dtdi

for high frequencies, an inductor is an open circuitdt


Mutual inductanceMutual inductance

P t f th ti flPart of the magnetic flux of L1 flows through L2

Part of the magnetic flux f L fl th h Lof L2 flows through L1

didi = mutual inductanceMdtdiM

dtdiLv 2

121

11 += Units: Henry [H]Typical values: pH…mH

M


Mutual inductance: dot conventionMutual inductance: dot convention

dididtdiM

dtdiLv 2

121

11 +=

dtdiL

dtdiMv 2

21

122 +=dtdt

dtdiM

dtdiLv 2

121

11 −=

diLdiMv

dtdt2

21

122 +−=dtdt 2122


HomeworkHomework

• Capacitance and inductance:– P3.13

– P3.53


Lecture 4Lecture 4

• Transients– First order RC circuits, ,

– RL circuits,

Second order circuits– Second order circuits

• Book chapter 4– 4.1/4.2/4.3

Discharging a charged capacitor with a current source

SC

C IdvCti −==)(SI SC Idt

Cti )(S

iC V0tv == )( CviV

S

i

ICVt =dttitv

t

CC = ∫ )()( SI

tIV S

0CC ∫ )()(

ttCIV S

i ⋅−=

RC circuits: discharging a capacitor through a resistor

?)( =tvOUT

vdv0ii RC =+

V0t )(0

Rv

dtdvC CC =+

iC V0tv == )(

0vdvRC C =+ tC Ketv α=)(0v

dtRC C =+ C Ketv =)(

RC circuits: discharging a capacitor through a resistor (2)

0vdt

dvRC CC =+

dtt

C Ketv α=)(

V0t )(

C Ketv )(

iC V0tv == )(

KKetv0tVv0t

0iC

==→=

=→=α+

+

)( iVK =KKetv0t C ==→= )(


dv 0vdt

dvRC CC =+

tC Ketv α=)(

V0t )(

C )(

iC V0tv == )(

0KeeRCK tt =+α αα

RC1

−=α


τ−− ==t

RCt

eVeVtv )( == iiC eVeVtv )(

V)(tvC

iV

tiV370 ⋅.

RC=τt


τ−==tiC eVtvti )()( ==R e

RRti )(

RV /)(tiR

RVi /

tRV370 i /. ⋅

RC=τt

Charging a capacitor with a current source

ddt

dvCtiI CCS == )(

tS dtIt ∫)(

0

SC

I

dtC

tv = ∫)()(tvC

iS Vt

CI

+⋅=

0Vi =

t

RC circuits: charging a capacitor through a resistor

?)( =tvOUT

RC

vVdvii =

00t )(

CSC

RvV

dtdvC −

=

00tvOUT == )(SC

C Vvdt

dvRC =+

tt21C eKKtv α+=)(

RC circuits: charging a capacitor through a resistor (2)

SCC Vv

dtdvRC =+

t21C eKKtv α+=)(

dt

St

21t

2 VeKKeKRC =++α ααS212

1α VKdRC

−=α S1 VK =and


021C eKK0tv α+== )( 21C eKK0tv +)(

0KK 21 =+

S2 VK −=


⎟⎞⎜⎛= τ−t

e1Vtv )( ⎟⎠⎞⎜

⎝⎛ −= τ

SC e1Vtv )(

SV)(tvout

SV

SV630 ⋅.

tRC=τ

t


τ−

=−

=tScS

R eRV

RtvVi )(

R RR

)(tiR

RVi /

RV370 i /. ⋅

RC=τt

RL circuitsRL circuits

diSL

L VRidtdiL =+

t21L eKKti α+=)(

tt VRKRKKL ααS

t21

t2 VeRKRKeKL =++α αα

LR

−=αRVK S

1 = RVK S

2 −=L R R

RL circuitsRL circuits

⎞⎛ −tV⎟⎠⎞⎜

⎝⎛ −= τ

tSL e1

RVti )(

τ−

=t

SOUT eVv

L )(tvOUT

R=τ

SV

SV370 ⋅.

RL /=τt

RLC circuitsRLC circuits

• Second‐order differential equations– (Damped) oscillations( p )

HomeworkHomework

• T4.1

Lecture 5Lecture 5

• Steady‐state Sinusoidal Analysis– Sinusoidal currents and voltagesg

– Phasors

Complex impedance– Complex impedance

– Circuit analysis with phasors and complex i dimpedances.

• Book chapter 5p– 5.1/5.2/5.3/5.4/5.6

Sinusoidal voltages and currentsSinusoidal voltages and currents

)cos()cos()( ϕ+π⋅=ϕ+ω⋅= ft2AtAtv)(tv

A2T π=ω

t

A f2π=ω

= period [s]= frequency [Hz]f

T

f1T =

frequency [Hz]= angular frequency [rad/s]= amplitude [v]

ωf

A

Power and RMSPower and RMS

v 2

Rvivp =⋅=

2T ⎤⎡

dtvT1

v11

T

0

2

T 2T ⎥⎥⎦

⎤

⎢⎢⎣

⎡⋅∫

∫∫ Rdt

Rv

T1dtp

T1P

0

00avg

⎥⎦⎢⎣=⋅=⋅= ∫∫

T1 V 2

∫ ⋅=0

2RMS dtv

T1V

RVP

2RMS

avg =

RMS for sinusoidal signalsRMS for sinusoidal signals

A11 TT

( )2

AdttAT1dtv

T1V

T

0

2T

0

2RMS =⋅ϕ+ω=⋅ ∫∫= )cos(

00

A2

R2AP

2

avg =

Complex numbersComplex numbers

ϕ⋅=+= jeAjyxSIm (imaginary part)g y p

y ϕ⋅= Ax )cos(ϕ⋅=ϕ

Ay )sin()(

Re (real part)

x

ϕ

⎟⎞

⎜⎛

+=

yyxA 22

x ⎟⎠⎞

⎜⎝⎛=ϕ

xyarctan

PhasorsPhasors

jyxeAVtAtv j11 +=⋅=⇔ϕ+ω⋅= ϕ)cos()(

is a complex number that represents the sinusoidal voltage

1V)(tv1represents the sinusoidal voltage

with amplitude A and phase ϕ)(tv1

Summation of sinusoidal voltages = vectorial sum of phasorsSummation of sinusoidal voltages = vectorial sum of phasors

Phasors represent sinusoidsPhasors represent sinusoids)(ts

t

y

x

PhasorsPhasors

)cos()( tAtv ϕ+ω j jyxeAV 1 +ϕ

)cos()()cos()(

222

111

tAtvtAtv

ϕ+ω⋅=ϕ+ω⋅=

22j

22

11j

11

jyxeAVjyxeAV

2

1

+=⋅=

+=⋅=ϕ

ϕ

Summation of sinusoidal voltages = vectorial sum of phasors

1V ( ) ( )212121 yyjxxVV +++=+

21 VV +

2V

Complex impedances: CComplex impedances: C

dtdvCi C

C =

)cos()( tAtvC ϕ+ω⋅=)/cos()sin()( 2tCAtCAtiC π−ϕ+ω⋅ω⋅−=ϕ+ω⋅ω⋅−=

( ) ϕπ−ϕπ−ϕ

ϕ

⋅ω⋅⋅=⋅⋅ω⋅−=⋅ω⋅−=

⋅=j2jj2j

C

jC

eCAjeeCAeCAIeAV

//C j

C Z1VLi (b t l ) l ti hiC

C

C ZCjI

=ω

= Linear (but complex) relationship between voltage and current (phasors)

Complex impedances: LComplex impedances: L

dtdiLv L

L =

)cos()( tAtiL ϕ+ω⋅=)/cos()sin()( 2tLAtLAtvL π−ϕ+ω⋅ω⋅−=ϕ+ω⋅ω⋅−=

( ) ϕπ−ϕπ−ϕ

ϕ

⋅ω⋅⋅=⋅⋅ω⋅−=⋅ω⋅−=

⋅=j2jj2j

L

jL

eLAjeeLAeLAVeAI

//L j

L ZLjVLi (b t l ) l ti hiL

L

L ZLjI

=ω= Linear (but complex) relationship between voltage and current (phasors)

Complex impedances: RComplex impedances: R

Rvi R

R =

)cos()( ϕ+ω⋅= tAtvR

)cos(/)( ϕ+ω⋅⋅−= tR1AtiR

ϕ

ϕ⋅=j

jR

R1AIeAV

/ ϕ⋅⋅= jR eR1AI /

RV Li d l l ti hi b tR

R

R ZRIV

== Linear and real relationship between voltage and current (phasors)

Circuit analysis with phasors and complex impedances

• KCL, KVL, series and parallel, Thévenin and Norton…are also valid for complex pimpedances!

Cj

Cj1ZC

−== Cj

Z1YC ω==

CCj ωω ZC

LjZL ω= Lj1YL ω

=jL Ljω


What is the amplitude and phase of the output voltage for C1=10nF, C2=20nF, R1=10kΩ and vIN a sinewave with frequency 1kHz and amplitude 1V?

dvdvC1vdCR INOUT2OUT2

⎟⎞

⎜⎛

dtdtC1

dtCR INOUT

1

22OUT

21 =⎟⎟⎠

⎜⎜⎝

++


• Assume vIN is sinusoidal

• Series and parallel still valid!Series and parallel still valid!

111 Cj

1RZω

+=

22

1

Cj1Z

j

ω=

2j


2INOUT

ZVV =

1

21INOUT

1CV

ZZVV

+

12121

1IN

RCC

CCj1

1CC

CV

+ω+

⋅+

⋅=

21 CC +

( )ϕjAVV O l l b !( )ϕ⋅= jINOUT AeVV Only one complex number!


What is the amplitude and phase of the output voltage for C1=10nF, C2=20nF, R1=10kΩ and vIN a sinewave with frequency 1kHz and amplitude 1V?

( ) 39670jININOUT e30750V11880j28360VV .... −=−⋅= ( ) ININOUT j

)sin( te12Av 3IN π=

).sin(. 39670te1230750Av 3OUT −π=

O t t i i ith lit d f 308 V d i iti lOutput is a sinewave with amplitude of 308mV and initial phase of ‐0.397 rad = ‐22.73 degrees.

Thévenin and NortonThévenin and Norton HomeworkHomework

• P5.44

• P5.46P5.46

(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011

1

Lecture 6

• Frequency response and bode‐plots– Fourier

– Transfer functions

– Bode plot

• Book chapter 6– 6 1 / 6 2 / 6 3 / 6 4 / 6 5– 6.1 / 6.2 / 6.3 / 6.4 / 6.5

Fourier

• Fourier series: – Periodic signal = sum of sinusoids


2

Transfer functions

121

2121

1

21

2

IN

OUT

RCC

CCj1

1CC

CZZ

ZV

V

+ω+

⋅+

=+

=

Example: low‐pass filter

11Cj1

V 1

C

IN

OUT

ffj1

1RCj1

1

Cj1R

CjV

V

+=

ω+=

ω+

ω=RC21fC π

=with


3

low‐pass filter: transfer function

1RC21fC π

=

( )1jH )( =ω

( )

C

IN

OUT

ffj1

1jHV

V

+=ω=

( )2Cff1j

/)(

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−=ω∠

Cff0jH arctan)(

low‐pass filter: magnitude

( )2ff11jH

/)(

+=ω 7070

21jH .)( ==ω⇒= Cff( )Cff1 /+ 2

j )(C

Normalized frequency


4

low‐pass filter: phase

4jH /)( π−=ω∠⇒= Cff⎟⎟⎠

⎞⎜⎜⎝

⎛−=ω∠

ff0jH arctan)( j )(C⎟⎠

⎜⎝ Cf

j )(

Normalized frequency

Low‐pass filter: magnitude


5

Logarithms

( ) x10 x10 =log ( )

( ) x10

110 x10x

10 −=⎟⎠⎞

⎜⎝⎛=− loglog

( ) ( )ABAB ll ( ) ( )ABA 10B

10 loglog =

Bell and deciBell

⎟⎟⎞

⎜⎜⎛

= OUT10

PBell log

⎟⎟⎞

⎜⎜⎛

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎠

⎜⎝

OUT2

OUT

IN

OUT10

IN10

RV10

PP10dB

P

/log

log

g

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎜⎜⎝

=

IN

OUT10

IN2

IN10

VV20

RV10

log

/log

If ROUT = RIN


6

deciBell

( ))(log ω= jH20H 10dBdBH)( ωjH( ))(g j10dB

1100.11000.012

0dB+20dB‐20dB+40dB‐40dB

240.5

+6dB+12dB‐6dB+3dB‐3dB

41412 .=707021 ./ =

Low‐pass filter

( )( )( )2

CdB2ff110H

ff11jH /log

/)( +−=⇒

+=ω

( )Cff1 /+

dB3H7070HffdB0H1Hff

dBC

dBC

. −=⇒=⇒==⇒≈⇒<<

( )f20f20HffHff CdBCC log)log(/ −=⇒≈⇒>>


7

Bode plot ‐magnitude

Bode plot ‐magnitude


8

Bode plot – magnitude: asymptotic

( ) 1jHω

=ω dB3H7070HffdB0H1Hff

dBC

dBC

. −=⇒=⇒==⇒≈⇒<<

Cj1ωω

+ ( )f20f20HffHff CdBCC

dBC

log)log(/ −=⇒≈⇒>>

πω

=2

f CC

Bode plot ‐ phase


9

Bode plot ‐ phase

Bode plot – phase: asymptotic

4Hff0Hff

C

C

/π−=∠⇒=≈∠⇒<<

⎟⎟⎠

⎞⎜⎜⎝

⎛−=ω∠

Cff0jH arctan)(

2Hff C /π−≈∠⇒>>

πω

=2

f CC

f

⎠⎝ Cf

10fC / Cf10


10

Bode plot – asymptotic

( )Cj1

1jHωω+

=ω/

( ) Cj1jH ωω+=ω /( ) KjH =ω

)log(K20

dBH dBH

f f

)(∠ jH )( ω∠ jH

dBH

f

)( ω∠ jH

CfCf

“pole” “zero”

‐20dB/dec

+20dB/dec

f f

)( ω∠ jH )( ω∠ jH

f

)( ω∠ jH

2/π−

2/π+

Bode plot – asymptotic

( )Cj

1jHωω

=ω/

( ) CjjH ωω=ω /( ) KjH =ω

)log(K20

dBH dBH

f f

)(∠ jH )( ω∠ jH

dBH

f

)( ω∠ jH

CfCf

‐20dB/dec+20dB/dec

f f

)( ω∠ jH )( ω∠ jH

f

)( ω∠ jH

2/π−

2/π+


11

Using Bode‐plots

( ) ( )21

jHjH20jH20HjHjHjH

)()(log)(log)()()(

ω⋅ω=ω=

ω⋅ω=ω

( ) ( )( ) ( )

dB2dB1

210110

211010dB

HHjH20jH20

jHjH20jH20H

,,

)(log)(log)()(log)(log

+=

ω+ω=

ω⋅ω=ω=

jH )(

dB2dB12

110dB

2

1

HHjHjH20H

jHjHjH

,,)()(log

)()()(

−=⎟⎟⎠

⎞⎜⎜⎝

⎛

ωω

=

ωω

=ω

Using Bode plots: example( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )ω⋅ω⋅ω⋅ω=⋅⋅ωω+⋅=

ωω+⋅ωω+ωω+

=ω

jHjHjHjHj11

j11j1A

j1j1j1AjH

43211C

3C2C

1C

///

///

( ) ( ) ( ) ( ) ( ) ( ) ( )ωω+ωω+

jjjjj1j1

j 43213C2C

1C //

dBH

f

1Cf

2Cf 3Cf


12


( ) ( )

( ) ( ) ( ) ( ) ( )

ωω+⋅ωω+ωω+

=ω

11j1j1

j1AjH3C2C

1C

///

( ) ( ) ( ) ( ) ( ) ( ) ( )ω⋅ω⋅ω⋅ω=ωω+

⋅ωω+

⋅ωω+⋅= jHjHjHjHj11

j11j1A 4321

3C2C1C //

/

dBH

)/log()log( 1C2C ff20A20 +

f

1Cf

2Cf 3Cf


( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )ω⋅ω⋅ω⋅ω=⋅⋅ωω+⋅=

ωω+⋅ωω+ωω+

=ω

jHjHjHjHj11

j11j1A

j1j1j1AjH

43211C

3C2C

1C

///

///

( ) ( ) ( ) ( ) ( ) ( ) ( )ωω+ωω+

jjjjj1j1

j 43213C2C

1C //

)( ω∠ jH

1Cf 2Cf 3Cf

90+

f

90−


13


( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )ω⋅ω⋅ω⋅ω=⋅⋅ωω+⋅=

ωω+⋅ωω+ωω+

=ω

jHjHjHjHj11

j11j1A

j1j1j1AjH

43211C

3C2C

1C

///

///

( ) ( ) ( ) ( ) ( ) ( ) ( )ωω+ωω+

jjjjj1j1

j 43213C2C

1C //

)( ω∠ jH

1Cf 2Cf 3Cf

90+

f

90−

‐20dB/dec, ‐40dB/dec( ) ( ) ( )

( ) ( )

ωω+⋅ωω+=ω

11j1j1

1jH2C1C //

( ) ( ) ( ) ( )ω⋅ω=ωω+

⋅ωω+

= jHjHj11

j11

212C1C //

dBH

f1Cf 2Cf

‐20dB/dec = ‐6dB/oct‐40dB/dec = ‐12dB/oct


14

Using Bode plots (1)

( ) ( ) ( )ω⋅ω=ωω+

⋅=ω jHjHj11AjH 21

1C/

dBH

f

)( ω∠ jH

dBH

f

)( ω∠ jH

1Cf1Cf 2dB1dBdB HHH ,, +=1dBH ,

2dBH ,

f

)( ω∠ jH

f

)( ω∠ jH

2/π− 2/π−

)()()( ω∠+ω∠=ω∠ jHjHjH 21

Using Bode plots (2)( ) ( ) ( ) ( )ω⋅ω=

ωω+⋅ωω+=

ωω+ωω+

=ω jHjHj11j1

j1j1jH 21

2C1C

2C

1C

//

//

dBH

f

)( ω∠ jH

dBH

f

)( ω∠ jH

2Cf2Cf 2dB1dBdB HHH ,, +=

2dBH ,

1Cf1Cf

f

)( ω∠ jH

f

)( ω∠ jH

2/π− 2/π−

)()()( ω∠+ω∠=ω∠ jHjHjH 21


15

Using Bode plots (3)( ) ( ) ( ) ( ) ( ) ( ) ( )ω⋅ω=

ωω+⋅

ωω+=

ωω+⋅ωω+=ω jHjH

j11

j11

j1j11jH 21

2C1C2C1C ////

dBH

f

)( ω∠ jH

dBH

f

)( ω∠ jH

2Cf2Cf 2dB1dBdB HHH ,, +=1Cf 1Cf

f

)( ω∠ jH

f

)( ω∠ jH

2/π− 2/π−

)()()( ω∠+ω∠=ω∠ jHjHjH 21

π−

Homework

• P6.25

• P6.52

• P6.65

• T6.3


Part 7 1

Lecture 7

• Amplifiers: specifications and external h t i ticharacteristics

• Book chapter 11– 11.1, 11.2, 11.4, 11.5, 11.6, 11.7, 11.10

Ideal voltage amplifier

INVOUT vAv ⋅=

AV = voltage gain (can be negative)

OUTv

v

t

INv


Part 7 2

Ideal voltage amplifier

12V34 vAv ⋅=

AV = voltage gain

Realistic amplifier model

INVOUT vGv ⋅=

AV = (internal) voltage gain (can be negative)RIN = input resistance (should be high)ROUT = output resistance (should be low)


Part 7 3

Input and output loading effect

LOUT RAv LINOUT RRAvOUTL

LV

IN

OUT

RRA

v +⋅=

RIN = should be large compared to RS

ROUT = should be small compared to RL

OUTL

L

SIN

INV

S

OUT

RRRRA

v +⋅

+⋅=

Voltage gain

ROUTL

LV

IN

OUTV RR

RAv

vG+

⋅==


Part 7 4

Current gain

RiL

INV

IN

OUTI R

RGi

iG ⋅==

RIN = should be large compared to RS

ROUT = should be small compared to RL

Power gain

RP ( )L

IN2VIV

IN

OUTP R

RGGGP

PG ⋅=⋅==


Part 7 5

Cascading amplifiers

2V1V2OUTL

L2V

1OUT2IN

2IN1V

1IN

2OUTV GG

ZRRA

ZZZ

Av

vG ,,,

,,

, ⋅=+

⋅⋅+

⋅==2OUTL1OUT2IN1IN ZRZZv ,,,, ++

2dBV1dBVdBV GGG ,,,,, +=

Voltage‐amplifier model

AV = (internal) voltage gainRIN = input resistance (HIGH)ROUT = output resistance (LOW)


Part 7 6

Current‐amplifier model

Ai = (internal) current gainR = input resistance (LOW)RIN = input resistance (LOW)ROUT = output resistance (HIGH)

Transconductance‐amplifier model

Gm = transconductance [S]RIN = input resistance (HIGH)ROUT = output resistance (HIGH)


Part 7 7

Transresistance‐Amplifier model

Rm = transresistance [Ohm]RIN = input resistance (LOW)ROUT = output resistance (LOW)

High or low input impedance ?

Electrodes = high impedance source high input impedance neededhigh input impedance needed

Current measurements = low input impedance needed

OUTL

L

SIN

INV

S

OUT

RRR

RRRA

vv

+⋅

+⋅=


Part 7 8

High or low output impedance

Driving a loudspeaker = low output impedance needed ( l d )(voltage drive)

Driving a laser‐diode = high output impedance needed (current drive)

Specific input/output impedance

Transmissionline needs to be terminated by the d flcorrect resistor to avoid reflections


Part 7 9

Frequency responseComplex voltage gain = ratio of phasors = transfer function

IN

OUTV V

VjG =ω)(

dBVG ,

High‐frequency

f

low‐frequency or midband region

region

Example

Calculate the transfer function vOUT/vIN and draw the Bode plot LL

LL CRj1

RZω+

=

OUTLC

CRR1

⋅=ω

LOUT

LV

IN

OUTV ZR

ZAV

VjG+

⋅==ω)(

C0v

LOUTL

OUTLLOUT

LV j1

1AC

RRRRj1

1RR

RAωω+

⋅=

+ω+

⋅+

⋅=/

LOUTL

CRR +


Part 7 10

DC‐ or AC‐ coupleddBVG ,

High‐frequency region

DC‐coupled amplifier

f

low‐frequency or midband region

region

dBVG ,

Low

f

midband region

High‐frequency region

Low‐frequency region

AC‐coupled amplifier

Example

Calculate the transfer function vOUT/vIN and draw the Bode plot

ZV

( )OUTLC

CLV

LOUT

LV

IN

OUTV

RRCj1CRjA

ZRZA

VVjG

+ω+ω

⋅=

+⋅==ω)(


Part 7 11

Narrowband amplifier

Wideband amplifier

dBVG ,

High‐frequency Low‐

frequency

dBVG ,

f

midband region

regionfrequency region

f

Narrowband or Tuned amplifier

Amplifier limits: distortion

OUTvINVOUT vAv ⋅=

t


Part 7 12

Amplifier limits: distortion

OUTv

INVOUT vAv ⋅=Maximum output voltage is limited

INv

Amplifier limits: noise

OUTv

INVOUT vAv ⋅=

INv

Noise at the output


Part 7 13

Funny circuitsRi

RR

R

ININ i

vZ =

RvAvi INvIN

R−

=V

IN A11RZ−

=

What happens if AV=1, ‐1, ‐10, … ?

Homework

• P11.11

• P11.12

• P11.51


8/2 1

Lecture 8: OpAmps

• Operational Amplifiers– Amplifiers

– Filters

– Schmitt‐Trigger

• Book chapter 14– 14 1 14 2 14 3 14 4 14 5 14 9– 14.1, 14.2, 14.3, 14.4, 14.5, 14.9


8/2 2

Ideal OpAmp

vNon‐inverting input

( )nINpINOLOUT vvAv ,, −⋅=

OUTvpINv ,

nINv ,

Non‐inverting input

inverting inputoutput

AOL is very large (105…107)

Ideal OpAmp

OUTvpINv ,

( )nINpINOL vvA ,, −⋅OUTvnINv ,

AOL is very large (105…107)Infinitely high input impedanceZero output impedance

OUTv

( )nINpIN vv ,, −


8/2 3

Non‐Inverting amplifier

( )nINpINOLOUT vvAv ,, −⋅=OpAmp:

Resistors:

21

1OUTnIN RR

Rvv+

⋅=,

1OUTOLINOLOUT RR

RvAvAv+

−=AOL is very large (105…107)Infinitely high input impedance

21 RR +

1

2

1

21

1OL

21

21

IN

OUT

RR1

RRR

RA

RRRR

vv

+=+

≈+

++

=

Infinitely high input impedanceZero output impedance

Feed‐back

OLAOUTv

pININ vv ,= +

( )A

21

1

RRR+

nINv ,

‐

R( )

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=

−=

21

1OUTINOL

nINpINOLOUT

RRRvvA

vvAv ,,

1

2

IN

OUTCL R

R1v

vA +==


8/2 4

Open‐loop gain, closed‐loop gain

OLAOUTv

pININ vv ,= +G

in out+ ε

21

1

RRR+

OL

nINv ,

‐

H

‐

( )

GH1G

inout

outHinGout

+=

⋅−=εε⋅== open‐loop gain

= loop‐gain

= closed‐loop gainGH1

G+

G

GH

Feed‐back

Gin out+

Gout ε⋅=ε

H

‐ ( )

GH1G

inout

outHin

+=

⋅−=ε

H1

inout1GH ≈⇒>>


8/2 5

Feed‐back

Gin out+ ε

H

‐ ( )

⎟⎠⎞

⎜⎝⎛

+=

⋅−=ε

GH11in

outHin

01GH ≈ε⇒>>

Virtual short‐circuit

nINpIN

nINpIN

vv0vv

,,

,,

≈⇒

≈−=ε

Virtual short circuit


8/2 6

Voltage follower

0R2 =

Rv v1

2

IN

OUT

RR1

vv

+≈ 1v

vIN

OUT ≈

Ideal voltage amplifier with gain=1Infinitely high input impedanceZero output impedance

Inverting amplifier

Simplified analysis:2Ri

2

OUT

1

IN2R1R

nINpIN

Rv

Rvii

0vv−

=⇒=

=≈ ,,

1Ri

1

2

IN

OUT

RR

vv

−=


8/2 7

Virtual ground

0vvi 0vv nINpIN =≈ ,,

1Ri

2Ri

1IN1

IN1R RZ

Rvi ≈⇒≈

Inverting amplifier

Exact analysis:

RR

2Ri

nINOLOUT

21

1OUT

21

2INnIN

vAvRR

RvRR

Rvv

,

,

−=+

++

=

RR

1Ri

1

2

1OL

21

2

IN

OUT

RR

RA

RRR

vv

−≈+

+−

=


8/2 8

Inverting amplifier

21

1OUT

21

2INnIN RR

RvRR

Rvv , ++

+=

21

2

RRR+

in out+

+

ε

nINOLOUT vAv ,−=

OLA−

+

21

1

RRR+

Inverting amplifier

( )+

−−+

==ε21

1OUT

21

2INnIN RR

RvRR

Rvv ,

21

2

RRR+

in out−+ ε

ε==− OLnINOLOUT AvAv ,

OLAout

1−‐

21

1

RRR+


8/2 9

Summation amplifier

RAi1IN

1R Rvi =

Virtual ground:

2IN2R R

vi = OUTRA R

vi −=

1Ri

RAi1R 2R

Infinitely high input impedance:

RA2R1R iii =+

AR

vvv

2RiA

OUT

2

2IN

1

1IN

Rv

Rv

Rv

−=+

⎟⎟⎠

⎞⎜⎜⎝

⎛+−= 2IN

2

A1IN

1

AOUT v

RRv

RRv

IntegratorCi

vi INR = dt

dvCi OUTC −=

Ri

RiR dtC

INOUT

RC vRC1

dtdvii −=⇒=

0tOUT

t

0INOUT vdtv

RC1v =+−= ∫ ,


8/2 10

DifferentiatorRi

vi OUTR −= dt

dvCi INC =

Ci

RiR dtC

dtdvRCvii IN

OUTRC −=⇒=

Active filters

1

2

1

2

IN

OUT

ZZ

RR

vv

−⇒−=


8/2 11

Example: Low‐pass filter

R2=10kR1=1kC2 100 FC2=100nFDraw the Bode‐plot

OpAmp: gain, bandwidth and GBW

dBOLA ,0AjA )(

f

dB0A ,

C

0OL

j1jA

ωω

+=ω)(

C0A2GBW ω⋅⋅π=

BWfC =GBWfA C0 =⋅


8/2 12

Feed‐back

)( ωjAOL

OUTvINv +

21

1

RRRH+

=

H

‐

21

C

0OL

j1

AjA

ωω

+=ω)(

H1

GH1G

vv

IN

OUT ≈+

= only if GH is large enough!

Closed‐loop Transfer function

)( ωjAOL

OUTvINv +

‐ C

0OL

j1

AjAG

ωω

+=ω= )(

21

1

RRRH+

=H

C

0

1A

G /

( ) C00

0

C0

0

C

0

C

IN

OUT

HA1j11

HA1A

jHA1A

Hj1A1

j1GH1

Gv

v

ω+ω+⋅

+=

ωω++=

ωω++

ωω+=

+=

/

//

/


8/2 13

Trading gain for bandwidth

)( ωjAOL

OUTvINv +

‐ 1Av

H

( ) C00

0

IN

OUT

HA1j11

HA1A

vv

ω+ω+⋅

+=

/

1A0 ( )HA1f

Low‐frequency gain: bandwidth

( ) H1

HA1A

0

0 ≈+

( )HA1f 0C +

( ) C00C0

0 fAHA1fHA1

AGBW =+⋅+

=GBW is the same!

Trading gain for bandwidth

)( ωjAOL

OUTvINv +

‐ C

0OL

j1

AjA

ωω

+=ω)(

H ( ) C00

0

IN

OUT

HA1j11

HA1A

vv

ω+ω+⋅

+=

/

0Aopen‐loop gain

fBWfC =

GBWfA C0 =⋅

H1

HA1A

0

0 ≈+

( )HA1f 0C +

open loop gain

closed‐loop gain


8/2 14

Voltage limits

OUTvDDV

pINv ,

nINv ,( )nINpIN vv ,, −

DDV

SSV

Positive feedback

OUTvDDV

( )nINpIN vv ,, −

DD

0v1v

OUT

IN

==

SSV


8/2 15

Positive feedback

0v

DDOUTIN21

1DDpINDDOUT Vvv

RRRVvVv =⇒>+

=⇒= ,

SSOUTIN21

1SSpINSSOUT Vvv

RRRVvVv =⇒<+

=⇒= ,

0v IN =

Hysteresis / Schmitt‐Trigger

OUTv21

1DDpIN RR

RVv+

=,

INv

DDV21 RR +

SSV

21

1SSpIN RR

RVv+

=,


8/2 16

Example: temp controlINv

t

OUTv

t

Temp control with hysteresisINv

t

OUTv

t


8/2 17

Homework

• P14.32

• P14.33

• P14.78


1

Lecture 9

• Diodes (Book Chapter 10)– 10.1, 10.2, 10.3, 10.6, 10.7, 10.8

Semiconductor basics

‐‐

‐ ‐N‐type doped Silicium

free charges = electrons (negative)

‐‐

‐

‐

‐

‐‐

‐ ‐‐‐

‐‐

‐

‐

P‐type doped Siliciumfree charges = holes (positive)

+++

+ +

++

+ ++

++

+

+

++ ++


2

junction

‐‐

‐ ‐+++

+ +‐

‐

‐

‐ ‐

‐ ‐‐‐

‐‐

‐

++

+

+ +

+ +++

++

+

Electrons and holes will recombine

Diode basics

‐‐

‐ ‐+++

+ Electric ‐

‐

‐

‐ ‐

‐‐‐

‐+

+

+

+ +

+ ++

+

+

Depleted region (no free charges)Electric

l

field

potential


3

Diode forward bias

‐‐

‐ ‐+++

+ Electric V+ V‐‐

‐

‐

‐ ‐

‐‐‐

‐+

+

+

+ +

+ ++

+

+

Electric l

fieldV+ V‐

current

potential

Diode reverse bias

‐‐

‐ ‐+++

+ Electric V‐ V+‐

‐

‐

‐ ‐

‐‐‐

‐+

+

+

+ +

+ ++

+

+

Electric l

fieldV‐ V+

current

potential


4

Diode symbol

cathodeanode

‐‐‐

‐

‐

‐

‐‐

‐‐‐

‐

‐+++

+

+

+

++

+ ++

+

+

Electric

Electric field

Electric potential

Diode curve

Di

100uA … 1A

Forward bias regionReverse breakdown

Dv0.5V…0.8V

Reverse bias region


5

Simplified diode curve

Dv∝SI area

DiTnV

SD eIi ≈ mV25q

kTVT ≈=

n depends on doping profile (1…2)

Dv

≈Di very small (nA)k=1.38e‐23 (Boltzmann Constant)T = temperature in Kelvinq = charge of an electron (1.61e‐19 C)

Solving network with diodes

DRDC

vRivvV+=+=

DD vRi +=

vVi DDC −=

T

DnVv

SD eIi ≈

RiD =

Non‐linear equation!


6

Load line analysis: diodeDi

RVi DC

D =

Dv

DCD Vv =

Dv

0iVvR

Vi0v

DDCD

DCDD

=⇒=

=⇒=

RvVi DDC

D−

=

T

DnVv

SD eIi ≈

Iteration: convergenceDi

DvDv

⎟⎟⎠

⎞⎜⎜⎝

⎛≈

S

DTD I

inVv ln 2. calculate vD

Start: chose vDR

vVi DDCD

−= 1. calculate iD


7

Iteration: divergenceDi

DvDv

T

DnVv

SD eIi ≈ 1. calculate iD

Start: chose vD

RvVi DDC

D−

= 2. calculate vD

Example

R = 10k Result:VDC = 5VIS = 5*10e‐14 An = 1

Result: vD = 573mViD = 443uA


8

(very) simplified diode curve

Di

V60vA0i

D

D

.=>

V60vA0i

D

D

.<=

Dv0.6V

Zener diode

Di

DvZV

Typical VZ values: 3.6V, 4.5V, 6.8V, 9.1V, …


9

Load line analysis: Zener diodeDi

Dv

DR

DRDC

vRivvV−=−=

DCD Vv −=

DD

DR

vRivRi−−=

RvVi DDC

D−−

= RVi DC

D −=

Rectifier circuits

)(tvIN )(tvR)(IN

t

)(R

~0.6V

Half‐wave rectifier


10

Rectifier circuits

Positive input voltage Negative input voltage

Rectifier circuits

)(tvIN )(tvR)(IN

t

)(R

~1.2V

Full‐wave rectifier


11

Filtering

t

Filtering

( ) ( )τ−≈−= τ /)( / t1Ve1Vtv 0t

0OUT

tfC2I

RC2TVv L0

OUT ≈≈Δ


12

Wave‐shaping circuits

OUTv

INv‐0.6V

Wave‐shaping circuits

OUTv

VZ+0.6V

INv‐0.6V


13

Voltage doubler and shifter

• Page 518

Diode: small‐signals

iDi

DI

DvdDDDD vVvVv +=Δ+=

dDDDD iIiIi +=Δ+=DV

DC Small‐signal AC


14


iDi

dDDDD vVvVv +=Δ+=dDDDD iIiIi +=Δ+=

DI

di ⎞⎛Dv

DVD

D

DD v

dvdii Δ⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛=Δ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

D

D

d dvdi

r1

small‐signal equivalent circuit

⎟⎟⎠

⎞⎜⎜⎝

⎛= D

dvdi

r1

T

DnVv

SD eIi = DnVv

SD

nVIe

nVI

dvdi

T

D

==⎟⎠

⎜⎝ Dd dvr TTD nVnVdv

D

Td I

nVr =


15


DiDd

dDvvii Δ

===Δ

ITnVr

2DIdd

dD rr

Dv1DI

1DV

D

Td I

r =

2DV

Larger DC current lower small‐signal resistance larger AC small‐signal current

Example

R1=1kRBIAS 10kRBIAS=10k

Calculate the ratio vout/vin for VBIAS=1V and 5V


16

Example: DC calculation

Assume VD=0.6VVBIAS = 1V ID = 40uAVBIAS = 5V ID = 440uA

Example: small‐signal calculation

Assume VD=0.6VVBIAS = 1V ID = 40uA rd1=625Ω AV= 0.384 = ‐8dBVBIAS = 5V ID = 440uA rd2=57Ω AV = 0.054 = ‐25dB


17

Homework

• P10.9

• P10.28

• P10.72


1

Lecture 10

• MOSFET transistors– Operating principles

– Amplifiers

– Logic gates

• Handbook: Chapter 12– Sections: 12 1 12 2 12 3 12 4 12 5 12 6 12 7– Sections: 12.1, 12.2, 12.3, 12.4, 12.5, 12.6, 12.7

NMOS transistor

MOS = Metal-Oxide-Semiconductor4 terminals:GateDrainSourceBulk


2

NMOS Transistor (2)

Cross-sectional view

Top view

NMOS Channel

drain

gate

drainsourceGate-oxide SiO2

no voltage no current can flow from drain to source


3

NMOS ChannelGATE: V++

DRAIN: 0VSOURCE: 0V Gate-oxide SiO2

Electrons are attracted by the positive voltage at the gate conductive channel between drain and source!

NMOS CurrentGATE: V++

DRAIN: V++SOURCE: 0V

current

Gate-oxide SiO2

Electrons move from source to drain current flows from drain to source


4

NMOS symbol

NMOS: Triode


5

NMOS: Triode

DSi

MOS = voltage-controlled resistor

DSv↓GSv

↑GSv

NMOS: Saturation

DSi↑GSv

MOS = voltage-controlled current source

DSv

↓GSv


6

NMOS: Triode equation( )[ ]( )[ ]DSTGS

2DSDSTGSDS

vVv2KvvVv2Ki

⋅−≈−⋅−=

( )TGSon

DS

DS

VvK21r

iv

−=≈

DSi↑GSv

( )[ ]DSTGS vVv2K≈ ( )TGSDS

DSv

↓GSv

NMOS: Saturation equation

( )2TGSDS VvKi −=( )2DSDS vKi =

DSi↑GSv

( )TGSDS VvKi =( )DSDS

DSv

↓GSv


7

K, W, L, KP

⎟⎠⎞

⎜⎝⎛⋅=

LW

2KPK OXCKP μ= OX

OX tC ε

=⎟⎠

⎜⎝ L2 OX

OXt

Moore’s Law


8

Example‐1

R=1kΩ 100WVuA100KP 2/=

V=5V

V=0.8V

R=1kΩ

V50Vum1L

um100W

T .===

What is the current through the transistor?

Load‐line approach

VDDRLDSDD

DS

DS

vVi

i−

=

= NMOS equations

VGSDSi ↑GSv

L

DD

RV

DSi LDS R

i

DSv

↓GSv

DDV


9

Load‐line approachDSi ...=GSv

...=GSv

DSv...=GSv

Load‐line approach (2)

Input signal = sinewave with 1V amplitude = 2Vptp


10


of the 1kOhm resistor


Input: 2Vptp

Output: 12Vptp

Gain = -6 = 15.6dB


11

Small‐signal equivalent circuit

( )2TGSDS VvKi −= in saturation region

gsGSGS vVv += DC bias voltage + small signal

Small‐signal equivalent circuit

( )2TGSDS VvKi −= in saturation region

gsGSGS vVv += DC bias voltage + small signal voltage

( ) ( )2gsgsmDS

2gsgsTGS

2TGSDS

KvvgI

KvvVVK2VVKi

++≈

+−+−=

gsgsmDS g

dsDSDS iIi += DC bias current + small signal current

gsmds vgi = LINEAR relationship between small-signal gate-source voltage and small-signal drain-source current!


12

Small‐signals

dsDSDS iIi +=

gsGSGS vVv +=

DSi

GSv

t

gsGSGS vVv + GSv

t gsmds vgi =

Transconductance( ) ( )

2gsgsmDS

2gsgsTGS

2TGSDS

KvvgI

KvvVVK2VVKi

++≈

+−+−=

gsgsmDS g

( )TGSm VVK2g −=

( )2TGSDS VVKI −=

DSm KI2g = DSm

⎟⎠⎞

⎜⎝⎛⋅=

LW

2KPK

Example: IDS=1mA, W=100um, L=0.5um gm = ?

What is VGS-VT for that transistor?


13

Small‐signal equivalent model

In the saturation region and forregion and for

small-signals only

gsmds vgi =

( ) DSTGSm KI2VVK2g =−=

Amplifier calculation

DDV

v

dsDS iI +

LR

outOUTOUT vVv +=

inv

BIASVtwo sources superposition!


14

Amplifier calculation: DCDDV

R R

DDV

inv

LR LROUTV

BIASV

DC calculation set all small-signal sources to zero

BIASV

Amplifier calculation: ACDDV

R R

inv

LR LR

inv

outv

BIASV

AC calculation set all DC sources to zero


15

Amplifier calculation: AC

outvinv

Amplifier calculation: AC

outv

inLmLgsmout vRgRvgv −=−=

inv

Lmin

outv Rg

vvA −==


16

Common‐source amplifier



17

Common‐source amplifier: DC

Common‐source amplifier: AC


18

Common‐source amplifier: AC

( )ings

DLgsmout

vvRRvgv

=

−= //2G1G1

2G1G1in RRR

RRvv//

//+

=

ings

( )DLmin

outv RRg

vvA //−==


Dout RR =2G1Gin RRR //=

Dout

( )DLmv RRgA //−=


19

Common‐drain amplifier

Common‐drain amplifier: DC


20

Common‐drain amplifier: AC

Common‐drain amplifier: AC

( )SLgsmout RRvgv //=

outings vvv −=

2G1G1

2G1G1in RRR

RRvv//

//+

= ( )( ) 1

RRg1RRg

vv

SLm

SLm

in

out ≈+

=//

//


21

Common‐drain amplifier

2G1Gin RRR //=

m

mDout

g1

g1RR

≈

= //

( )( )SLm

SLmv RRg1

RRgA//

//+

=

Common‐drain output impedance

gsmS

testtest

gstest

vgR

vi

vv

−=

−=

mtest

testout g

1ivR ≈=


22

PMOS transistors

Logic gates


23

Homework

• P12.22

• P12.40

• P12.27

• P12.53


1

Lecture 11

• Bipolar transistors– Operating principles

– Amplifiers

• Handbook: Chapter 13

NPN structure and symbol


2

NPN: biasing

NPN: Currents

BC ii β=

⎟⎞

⎜⎛ v

100020...=β

⎟⎟⎠

⎞⎜⎜⎝

⎛

= T

BEVv

SB eIi

CCBE iiii ≈+=


3

NPN transistor curves


Linear region: BECE ii β=


4


Saturation region: 3020vv satCECE ....., ==

NPN common‐emitter amplifier


5

Voltage gain


6

distortion

PNP transistor


7

PNP transistor curves

biasing

( ) EB

BEBB R1R

VVI+β+

−=


8

Practical common‐source amplifier

Small‐signal circuit


9

Small‐signal circuit

t

B

BE

BE

VI

vi

r1

=∂∂

=π

Small‐signal analysis

Documents

Electric current Electric current ‐ 2 · Electric current • Electric current = movement of charges dq(t) Electric current tells us how much dt i(t) = ect ccu e tte s charges per