Chapter 10Chapter 10 Diodes Diodes1. Understand diode operation and select diodes for various applications.

2. Analyze nonlinear circuits using the graphical load-line technique.

3. Analyze and design simple voltage-regulator circuits.

4. Solve circuits using the ideal-diode model andpiecewise-linear models.

5. Understand various rectifier and wave-shaping circuits.

6. Understand small-signal equivalent circuits.

2

10. Diodes – Basic Diode Concepts10.1 Basic Diode Concepts10.1.1 Intrinsic Semiconductors* Energy Diagrams – Insulator, Semiconductor, and Conductor the energy diagram for the three types of solids

3

10. Diodes – Basic Diode Concepts10.1.1 Intrinsic Semiconductors* Intrinsic (pure) Si Semiconductor: Thermal Excitation, Electron-Hole Pair, Recombination, and Equilibrium

)cm 105~ is

density atom crystal Si:Note (K 300 at crystal Siintrinsic for

cm 101.5pn

density hole density electron: reached is

ionrecombinat and excitation between mequilibriu When

3-22

3-10ii

4

10. Diodes – Basic Diode Concepts10.1.1 Intrinsic Semiconductors*Apply a voltage across a piece of Si: electron current and hole current

5

10. Diodes – Basic Diode Concepts10.1.2 N- and P- Type Semiconductors* Doping: adding of impurities (i.e., dopants) to the intrinsic semi-con

ductor material.* N-type: adding Group V dopant (or donor) such as As, P, Sb,…

carrier cahage minor the holecarrier charge major the electron

call Wepp ,nNn

nconceratio donor the Nnmaterial type-n In

101.5pnpn

300K at SiForctor semicondua for constantpn

iid

d

2102i

2i

6

10. Diodes – Basic Diode Concepts

10.1.2 N- and P- Type Semiconductors* Doping: adding of impurities (i.e., dopants) to the intrinsic semi-con

ductor material.* P-type: adding Group III dopant (or acceptor) such as Al, B, Ga,…

carrier cahage minor the electroncarrier charge major the hole

call Wenn ,pNp

nconceratio acceptor the Npmaterial type-p In

101.5pnpn

300K at SiForctor semicondua for constantpn

iia

a

2102i

2i

7

10. Diodes – Basic Diode Concepts10.1.3 The PN-Junction* The interface in-between p-type and n-type material is called a pn-junction.

. V ,T as :300K at

Ge for 0.3V and Sifor 0.7V6.0V potential barrier The

B

B

8

10. Diodes – Basic Diode Concepts10.1.4 Biasing the PN-Junction* There is no movement of charge

through a pn-junction at equilibrium.

* The pn-junction form a diode which allows current in only one direction and prevent the current in the other direction as determined by the bias.

9

10. Diodes – Basic Diode Concepts10.1.4 Biasing the PN-Junction*Forward Bias: dc voltage positive terminal connected to the p regi

on and negative to the n region. It is the condition that permits current through the pn-junction of a diode.

10

10.1.4 Biasing the PN-Junction*Forward Bias: dc voltage positive terminal connected to the p regi

on and negative to the n region. It is the condition that permits current through the pn-junction of a diode.

11

10. Diodes – Basic Diode Concepts10.1.4 Biasing the PN-Junction*Forward Bias:

12

10. Diodes – Basic Diode Concepts*Reverse Bias: dc voltage negative terminal connected to the p

region and positive to the n region. Depletion region widens until its potential difference equals the bias voltage, majority-carrier current ceases.

13

10. Diodes – Basic Diode Concepts*Reverse Bias: majority-carrier current ceases.* However, there is still a very

small current produced by minority carriers.

14

10. Diodes – Basic Diode Concepts10.1.4 Biasing the PN-Junction* Reverse Breakdown: As reverse voltage reach certain

value, avalanche occurs and generates large current.

15

10. Diodes – Basic Diode Concepts10.1.5 The Diode Characteristic I-V Curve

16

10. Diodes – Basic Diode Concepts10.1.6 Shockley Equation* The Shockley equation is a theoretical

result under certain simplification:

0v whenapplicable not is equation ThisVn

vexpIi 0.1V,v when

C101.60q constant, sBoltzman' the is k

voltage thermal the is 300K at 0.026V qTkV

t,coefficien emission the is 2 to 1 n current, n saturatio(reverse) the is 300K at A10I where

1Vn

vexpIi

D

T

DsDD

19-

T

14-s

T

DsD

17

10. Diodes – Load-Line Analysis of Diode Circuits10.2 Load-Line Analysis of Diode Circuit

equations. KVL or KCL writeto difficult is It

1Vn

vexpIi: diode a is there whenbut

,...dtdiLv ,

dtdvCi iR,v use can We

T

DsD

Analysis" Line-Load"the perform can we

given, is diode theof curve V-I the If

viRV :gives KVL

shown,circuit the For

DDSS

18

10. Diodes – Load-Line Analysis of Diode CircuitsExample 10.1- Load-Line Analysis

point)-(Q point operating the at voltage and current diode the :Find

diode the of curve V-I the ,Ω1kR 2V,V :Given

shown,circuit the For

SS

mA 1.3i V, 0.70V point operating the at

analysis line-load performvi 10002

i.e., ,viRV

DQDQ

DD

DDSS

19

10. Diodes – Load-Line Analysis of Diode CircuitsExample 10.2 - Load-Line Analysis

point operating the at voltage and current diode the :Find

diode the of curve V-I the ,k 10R V, 10Vss :Given

shown,circuit the For

mA 0.93i V, 0.68V point operating the at

analysis line-load performvi 10k10

i.e., ,viRV

DQDQ

DD

DDSS

20

10. Diodes – Zener Diode Voltage-Regulator Circuits10.3 Zener-Diode Voltage-Regulator Circuits10.3.1 The Zener Diode* Zener diode is designed for operation in the reverse-breakdown

region.* The breakdown voltage is controlled by the doping level (-1.8 V to

-200 V).* The major application of Zener diode is to provide an output

reference that is stable despite changes in input voltage – power supplies, voltmeter,…

21

10. Diodes – Zener-Diode Voltage-Regulator Circuits10.3.2 Zener-Diode Voltage-Regulator Circuits* Sometimes, a circuit that produces constant output voltage while

operating from a variable supply voltage is needed. Such circuits are called voltage regulator.

* The Zener diode has a breakdown voltage equal to the desired output voltage.

* The resistor limits the diode current to a safe value so that Zener diode does not overheat.

22

10. Diodes – Zener-Diode Voltage-Regulator CircuitsExample 10.3 – Zener-Diode Voltage-

Regulator Circuits

Actual Zener diodeperforms much better!

V 20V and V 15V for voltage output the :Find1kR curve, V-I diode Zenerthe :Given

SS

SS

o

SSo

SSo

DDSS

v in change 0.5V input in change 5V

V 20V for V 10.5vV 15V for V 10.0v

:have wepoint-Q the From0viRV

:line load the gives KVL

23

10. Diodes – Zener-Diode Voltage-Regulator Circuits10.3.3 Load-Line Analysis of Complex Circuits* Use the Thevenin Equivalent

24

10. Diodes – Zener-Diode Voltage-Regulator CircuitsExample 10.4 – Zener-Diode Voltage-Regulator with a Load

SL

LSS

I currents sourceand v voltage load the :Find6kR ,1.2kR 24V,V curve, V-I diode Zener:Given

mA 11.67 )/Rv-(VI V 10.0-vv

0viRV

k1RR

RRR ,V20

RRR

VV Equivalent Thevenin Applying

LSSS

DL

DDTT

L

LT

L

LSST

25

10. Diodes – Zener-Diode Voltage-Regulator CircuitsQuiz – Exercise 10.5

100mAi and 20mA,i 0,i for v voltage output the :Find shown.as curve V-I doide Zenerthe and circuit the :Given

LLLo

26

10. Diodes – Ideal-Diode Model10.4 Ideal-Diode Model* Graphical load-line analysis is too cumbersome for complex circuits,* We may apply “Ideal-Diode Model” to simplify the analysis:(1) in forward direction: short-circuit assumption, zero voltage drop;(2) in reverse direction: open-circuit assumption.* The ideal-diode model can be used when the forward voltage drop and

reverse currents are negligible.

27

10. Diodes – Ideal-Diode Model10.4 Ideal-Diode Model* In analysis of a circuit containing diodes, we may not know in

advance which diodes are on and which are off.* What we do is first to make a guess on the state of the diodes in

the circuit:

YES" ALL" until iterates guess.... another make YES ALL not

BINGO! YES ALL negative is v if check :diodes" off assumed" For (2)

positive; is i if check :diodes" on assumed" (1)For

D

D

28

10. Diodes – Ideal-Diode ModelExample 10.5 – Analysis by Assumed Diode States

on D off,Dassume (1)

21

on D and off isD assumingby circuit the Analysis 21

OK! not 7VvOK! 0.5mAi

D1

D2

off D on, D assume (2)

21

OK! V -3v OK! mA 1i

D2

D1

29

10. Diodes – Ideal-Diode ModelQuiz – Exercise 10.8c* Find the diode states by using ideal-diode model. Starting by

assuming both diodes are on.

on D on D

assume (1)

4

3

OK mA, 6.7iOK not mA, -1.7i

4 D

3 D

on D and off D assume (2) 43

OK V, -5v OK mA, 5i

3 D

4 D

30

10. Diodes – Piecewise-Linear Diode Models10.5 Piecewise-Linear Diode Models10.5.1 Modified Ideal-Diode Model

* This modified ideal-diode model is usually accurate enough in most of the circuit analysis.

31

10. Diodes – Piecewise-Linear Diode Models10.5.2 Piecewise-Linear Diode Models

aa ViRv

32

10. Diodes – Rectifier Circuits10.6 Rectifier Circuits* Rectifiers convert ac power to dc power.* Rectifiers form the basis for electronic power suppliers and battery

charging circuits.10.6.1 Half-Wave Rectifier

33

10. Diodes – Rectifier Circuits* Battery-Charging Circuit

* The current flows only in the direction that charges the battery.

34

10. Diodes – Rectifier Circuits* Half-Wave Rectifier with Smoothing Capacitor

* To place a large capacitance across the output terminals:

35

10. Diodes – Rectifier Circuits10.6.2 Full-Wave Rectifier Circuits

* Center-Tapped Full-Wave Rectifier – two half-wave rectifier with out-of-phase source voltages and a common ground.

* When upper source supplies “+” voltage to diode A, the lower source supplies “-” voltage to diode B; and vice versa.* We can also smooth the output by using a large capacitance.

36

10. Diodes – Rectifier Circuits10.6.2 Full-Wave Rectifier Circuits* The Diode-Bridge Full-Wave Rectifier:

A,B C,D

37

10. Diodes – Wave-Shaping Circuits10.7 Wave-Shaping Circuits10.7.1 Clipper Circuits* A portion of an input signal waveform is “clipped” off.

38

10. Diodes – Wave-Shaping Circuits10.7 Wave-Shaping

Circuits10.7.2 Clamper Circuits* Clamp circuits are used to

add a dc component to an ac input waveform so that the positive (or negative) peaks are “clamped” to a specified voltage value.

39

10. Diodes – Linear Small-Signal Equivalent Circuits10.8 Linear Small-Signal Equivalent Circuits* In most of the electronic circuits, dc supply voltages are used to

bias a nonlinear device at an operating point and a small signal is injected into the circuits.

* We often split the analysis of such circuit into two parts:(1) Analyze the dc circuit to find operating point,(2) Analyze the small signal ( by using the “linear small- signal equivalent circuit”.)

40

10. Diodes – Linear Small-Signal Equivalent Circuits10.8 Linear Small-Signal Equivalent

Circuits* A diode in linear small-signal equivalent circuit is simplified to a resistor.* We first determine the operating point (or the “quiescent point” or Q point) by dc bias.* When small ac signal injects, it swings

the Q point slightly up and down.* If the signal is small enough, the

characteristic is straight.

voltage diode in change smallthe is vcurrent diode in change smallthe is i

vvdid

i

D

D

DQD

DD

QD

D

vdid

41

10. Diodes – Linear Small-Signal Equivalent Circuits10.8 Linear Small-Signal Equivalent

CircuitsQD

D

vdid

QD

Td

d

dd

ddDD

d

DDD

QD

DD

1

QD

Dd

IVn

r :have we

equation, Shockley the applyingby e,Furthermorrv

i

: signalsac for have wechanges, small denoting v and iby v and i Replace

rv

i vvdid

i

:have willWe vdid

r

:as diode the of resistance dynamic the Define

42

10. Diodes – Linear Small-Signal Equivalent Circuits10.8 Linear Small-Signal

Equivalent Circuits

* By using these two equations, we can treat diode simply as a linear resistor in small ac signal analysis.

* Note: An ac voltage of fixed amplitude produces different ac current change at different Q point.

QD

Td

d

dd I

Vnr ,

rv

i

43

10. Diodes – Linear Small-Signal Equivalent Circuits10.8 Linear Small-Signal Equivalent Circuits

current. andvoltage diode ousinstantane totalthe represent i and v (3)

signals.sc smallthe represent i and v (2)point.Q the at

signalsdc the represent I and V (1)

DD

dd

DQDQ

dDQD

dDQD

vVv

iIi

QD

Td

d

dd I

Vnr ,

rv

i

44

10. Diodes – Linear Small-Signal Equivalent CircuitsVoltage-Controlled Attenuator

* The function of this circuit is to produce an output signal that is a variable fraction of the ac input signal.

* Two large coupling capacitors: behave like short circuit for ac signal and open circuit for dc, thus the Q point of the diode is unaffected by the ac input and the load.

Cj1ZC

45

10. Diodes – Linear Small-Signal Equivalent CircuitsVoltage-Controlled Attenuator

QD

TddDQ I

Vnr :diode the of r the then ,I determine

point, Q diode the find to analysis dcapply First

1RR

Rvv

A :divider voltage on based ,r1R1R1

1R

signal.)ac for circuit shorta to equivalent is sourcevoltage dc thevoltage, ac no but current of component ac an has sourcevoltage dc the :(note

: analysis signalac smallperform weNext,

p

p

in

ov

dLCp

46

10. Diodes – Linear Small-Signal Equivalent CircuitsExercise 10.20 - Voltage-Controlled Attenuator

0.026VV withIVn

r ,R

0.6-VI

point,Q diode the find to analysisdc apply First10.6V and 1.6V for A and

0.6VVassuming values point-Qthe :Find 300Kat 1ndiode ,Ω2kRR ,Ω100R :Given

TQD

Td

C

CDQ

Cv

f

LC

p

p

in

ov

dLCp RR

Rvv

A ,r1R1R1

1R

: analysis signalac small perform we Next,

Chapter 11: Amplifiers:Chapter 11: Amplifiers:Specifications and External CharacteristicsSpecifications and External Characteristics

48

11. Amplifiers – Basic Amplifier Concepts11.1 Basic Amplifier Concepts11.1.1 For Starting* Ideally, an amplifier produces an output signal with identical wave-

shape as the input signal but with a larger amplitude.

GainVoltage the is A

tvAtv

v

ivo

49

11. Amplifiers – Basic Amplifier Concepts11.1.1 For Starting* inverting and non-inverting amplifiers

50

11. Amplifiers – Basic Amplifier Concepts11.1.1 For Starting* Often, one of the amplifier input

terminals and one of the output terminals are connected to a common ground.

* The common ground serve as the return path for signal and the dc power supply currents.

51

11. Amplifiers – Basic Amplifier Concepts11.1.1 For Starting* Another example for common ground

52

11. Amplifiers – Basic Amplifier Concepts11.1.2 Voltage-Amplifier Model* Amplification can be modeled by a controlled source.

Amplifier

)A tham smaller is gain realthe :(note gainvoltage circuit-openthe : /vvA

impedance) (or resistance outputthe is terminals, outputthe withseries in :R

terminals. inputthe intolooking when seen resistance equivalentthe is ,impedance) (or resistance inputthe :R

vo

iocvo

o

i

53

11. Amplifiers – Basic Amplifier Concepts11.1.3 Current and Voltage Gains

load.the through flowing currentthe is i

amplifier;the of terminals inputthe into

delivered currentthe is i

o

i

vov

i

ov

L

iv

ii

Lo

i

oi

i

oii

A gainvoltage circuit-openthe than smallerusually is A gainvoltage The

gainvoltage the is vv

A where ,RR

ARvRv

ii

A e,Furthermor

ii

A :currents input and output between ratiothe is A gain currentThe

54

11. Amplifiers – Basic Amplifier Concepts11.1.3 Power Gain

L

i2viv

i rmsi rms

o rmso rms

i

o

rmsrms

i

o

RR

AAAIVIV

PP

G :havewe

,I and V values rmsthe of productthe is poweraverage the Since PP

G :power inputthe to power outputthe of ratiothe is gain powerThe

55

11. Amplifiers – Basic Amplifier ConceptsExample 11.1 Calculating Amplifier Performance

gain powerthe andgain currentthe find Also,

/VVA and /VVAgainsvoltage the Find

iovsovs

12iv

9

L

ivi

ov,vsv

si

iv

isii

o

s

osv

4

Lo

Lov

i

LoLivo

i

ov

1016AAG ,102RR

AA

AAA effect,loading the todue :Note

5333RR

RA

)/RR(RVV

VV

A

800082

810

RRR

AV

)R/(RRVAVV

A

56

11. Amplifiers – Cascade Amplifiers11.2 Cascade Amplifiers

212i1ii

2ov1vov2i

1o2ov

1i

2i2ov

1i

2coov

2v1vv2i

2o

1i

1o

01

2o

1i

1o

1i

2ov

GGG , AAA addition, In

AAA vv

Av

vAvv

A

AAA vv

vv

vv

vv

vv

A

57

11. Amplifiers – Cascade AmplifiersExample 11.2 – Calculating Performance of Cascade Amplifier

Find the current gain, voltage gain, and power gain

1121

42i2v2

71i1v1

62i1ii

L

2i2v2i

5

2i

1i1v1i

2v1vv2oL

L2vo2v

1o2i

2i1vo1v

10625.5GGG

1075.3AAG ,105.1AAG

1075AAA 750RR

AA ,10RR

AA

7500AAA 50RR

RAA ,150

RRR

AA

1L2i RR

58

11. Amplifiers – Cascade AmplifiersExample 11.3– Simplified Model for Amplifier Cascade

32ov1vov

2ov

1o2i

2i1ov1v

1015AAA

100A

150RR

RAA

59

11. Amplifiers – Power Supplies and Efficiency11. 3 Power Supplies and Efficiency* The power gain of an amplifier is usually large, the additional

power is taken from the power supply.

100%PP

η

:amplifier an of efficiencyThe carcuits internal

in dissipated powerthe is Psignal output of powerthe is P

signal input of powerthe is PPPPP

:energy of onConservatiIVIVP

is amplifierthe to suppliedpoweraverage totalThe

s

o

d

o

i

dosi

BBBAAAs

60

11. Amplifiers – Power Supplies and EfficiencyExample 11.4 Amplifier Efficiency

%6.35PP

η

W5.14PPPPW5.22IVIVP

W8RV

P

rmsV8RR

RVAV

pW10W10R

VP

s

o

oisd

BBBAAAs

L

2o

o

oL

Livoo

11

i

2i

i

61

11. Amplifiers – Additional Amplifier Models11.4 Additional Amplifier ModelsCurrent-Amplifier Model

models. twothe for samethe are R and R

RR

Ai

iA

:gain current circuit shortthe R

vAiand

Rv

i

approach Norton vs. Thevenin

oi

o

ivo

i

oscisc

o

ivoosc

i

ii

62

11. Amplifiers – Additional Amplifier ModelsExample 11.5 – Transform Voltage-Amplifier to Current-Amplifier Model

3

o

ivo

i

oscisc

o

ivoosc

i

ii

10RR

Ai

iA

RvA

iandRv

i

63

11. Amplifiers – Additional Amplifier Models11.4 Additional Amplifier ModelsTransconductance-Amplifier Model

Transresistance-Amplifier Model

i

oscmsc v

iG

i

oscmsc i

vR

64

11. Amplifiers – Ideal Amplifiers11.6 Ideal Amplifiers

Gain!Voltage Maximum max vAv 0, R

max. vv , R:AmplifierVoltage

ivooo

sii

Gain! Current Maximum

max. iAi , R

max. ii 0, R:Amplifier Current

iscioo

sciii

65

11. Amplifiers – Frequency Response11.7 Frequency Response* The gain of an amplifier is a function of frequency.* Both amplitude and phase are affected.

Example 11.8

i

ov V

VA

dB40)100log(20Alog20Ashift.phase 45a isthere :Note

45100301.0

1510VV

A 1510V , 301.0V

dB inamplitude the express and gainvoltage complex the Find15tπ2000cos10tv :is outpotthe

30tπ2000cos1.0tv:is amplifier certaina for inputThe

vdBv

i

ovoi

0

i

66

11. Amplifiers – Frequency ResponseGain as a Function of Frequency

67

11. Amplifiers – Frequency ResponseAC Coupled Amplifiers

The gain drops to zero at dc (low frequency).

68

11. Amplifiers – Frequency ResponseDC Coupled Amplifiers

Amplifiers that are realized as integrated circuits are often dc coupled, because capacitors or transformers can not be fabricated in integrated form.

69

11. Amplifiers – Frequency ResponseHigh-Frequency Drop Off* The small amount of capacitance in parallel or inductances in

series with the signal path in the amplifier circuit will cause the gain of the amplifier to drop at high frequencies.

f as short fC,π1/j2 f as open fL,πj2

70

11. Amplifiers – Frequency ResponseHalf-Power Frequencies and

Bandwidth* Bandwidth is the distance bet

ween the half-power frequencies.

* Half-power frequencies:

* Wideband (Baseband) Amp.* Narrowband (Bandpass) Amp.

dB -3.01)220log(1/

A from 3dB 2/AA midmid

71

11. Amplifiers – Linear Waveform Distortion11.8 Linear Waveform Distortion* Distortion may occur even though the amplifier is linear (i.e., obeys

superposition principle).Amplitude DistortionIf a signal contains components of various frequencies, the output

waveform may be distorted due to the frequency response of the amplifier gain.

72

11. Amplifiers – Linear Waveform DistortionPhase Distortion* If the phase shift of an amplifier is not

proportional to the frequency, phase distortion may occur.

:gainsthe having samplifiierthree and

tπ6000costπ2000cos3tv:like input anhave Assume we

i

45tπ6000cos1045tπ2000cos30tv135tπ6000cos1045tπ2000cos30tv

tπ6000cos10tπ2000cos30tv:like look wouldoutputThe

oC

oB

oA

73

11. Amplifiers – Linear Waveform Distortion

11.8 Linear Waveform Distortion* In order to avoid distortion:(1) the magnitude of the gain must be constant against frequency.(2) The phase response must be proportional to the frequency.

delay!time same The 1:1

)ω/ω(Tπ2

)ω/ω(θ:T

π2θ

Tπ2

θ:T

π2θ

tΔ:tΔ

T:Tω:ωθ:θ

121211

11

22

11

21

122121

74

11. Amplifiers – Pulse Response11.9 Pulse Response

75

11. Amplifiers – Nonlinear Distortion11.10 Nonlinear Distortion

76

11. Amplifiers – Differential Amplifiers11.11 Differential Amplifiers* A Differential amplifier has two input sources, the output voltage is

proportional to the difference between the two input voltages.

Non-inverting input Inverting input

tvAtvA

tvtvAtv

2id1id

2i1ido

iddo

d

2i1iid

vAv:as amplifier ildifferentaa

of outputthe write canWe gain ildifferentathe A and

vvv:signal aldifferentithe Define

77

11. Amplifiers – Differential AmplifiersElectrocardiogram* The desired waveform is given by the difference between the

potentials measured by the two electrodes, i.e., the output of an ideal differential amplifier.

* While both electrodes (also act like antennas) pick a common-mode signal (noise) from the 60 Hz power line:

)v-(vAvAv i2i1diddo

)φcos(377tVv ),φcos(377tVv nni2nni1

78

11. Amplifiers – Differential AmplifiersElectrocardiogram

* Ideally, the common-mode signal was nullified by the differential amplifier.* However, real differential amplifier responds to both differential-mode and

common-mode signals:

Common-Mode Rejection Ratio (CMRR):

* At 60 Hz, CMRR of 120 dB is considered good.

)v-(vAvAv i2i1diddo

)φcos(377tVv ),φcos(377tVv

nni2

nni1

signalmode -commonthe v and signalmode -aldifferentithe is v gain,mode -commonthe is A where vAvAv

icmid

cmicmcmiddo

cm

d

AA

log20CMRR

79

11. Amplifiers – Differential AmplifiersExample 11.12 Determination of Minimum CMRR Specification* Find the minimum CMRR for an electrocardiogram amplifier if the

differential gain is 1000, the desired differential input signal has a peak amplitude of 1 mV, the common-mode signal is 100 V-peak 60 Hz sine wave, and it is desired that output contain a peak common-mode signal that is 1% or less of the peak output caused by the differential signal.

dB140 CMRR :Ans

dB140101000

log20AA

log20CMRR

10V100V01.0

A V 0.011%V 1 v peak

V 1v 1000 A mV, 1v peak

4cm

d

4cmcmo

doddi