SECTION 5 Advanced Opamp Circuits

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


3


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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


5


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Comparators


6

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


7

Theoutputofacomparatorswitchesbetweentwostates theupperandlower

outputlimits dependingontherelativevaluesoftheinputvoltages:

ComparatorInputs

ComparatorOutputs


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Comparators ExampleUses


8

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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400hystV mV

Hysteresisismeasuredas

thefullpeaktopeak

swingofthethreshold

voltage

Hysteresisistwicethe

magnitudeofthefeedbacksignal:

2

1 2

2maxhyst o

RV V

R R


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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


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


16


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SchmittTriggerExample


17

DesignaSchmitttriggertosatisfythecriteriaonthepreviousslide.Theentirecircuit

isshownbelow.

WhentheRTDoutputexceeds0VtheSchmitttriggeroutputwillgolowand

theLEDwillbebiasedwith9mAandwillilluminate

NotethattheRTDitselfisjustaresistor;itsbiasnetworkisnotshown.

Currentflowingthroughthehysteresisfeedbacknetworkmustbelimitedto

1mA


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18

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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19

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


21


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OpampRelaxationOscillator


22

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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23

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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24

AstheoutputswitchesbackandforthbetweenV,twothingshappen

Thenoninvertinginput,V+,switchesbetweenV

Thecapacitorvoltageattheinvertinginputchargesanddischargesbetween

VataratedeterminedbytheRCtimeconstantofthenegativefeedback

network

OutputswitchingoccurseachtimeV reachesV

EachtimeVoswitches,V ischarging/dischargingtowardV


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OpampOscillatorfosc


25

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


26

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


27

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


29


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OpampIntegrator


30

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


31

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


32

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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33

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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34

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


35

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


36

AtDC(=0),thecapacitoris

anopencircuit,andthegainissetbythetworesistors

As,thegain0

At=1/(RfC),thegainis3dB

Thisisalowpassfilter!

Aninvertingamplifierforfrequenciesbelowfc.

Anintegratorforfrequencies

wellabovefc.

IntegratorAmplifier

fc=159Hz


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OpampDifferentiator


37

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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39/104

OpampDifferentiator Example


39

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


40

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


42


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FirstOrderLowPassFilter


43

Wevealreadyseenthatafirstorderlowpassresponse

canbeobtainedbyaddingafeedbackresistortoan

opampintegratorcircuit:

1

1

f

f

RH

jR R C

Thecornerfrequencyis:

2

1c

fC

fR

Magnituderesponsetothe

rightisforRf=R.Responsewouldshiftupordownby

20log fR

R


44/104

FirstOrderHighPassFilter


44

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


45/104

HigherOrderActiveFilters

Higherorderactivefilterscanbeconstructedinacoupleofdifferentways:CascadingfirstorderactivefiltersUsingsecondorderactivefilterstages(nexttopic)Cascadingsecondandfirstorderstages

Cancreatehigherorderbandpass/stopfilters

similarly:Cascadefirstorderhigh/lowpassfiltersUseand/orcascadesecondorderbandpass/stopstages

Therearemanydifferentsecondorderactivefilter

topologies.Welllookatoneofthemorepopularones:theSallenKeytopology.


45


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SallenKeyFilter GeneralizedForm


46

TheSallenKeyfiltertopologycanbeusedtoconstructlowpassandhighpassfilters,aswellbandpass,andnotchfilterswithslightmodificationstothetopology(additionofafewmorecomponents).

Wellfirsttakealookatthefilterinitsmostgeneralizedform,thenconsiderthespecificlowpassandhighpassfilterforms.

Typeoffilterdependsonthelocationofcomponents resistorsandcapacitors.


47/104

SallenKeyFilter GeneralizedForm


47

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.


48/104

SallenKeyLowPassFilter


48

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


49/104

SallenKeyLowPassFilter


49

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.


50/104

SallenKeyLowPass Simplified


50

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.


51/104

SallenKeyLowPass Simplified


51

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.


52/104

SallenKeyLowPass BodePlot


52

NotethatthefiltersDC

gainandQvalueare

dependentonone

another.

QvalueandDCgainare

bothsetbytheratioof

resistorsinthenegativefeedbacknetwork.

Qvalue(andDCgain)is

independentofo.

oissetbytheresistor

andcapacitorvalues.

Gainrollsoffat

40dB/dec

ll h l


53/104

SallenKeyHighPassFilter


53

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


54/104

SallenKeyHighPass BodePlot


54

NotethatthefiltersHF

gainandQvalueare

dependentonone

another.

QvalueandHFgainare

bothsetbytheratioof

resistorsinthenegativefeedbacknetwork.

Qvalue(andHFgain)is

independentofo.

oissetbytheresistor

andcapacitorvalues.

Gainrollsoffat

40dB/dec

S ll K Filt St bilit


55/104

SallenKeyFilter Stability


55

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


56/104

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.


56

Filt F ili F R


57/104

FilterFamilies FrequencyResponse


57

Eachfilterfamilyhas

advantagesanddisadvantages.

Butterworth: Maximallyflatpassband

Slowrolloff

Chebyshev:

Steeperrolloff

Passbandripple

Elliptic:

Verysteeprolloff

Passbandripple

Stopbandripple Aswithallengineeringdesign,

filterdesignisaboutmaking

tradeoffs.

Filt F ili S t P l


58/104

FilterFamilies SystemPoles


58

Polelocationstellalotabout

thebehaviorofasystem.

Youlllearnmuchmoreaboutthisinfuturecourses.

Butterworth:

Polesareevenlyspaced

alongasemicircleinthe

lefthalfplane

Chebyshev:

Poleslieonasemiellipse

inthelefthalfplane

Rememberthesepolesaretherootsofthetransferfunctions

denominatorpolynomial(the

characteristicequation)

Sallen Key Filter Example


59/104

SallenKeyFilter Example

DesignaButterworth(maximallyflat)lowpass

activefiltertosatisfythefollowingspecifications: Cornerfrequency:fc=1MHz

Frequencyresponserolloffbeyondfc:80dB/dec

Passband(DC)gain:12dB(4)

Rolloffspecof80dB/dec tellsusweneedafourth

orderfilter cascadetwoSallenKeystages

Addaconstantgainstageifnecessarytomeetgain

specification


59



60/104



60

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



61/104



61

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



62/104



62

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



63/104



63

Thecomplete4thorderSallenKeyButterworthlowpassfilter:

SallenKey Filter Example


64/104



64

DCgainis12dB

fc =1MHz

Gainrolloffabovefc is

80dB/dec

Firststageisoverdamped

Secondstageisunderdamped

peaked Thirdstageisconstantgain

Overallresponseistheproduct

oftheindividualresponseson

alinearscale

Overallresponseisthesumof

theindividualresponsesona

dB(log)scale


65/104

InstrumentationAmplifiers65


SingleEnded Measurement


66/104

Single EndedMeasurement

Considerthefollowingscenario: Yourerunninganexperimentinthelab,inwhichyou

aremeasuringthepressureinsideatube Youareamplifyingthepressuresensoroutputbefore

measuringitwiththedataacq.System Thesensorhasasignaloutputandagroundoutput Thesensorisalongdistancefromthemeasurement

systemandisconnectedwithasinglelongwire Thesensorisgroundedlocally

Alotofnoise(frompowerline,pumps,motors,lights,etc.)couplesontothesignalwire,corruptingthemeasurement


66

SingleEnded Measurement


67/104

Single EndedMeasurement


67

Sensorgroundandamplifiergroundmaynot

beatthesamepotential(DCorAC)

Thissituationiswhatwecallasingleendedmeasurement:

Amplifierismeasuringandamplifyingthesinglevoltagesignalfromthesensor

referencedtotheamplifiersground. Sensoroutputisreallydifferential voltagedifferencebetweensignalandground

terminals

Amplifierinputisreally

differentialalso voltage

differencebetweennoninvertingterminalandground

Twoproblems:

1) Groundsmaynotbeat

thesamepotential

2) Anynoisepickedupbythesignalwiregets

amplifiedalongwiththe

signal

Differential Measurement


68/104

DifferentialMeasurement

Singleendedmeasurement:Sensorandamplifier/measurementsystemaregrounded

separatelySingle,longsignalwirerunstotheamplifier,pickingupnoise

Differentialmeasurement:Sensorsignalandgroundterminalsbothwiredtoamplifier

inputVoltagedifferentialbetweensensoroutputterminalsis

amplifiedSolvestwoproblems:

1)Bothwirespickupthesamenoisecommonmode

noise

2)GroundpotentialdifferencesnolongeramplifiedwithsignalHowdowedothis?Differentialamplifier


68

Opamp Differential Amplifier


69/104

OpampDifferentialAmplifier


69

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


70/104

e e t a a d Co o ode puts


70

Inputsignalstothedifferentialamplifier,v1andv2,canbeexpressedintermsoftheir

differentialandcommonmodecomponents.

Differentialcomponentisthedifference

betweenthetwoinputsignals:

Commonmodecomponentistheaverage

ofthetwoinputsignals:

1 2idv v v

1 22icm

v v

v

Thinkofthecommonmodesignalasany

signalthatappearsequallyatbothinputs.

DifferentialandCommonModeGains


71/104


71

Foreachofthetwoinputcomponents differentialandcommonmode an

amplifierwill,ingeneral,havedifferentgainvalues.Thatis,itwillamplifydifferential

signalsandcommonmodesignalsbydifferentamounts.

Differentialgain:

Gainoftheamplifierwhendriven

byapurelydifferentialinputsignal.

od

id

vA

v

Commonmodegain:

Gainoftheamplifierwhendriven

byacommonmodeinputsignal.

ocm

icm

vA

v

CommonModeRejectionRatio


72/104

j


72

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


73/104

p p p d cm


73

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


74/104

cm


74

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


75/104

cm


75

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


76/104


76

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

OpampDifferentialAmplifier Problems


77/104

Theopampdifferentialamplifiercanideallyprovidehighdifferentialgain,lowcommonmode

gainandveryhighCMRR,butithasafewdrawbacks: HighCMRRisheavilydependentoncloseresistor

matching GainandCMRRaredependentonsourceresistances.

CMRRonsourceresistancematching,gainonabsolutevalues

Thereisabetteralternative:theinstrumentation

amplifier


77

Instrumentationamplifier


78/104


78

Stage1 Stage2

Thefollowingcircuitisaclassicthreeopampinstrumentationamplifier.

Bestunderstoodandanalyzed

bytreatingitastwoseparatestages.

Firststagehashigh

impedanceinputs,highAd,

whichisindependentof

sourceimpedance,andlowAcm,whichreliesonlyonthe

matchingoftheR2resistors.

Secondstageisdiff.amp

wevealreadylookedat.

HighCMRRoffirststage

relaxesmatching

requirementsofsecondstage.

Instrumentationamplifier Stage1


79/104


79

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.



80/104


80

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



81/104


81

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.



82/104


82

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


83/104


83

TheCMRRofthesecondstageisnominallyinfinite,

howeverresistormismatcheswillresultinnonzero

commonmodegainand,therefore,finiteCMRR.

Assumeresistancematchingto(e.g.for5%

matching,=0.05).

TheworstcaseCMRRofthecircuitisoccursforthe

resistorvaluesshown.Itis:

Thisisanimportantresult:itprovidesalinkbetweenaCMRRrequirementand

resistortolerancesforathreeopampinstrumentationamplifier.

InstrumentationAmplifier


84/104

TheoverallCMRRoftheInAmp istheproductof

theCMRRofeachstage(sum,ifindB).

ThehighCMRRofthefirststagesignificantlyeases

thematchingrequirementsofthesecondstageto

achievesimilar(ormuchhigher)overallCMRR.

AdvantagesofInAmp overdifferentialamplifier: Highimpedanceinputs:sourceresistancesdonot

affectgainorCMRR.

Resistormatchingislesscriticalbecausethereare

twocascadedhighCMRRstages


84

InstrumentationAmplifier Example


85/104

DesignanInAmp forthefollowingapplication:MeasuringstrainwithaWheatstonebridgecircuit

Strainsignalis100mVdifferentialatmax/minstrainWanttoamplifybridgeoutputtousefull5Vdynamic

rangeofdataacq.system.

Commonmodenoiseonthebridgeoutputsignalis

200mVpp.Thatsthefulldynamicrangeofthestraingauge!

Assumeresistorsinstage1arematched.

Assume1%matchingofresistorsinstage2.

WhatisworstcaseCMRR? HowmuchnoiseisthereattheoutputoftheInAmp?


85



86/104


86

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



87/104


87

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



88/104


88

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


89/104

DCOpampNonIdealities89


RealOpamps


90/104

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.


90

OpampDCOffsets


91/104

TherearethreeprimaryDCimperfections

associatedwithrealopamps: Bias

current:

DCcurrentsthatflowintotheopamp

inputs

Offsetcurrent:thedifferencebetweenbiascurrents

atthenoninvertingandinvertinginputs Offsetvoltage:differenceininputterminalvoltagesin

thepresenceofnegativefeedback.Or,theequivalent

DCsourcethat,whenconnectedinserieswithoneof

theinputs,wouldexplainanonzerooutputvoltage

forzeroinputvoltage.


91

BiasCurrent


92/104

Theinternaldevices(transistors)attheopampinputterminalsrequireasmallamountofDCbiascurrentto

function. ThatcurrentmustflowinthroughtheinputterminalsOpampswithBJT(BipolarJunctionTransistor)input

deviceshavehigherbiascurrents.OpampswithFET(FieldEffectTransistor)inputdeviceswill

havemuchlowerbiascurrents.

BiascurrentistheaverageDCinputcurrent

Biascurrentcanbeaccountedforbyaddingcurrentsourcestotheidealopampmodel.


92

OffsetCurrent93


93/104

Duetomismatchbetweeninternalopampinput

devices,biascurrentsateachinputmaynotbe

equal.

Thedifferencebetweenthebiascurrentsateach

inputterminalistheoffsetcurrent.

Offsetcurrentcanbeaccountedforbyaddinga

currentsource,inadditiontothebiascurrent

sources,totheidealopampmodel.


93

ModelingBiasandOffsetCurrents94


94/104


94

Biascurrentistheaverageofthe

DCinputcurrentsateach

terminal:

2B B

B

I II

Offsetcurrentisthedifference

betweenthebiascurrentsateach

inputterminal:

off B BI I I

Opampmodelaccountingforbias

andoffsetcurrents:

IdealOpamp

OffsetVoltage95


95/104

Duetomismatchesandimperfectionsinthe

internalopampcircuitry,theoutputvoltageofan

opampmaynotbezeroeveniftheinputvoltage

(differential)iszero.

Thisoutputerrorvoltagecanbetreatedasan

inputreferrederrorvoltage,modeledasaDC

voltagesourceinserieswithoneoftheinputs.

Thevalueofthisinputreferredsourceistheoffset

voltage.


95

ModelingOffsetVoltage96


96/104


96

Offsetvoltageismodeledasasourceinserieswithoneoftheinputs.

Usuallyspeced as somevalue,sopolarity(orwhatterminalits

connectedto)isnotcritical.

Anotherwaytothinkofoffsetvoltage:itisthevoltagerequiredacrossthe

inputterminalstoforcetheoutputvoltagetozero.

IdealOpamp

ModelingOpampDCNonIdealities97


97/104


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


98/104


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



99/104


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.



100/104


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



101/104


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.



102/104


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.

BiasCurrentCancellation103


103/104


Itispossibletocanceltheeffectofbiascurrentbyaddingasingleresistortoboth

theinvertingandnoninvertingopampamplifiertopologies:

InvertingAmplifier: NonInvertingAmplifier:

Inbothcases: 1 2

1 2

1 2

||biasR R

R R RR R

OffsetVoltageNulling104


104/104


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.

SECTION 5 Advanced Opamp Circuits

Documents

Transcript of SECTION 5 Advanced Opamp Circuits