opamp pratik
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Table of ContentsTable of ContentsThe Operational Amplifier______________________________slides 3The Operational Amplifier______________________________slides 3--44
The Four Amplifier Types______________________________slide 5The Four Amplifier Types______________________________slide 5
VCVS(Voltage Amplifier) Summary:VCVS(Voltage Amplifier) Summary:
Noninverting Configuration____________slides 6Noninverting Configuration____________slides 6--99
Inverting Configuration________________slides 10Inverting Configuration________________slides 10--1212
ICIC(Current Amplifier) Summary________________________slide 13ICIC(Current Amplifier) Summary________________________slide 13
VCIS (Transconductance Amplifier) Summary_____________slides 14VCIS (Transconductance Amplifier) Summary_____________slides 14--1515
ICVS (Transresistance Amplifier) Summary_______________slides 16ICVS (Transresistance Amplifier) Summary_______________slides 16--1818Power Bandwidth_____________________________________slide 19Power Bandwidth_____________________________________slide 19
Slew Rate____________________________________________slide 20Slew Rate____________________________________________slide 20
Slew Rate Output Distortion____________________________slide 21Slew Rate Output Distortion____________________________slide 21
Noise Gain___________________________________________slide 22Noise Gain___________________________________________slide 22
GainGain--Bandwidth Product_______________________________slide 23Bandwidth Product_______________________________slide 23Cascaded AmplifiersCascaded Amplifiers -- Bandwidth________________________slide 24Bandwidth________________________slide 24
Common Mode Rejection Ratio__________________________slides 25Common Mode Rejection Ratio__________________________slides 25--2626
Power Supply Rejection Ratio___________________________slide 27Power Supply Rejection Ratio___________________________slide 27
Sources_____________________________________________slide 28Sources_____________________________________________slide 28
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The Operational AmplifierThe Operational Amplifier
Usually Called Op AmpsUsually Called Op Amps
An amplifier is a device that accepts a varying input signal andAn amplifier is a device that accepts a varying input signal andproduces a similar output signal with a larger amplitude.produces a similar output signal with a larger amplitude.
Usually connected so part of the output is fed back to the input.Usually connected so part of the output is fed back to the input.
(Feedback Loop)(Feedback Loop)
Most Op Amps behave like voltage amplifiers. They take an inputMost Op Amps behave like voltage amplifiers. They take an input
voltage and output a scaled version.voltage and output a scaled version.
They are the basic components used to build analog circuits.They are the basic components used to build analog circuits.
The name operational amplifier comes from the fact that they wereThe name operational amplifier comes from the fact that they were
originally used to perform mathematical operations such asoriginally used to perform mathematical operations such as
integration and differentiation.integration and differentiation.
Integrated circuit fabrication techniques have made highIntegrated circuit fabrication techniques have made high--
performance operational amplifiers very inexpensive in comparisonperformance operational amplifiers very inexpensive in comparison
to older discrete devices.to older discrete devices.
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ii(+)(+), i, i((--)) : Currents into the amplifier on the inverting and noninverting lines: Currents into the amplifier on the inverting and noninverting lines
respectivelyrespectively
vvidid : The input voltage from inverting to non: The input voltage from inverting to non--inverting inputsinverting inputs +V+VSS ,, --VVSS : DC source voltages, usually +15V and: DC source voltages, usually +15V and 15V15V
RRii : The input resistance, ideally infinity: The input resistance, ideally infinity
A : The gain of the amplifier. Ideally very high, in the 1x10A : The gain of the amplifier. Ideally very high, in the 1x101010 range.range.
RROO: The output resistance, ideally zero: The output resistance, ideally zero
vvOO: The output voltage; v: The output voltage; vOO = A= AOLOLvvidid where Awhere AOLOL is the openis the open--loop voltage gainloop voltage gain
The Operational AmplifierThe Operational Amplifier+V+VSS
--VVSS
vvidid
InvertingInverting
NoninvertingNoninverting
OutputOutput
++
__ii((--))
ii(+)(+)
vvOO = A= Addvvidid
RROOAA
RRii
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The Four Amplifier TypesThe Four Amplifier Types
Description
Gain
Symbol
Transfer
Function
Voltage Amplifier
or
Voltage Controlled Voltage Source (VCVS)
Av vo/vin
Current Amplifier
or
Current Controlled Current Source (ICIS)
Ai io/iin
Transconductance Amplifier
or
Voltage Controlled Current Source (VCIS)
gm
(siemens)io
/vin
Transresistance Amplifier
or
Current Controlled Voltage Source (ICVS)
rm(ohms)
vo/iin
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VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryNoninverting ConfigurationNoninverting Configuration
++
__
vvinin
++++
--
vvOO
vvidid
ii(+)(+)
ii((--))
iiOO
iiFF
RRFF RRLL
RR11
ii11
vvidid
= v= voo/A/A
OLOLAssuming AAssuming AOLOL
vvidid =0=0
Also, with theAlso, with the
assumption that Rassumption that Rinin ==
ii(+)(+) = i= i((--)) = 0= 0
__vvFF
++
__
vv11
++
__
vvLL
++
__
iiLL
Applying KVL theApplying KVL the
following equationsfollowing equations
can be found:can be found:
vv11 = v= vinin
vvOO = v= v11 + v+ vFF = v= vinin+ i+ iFFRRFF
This means that,This means that,
iiFF = i= i11
Therefore: iTherefore: iFF = v= vinin/R/R11
Using the equation to the left the outputUsing the equation to the left the output
voltage becomes:voltage becomes:
vvoo = v= vinin + v+ vininRRFF = v= vinin RRFF + 1+ 1
RR11 RR11
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VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryNoninverting Configuration ContinuedNoninverting Configuration Continued
The closedThe closed--loop voltage gain is symbolized by Aloop voltage gain is symbolized by Avv and is found to be:and is found to be:AAvv = v= voo = R= RFF + 1+ 1
vvinin RR11
The original closed loop gain equation is:The original closed loop gain equation is:
AAvv = A= AFF = A= AOLOL
1 + A1 + AOLOLFF
Ideally AIdeally AOLOL , Therefore A, Therefore Avv = 1= 1
FF
Note: The actual value of ANote: The actual value of AOLOL is given for the specific device andis given for the specific device and
usually ranges from 50kusually ranges from 50k 500k.500k.
FF is the feedback factor and by assuming openis the feedback factor and by assuming open--loop gain is infinite:loop gain is infinite:
FF = R= R11
RR11 + R+ RFF
AAFF is the amplifieris the amplifier
gain withgain withfeedbackfeedback
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VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryNoninverting Configuration ContinuedNoninverting Configuration Continued
Input and Output ResistanceInput and Output ResistanceIdeally, the input resistance for this configuration is infinity, but the aIdeally, the input resistance for this configuration is infinity, but the a
closer prediction of the actual input resistance can be found with thecloser prediction of the actual input resistance can be found with the
following formula:following formula:
RRinFinF = R= Rinin (1 +(1 +FFAAOLOL)) Where RWhere Rinin is given for theis given for the
specified device. Usually Rspecified device. Usually Rinin isisin the Min the M;; range.range.
Ideally, the output resistance is zero, but the formula below gives aIdeally, the output resistance is zero, but the formula below gives a
more accurate value:more accurate value:
RRoFoF = R= Roo Where RWhere Roo is given for theis given for theFFAAOLOL + 1+ 1 specified device. Usually Rspecified device. Usually Roo is inis in
the 10the 10ss ofof;;ss range.range.
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VCVS (Voltage Amplifier)VCVS (Voltage Amplifier)Noninverting Configuration ExampleNoninverting Configuration Example
++
__
vvinin++
++
--
vvOO
vvidid
ii(+)(+)
ii((--))
iiOO
iiFF
RRFF RRLL
RR11
ii11
__vvFF
++
__
vv11
++
__
vvLL++
__
iiLL
Given:Given: vvinin = 0.6V, R= 0.6V, RFF = 200 k= 200 k;;RR11 = 2 k= 2 k;; , A, AOLOL = 400k= 400k
RRinin = 8 M= 8 M ;; , R, Roo = 60= 60 ;;
Find: vFind: voo , i, iFF , A, Avv ,, FF , R, RinFinF and Rand RoFoF
Solution:Solution:
vvoo = v= vinin + v+ vininRRFF = 0.6 + 0.6*2x10= 0.6 + 0.6*2x1055 == 60.6 V60.6 V iiFF = v= vinin = 0.6 == 0.6 = 0.3 mA0.3 mA
RR11 20002000 RR11 20002000
AAvv = R= RFF + 1 = 2x10+ 1 = 2x1055
+ 1 =+ 1 = 101101 FF = 1 = 1 == 1 = 1 = 9.9x109.9x10--33
RR11 20002000 AAOLOL 101101
RRinFinF = R= Rinin (1 +(1 +FFAAOLOL) = 8x10) = 8x1066 (1 + 9.9x10(1 + 9.9x10--33*4x10*4x1055) =) = 3.1688x103.1688x101010 ;;
RRoFoF = R= Roo = 60= 60 == 0.0150.015 ;;
FFAAOLOL + 1 9.9x10+ 1 9.9x10--33
*4x10*4x1055
+ 1+ 1
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VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryInverting ConfigurationInverting Configuration
++
__
RRLL
vvOO
++
--
vvinin
++
__
RR11ii11
RRFFiiFF
The sameThe same
assumptions used toassumptions used tofind the equations forfind the equations for
the noninvertingthe noninverting
configuration areconfiguration are
also used for thealso used for the
invertinginvertingconfiguration.configuration.
General Equations:General Equations:
ii11 = v= vinin/R/R11
iiFF = i= i11
vvoo == --iiFFRRFF == --vvininRRFF/R/R11
AAvv = R= RFF/R/R11 FF = R= R11/R/RFF
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Input and Output ResistanceInput and Output ResistanceIdeally, the input resistance for this configuration is equivalent to RIdeally, the input resistance for this configuration is equivalent to R11..
However, the actual value of the input resistance is given by theHowever, the actual value of the input resistance is given by the
following formula:following formula:
RRinin = R= R11 + R+ RFF
1 + A1 + AOLOL
Ideally, the output resistance is zero, but the formula below gives aIdeally, the output resistance is zero, but the formula below gives a
more accurate value:more accurate value:
RRoFoF = R= Roo
1 +1 +FFAAOLOL
Note:Note: FF = R= R11 This is different from the equation usedThis is different from the equation used
RR11 + R+ RFF on the previous slide, which can be confusing.on the previous slide, which can be confusing.
VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryInverting Configuration ContinuedInverting Configuration Continued
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VCVS (Voltage Amplifier)VCVS (Voltage Amplifier)Inverting Configuration ExampleInverting Configuration Example
++
__
RRLL
++
--
vvinin
++
__
RR11ii11
RRFFiiFF
Given:Given: vvinin = 0.6 V, R= 0.6 V, RFF = 20 k= 20 k;;RR11 = 2 k= 2 k;; , A, AOLOL = 400k= 400k
RRinin = 8 M= 8 M ;; , R, Roo = 60= 60 ;;
Find: vFind: voo , i, iFF , A, Avv ,, FF , R, RinFinF and Rand RoFoFvvOO
Solution:Solution:
vvoo == --iiFFRRFF == --vvininRRFF/R/R11 == --(0.6*20,000)/2000 =(0.6*20,000)/2000 = 12 V12 V
iiFF = i= i11 == vvinin/R/R11 = 1 / 2000 == 1 / 2000 = 0.50.5 mAmA
AAvv = R= RFF/R/R11 = 20,000 / 2000 == 20,000 / 2000 = 1010 FF = R= R11/R/RFF = 2000 / 20,000 == 2000 / 20,000 = 0.10.1
RRinin = R= R11 + R+ RFF = 2000 + 20,000= 2000 + 20,000 == 2,000.052,000.05 ;;
1 + A1 + AOLOL 1 + 400,0001 + 400,000
RRoFoF = R= Roo = 60= 60 == 1.67 m1.67 m ;;
1 +1 +FFAAOLOL 1 + 0.09*400,0001 + 0.09*400,000 Note:Note: FF is 0.09 because usingis 0.09 because usingdifferent formula than abovedifferent formula than above
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ICIS (Current Amplifier) SummaryICIS (Current Amplifier) Summary Not commonly done using operational amplifiersNot commonly done using operational amplifiers
++
__LoadLoad
iiinin
iiLL
Similar to the voltageSimilar to the voltage
follower shown below:follower shown below:
Both these amplifiers haveBoth these amplifiers have
unity gain:unity gain:
AAvv = A= Aii = 1= 1
++
__
iiinin = i= iLL
vvinin = v= voovvinin++
__ ++
--
vvOO
Voltage FollowerVoltage Follower
1 Possible1 Possible
ICISICIS
OperationalOperational
AmplifierAmplifier
ApplicationApplication
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VCIS (Transconductance Amplifier) SummaryVCIS (Transconductance Amplifier) SummaryVoltage to Current ConverterVoltage to Current Converter
++
__
LoadLoad
iiLL
RR11ii11
vvinin
++
__
OROR++
__
LoadLoad
iiLL
RR11ii11
vvinin
++
__
vvinin++
__
General Equations:General Equations:
iiLL = i= i11 = v= v11/R/R11
vv11 = v= vinin
The transconductance, gThe transconductance, gmm = i= ioo/v/vinin = 1/R= 1/R11
Therefore, iTherefore, iLL = i= i11 = v= vinin/R/R11 = g= gmmvvinin
The maximum load resistance is determined by:The maximum load resistance is determined by:
RRL(max)L(max) = v= vo(max)o(max)/i/iLL
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VCIS (Transconductance Amplifier)VCIS (Transconductance Amplifier)Voltage to Current Converter ExampleVoltage to Current Converter Example
++
__
LoadLoad
iiLL
RR11ii11
vvinin
++
__
Given: vGiven: vinin = 2 V, R= 2 V, R11 = 2 k= 2 k;;vvo(max)o(max) == 10 V10 V
Find: iFind: iLL , g, gmm and Rand RL(max)L(max)
Solution:Solution:
iiLL = i= i11 = v= vinin/R/R11 = 2 / 2000 == 2 / 2000 = 1 mA1 mA
ggmm = i= ioo/v/vinin = 1/R= 1/R11 = 1 / 2000 == 1 / 2000 = 0.5 mS0.5 mS
RRL(max)L(max) = v= vo(max)o(max)/i/iLL = 10 V / 1 mA= 10 V / 1 mA
== 10 k10 k ;;
Note:Note:
If RIf RLL > R> RL(max)L(max) the op ampthe op ampwill saturatewill saturate
The output current, iThe output current, iLL isis
independent of the loadindependent of the load
resistance.resistance.
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VCIS (Transresistance Amplifier) SummaryVCIS (Transresistance Amplifier) SummaryCurrent to Voltage ConverterCurrent to Voltage Converter
General Equations:General Equations:
iiFF = i= iinin
vvoo == --iiFFRRFF
rrmm = v= voo/i/iinin = R= RFF
++
__
iiFF
iiinin
RRFF
vvOO
++
--
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VCIS (Transresistance Amplifier) SummaryVCIS (Transresistance Amplifier) SummaryCurrent to Voltage ConverterCurrent to Voltage Converter
Transresistance Amplifiers are used for lowTransresistance Amplifiers are used for low--powerpowerapplications to produce an output voltage proportional toapplications to produce an output voltage proportional to
the input current.the input current.
Photodiodes and Phototransistors, which are used in thePhotodiodes and Phototransistors, which are used in the
production of solar power are commonly modeled asproduction of solar power are commonly modeled as
current sources.current sources.
Current to Voltage Converters can be used to convert theseCurrent to Voltage Converters can be used to convert these
current sources to more commonly used voltage sources.current sources to more commonly used voltage sources.
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VCIS (Transresistance Amplifier)VCIS (Transresistance Amplifier)Current to Voltage Converter ExampleCurrent to Voltage Converter Example
++
__
iiFF
iiinin
RRFF
vvOO
++
--
Given: iGiven: iinin = 10 mA= 10 mA
RRFF = 200= 200 ;;
Find: iFind: iFF , v, voo and rand rmm
Solution:Solution:
iiFF = i= iinin == 10 mA10 mA
vvoo == --iiFFRRFF = 10 mA * 200= 10 mA * 200 ;; == 2 V2 V
rrmm = v= voo/i/iinin = R= RFF == 200200
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Power BandwidthPower BandwidthThe maximum frequency at which a sinusoidal output signal can beThe maximum frequency at which a sinusoidal output signal can be
produced without causing distortion in the signal.produced without causing distortion in the signal.
The power bandwidth, BWThe power bandwidth, BWpp is determined using the desiredis determined using the desired
output signal amplitude and the the slew rate (output signal amplitude and the the slew rate (see next slidesee next slide))
specifications of the op amp.specifications of the op amp.
BWBWpp = SR= SR
22TTVVo(max)o(max)
SR = 2SR = 2TTffVVo(max)o(max) where SR is the slew ratewhere SR is the slew rate
Example:Example:
Given: VGiven: Vo(max)o(max) = 12 V and SR = 500 kV/s= 12 V and SR = 500 kV/s
Find: BWFind: BWpp
Solution:Solution: BWBWpp == 500 kV/s500 kV/s = 6.63 kHz= 6.63 kHz
22TT * 12 V* 12 V
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Slew RateSlew RateA limitation of the maximum possible rate of change of theA limitation of the maximum possible rate of change of the
output of an operational amplifier.output of an operational amplifier.
As seen on the previous slide,As seen on the previous slide, This is derived from:This is derived from:
SR = 2SR = 2TTffVVo(max)o(max) SR =SR = vvoo//ttmaxmax
Slew Rate is independent of theSlew Rate is independent of the
closedclosed--loop gain of the op amp.loop gain of the op amp.
Example:Example:
Given: SR = 500 kV/s andGiven: SR = 500 kV/s and vvoo = 12 V (Vo(max) = 12V)= 12 V (Vo(max) = 12V)
Find: TheFind: The t and f.t and f.
Solution:Solution: t =t = vo / SR = (10 V) / (5x10vo / SR = (10 V) / (5x1055 V/s) = 2x10V/s) = 2x10--55 ss
f = SR /f = SR / 22TTVVo(max)o(max) == (5x10(5x1055 V/s) / (V/s) / (22TT * 12) = 6,630 Hz* 12) = 6,630 Hz
f is thef is the
frequency infrequency in
HzHz
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Slew Rate DistortionSlew Rate Distortion
vv
tt
desired outputdesired output
waveformwaveform
actual outputactual output
because ofbecause of
slew rateslew rate
limitationlimitation
tt
vv
The picture above shows exactly what happens when theThe picture above shows exactly what happens when the
slew rate limitations are not met and the output of theslew rate limitations are not met and the output of the
operational amplifier is distorted.operational amplifier is distorted.
SR =SR = v/v/t = m (slope)t = m (slope)
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Noise GainNoise GainThe noise gain of an amplifier is independent of the amplifiersThe noise gain of an amplifier is independent of the amplifiers
configuration (inverting or noninverting)configuration (inverting or noninverting)
The noise gain is given by the formula:The noise gain is given by the formula:
AANN = R= R11 + R+ RFF
RR11
Example 1: Given a noninverting amplifier with the resistanceExample 1: Given a noninverting amplifier with the resistance
values, Rvalues, R11 = 2 k= 2 k;; and Rand RFF = 200 k= 200 k;;Find: The noise gain.Find: The noise gain.
AANN == 2 k2 k;; + 200 k+ 200 k;; = 101= 101 Note: For theNote: For the
2 k2 k;; noninverting amplifier Anoninverting amplifier ANN = A= AVV
Example 2: Given an inverting amplifier with the resistanceExample 2: Given an inverting amplifier with the resistancevalues, Rvalues, R11 = 2 k= 2 k;; and Rand RFF = 20 k= 20 k;;
Find: The noise gain.Find: The noise gain.
AANN == 2 k2 k;; + 20 k+ 20 k;; = 12= 12 Note: For theNote: For the
2 k2 k;; inverting amplifier Ainverting amplifier ANN > A> AVV
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GainGain--Bandwidth ProductBandwidth Product
In most operational amplifiers, the openIn most operational amplifiers, the open--loop gain beginsloop gain begins
dropping off at very low frequencies. Therefore, to make thedropping off at very low frequencies. Therefore, to make theop amp useful at higher frequencies, gain is traded forop amp useful at higher frequencies, gain is traded for
bandwidth.bandwidth.
The GainThe Gain--Bandwidth Product (GBW) is given by:Bandwidth Product (GBW) is given by:
GBW = AGBW = ANNBWBW
Example: For a 741 op amp, a noise gain of 10 k correspondsExample: For a 741 op amp, a noise gain of 10 k corresponds
to a bandwidth of ~200 Hzto a bandwidth of ~200 Hz
Find: The GBWFind: The GBW
GBW = 10 k * 200 Hz = 2 MHzGBW = 10 k * 200 Hz = 2 MHz
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Cascaded AmplifiersCascaded Amplifiers -- BandwidthBandwidth
Quite often, one amplifier does not increase the signal enoughQuite often, one amplifier does not increase the signal enough
and amplifiers are cascaded so the output of one amplifier is theand amplifiers are cascaded so the output of one amplifier is theinput to the next.input to the next.
The amplifiers are matched so:The amplifiers are matched so:
BWBWSS = BW= BW11 = BW= BW22 == GBWGBW where, BWwhere, BWSS is the bandwidth of allis the bandwidth of all
AANN the cascaded amplifiers and Athe cascaded amplifiers and ANN isis
the noise gainthe noise gain
The Total Bandwidth of the Cascaded Amplifiers is:The Total Bandwidth of the Cascaded Amplifiers is:
BWBWTT = BW= BWss(2(21/n1/n 1)1)1/21/2 where n is the number of amplifierswhere n is the number of amplifiers
that are being cascadedthat are being cascaded
Example: Cascading 3 Amplifiers with GBW = 1 MHz and AExample: Cascading 3 Amplifiers with GBW = 1 MHz and ANN = 15,= 15,Find: The Total Bandwidth, BWFind: The Total Bandwidth, BWTT
BWBWSS = 1 MHz / 15 = 66.7 kHz= 1 MHz / 15 = 66.7 kHz
BWBWTT = 66.7 kHz (2= 66.7 kHz (21/31/3 1)1)1/21/2 = 34 kHz= 34 kHz
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CommonCommon--Mode Rejection RatioMode Rejection Ratio
The commonThe common--mode rejection ratio (CMRR) relates to the ability ofmode rejection ratio (CMRR) relates to the ability of
the op amp to reject commonthe op amp to reject common--mode input voltage. This is verymode input voltage. This is very
important because commonimportant because common--mode signals are frequentlymode signals are frequentlyencountered in op amp applications.encountered in op amp applications.
CMRR = 20 log|ACMRR = 20 log|ANN/ A/ Acmcm||
AAcmcm == AANN
loglog--11
(CMRR / 20)(CMRR / 20)We solve for AWe solve for Acmcm because Op Amp data sheets list the CMRR value.because Op Amp data sheets list the CMRR value.
The commonThe common--mode input voltage is an average of the voltages thatmode input voltage is an average of the voltages that
are present at the nonare present at the non--inverting and inverting terminals of theinverting and inverting terminals of the
amplifier.amplifier.vvicmicm = v= v(+)(+) + v+ v((--))
22
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CommonCommon--Mode Rejection RatioMode Rejection RatioExampleExample
Given: A 741 op amp with CMRR = 90 dB and a noise gain,Given: A 741 op amp with CMRR = 90 dB and a noise gain,AANN = 1 k= 1 k
Find: The common mode gain, AFind: The common mode gain, Acmcm
AAcmcm == AANN = 1000= 1000
loglog--11 (CMRR / 20)(CMRR / 20) loglog--11 (90 / 20)(90 / 20)
= 0.0316= 0.0316
It is very desirable for the commonIt is very desirable for the common--mode gain to be small.mode gain to be small.
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Power Supply Rejection RatioPower Supply Rejection Ratio
One of the reasons op amps are so useful, is that they canOne of the reasons op amps are so useful, is that they can
be operated from a wide variety of power supply voltages.be operated from a wide variety of power supply voltages.
The 741 op amp can be operated from bipolar suppliesThe 741 op amp can be operated from bipolar supplies
ranging fromranging from 5V to5V to 18V with out too many changes to the18V with out too many changes to the
parameters of the op amp.parameters of the op amp.
The power supply rejection ratio (SVRR) refers to the slightThe power supply rejection ratio (SVRR) refers to the slight
change in output voltage that occurs when the powerchange in output voltage that occurs when the power
supply of the op amp changes during operation.supply of the op amp changes during operation.
SVRR = 20 log (SVRR = 20 log (VVss// VVoo))
The SVRR value is given for a specified op amp. For theThe SVRR value is given for a specified op amp. For the
741 op amp, SVRR = 96 dB over the range741 op amp, SVRR = 96 dB over the range 5V to5V to 18V.18V.
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OpenOpen--Loop Op Amp CharacteristicsLoop Op Amp CharacteristicsTable 12.1Table 12.111
Device LM741C LF351 OP-07 LH0003 AD549K
Technology BJT BiFET BJTHybrid
BJTBiFET
AOL(typ)
200 k 100 k 400 k 40 k 100 k
Rin 2 M; 1012 ; 8 M; 100 k; 1013; || 1 pF
Ro 50 ; 30 ; 60 ; 50 ; ~100 ;
SR 0.5 V/Qs 13 V/Qs 0.3 V/Qs 70 V/Qs 3 V/Qs
CMRR 90 dB 100 dB 110 dB 90 dB 90 dB
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SourcesSources
Dailey, Denton.Dailey, Denton. Electronic Devices and Circuits, Discrete and Integrated.Electronic Devices and Circuits, Discrete and Integrated. Prentice Hall, NewPrentice Hall, New
Jersey: 2001. (pp 456Jersey: 2001. (pp 456--509)509)
11Table 12.1: Selected Op Amps and Their Open Loop Characteristics, pg 457Table 12.1: Selected Op Amps and Their Open Loop Characteristics, pg 457
Liou, J.J. and Yuan, J.S. Semiconductor Device Physics and Simulation. Plenum Press,Liou, J.J. and Yuan, J.S. Semiconductor Device Physics and Simulation. Plenum Press,
New York: 1998.New York: 1998.
Neamen, Donald.Neamen, Donald. Semiconductor Physics & Devices. Basic Principles.Semiconductor Physics & Devices. Basic Principles. McGrawMcGraw--Hill,Hill,
Boston: 1997. (pp 351Boston: 1997. (pp 351--357)357)
Web SourcesWeb Sources
www.infoplease.com/ce6/sci/A0803814.htmlwww.infoplease.com/ce6/sci/A0803814.html
http://www.infoplease.com/ce6/sci/A0836717.htmlhttp://www.infoplease.com/ce6/sci/A0836717.html
http://people.msoe.edu/~saadat/PSpice230Part3.htmhttp://people.msoe.edu/~saadat/PSpice230Part3.htm
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