Chapter 10 Diodes - libvolume3.xyzlibvolume3.xyz/computers/bca/semester1/basic... · 10. Diodes –...

Chapter 10 Diodes 1. 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 and piecewise-linear models. 5. Understand various rectifier and wave-shaping circuits. 6. Understand small-signal equivalent circuits.

2

10. Diodes – Basic Diode Concepts

10.1 Basic Diode Concepts

10.1.1 Intrinsic Semiconductors

* Energy Diagrams – Insulator, Semiconductor, and Conductor

the energy diagram for the three types of solids

3



* 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

3-10

ii

×

×==

=

4



*Apply a voltage across

a piece of Si:

electron current

and hole current

5


10.1.2 N- and P- Type Semiconductors

* Doping: adding of impurities (i.e., dopants) to the intrinsic semi-

conductor material.

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

( )

carrier cahage minor the hole

carrier charge major the electron

call We

pp ,nNn

nconceratio donor the Nn

material type-n In

101.5pnpn

300K at SiFor

ctor semicondua for constantpn

iid

d

2102

i

2

i

<<>>=

≅

×===⋅

=⋅

6


10.1.2 N- and P- Type Semiconductors

* Doping: adding of impurities (i.e., dopants) to the intrinsic semi-

conductor material.

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

( )

carrier cahage minor the electron

carrier charge major the hole

call We

nn ,pNp

nconceratio acceptor the Np

material type-p In

101.5pnpn

300K at SiFor

ctor semicondua for constantpn

iia

a

2102

i

2

i

<<>>=

≅

×===⋅

=⋅

7


10.1.3 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


10.1.4 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



*Forward Bias: dc voltage positive terminal connected to the p

region and negative to the n region. It is the condition that

permits current through the pn-junction of a diode.

10


*Forward Bias: dc voltage positive terminal connected to the p

region and negative to the n region. It is the condition that

permits current through the pn-junction of a diode.

11



*Forward Bias:

12


*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


*Reverse Bias:

majority-carrier current ceases.

* However, there is still a very

small current produced by

minority carriers.

14



* Reverse Breakdown: As reverse voltage reach certain value,

avalanche occurs and generates large current.

15


10.1.5 The Diode Characteristic I-V Curve

16


10.1.6 Shockley Equation

* The Shockley equation is a theoretical result

under certain simplification:

0v whenapplicable not is equation This

Vn

vexpIi 0.1V,v when

C101.60q constant, sBoltzman' the is k

voltage thermal the is 300K at 0.026V q

TkV

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 Circuits

10.2 Load-Line Analysis of Diode Circuit

equations. KVL or KCL writeto difficult is It

1Vn

vexpIi: diode a is there whenbut

,...dt

diLv ,

dt

dvCi iR,v use can We

T

D

sD

−

=

===

Analysis" Line-Load"

the perform can we

given, is diode the

of curve V-I the If

viRV

:gives KVL

shown,circuit the For

DDSS +=

18


Example 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


SS ==

mA 1.3i V, 0.70V

point operating the at

analysis line-load perform

vi 10002

i.e., ,viRV

DQDQ

DD

DDSS

≅≅

⇒

⇒

+=

+=

19


Example 10.2 - Load-Line Analysis


voltage and current diode the :Find

diode the of curve V-I the

,k 10R V, 10Vss :Given


Ω==

mA 0.93i V, 0.68V


analysis line-load perform

vi 10k10

i.e., ,viRV

DQDQ

DD

DDSS

≅≅

⇒

⇒

+=

+=

20

10. Diodes – Zener Diode Voltage-Regulator Circuits

10.3 Zener-Diode Voltage-Regulator Circuits

10.3.1 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 Circuits

10.3.2 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


Example 10.3 – Zener-Diode Voltage-

Regulator Circuits

Actual Zener diode

performs much better!

V 20V

and V 15V for voltage output the :Find

1kR curve, V-I diode Zenerthe :Given

SS

SS

=

=

= Ω

o

SSo

SSo

DDSS

v in change 0.5V

input in change 5V

V 20V for V 10.5v

V 15V for V 10.0v

:have wepoint-Q the From

0viRV

:line load the gives KVL

⇒

==

==

=++

23


10.3.3 Load-Line Analysis of Complex Circuits

* Use the Thevenin Equivalent

24


Example 10.4 – Zener-Diode Voltage-Regulator with a Load

SL

LSS

I currents sourceand v voltage load the :Find

6kR ,1.2kR 24V,V curve, V-I diode Zener:Given ΩΩ ===

mA 11.67 )/Rv-(VI

V 10.0-vv

0viRV

k1RR

RRR ,V20

RR

RVV Equivalent Thevenin Applying

LSSS

DL

DDTT

L

LT

L

LSST

==

==⇒

=++⇒

=+

==+

=⇒ Ω

25


Quiz – 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 Model

10.4 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


10.4 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


Example 10.5 – Analysis by Assumed Diode States

⇒ on D off,D

assume (1)

21

on D and off isD assumingby circuit the Analysis 21

OK! not 7Vv

OK! 0.5mAi

D1

D2

=

=

⇓

off D on, D

assume (2)

21

⇓

OK! V -3v

OK! mA 1i

D2

D1

=

=⇒

29


Quiz – 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, 6.7i

OK 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 Models

10.5 Piecewise-Linear Diode Models

10.5.1 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 Models

10.5.2 Piecewise-Linear Diode Models

aa ViRv +=

32

10. Diodes – Rectifier Circuits

10.6 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


* Battery-Charging Circuit

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

34


* Half-Wave Rectifier with Smoothing Capacitor

* To place a large capacitance across the output terminals:

35


10.6.2 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


10.6.2 Full-Wave Rectifier Circuits

* The Diode-Bridge Full-Wave Rectifier:

A,B C,D

37

10. Diodes – Wave-Shaping Circuits

10.7 Wave-Shaping Circuits

10.7.1 Clipper Circuits

* A portion of an input signal waveform is “clipped” off.

38

10. Diodes – Wave-Shaping Circuits

10.7 Wave-Shaping

Circuits

10.7.2 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 Circuits

10.8 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


10.8 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 v

current diode in change smallthe is i

vvd

idi

D

D

D

QD

DD

∆

∆

∆∆

≅

QD

D

vd

id

41


10.8 Linear Small-Signal Equivalent

Circuits

QD

D

vd

id

QD

T

d

d

dd

ddDD

d

D

DD

QD

D

D

1

QD

Dd

I

Vnr :have we

equation, Shockley the applyingby e,Furthermor

r

vi

: signalsac for have wechanges, small

denoting v and iby v and i Replace

r

vi v

vd

idi

:have willWe vd

idr

:as diode the of resistance dynamic the Define

=

=

≅⇒

≅

≅

−

∆∆

∆∆∆∆

42


10.8 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 ,

r

vi ==

43


10.8 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 ,

r

vi ==

44


Voltage-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.

Cj

1Z

C ω=

45


Voltage-Controlled Attenuator

QD

TddDQ

I

Vnr :diode the of r the then ,I determine

point, Q diode the find to analysis dcapply First

=

1RR

R

v

vA :divider voltage on based ,

r1R1R1

1R

signal.)ac for circuit shorta to equivalent is sourcevoltage dc the

voltage, ac no but current of component ac an has sourcevoltage dc the :(note

: analysis signalac smallperform weNext,

p

p

in

ov

dLC

p <+

==++

=

46


Exercise 10.20 - Voltage-Controlled Attenuator

0.026VV withI

Vnr ,

R

0.6-VI

point,Q diode the find to analysisdc apply First

10.6V and 1.6V for A and

0.6VVassuming values point-Qthe :Find

300Kat 1ndiode ,Ω2kRR ,Ω100R :Given

T

QD

Td

C

C

DQ

Cv

f

LC

============

====

====

================

p

p

in

o

v

dLC

pRR

R

v

vA ,

r1R1R1

1R

: analysis signalac small perform we Next,

++++========

++++++++====

Chapter 11: Amplifiers: Specifications and External Characteristics

48

11. Amplifiers – Basic Amplifier Concepts

11.1 Basic Amplifier Concepts

11.1.1 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


11.1.1 For Starting

* inverting and

non-inverting amplifiers

50


11.1.1 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


11.1.1 For Starting

* Another example for common ground

52


11.1.2 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


11.1.3 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

o

v

L

i

v

ii

Lo

i

o

i

i

o

ii

A gainvoltage circuit-openthe than smallerusually is A gainvoltage The

gainvoltage the is v

vA where ,

R

RA

Rv

Rv

i

iA e,Furthermor

i

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

================

====

54


11.1.3 Power Gain

(((( ))))L

i2

viv

i rmsi rms

o rmso rms

i

o

rmsrms

i

o

R

RAAA

IV

IV

P

PG :havewe

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

P

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

================

====

55


Example 11.1 Calculating Amplifier Performance

gain powerthe and

gain currentthe find Also,

/VVA and /VVA

gainsvoltage the Find

iovsovs ========

(((( ))))

(((( ))))(((( ))))

12

iv

9

L

i

vi

ov,vsv

si

i

v

isii

o

s

o

sv

4

Lo

L

ov

i

LoLivo

i

o

v

1016AAG ,102R

RAA

AAA effect,loading the todue :Note

5333RR

RA

)/RR(RV

V

V

VA

800082

810

RR

RA

V

)R/(RRVA

V

VA

××××========××××========

<<<<<<<<

====++++

====++++

========

====++++

====++++

====++++

========

56

11. Amplifiers – Cascade Amplifiers

11.2 Cascade Amplifiers

212i1ii

2ov1vov

2i

1o

2ov

1i

2i2ov

1i

2co

ov

2v1vv

2i

2o

1i

1o

01

2o

1i

1o

1i

2o

v

GGG , AAA addition, In

AAA v

vA

v

vA

v

vA

AAA v

v

v

v

v

v

v

v

v

vA

========

====⇒⇒⇒⇒============

====⇒⇒⇒⇒××××====××××========

57


Example 11.2 – Calculating Performance of Cascade Amplifier

Find the current gain, voltage gain, and power gain

11

21

4

2i2v2

7

1i1v1

6

2i1ii

L

2i

2v2i

5

2i

1i

1v1i

2v1vv

2oL

L

2vo2v

1o2i

2i

1vo1v

10625.5GGG

1075.3AAG ,105.1AAG

1075AAA 750R

RAA ,10

R

RAA

7500AAA 50RR

RAA ,150

RR

RAA

××××========⇒⇒⇒⇒

××××========××××========

××××========⇒⇒⇒⇒================

========⇒⇒⇒⇒====++++

========++++

====

1L2i RR ====

58


Example 11.3– Simplified Model for Amplifier Cascade

3

2ov1vov

2ov

1o2i

2i

1ov1v

1015AAA

100A

150RR

RAA

××××========

====

====++++

====

59

11. Amplifiers – Power Supplies and Efficiency

11. 3 Power Supplies and Efficiency

* The power gain of an amplifier is usually large, the additional

power is taken from the power supply.

100%P

Pη

:amplifier an of efficiencyThe

carcuits internal

in dissipated powerthe is P

signal output of powerthe is P

signal input of powerthe is P

PPPP

:energy of onConservati

IVIVP

is amplifierthe to supplied

poweraverage totalThe

s

o

d

o

i

dosi

BBBAAAs

××××====

++++====++++

++++====

60

11. Amplifiers – Power Supplies and Efficiency

Example 11.4 Amplifier Efficiency

%6.35P

Pη

W5.14PPPP

W5.22IVIVP

W8R

VP

rmsV8RR

RVAV

pW10W10R

VP

s

o

oisd

BBBAAAs

L

2

o

o

oL

L

ivoo

11

i

2

i

i

========

====−−−−++++====

====++++====

========

====++++

====

============ −−−−

61

11. Amplifiers – Additional Amplifier Models

11.4 Additional Amplifier Models

Current-Amplifier Model

models. twothe for

samethe are R and R

R

RA

i

iA

:gain current circuit shortthe

R

vAiand

R

vi

approach Norton vs. Thevenin

oi

o

i

vo

i

osc

isc

o

ivo

osc

i

i

i

========

========

62


Example 11.5 – Transform Voltage-Amplifier to

Current-Amplifier Model

3

o

i

vo

i

osc

isc

o

ivo

osc

i

i

i

10R

RA

i

iA

R

vAiand

R

vi

============

========

63


11.4 Additional Amplifier Models

Transconductance-Amplifier Model

Transresistance-Amplifier Model

i

oscmsc

v

iG =

i

osc

msci

vR =

64

11. Amplifiers – Ideal Amplifiers

11.6 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 Response

11.7 Frequency Response

* The gain of an amplifier is a function of frequency.

* Both amplitude and phase are affected.

Example 11.8

i

o

vV

VA ====

(((( )))) (((( ))))(((( )))) (((( ))))

dB40)100log(20Alog20A

shift.phase 45a isthere :Note

45100301.0

1510

V

VA 1510V , 301.0V

dB inamplitude the express and gainvoltage complex the Find

15tπ2000cos10tv :is outpotthe

30tπ2000cos1.0tv:is amplifier certaina for inputThe

vdBv

i

o

voi

0

i

============

°°°°

°°°°∠∠∠∠====°°°°−−−−∠∠∠∠

°°°°∠∠∠∠========⇒⇒⇒⇒°°°°∠∠∠∠====°°°°−−−−∠∠∠∠====

°°°°++++====

°°°°−−−−====

66


Gain as a Function of Frequency

67


AC Coupled Amplifiers

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

68


DC 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


High-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


Half-Power Frequencies and

Bandwidth

* Bandwidth is the distance

between 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 Distortion

11.8 Linear Waveform Distortion

* Distortion may occur even though the amplifier is linear (i.e., obeys

superposition principle).

Amplitude Distortion

If a signal contains components of various frequencies, the output

waveform may be distorted due to the frequency response of the

amplifier gain.

72


Phase 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−−−−====

(((( )))) (((( )))) (((( ))))(((( )))) (((( )))) (((( ))))(((( )))) (((( )))) (((( ))))°°°°−−−−−−−−°°°°−−−−====

°°°°−−−−−−−−°°°°−−−−====

−−−−====

45tπ6000cos1045tπ2000cos30tv

135tπ6000cos1045tπ2000cos30tv

tπ6000cos10tπ2000cos30tv

:like look wouldoutputThe

oC

oB

oA

73


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

π2

θ

Tπ2

θ:T

π2

θtΔ:tΔ

T:Tω:ωθ:θ

121

211

1

1

2

2

1

1

21

122121

====

××××××××====

××××××××====

========

74

11. Amplifiers – Pulse Response

11.9 Pulse Response

75

11. Amplifiers – Nonlinear Distortion

11.10 Nonlinear Distortion

76

11. Amplifiers – Differential Amplifiers

11.11 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


Electrocardiogram

* 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-(vAvAvi2i1diddo

========

)φcos(377tVv ),φcos(377tVv nni2nni1 ++++++++++++++++

78


Electrocardiogram

* 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

A

Alog20CMRR ====

79


Example 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

dB14010

1000log20

A

Alog20CMRR

10V100

V01.0A V 0.011%V 1 v peak

V 1v 1000 A mV, 1v peak

4

cm

d

4

cmcmo

doddi

>>>>

====>>>>====

====<<<<⇒⇒⇒⇒====××××<<<<

====⇒⇒⇒⇒========

−−−−

−−−−

Chapter 10 Diodes - libvolume3.xyzlibvolume3.xyz/computers/bca/semester1/basic... · 10. Diodes –...

Documents

Transcript of Chapter 10 Diodes - libvolume3.xyzlibvolume3.xyz/computers/bca/semester1/basic... · 10. Diodes –...