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7/23/2019 SECTION 5 Advanced Opamp Circuits
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SECTION5:ADVANCEDOPAMPCIRCUITS
MAE3055 MechetronicsII
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Introduction2
K.Webb MAE3055 MechetronicsII
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MoreOpampCircuits Opampcircuitscanbebroadlygroupedintotwocategories:linear
circuitsandnonlinearcircuits.
Inthefirstsectionofnotes,welookedatlinearopampcircuitsthoseemployingnegativefeedbacktoprovidelinearamplificationofaninputsignal.
Inthissectionofnotes,wewilltakealookatacoupleofexamples
ofnonlinear
opamp
circuits
thosethatemploypositivefeedback
(ornofeedback)toproduceoutputsthatswitchbetweenthepositiveandnegativelimits.
Thefirstandmostimportantofthesearecomparators.
Wewillalsolookatafewothertypesofopampcircuits: Activefilters
Instrumentationamplifiers
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Comparators4
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OpenLoopOpampBehavior Inthefirstsectionofnotes,welookedatopampamplifiercircuits
Closedloopconfiguration
Negativefeedback Outputremainswithintheopamps linearoutputrange
Wellnowtakealookatwhathappenswhenyouuseanopampopenloop withoutfeedback orwithpositivefeedback Idealopamphasinfinitegain,soforanynonzerodifferentialinput
voltage,theoutputwanttobeV CantgotoV limits,orsaturates,somewherenearthesupply
voltages
Saturationvoltageisnotatthesupplyvoltages,butforsimplicityinthissetofnotes,welltypicallyassumethattheopampoutputcanswingbetweenthepositiveandnegativesupplyvoltages
Opampswhoseoutputscanswingalltheway(almost)tothesuppliesarecalledrailtorailopamps
Gainissohigh(ideally,infinite)thatoutputwillalwaysbesaturatedhighorlow
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Comparators
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Inanopenloopconfiguration,theoutputofanopampisdeterminedbytherelative
valueofthetwoinputs.
V+,V Vid Vo
V+ V > 0V +V
Thedifferentialopampinputisthedifferencebetweenthevoltageatthetwoinput
terminals:idV V V
If ,V V then 0 ,idV V and oV V
If ,V V then 0 ,idV V and oV V
Theopampcomparesthetwoinputvoltagesandgeneratesanoutputbasedon
whichinputishigher itisactingasacomparator.
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Comparators
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Theoutputofacomparatorswitchesbetweentwostates theupperandlower
outputlimits dependingontherelativevaluesoftheinputvoltages:
ComparatorInputs
ComparatorOutputs
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Comparators ExampleUses
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Comparatorsareusedtocompareonevoltagewithanother.
Forexample,thermostat,usedtoturnaheatingsystemonandoff: Invertinginputconnected toatemperaturesensor
Noninvertinginputconnectedtoavariablereferencevoltagedeterminedby
thetemperaturesetpoint Outputishighwhentheroom
temperatureisbelowthesetpoint temperature
Outputislowwhenroom
temperatureisabovethe
setpoint temperature
Anotherexampleoftheuseofacomparatorisamotionsensinglight:
Oneinputcomesfromamotionsensoroutput(analogvoltage)
Otherinputisathresholdvoltagesetbythesensitivitysetting
Wantlighttoturnonforpeople,cars,notinsects,birds
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ComparatorsandNoise9
Considerthefollowingcomparatorcircuit:
Invertinginput
(V)
is
driven
by
a
noiseless,1KHzsinusoidalsignal
Noninvertinginput(V+),thethreshold
voltage,isconnectedtoground(0V)
Output,Vo,switchesasexpected Wheninputishigherthanthe
thresholdvoltage(V >V+),outputis
low,Vo=12V,andviceversa
Zoomedinviewofasinglethresholdcrossing
Voswitchescleanlyatthepointwhere
theinputcrossesthreshold
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ComparatorsandNoise10
Samecircuitandsourceasbefore
Now,theinputsignaliscorruptedbynoise
Thisisamorerealisticscenario thereis
alwaysnoise
Inputtraceisvisiblyfatternoisier Outputedgeslookfattertoo
Somethingishappeningattheoutput
transitions takeacloserlook
Zoomingin,noiseoninputisvisible Noisyinputtransitionsbackandforth
acrossthethreshold
Voswitchesateachthresholdcrossing
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Hysteresis SchmittTrigger11
Thiscomparatorcircuitusesfeedbacktogenerate
hysteresis aphenomenainwhichthecharacteristics
of
a
system
are
dependent
on
its
previous
states.
Lookcloselynotanoninvertingamplifier
feedbackispositive
Thresholdvoltage,V+,isnolongerconstant
dependentontheoutputvoltage:
Theoutputcanassumeoneoftwostates,
+12Vor12V,sothethresholdvoltagewillbe:
2
1 2o
R
V V R R
2
1 2
122
hystVR
V VR R
Iftheinputislow,theoutputwillbehigh,andthethresholdvoltagewillbea
(relativelysmall,typically)positivevalue:V+ =+Vhyst/2 Astheinputrisesandtransitionsthroughthethresholdvoltage,+Vhyst/2,the
outputwillswitchto12V,andthethresholdvoltagewillmoveawayfromthe
risinginputsignaltoVhyst/2.theoppositeoccursfornegativeinputtransitions.
Themagnitudeof+Vhyst/2willdeterminethecomparatorssensitivitytonoise.
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ComparatorsandNoiseHysteresis12
Nowweveaddedhysteresistodesensitizethe
comparatortothenoiseontheinputsignal
Acomparatorwithhysteresisiscalleda
Schmitttrigger
R1andR2selectedtoprovideadequate
hysteresisfortheamountofnoisepresent
Inputisstillnoisy
Outputedgesnowlookcleaner
ThresholdvoltageswitcheswithVo
Single,cleanoutputtransition
Thresholdvoltageswitchesbetween
Vhyst/2movesawayfromnoisyinput
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ComparatorsandNoiseHysteresis13
K.Webb MAE3055 MechetronicsII
400hystV mV
Hysteresisismeasuredas
thefullpeaktopeak
swingofthethreshold
voltage
Hysteresisistwicethe
magnitudeofthefeedbacksignal:
2
1 2
2maxhyst o
RV V
R R
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ComparatorsandNoiseHysteresis14
K.Webb MAE3055 MechetronicsII
AnotherwaytoquantifyhysteresisistoperformabidirectionalDCsweepofthe
input lowtohigh,thenhightolow whilemonitoringtheoutput:
Vin
increasing
Vin
decreasing
Outputtracesadifferent
path,dependentupon
directionofVin increasing
ordecreasing Thresholdvoltagedepends
onhistoryofVin
Thresholdvoltagesindicated
byverticalportionsoftheVo
trace Hysteresisgivenby
differencebetweentwo
thresholdlevels
Vhyst=400mV
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ComparatorwithAdjustableHysteresis15
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Ifthelevelofnoiseattheinputtoacomparatorisnotknownaheadoftime,
amountofhysteresiscanbemadeadjustablebyincludingapotentiometer(variableresistor)inthefeedbackpath.
Magnitudeofthefeedbacksignal portionoftheoutputthatisfedback isvaried
byvaryingtheresistanceofthepotentiometer
R2
isadjustablebetween0and
somemaximumvalue,Rmax.
WhenR2=0:
WhenR2=Rmax:
2
1 22maxhyst o
R
V V R R
1
02 0
0maxhyst oV V
R
max
1 max
2maxhyst o
RV V
R R
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SchmittTriggerExampleConsiderthefollowingscenario:
Youhaveatestengineinstrumentedandrunningonadynamometerinthelab.
Youwanttogenerateasignalthatwillilluminateawarninglight(LED)iftheenginetemperatureexceedssomeupperlimit.
YouhaveanRTD(resistivetemperaturedetectororresistivethermaldevice)installedtomeasurecoolanttemperature.
TheRTDisbiasedsuchthata0Voutputcorrespondstothethresholdtemperaturewewishtodetect.
NoiseontheRTDsignalisapproximately100mVpp
Anopampisavailableforuseasacomparator
Opampsuppliesare5V
Opampoutputssaturateat+VCC 500mVandVCC+500mV
Maximumopampoutputcurrentisonly10mA
Whenon,theLEDsinks9mA
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SchmittTriggerExample
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DesignaSchmitttriggertosatisfythecriteriaonthepreviousslide.Theentirecircuit
isshownbelow.
WhentheRTDoutputexceeds0VtheSchmitttriggeroutputwillgolowand
theLEDwillbebiasedwith9mAandwillilluminate
NotethattheRTDitselfisjustaresistor;itsbiasnetworkisnotshown.
Currentflowingthroughthehysteresisfeedbacknetworkmustbelimitedto
1mA
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SchmittTriggerExample
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DesignaSchmitttriggertosatisfythecriteriaonthepreviousslide.
Requiredhysteresisvalueisatleastthepeaktopeaknoisevalue.Wellset
hysteresistobe150mV:
Theoutputsaturatesat4.5V,so
Opampoutputcurrent,excludingLEDbiascurrent,willbe
Choosingastandardresistorvalueof5.1KforR1,wecanthencalculateR2:
2
1 2
2 150maxhyst o
RV V mV
R R
3 32 2 1
1 2
15016.67 10 16.95 102 4.5
R mVR R
R R V
1 2
1 2
4.51 4.5oI mA R R K
R R
3 32 116.95 10 16.95 10 5.1 86.4R R K
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SchmittTriggerExample
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Choosingthenearest(larger,soasnottodecreasehysteresisvalue)standard
resistorvalueforR2gives:
Recalculatingtheexpectedhysteresis:
Andcheckingthatthecurrentthroughthefeedbacknetworkdoesnotexceed1mA:
TheresultingSchmitttriggercircuit:
1 25.1 91,R K R
2
1 2
912 2 4.5 157
5.1 91maxhyst o
RV V V mV
R R K
1 2
4.5867 1
5191maxo
o
V VI mAA
R R
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OpampRelaxationOscillator20
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Oscillators Oscillatorsarecircuitsorcomponentsthatprovideanoutputvoltagethat
oscillatesatacertain(possiblyadjustable)frequency
Oscillatingoutputmaybesinusoidal orsomethingapproximating
sinusoidal oritmaybeasquarewave
Crystaloscillatorsareusedforhighperformanceapplications(e.g.clock
onthemotherboardinyourPC,oscillatorinmobilephonetransceiver,
etc.)
LowperformanceapplicationsmayuseoscillatorICs,suchasthe555
timerIC
Canalsobuildanoscillatorusinganopamp possiblyusefulinsome
circumstances mostlyjustaninterestingcircuitthatwillaidinyou
understandingofopamps,feedback,stepresponseofRCcircuits,andthe
basicprinciplesbehindthefunctioningofmanytypesofoscillators
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OpampRelaxationOscillator
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Thefollowingcircuitiscalledarelaxationoscillator relaxationbecauseits
oscillationisduetothecharginganddischargingofacapacitor.
Thisisafairlytrickycircuittoanalyze.Letsbeginbynotingafewimportant
propertiesandcharacteristicsofthecircuit:
ThereispositivefeedbackprovidedbyR1andR2,so
wellassumetheoutputisalwayssaturatedatV
ThereisalsoanegativefeedbackpathfromVotoV Thevoltageatthenoninvertinginputisafunctionof
theoutputvoltage.Itstheoutputvoltagescaledby
thegainofthefeedbacknetwork.
Thevoltageattheinvertinginputisthevoltageacross
acapacitor.Itwantstochargeuptotheleveloftheoutputvoltage
IfV V+,thentheoutputwillbelow,Vo=V
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OpampRelaxationOscillator
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Toanalyzethebehaviorofthecircuit,letusbeginwiththeassumptionthatthe
outputwillswitchbackandforthbetweenV(itisanoscillator,afterall).Well
assumeanoutputstateandworkourwaythroughthedifferentcircuitnodes:
Assumethatatt=0theoutputhasjustswitchedfrom
low,V,tohigh,+V
WhenVowaslow,thevoltageatthenoninverting
inputwas:
ThetransitionmusthaveoccurredbecauseV
transitionedfrombeinghigherthantolowerthanV+.
So,attheswitchinginstant:
AssoonasVoswitchesfromlowtohigh,thecapacitor
voltage,V,beginstochargetoward+V,and:
2
1 2
RV VR R
V
V V
V V
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OpampRelaxationOscillator
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AstheoutputswitchesbackandforthbetweenV,twothingshappen
Thenoninvertinginput,V+,switchesbetweenV
Thecapacitorvoltageattheinvertinginputchargesanddischargesbetween
VataratedeterminedbytheRCtimeconstantofthenegativefeedback
network
OutputswitchingoccurseachtimeV reachesV
EachtimeVoswitches,V ischarging/dischargingtowardV
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OpampOscillatorfosc
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Bynowitshouldbeclearthatthefrequencyofoscillationwillberelatedtotherate
atwhichthecapacitorchargesanddischargesbetweentheswitchingpointsatV.
WewillnowcalculatethefrequencyofoscillationtofindthatitisafunctionofboththeRCtimeconstantofthenegativefeedbacknetworkandthegainofthepositive
feedbacknetwork,.
Againassumethatatt=0,VoswitchesfromVto+V.
RecallthatthestepresponseofafirstorderRCcircuit
isgivenby:
Att=0,V= V,andV thenbeginschargingtoward
+V,so:
Thechargingofthecapacitorfort>0isthengivenby:
t
RCF I FV t V V V e
,I FVV V V
1t t
RC RCV t V V V e V eV
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OpampOscillatorfosc
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Thecapacitorvoltageneverreachesitsdestinationof+V,becauseonceitreaches
+Vtheoutputswitchesagain,andthecapacitorbeginsdischarging
Thecapacitorvoltagereaches+Vatt=T/2,whereTistheperiodof
oscillation.So,
212
T
RCVT
V V V e
T/2
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OpampOscillatorfosc
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Solvingfortheperiodofoscillation,T:
21 1 1
21 2 1
ln ln1
T
RC T
RCC
e TR
T/2
Thefrequencyofoscillation,fosc,is: 1
l1
n 1
2
osc
f RC
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OpampIntegratorsand
Differentiators28
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IntegratorsandDifferentiators Opampscanbeusedtobuildcircuitsthatperform
manydifferentmathematicaloperations hencethenameoperationalamplifiers
Wevealreadyseenopampscanbeusedtoaddandsubtractelectricalsignals
Theycanalsobeusedtoperformintegrationanddifferentiationofelectricalsignals Especiallyimportantwhenbuildingfeedbackcontrol
systems theverycommonproportionalintegral
derivative(PID)controllercanbeimplementedwithasimpleopampcircuit
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OpampIntegrator
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Analyzetheopampintegrator:
Thereisnegativefeedbackaroundthe
opamp,sowecanassumethattheinputterminalvoltagesareequal:
Thecurrentthroughtheresistoris:
0V V V
( )
( )
iv t
i t R Theopamphasinfiniteinputimpedance,sonocurrentflowsintotheinverting
terminal,andallofthecurrentthroughtheresistorflowsontothecapacitor.
Theoutputvoltage,Vo,isthevoltageacrossthecapacitor:
0 0 0( )1 1 1 ( )( ( )) t t
io i
t
vv t i d d v d C C R RC
0
1( )) ( o i
t
v t v d RC
Theoutputisthe(invertedandscaled)
integraloftheinput:
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OpampIntegrator
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NowinsteadofbothZ1andZ2beingresistive,Z2iscapacitive:
Thefrequencyresponsefunctionofthecircuitis:
Acoupleofthingstonotice:1)thephaseisalways90,and2)gaindecreaseswith
frequencyandisinfiniteatDC! Mightthisposeaproblem?
2
1
o
i
v Z
v Z
1 2
1,Z Z
CR
j
21
1
190
C
R RC
Z jjH
Z RC
Wehavejustanalyzedtheopampintegratorcircuitin
thetimedomain.Itisalsopossibletoanalyzethe
circuitinthefrequencydomain:
Thiscircuitlooksalotlikethesimpleinverting
opampwelookedatinaprevioussectionofnotes.
Itcanbetreatedassuch:
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OpampIntegrator Example
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310 10
90
j
CH R
Theintegratorcircuittotherighthasthefollowing
freq.responsefunction:
anditsBodeplotis:
Thegainoftheintegratorcircuit
decreaseswithincreasing
frequency.
GainisinfiniteatDC capacitor
lookslikeanopencircuit(infinite
impedance)atDC. Phaseisalways90.Thinkof
integratingacosine resultisa
sine:a90 phaseshift,plusthe
inversiongives+90.
20dB/dec
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OpampIntegrator Example
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NowthatweveseentheBodeplotfortheintegrator,
letstakealookatitstimedomainresponsefora
coupleofdifferentinputsignals.First,a1KHzsinusoid:
0 0
3
3
110 10 cos 2
10 10sin 2 1.59si
12
1
n 21 1
t t
o i
o
KHz
KHz KHz
KH
v v dt t dt RC
v
z
t t
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OpampIntegrator Example
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Next,letstakealookattheintegratorsresponsetoa
DC(f=0Hz)input.Wellassumethattheinputis
switchedonatt=0,sothisisreallythecircuitsstepresponse.
fort0: ( ) 1 ( )i
v t V u t ( 1
( ) 11
)iv Vi t mAR K
t
and
Aconstantcurrentflowsontothecapacitor,andtheoutputisthe(negativeofthe)
voltageacrossthecapacitor: 31( ( )0
0 1.
) 1 01
o c
mI tt
Av t v
Ct t
F
Theoutputincreaseslinearlywithtime!Thisintegratorcircuitwillquicklysaturateif
aDCvoltageisappliedtotheinput.Thisiswhatwewouldexpectfromacircuitwith
infinitegainatDC.Thismaybeaproblem: SaywewantacircuitthatbehavesasanintegratorforsignalsintheKHzregion
IfthesignalshaveanyDCoffset(oriftheopampisnotideal,andhasanonzero
offsetvoltage morelater)theoutputwillsaturate.
Fortunately,thereisasolution
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ABetterOpampIntegrator
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Mainproblemwiththeidealintegratorcircuitisthe
veryhigh(infinite)DCgain
ThecircuitisessentiallyoperatingopenloopforDCthereisnoDCfeedback
Addaresistorinparallelwiththeintegratingcapacitor
DCgainislimitedtoRf/R
Stillbehavesasanintegratorathigherfrequencies
Treatthecircuitasasimpleinvertingamplifiertodeterminethefrequencyresponse:
21
ZH
Z
1Z R 2 1 1
f
f
ff
R
RjZ
j
j
C
R CR
C
where and
Thefreq.responsefunctionis:
1
1
f
f
RH
jR R C
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ABetterOpampIntegrator
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AtDC(=0),thecapacitoris
anopencircuit,andthegainissetbythetworesistors
As,thegain0
At=1/(RfC),thegainis3dB
Thisisalowpassfilter!
Aninvertingamplifierforfrequenciesbelowfc.
Anintegratorforfrequencies
wellabovefc.
IntegratorAmplifier
fc=159Hz
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OpampDifferentiator
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Analyzetheopampdifferentiator:
Thereisnegativefeedbackaroundthe
opamp,sowecanassumethattheinputterminalvoltagesareequal:
Thecurrentthroughthecapacitoris:
0V V V
( ) i
dv
i t C dt Nocurrentflowsintotheinvertingterminal,andallofthecurrentthroughthe
capacitorflowsthroughthefeedbackresistor.
Theoutputvoltage,Vo,isthevoltageacrossthefeedbackresistor:
( ) ( ) io
dvv t i t R RC dt
( ) iodv
v t RC dt
Theoutputisthe(invertedandscaled)
derivativeoftheinput:
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OpampDifferentiator Example
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Theintegratorcircuittotherighthasthefollowing
freq.resp.function:
anditsBodeplotis:
Thegainofthedifferentiator
circuitincreaseswithincreasing
frequency.
GainiszeroatDC capacitor
lookslikeanopencircuit(infinite
impedance)atDC.
Phaseisalways90.Thinkof
differentiatingasine resultisa
cosine:a+90 phaseshift,plus
theinversiongives90.
+20dB/dec
6100 10 90j RCH
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OpampDifferentiator Example
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NowthatweveseentheBodeplotforthe
differentiator,letstakealookatitstimedomain
responsetoa1KHzsinusoidalinputsignal:
6
6
100 10 cos 2
100 10 2 sin 2 0.628cos
1
1 91 2 01
io
o
KHdv d
v RC t dt
z
KHz
d
KHz H t z
t
v t K
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OpampActiveFilters41
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ActiveFilters
Inaprevioussectionofnoteswelookedatfirst andsecondorderpassivefiltersofdifferentvarietiesPassive,
becausetheycontainedonlypassivecomponents resistors,capacitors,andinductors
Itisalsopossibletoconstructfiltersusingopamp
circuits wecalltheseactivefilters
Activefiltershaveseveraladvantagesoverpassivefilters:CanbuildhighQfilterswithoutinductors
Lowoutput
impedance
Easilyadjustable fc,QFiltercanprovidepositive(>0dB)gain
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FirstOrderLowPassFilter
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Wevealreadyseenthatafirstorderlowpassresponse
canbeobtainedbyaddingafeedbackresistortoan
opampintegratorcircuit:
1
1
f
f
RH
jR R C
Thecornerfrequencyis:
2
1c
fC
fR
Magnituderesponsetothe
rightisforRf=R.Responsewouldshiftupordownby
20log fR
R
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FirstOrderHighPassFilter
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Justastheintegratorwastransformedintoalow
passfilterwiththeadditionofasingleresistor,a
differentiatorcircuitcanbesimilarlytransformedtoahighpassfilter.
Onceagainwelldeterminethiscircuitsfrequency
responsebytreatingitasaninvertingamplifier:
21
ZHZ
1 1 1Rj C
Z
2 2Z Rwhere and
Thefreq.responseis:
21 1
C
C
j R
RH
j
withcorner(3dB)frequency:
12
1c
Cf
R
andpassbandgain:
21
c
R
RH
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HigherOrderActiveFilters
Higherorderactivefilterscanbeconstructedinacoupleofdifferentways:CascadingfirstorderactivefiltersUsingsecondorderactivefilterstages(nexttopic)Cascadingsecondandfirstorderstages
Cancreatehigherorderbandpass/stopfilters
similarly:Cascadefirstorderhigh/lowpassfiltersUseand/orcascadesecondorderbandpass/stopstages
Therearemanydifferentsecondorderactivefilter
topologies.Welllookatoneofthemorepopularones:theSallenKeytopology.
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SallenKeyFilter GeneralizedForm
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TheSallenKeyfiltertopologycanbeusedtoconstructlowpassandhighpassfilters,aswellbandpass,andnotchfilterswithslightmodificationstothetopology(additionofafewmorecomponents).
Wellfirsttakealookatthefilterinitsmostgeneralizedform,thenconsiderthespecificlowpassandhighpassfilterforms.
Typeoffilterdependsonthelocationofcomponents resistorsandcapacitors.
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SallenKeyFilter GeneralizedForm
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Toderivethefrequencyresponse
functionforthiscircuit,performanodal
analysis,applyingKCLatnodesvfandV+,
andapplythefactthatavirtualshort
existsbetweenV+ andV.
Afterseveralpagesofreallyuglyalgebra,
theresultingfrequencyresponse
is:
1 2 2 1 1
3 4 4 4 3
1
1H
Z Z
Z Z
Z Z Z
Z Z Z
where 1
1 2
f
f f
R
R R
isthegainofthenegativefeedbacknetwork.
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SallenKeyLowPassFilter
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ForalowpassfilterZ1andZ2areresistors,
andZ3andZ4arecapacitors.
Thegeneralizedfrequencyresponsefunction
thenbecomesalowpassfrequencyresponse:
1 2 1 2 2 2 1 2 1 11
1H
R jR C C C R C Rj R j j j C
Puttingthisintostandardform
1 2 1 2
2
1 1 2 1 2 2 1 2 1 2
11 1 1
1R C C
R C C C R C
RH
j jR R CR
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SallenKeyLowPassFilter
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Thefrequencyresponseofanysecondorderlowpasssystemcanbewrittenina
generalizedformas:
1 2 1 2
1o
RR C C
20
2 200
H
j
K
jQ
whereKistheDCgainofthesystem.ComparingtheSallenKeyfrequencyresponse
tothegeneralfrequencyresponseprovidessomeinsightintothebehaviorofthis
activefiltercircuit.Afewthingstonote:
Theresonantfrequency
is:
Thequalityfactoris: 1 2 1 2
2 2 1 1 1 1
1
R C CQ
C R C R
R
R C
TheDCgainis1/.Thiscanbeseenbyreplacingthefiltercapacitorswiththeir
DCequivalents(opencircuits) filterbecomesasimplenoninvertingamplifier.
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SallenKeyLowPass Simplified
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Thebandwidth(determinedbyoandQ)andtheQofthefiltercanbothbesetto
thedesiredvaluebyproperlyselectingcomponentvalues.Thereare,however,more
degreesoffreedomthanweneed,andthefrequencyresponsefunctionisabitmorecomplicatedthanwedlike.
Thecircuitanditsfrequencyresponsecanbesimplifiedbysettingthecomponent
valuesequal.Thecircuitanditsfrequencyresponsethenbecome:
2
2
2
1
1
3 1RCH
j jRC RC
where
1o
RC
1
13Q
and
Now,oandQaresimplyandindependentlydeterminedbyfourcomponentvalues.
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SallenKeyLowPass Simplified
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Thesimplifiedfiltersfrequencyresponsehasbeengivenintermsofthegainofthe
negativefeedbacknetwork,.ItcanalsobeexpressedintermsofthefiltersDC
gain,K,asisdoneinthetext,byrecognizingthat,foranidealopampK=1/.
2
2
2
3 1
K
RCH
j j
R C
K
C R
where
1o
RC
1
3Q
K
and
Lookingatthingsthiswaybringsupanimportantpoint:theDCgainofthefilteris
dependentonthefiltersQvalue,andviceversa.IfaDCgaindifferentthanthat
givenbythedesiredQvalueisrequired,anadditionalgainstagemaybenecessary.
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SallenKeyLowPass BodePlot
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NotethatthefiltersDC
gainandQvalueare
dependentonone
another.
QvalueandDCgainare
bothsetbytheratioof
resistorsinthenegativefeedbacknetwork.
Qvalue(andDCgain)is
independentofo.
oissetbytheresistor
andcapacitorvalues.
Gainrollsoffat
40dB/dec
ll h l
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SallenKeyHighPassFilter
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ASallenKeyhighpassfilterlooksverymuch
likethelowpassversion,butwiththe
positionoftheresistorsandcapacitorsswapped.
Againmakingthesimplificationthatthe
resistorandcapacitorvaluesareequal:
2
2
2
3 1
K jH
j jRC C
K
R
where
1o
RC
1
3Q
K
and
ThenaturalfrequencyandQvalueare
thesameasinthelowpasscase.
Notethatnow,aswedexpectfroma
highpassfilter,gaingoesto0atDC. Kisstillthepassbandgain,butnowit
it thethe highfrequency(asopposed
toDC)gain.
S ll K Hi h P B d Pl
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SallenKeyHighPass BodePlot
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NotethatthefiltersHF
gainandQvalueare
dependentonone
another.
QvalueandHFgainare
bothsetbytheratioof
resistorsinthenegativefeedbacknetwork.
Qvalue(andHFgain)is
independentofo.
oissetbytheresistor
andcapacitorvalues.
Gainrollsoffat
40dB/dec
S ll K Filt St bilit
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SallenKeyFilter Stability
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Positivefeedbackpath
Negative
feedback
path
TheSallenKeyfilterisafeedbacksystem,andaswithallfeedbacksystems,
stabilityisaconcern.
Ithastwofeedbackpaths:apositivefeedbackpathandanegative
feedback
path
Negativefeedbackgenerallyhasa
stabilizingeffect.
Positivefeedbackisdestabilizing.
Negativefeedbackpathgaindeterminesratioofnegativetopositivefeedback.
As,negativefeedback,K,and
positivefeedback.
ThereisanupperlimitonK:whenK=3,Q=.Thisistheborderbetweenstability(netnegativefeedback)and
instability(netpositivefeedback).
Forstability:K 3.
Filt F ili
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FilterFamilies
Higherorderfiltersofalltypes(i.e.LP,HP,BP,BR)canbedesignedtohave
frequencyresponsefunctionsthatfitintooneofseveralfamiliesof
filters.Forexample Butterworth(introducedinthetext)
Chebyshev
Elliptic
Bessel
Eachfilterfamilyisdefinedbythenatureofthepolynomialinthe
denominatorofitsfrequencyresponse(itscharacteristicequation).
Equivalently,eachfilterisdefinedbytherelativelocationsinthecomplex
planeoftherootsofthedenominatorpolynomial(rootsofthe
characteristicequation systempoles) Forexample,Butterworthpoleslieevenlyspacedonacircleinthelefthalfofthe
complexplane.
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Filt F ili F R
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FilterFamilies FrequencyResponse
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Eachfilterfamilyhas
advantagesanddisadvantages.
Butterworth: Maximallyflatpassband
Slowrolloff
Chebyshev:
Steeperrolloff
Passbandripple
Elliptic:
Verysteeprolloff
Passbandripple
Stopbandripple Aswithallengineeringdesign,
filterdesignisaboutmaking
tradeoffs.
Filt F ili S t P l
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FilterFamilies SystemPoles
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Polelocationstellalotabout
thebehaviorofasystem.
Youlllearnmuchmoreaboutthisinfuturecourses.
Butterworth:
Polesareevenlyspaced
alongasemicircleinthe
lefthalfplane
Chebyshev:
Poleslieonasemiellipse
inthelefthalfplane
Rememberthesepolesaretherootsofthetransferfunctions
denominatorpolynomial(the
characteristicequation)
Sallen Key Filter Example
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SallenKeyFilter Example
DesignaButterworth(maximallyflat)lowpass
activefiltertosatisfythefollowingspecifications: Cornerfrequency:fc=1MHz
Frequencyresponserolloffbeyondfc:80dB/dec
Passband(DC)gain:12dB(4)
Rolloffspecof80dB/dec tellsusweneedafourth
orderfilter cascadetwoSallenKeystages
Addaconstantgainstageifnecessarytomeetgain
specification
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Sallen Key Filter Example
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SallenKeyFilter Example
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ChooseRandCtogivethedesiredfc.
ArbitrarilychooseC=1nF:
1 1
12 12cf MHRC R F zn
1
1 1159
2 2 1cR
C MHz nF f
Assuming1%resistors,choosethe
standardvalueresistorof158.
AllfourresistorslabeledR,andfour
capacitorslabeledC,willhavethesamevalues:
1 , 158C nF R
Sallen Key Filter Example
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SallenKeyFilter Example
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Todeterminethegainrequiredforeach
stagetogiveaButterworthresponse,
consultTable14.1inthetext:
1 21.152, 2.235K K
ArbitrarilysetRf1=5.11K.
Thesegainvalueswillbeachievedby
settingtheratiooffeedbackresistors.
2 1 1 1 5.11 0.152 777f fR KR K
3 1 2 1 5.11 1.235 6.3f fR K K KR Choosingstandard1%resistorvalues:
1 2787 , 6.34f fR R K
Sallen Key Filter Example
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SallenKeyFilter Example
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Thefinalcomponentvaluetodetermine
isRf4,whichischosentosatisfythe
overallgainspecification:
1 2 3 31.152 2.235 4K K K K Choosingthecloseststandardvalued1%resistor:
4 2.8fR K 1 4
3
1
41.554
2.575
f f
f
RK
R
R
4 1 3 1 5.11 0.554 2.83f fR KKR K
Sallen Key Filter Example
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SallenKeyFilter Example
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Thecomplete4thorderSallenKeyButterworthlowpassfilter:
SallenKey Filter Example
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SallenKeyFilter Example
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DCgainis12dB
fc =1MHz
Gainrolloffabovefc is
80dB/dec
Firststageisoverdamped
Secondstageisunderdamped
peaked Thirdstageisconstantgain
Overallresponseistheproduct
oftheindividualresponseson
alinearscale
Overallresponseisthesumof
theindividualresponsesona
dB(log)scale
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InstrumentationAmplifiers65
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Single EndedMeasurement
Considerthefollowingscenario: Yourerunninganexperimentinthelab,inwhichyou
aremeasuringthepressureinsideatube Youareamplifyingthepressuresensoroutputbefore
measuringitwiththedataacq.System Thesensorhasasignaloutputandagroundoutput Thesensorisalongdistancefromthemeasurement
systemandisconnectedwithasinglelongwire Thesensorisgroundedlocally
Alotofnoise(frompowerline,pumps,motors,lights,etc.)couplesontothesignalwire,corruptingthemeasurement
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SingleEnded Measurement
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Single EndedMeasurement
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Sensorgroundandamplifiergroundmaynot
beatthesamepotential(DCorAC)
Thissituationiswhatwecallasingleendedmeasurement:
Amplifierismeasuringandamplifyingthesinglevoltagesignalfromthesensor
referencedtotheamplifiersground. Sensoroutputisreallydifferential voltagedifferencebetweensignalandground
terminals
Amplifierinputisreally
differentialalso voltage
differencebetweennoninvertingterminalandground
Twoproblems:
1) Groundsmaynotbeat
thesamepotential
2) Anynoisepickedupbythesignalwiregets
amplifiedalongwiththe
signal
Differential Measurement
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DifferentialMeasurement
Singleendedmeasurement:Sensorandamplifier/measurementsystemaregrounded
separatelySingle,longsignalwirerunstotheamplifier,pickingupnoise
Differentialmeasurement:Sensorsignalandgroundterminalsbothwiredtoamplifier
inputVoltagedifferentialbetweensensoroutputterminalsis
amplifiedSolvestwoproblems:
1)Bothwirespickupthesamenoisecommonmode
noise
2)GroundpotentialdifferencesnolongeramplifiedwithsignalHowdowedothis?Differentialamplifier
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Opamp Differential Amplifier
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OpampDifferentialAmplifier
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Thisopampdifferentialamplifierhastwoinputs.Wecanapplysuperpositionto
determinetheoutput.Lookslikeavoltagedivider
andanoninvertingamplifiertoinputv1:
1
2 1 2 21 1
1 2 1 1
o v
R R R Rv v v
R R R R
Lookslikeaninvertingamplifiertov2:
2
22
1
o v
Rv v
R
Theamplifieroutputis:
1 22
1 21
o o ov v
R
v v v v vR
Theoutputvoltageisthe(scaled)differencebetweenthetwoinputvoltages.
DifferentialandCommonModeInputs
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e e t a a d Co o ode puts
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Inputsignalstothedifferentialamplifier,v1andv2,canbeexpressedintermsoftheir
differentialandcommonmodecomponents.
Differentialcomponentisthedifference
betweenthetwoinputsignals:
Commonmodecomponentistheaverage
ofthetwoinputsignals:
1 2idv v v
1 22icm
v v
v
Thinkofthecommonmodesignalasany
signalthatappearsequallyatbothinputs.
DifferentialandCommonModeGains
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Foreachofthetwoinputcomponents differentialandcommonmode an
amplifierwill,ingeneral,havedifferentgainvalues.Thatis,itwillamplifydifferential
signalsandcommonmodesignalsbydifferentamounts.
Differentialgain:
Gainoftheamplifierwhendriven
byapurelydifferentialinputsignal.
od
id
vA
v
Commonmodegain:
Gainoftheamplifierwhendriven
byacommonmodeinputsignal.
ocm
icm
vA
v
CommonModeRejectionRatio
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j
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Theoutputoftheamplifieristhesumoftheinputcomponents,eachamplifiedby
theirrespectivegain:
o id d icm cm
v v A v A
Ideally,adifferentialamplifierwouldamplifyonlydifferentialinputsignals,andthe
commonmodegainwouldbezero.Inthatcase:
Theratioofthedifferentialgaintothecommonmodegain(typicallyexpressedin
dB)iscalledthecommonmoderejectionratio(CMRR):
20log d
cm
A
CMRR A
o id d v v A and CMRR
d
cmCMRR or
OpampDifferentialAmplifier Ad&Acm
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p p p d cm
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Fortheopampdifferentialamplifier,aslongastheratiooffeedbackresistorsis
equaltotheratioofvoltagedividerresistors(as
wevedefinedthemtobe)thenthecommonmodegainoftheamplifieris:
Thedifferentialgainis:
Inreality,anymismatchbetweenresistorratioswillresultinnonzerocommon
modegainandfiniteCMRR.
Anothersourceofnonzerocommonmodegainwouldbenonzero,andmismatchedsourceimpedances.
Ifsourceresistancesaremismatchedbutarebothknownandstable(oftennotthe
case)thenwecouldcompensateforthembyadjustingtheamplifierresistors.
0cm
A
2
1
dR
AR
Thecommonmoderejectionratiois:
CMRR
Acm MismatchedSourceResistances
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cm
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Herethedifferentialamplifierisdrivenbytwosourceswithmismatchedsource
resistances,Rs1andRs2where
Theoutputduetov1is:
2'
1 1sR R R and
1
' '1 1
' ' ' '1 1 1
2 2 2 21 1
2 21
o v
R R R R R Rv v v
R R R R RR R R
2
2'
1
2o v
Rv v
R
1 2s sR R R Reanalyzethecircuit,makingthefollowing
substitutions:
1'
1 1sRR R R
Theoutputduetov2is:
Acm MismatchedSourceResistances
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cm
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1 2
'1
' ' '
1
2 2 21 2
2
2 21
1 1
'1
' '1 1
2
2
o o ov v
o
R R R Rv v v v v
R R R RR R R
v v vRRR R
R
Theoveralloutputis:
wherenow,thev1inputisscaledbyanadditionalterm
containingR.
Foracommonmodeinput,bothinputsignalsarethesame:
Theoutputduetoacommonmodeinputsignalis:
1 2 icmv v v
'1
' '2 2
' '
1 1
2
1 12 2
1o icm icmR R R R
v v v
R R R R
R
R RR R
Thecommonmodegainis
nolongerzero: 2
2'
1 1 2' '
1
cm d
RA A
R R
R
R R
R
R R R
CMRR MismatchedResistances
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Whetherduetomismatchedsourceresistances,orto
mismatchedresistorsinthediff.ampitself,theresult
ofresistormismatchisareductioninCMRR.
Assumeresistancematchingto(e.g.for5%
matching,=0.05).
Notetherenumberingofresistorstoreflecttheir
individualvariation.
TheactualvaluesoftheworstcasecommonmodegainandCMRRdependonthe
relativenominalvaluesofallresistors.Wewillconsiderasimplifiedcasewhereall
resistorsarenominallyequalshortly.
Theworstcasecommonmodegainand,therefore,theworstcaseCMRRofthe
circuitoccurswhen:
1 , 1 , 1 , 1
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Theopampdifferentialamplifiercanideallyprovidehighdifferentialgain,lowcommonmode
gainandveryhighCMRR,butithasafewdrawbacks: HighCMRRisheavilydependentoncloseresistor
matching GainandCMRRaredependentonsourceresistances.
CMRRonsourceresistancematching,gainonabsolutevalues
Thereisabetteralternative:theinstrumentation
amplifier
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Instrumentationamplifier
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Stage1 Stage2
Thefollowingcircuitisaclassicthreeopampinstrumentationamplifier.
Bestunderstoodandanalyzed
bytreatingitastwoseparatestages.
Firststagehashigh
impedanceinputs,highAd,
whichisindependentof
sourceimpedance,andlowAcm,whichreliesonlyonthe
matchingoftheR2resistors.
Secondstageisdiff.amp
wevealreadylookedat.
HighCMRRoffirststage
relaxesmatching
requirementsofsecondstage.
Instrumentationamplifier Stage1
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Wevealreadyanalyzedstage2.Letstakealookatstage1.Wellconsidertwo
separatecases:first,apurelydifferentialinputand,second,apurelycommonmode
input Foradifferential
input,v1andv2moveoutof
phasewithoneanother:1 2
2idvv v
Thereisnegativefeedbackaroundbothopamps,so
wecanassumethatthevoltageateachoftheir
inputterminalsareequal:
1 1 1 2V V v v 2 2 2 1V V v v
and
Duetosymmetry,thenodelabeledvx,betweenthe
twoR1resistors,mustbesittingatground:
0xv V
Wecanassumethatnodevxisgrounded,andtreat
eachamplifierasaseparatenoninvertingamplifier.
Instrumentationamplifier Stage1
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Theoutputofstage1foradifferentialinputisthedifferencebetweentheoutputsof
twononinvertingamplifiers: 1 2 1 2 1 21 1 2 1 2
1 1 1
1 2 1 21
1 12 2
o d
id id o d id
R R R R R Rv v v v v
R R R
v vR R R Rv v
R R
Thedifferentialgainisequaltothegainofeachof
thenoninvertingamplifiers,consideredseparately.
Thevalueofthedifferentialgaincaneasilybemadeverylargewithreasonableresistorvalues.
Theoutputduetoadifferentialinputispurely
differentialaswell.
1 2
11
o d id
R R
v v R
1 1 2
11
o d
did
v R R
A v R
Instrumentationamplifier Stage1
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Nowletsconsidertheresponseofstage1toapurelycommonmodeinput:
Foracommonmodeinput,v1andv2areequal:
1 2 icmv v v Again,thereisavirtualshortattheopampinputs,
butnow:
1 1 1 2V V v v 2 2 2 1V V v v
and
NowthevoltageoneithersideofthepairofR1
resistorsisthesame.
Thereisnovoltagedropacrosstheseresistors,so
nocurrentflows.
IfnocurrentflowsthroughtheR1resistors,wecanremovethemfromthecircuitforthepurposeofthis
analysis.
Instrumentationamplifier Stage1
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WiththeR1resistorsremoved,nocurrentflowsthroughtheR2resistors,because
nocurrentcanflowintotheopampinputterminals.
Thecommonmodegainofthefirststageisunity,
whichcanbemuchsmallerthanthedifferentialgain.
1o cm icmv v
Eachopampcannowbetreatedasaseparateunitygainbuffer.
Thecommonmodeoutputofstage1inresponseto
acommonmodeinputisthevoltageattheoutput
oftheunitygainbuffers:
11 1
o cmcm
icm
vA
v
CMRRforstage1is: 1 1 2
1 1
1 1
dd
cm
A R RCMRR A
A R
CMRR MismatchedResistances
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TheCMRRofthesecondstageisnominallyinfinite,
howeverresistormismatcheswillresultinnonzero
commonmodegainand,therefore,finiteCMRR.
Assumeresistancematchingto(e.g.for5%
matching,=0.05).
TheworstcaseCMRRofthecircuitisoccursforthe
resistorvaluesshown.Itis:
Thisisanimportantresult:itprovidesalinkbetweenaCMRRrequirementand
resistortolerancesforathreeopampinstrumentationamplifier.
InstrumentationAmplifier
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TheoverallCMRRoftheInAmp istheproductof
theCMRRofeachstage(sum,ifindB).
ThehighCMRRofthefirststagesignificantlyeases
thematchingrequirementsofthesecondstageto
achievesimilar(ormuchhigher)overallCMRR.
AdvantagesofInAmp overdifferentialamplifier: Highimpedanceinputs:sourceresistancesdonot
affectgainorCMRR.
Resistormatchingislesscriticalbecausethereare
twocascadedhighCMRRstages
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InstrumentationAmplifier Example
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DesignanInAmp forthefollowingapplication:MeasuringstrainwithaWheatstonebridgecircuit
Strainsignalis100mVdifferentialatmax/minstrainWanttoamplifybridgeoutputtousefull5Vdynamic
rangeofdataacq.system.
Commonmodenoiseonthebridgeoutputsignalis
200mVpp.Thatsthefulldynamicrangeofthestraingauge!
Assumeresistorsinstage1arematched.
Assume1%matchingofresistorsinstage2.
WhatisworstcaseCMRR? HowmuchnoiseisthereattheoutputoftheInAmp?
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InstrumentationAmplifier Example
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DifferentialgainoftheInAmp shouldbe: 5 50100
d
VA
mV
ArbitrarilysetR2=10K.ChooseresistorR1togivetherequiredgain:1 2 2
1 1
1
10204 205
1 150d
d
R R R KA R R
R A
CMRRofthefirststageisthedifferentialgainofthatstage: 1 50dCMRR A
WorstcaseCMRRofthesecondstageis:
WorstcaseCMRRoftheoverallInAmp is:
21 50 50 2500 wc wc wcCMRR CMRR CMRR
2500 68 wcCMRR dB
Arbitrarilysetallresistorsinstage2tobeR=10K.
12
50
InstrumentationAmplifier Example
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Theresultingcircuitis:
Determinetheamountofnoisepresentonthe
outputsignal:
Commonmodegainof
theInAmp underworst
casemismatchconditionsis: 1 2 21cm cm cm cmwc wc wcA A A A
Thecommonmodegainofstage2underworstcasematchingconditionsis:
Thenoiseattheoutputis: 32 200 19.8 10 4ocm icm cm pp ppwc wcv v A mV mV
2
1 19.8 10
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Attheinputthefullscale
signalis200mVpp.
Noiseattheinputis200mVpp.
Inputsignaltonoiseratio
(SNR)is:
200
1 0200
pp
ipp
mV
SNR dBmV
Attheoutputthefullscalesignalis10Vpp,whilethenoisehasbeenreducedtoonly
4mVpp,resultinginanoutputSNRof:
TheimprovementinSNRisequaltotheCMRR.
Thisisabigdeal theInAmp hasallowedustoaccuratelymeasureadifferential
signalthatwascompletelyburiedincommonmodenoise.
102500 68
4
pp
o
pp
VSNR dB
mV
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DCOpampNonIdealities89
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RealOpamps
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Recallthecharacteristicsofanidealopamp:
1) Infiniteinputimpedance zeroinputcurrent
2) Infinite
differential
mode
gain3) Zerocommonmodegain
4) Zerooutputimpedance
5) Infinitebandwidth
6) Virtualshortatinputterminals(w/negativefeedback)(notpartoftheoriginallist followsfrom#2)
Inpractice,realopampsarenonideal.
Thetextcoversthreecategoriesofopampnonidealities.
Welltakeabrieflookatafewnonidealitiesthatfallintojustoneofthesecategories:DCoffsets.
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OpampDCOffsets
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TherearethreeprimaryDCimperfections
associatedwithrealopamps: Bias
current:
DCcurrentsthatflowintotheopamp
inputs
Offsetcurrent:thedifferencebetweenbiascurrents
atthenoninvertingandinvertinginputs Offsetvoltage:differenceininputterminalvoltagesin
thepresenceofnegativefeedback.Or,theequivalent
DCsourcethat,whenconnectedinserieswithoneof
theinputs,wouldexplainanonzerooutputvoltage
forzeroinputvoltage.
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BiasCurrent
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Theinternaldevices(transistors)attheopampinputterminalsrequireasmallamountofDCbiascurrentto
function. ThatcurrentmustflowinthroughtheinputterminalsOpampswithBJT(BipolarJunctionTransistor)input
deviceshavehigherbiascurrents.OpampswithFET(FieldEffectTransistor)inputdeviceswill
havemuchlowerbiascurrents.
BiascurrentistheaverageDCinputcurrent
Biascurrentcanbeaccountedforbyaddingcurrentsourcestotheidealopampmodel.
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OffsetCurrent93
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Duetomismatchbetweeninternalopampinput
devices,biascurrentsateachinputmaynotbe
equal.
Thedifferencebetweenthebiascurrentsateach
inputterminalistheoffsetcurrent.
Offsetcurrentcanbeaccountedforbyaddinga
currentsource,inadditiontothebiascurrent
sources,totheidealopampmodel.
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ModelingBiasandOffsetCurrents94
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94
Biascurrentistheaverageofthe
DCinputcurrentsateach
terminal:
2B B
B
I II
Offsetcurrentisthedifference
betweenthebiascurrentsateach
inputterminal:
off B BI I I
Opampmodelaccountingforbias
andoffsetcurrents:
IdealOpamp
OffsetVoltage95
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Duetomismatchesandimperfectionsinthe
internalopampcircuitry,theoutputvoltageofan
opampmaynotbezeroeveniftheinputvoltage
(differential)iszero.
Thisoutputerrorvoltagecanbetreatedasan
inputreferrederrorvoltage,modeledasaDC
voltagesourceinserieswithoneoftheinputs.
Thevalueofthisinputreferredsourceistheoffset
voltage.
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ModelingOffsetVoltage96
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96
Offsetvoltageismodeledasasourceinserieswithoneoftheinputs.
Usuallyspeced as somevalue,sopolarity(orwhatterminalits
connectedto)isnotcritical.
Anotherwaytothinkofoffsetvoltage:itisthevoltagerequiredacrossthe
inputterminalstoforcetheoutputvoltagetozero.
IdealOpamp
ModelingOpampDCNonIdealities97
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97
IdealOpamp
ThefollowingmodelaccountsforallDCopampimperfections.
TheresultofeacherrorsourceisaDCerrorvoltageattheoutput.
Thismodelcanbeused
in
place
of
an
ideal
opamp
modelforcircuitanalysisto
determinetheeffectofeachnonidealityoncircuitperformance.
Totaloutputerrorcanbe
determinedbyapplying
superpositiontodeterminethe
errorduetoeachindividual
source.
Absolutevaluesforeacherror
sourcewill,ingeneral,notbe
known.Arangeforthevaluesistypicallyspecified,allowingfora
worstcaseerroranalysis.
DCErrorAnalysisExample98
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K.Webb MAE3055 MechetronicsII
Forthenoninvertingamplifiershown,determinetheworst
caseDCoutputerror(atanambienttemperatureofTA=25C),
giventhefollowing:
R1=20K,R2=100K.
OpampisanLM741.
|Voff| 5mV.
Ibias 500nA. |Ioff| 200nA.
Usesuperpositiontodeterminetheworstcaseoutputerrorduetoeacherror
source,thensumtheindividualcontributionstodeterminethetotalworstcase
outputerror.
err off B off o o o oV I I
V V V V
DCErrorAnalysisExample99
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K.Webb MAE3055 MechetronicsII
First,determinetheoutputduetotheinputoffsetvoltage:
Here,theoffsetvoltageappearsasthe
inputtoanoninvertingamplifier,with
gain
1 2
1
206
2
100
0o
i
v R R K
v R K
K
1 2
1
5 6 30off
o off V
R RV V mV mV
R
Theoutputduetotheinputoffsetvoltageis:
30offo V
V mV
Theoffsetvoltagealwaysappearsattheoutputscaledbythenoninvertinggain,
evenforinvertingamplifiers.
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K.Webb MAE3055 MechetronicsII
Next,determinetheoutputduetotheinputbiascurrent:
Thebiascurrentatthenoninvertingterminal
comesdirectlyfromgroundand,thereforehasnoeffectontheoutput.
ApplyingKCLattheinvertinginput:
2 500 100 50B
o BIV R nA mI K V
Becausetheopampmodelitselfisideal,there
isavirtualshortattheinputterminals,and
Vo=V =0V,whichgives:
50B
o IV mV
1 2
1 2
oB
V VVI I I
R R
Theoutputvoltageduetothebiascurrentis
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Finally,determinetheoutputduetotheinputoffsetcurrent:
Theanalysisfordeterminingtheoutputvoltage
duetotheoffsetcurrentisessentiallyidenticaltotheanalysisforthebiascurrent,so
2 2002
100 20off
off
o I
IV R nA K mV
20off
o IV mV
Theoutputvoltageduetotheoffsetcurrentis:
NotethatbotheffectofboththebiasandoffsetcurrentsisproportionaltoR2.ThistellsusthatwecanreducetheireffectsbyreducingthevalueofR2.The
tradeoffisincreasedpowerdissipation.Wellseeshortlythatitsactuallypossible
tocanceltheeffectofbiascurrent.
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Theoveralloutputerrorvoltageisgivenbythesumofthe
individualerrorvoltagecontributions:
30 50 20 100
err off B off
err
o o o oV I Imax maxmax max
omax
V V V V
V mV mV mV mV
Themaximumoutputerrorvoltageduetothese
threeerrorsourcesis:
100erro max
V mV
Notethatthepolarityofthebiascurrentisknown inthiscaseIBflowsintothe
opampinputterminals whereasthepolaritiesoftheoffsetvoltageandthe
offsetcurrent
are
not
known
and
never
are.
Thisisnotalwaysentirelyobviousfromthedatasheet,butthesignofthebias
currentshouldindicateitspolarity.
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Itispossibletocanceltheeffectofbiascurrentbyaddingasingleresistortoboth
theinvertingandnoninvertingopampamplifiertopologies:
InvertingAmplifier: NonInvertingAmplifier:
Inbothcases: 1 2
1 2
1 2
||biasR R
R R RR R
OffsetVoltageNulling104
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Manyopamps,includingthebasic741,provideameansfornulling theiroffset
voltage.
Thisistypicallydonebyinsertingapotentiometerbetweentwoadditionalopamp
pins,typicallybothcalledoffset
null,orsomethingtothateffect.
Onthe741opampthesearepins1and5.
Connecta10K(forthe741)potentiometerbetweentheoffsetnullpins,withthe
wiperterminalconnectedtothenegativesupply.
Shorttheinputs(oftheamplifierwithfeedbackconnected)togroundthenadjust
thepotuntiltheoutputis0V.
Thepotentiometerisoftenreferredtoasatrim
pot,becauseitisusedfortrimmingtheoffset
voltage. Trimpotconnectionsandrecommendedresistance
valuewillvaryfromopamptoopamp checkthe
datasheet.