ess2220-10to11
TRANSCRIPT
Chapter 10Chapter 10 Diodes 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 andpiecewise-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
10. Diodes – Basic Diode Concepts
10.1.1 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 Concepts
10.1.1 Intrinsic Semiconductors
*Apply a voltage across
a piece of Si:
electron current
and hole current
5
10. Diodes – Basic Diode Concepts
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
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-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
2102i
2i
7
10. Diodes – Basic Diode Concepts
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. Diodes – Basic Diode Concepts
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
10. Diodes – Basic Diode Concepts
10.1.4 Biasing the PN-Junction
*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
10.1.4 Biasing the PN-Junction
*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
10. Diodes – Basic Diode Concepts
10.1.4 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 Concepts
10.1.4 Biasing the PN-Junction
* Reverse Breakdown: As reverse voltage reach certain value, avalanche occurs and generates large current.
15
10. Diodes – Basic Diode Concepts
10.1.5 The Diode Characteristic I-V Curve
16
10. Diodes – Basic Diode Concepts
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
DsD
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
10. Diodes – Load-Line Analysis of Diode Circuits
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
shown,circuit the For
SS
mA 1.3i V, 0.70V
point operating the at
analysis line-load perform
vi 10002
i.e., ,viRV
DQDQ
DD
DDSS
19
10. Diodes – Load-Line Analysis of Diode Circuits
Example 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 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
10. Diodes – Zener-Diode Voltage-Regulator Circuits
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
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. Diodes – Zener-Diode Voltage-Regulator Circuits
10.3.3 Load-Line Analysis of Complex Circuits
* Use the Thevenin Equivalent
24
10. Diodes – Zener-Diode Voltage-Regulator Circuits
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
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
25
10. Diodes – Zener-Diode Voltage-Regulator Circuits
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. Diodes – Ideal-Diode Model
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
10. Diodes – Ideal-Diode Model
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
10. Diodes – Ideal-Diode Model
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
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 Circuits
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. Diodes – Rectifier Circuits
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. Diodes – Linear Small-Signal Equivalent Circuits
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. Diodes – Linear Small-Signal Equivalent Circuits
10.8 Linear Small-Signal Equivalent Circuits
QD
D
vd
id
QD
Td
d
dd
ddDD
d
DDD
QD
DD
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. Diodes – Linear Small-Signal Equivalent Circuits
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. 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 ,
r
vi
44
10. Diodes – Linear Small-Signal Equivalent Circuits
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
1ZC
45
10. Diodes – Linear Small-Signal Equivalent Circuits
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
dLCp
46
10. Diodes – Linear Small-Signal Equivalent Circuits
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
TQD
Td
C
CDQ
Cv
f
LC
p
p
in
ov
dLCp RR
R
v
vA ,
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 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. Amplifiers – Basic Amplifier Concepts
11.1.1 For Starting
* inverting and
non-inverting amplifiers
50
11. Amplifiers – Basic Amplifier Concepts
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. Amplifiers – Basic Amplifier Concepts
11.1.1 For Starting
* Another example for common ground
52
11. Amplifiers – Basic Amplifier Concepts
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. Amplifiers – Basic Amplifier Concepts
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
ov
L
iv
ii
Lo
i
oi
i
oii
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. Amplifiers – Basic Amplifier Concepts
11.1.3 Power Gain
L
i2viv
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
11. Amplifiers – Basic Amplifier Concepts
Example 11.1 Calculating Amplifier Performance
gain powerthe and
gain currentthe find Also,
/VVA and /VVA
gainsvoltage the Find
iovsovs
12iv
9
L
ivi
ov,vsv
si
iv
isii
o
s
osv
4
Lo
Lov
i
LoLivo
i
ov
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
2ov1vov2i
1o2ov
1i
2i2ov
1i
2coov
2v1vv2i
2o
1i
1o
01
2o
1i
1o
1i
2ov
GGG , AAA addition, In
AAA v
vA
v
vA
v
vA
AAA v
v
v
v
v
v
v
v
v
vA
57
11. Amplifiers – Cascade Amplifiers
Example 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 750R
RAA ,10
R
RAA
7500AAA 50RR
RAA ,150
RR
RAA
1L2i RR
58
11. Amplifiers – Cascade Amplifiers
Example 11.3– Simplified Model for Amplifier Cascade
32ov1vov
2ov
1o2i
2i1ov1v
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
2o
o
oL
Livoo
11
i
2i
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
ivo
i
oscisc
o
ivoosc
i
ii
62
11. Amplifiers – Additional Amplifier Models
Example 11.5 – Transform Voltage-Amplifier to
Current-Amplifier Model
3
o
ivo
i
oscisc
o
ivoosc
i
ii
10R
RA
i
iA
R
vAiand
R
vi
63
11. Amplifiers – Additional Amplifier Models
11.4 Additional Amplifier Models
Transconductance-Amplifier Model
Transresistance-Amplifier Model
i
oscmsc v
iG
i
oscmsc i
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
ov V
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
ovoi
0
i
66
11. Amplifiers – Frequency Response
Gain as a Function of Frequency
67
11. Amplifiers – Frequency Response
AC Coupled Amplifiers
The gain drops to zero at dc (low frequency).
68
11. Amplifiers – Frequency Response
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
11. Amplifiers – Frequency Response
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
11. Amplifiers – Frequency Response
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
11. Amplifiers – Linear Waveform Distortion
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. 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
π2
θ
Tπ2
θ:T
π2
θtΔ:tΔ
T:Tω:ωθ:θ
121211
11
22
11
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
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
A
Alog20CMRR
79
11. Amplifiers – Differential Amplifiers
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
4cm
d
4cmcmo
doddi