CHAPTER 10 Introduction803 10.1TheGeneralFeedback Structure804 10.8Summary of the Feedback AnalysisMethod863 10.9Determining the Loop Gain863 10.2SomeProperties of Negative Feedback80910.10The Stabili ty Problem868 10.3TheFour BasIcFeedback Topologies814 10.4TheFeedback VoltageAmplifier (Series-Shunt)823 10.5TheFeedback Transconductance Amplifier (Series-Series)834 10.6TheFeedback Transresistance Amplifier (Shunt-Shunt)846 10.7TheFeedback Current Amplifier (Shunt-Series)855 10.11Effect of Feedback on theAmplifier Poles870 10.12Stability Study Using80dePlots87910.13Frequency Compensation884 Summary890 Problems890 INTHISCHAPTERYOU WILL LEARN 1.Thegeneralstructure ofthenegati ve-feedbackamplifier andthebasic principle t hatunderl iesit s operation. 2. Theadvant agesofnegat ivefeedback.howthesecomeabout,andat whatcost. 3. The appropriate feedback t opology to employ with each of the four am-plifiertypes:vol t age,current ,t ransconductance,andtransresistance amplifiers. 4. Anintui t ive and insrght ful approach for the analysis of practicalfeedback-amplifier circuits. S. Whyandhownegat ive-feedbackamplifierscanbecomeunstable(i.e. , oscillat e) andhow t odesign the ci rcuit to ensure stable performance. Introduction Mostphysicalsystemsincorporateorneformof feedback. It isimeresting to note,though, that thetheoryof negativefeedbackhasbeen developedbyelectroni csengineers.Inhi s searchformethodsforthedesignofamplifierwithstablegai nforuseintelephone repeaters,HaroldBl ack,anelectronicsengi neerwiththeWesternElectri cCompany, invented the feedback amplifier in1928. Si ncethen the technique has been so widel y used that itis almost impossible to think of electroni c circuits wi thoutsomeformof feedback, ei ther impl icit or expl icit.Furthermore,the conceptof feedback andits associatedtheory are currentl yusedinareasotherthan engineering,such asinthemodeling of bi ological systems. Feedback can be ei ther negative(degenerative) or positive (regenerative). Inamplifier design, negati vefeedbackisapplied to effect one or more of the following properties: 1.Desensitize the gain: that is,make the value of the gainlesssensiti ve tovariations in the valuesof circuitcomponents,such as mightbe causedby changes intemper ature. 2.Reducenonlmear distortion:thatis, make the outputproporti onaltothemput(in other words,makethe gainconstant , independentof signal level). 3.Reducetheeffectof noise:thatis,minimi zethecontributiontotheoutputof unwanted electri c signals generated, either by the circuit components themselves, or by extraneous interfe rence. 803 804Chapter 10Feedback 4Control theinpllt and Olltplltresistancesthatis,raise or lower theInput and ...fidb resistancesbythe selection of anappropnateeeacktopology. 5.Extend the bandwidth of theamplifier. Allof the deSirable properti esaboveareobtainedattheexpense of a reduction In gam. It wi llbeshownthatthe gam-reductIOnfactor, call edtheamountof feedback, isthefactor b) whichtheCirCUIt IS desensIti zed,bywhichtheInputresistanceofavoltageamplifierisincreased, by which thebandWidthIS extended:andsoon.In short, the basicidea a/nega. tive feedbackisto tradeoff gam forother deSirablepropertres.Thlchapter isdevoted t the studyof negati ve-feedback amplifiers:their analysis,design,andcharacteristics.0 Under certain conditions.the negative feedback Inanampli fier can become positive and of such a magnitude asto cause oscill ation. In fact,InChapter17 wewillstudy the use of posltilefeedback inthe designof OSCillatorsandbistablecircutts.Here,Inthischapter.however. weareInterestedinthe designof stableamplifiers.Weshallthereforestudy thestabilityprob-lem of negati ve-feedback amplifiers and thetr potentialforoscillation. It shouldnot beImplied.however, thatposi tivefeedbackalwaysleadstoinstability. Infact, positi vefeedbackisqUIteuseful inanumber of nonregenerativeapplicatIOns.such asthedesign of acti vefi lters,which arestudied inChapter16. Beforewebeginourstudyof negativefeedback.wewishtoremindthereader that wehavealready encountered negati vefeedback in a number of applications.Almost all opampcircuits(Chapter2)empl oynegati vefeedback.Anotherpopularapplicationofnegativefeedback is the useof the emitter resistanceRE tostabilizethebiaspoi ntof bipolar transistorsandtoIncreasetheinputresistance,bandwidth,andlinearityofaBJT amplifier.Inadditi on,the sourcefollower and the emitter follower both employa largeamountof negativefeedback. The questi on then arises aboutthe need fora formalstudy of negatt ve feedback.Aswillbeappreciatedbytheendof thischapter,thefOlmalstudyoffeedbackprovidesanInvaluable tool for the analysis and design of electroniccircuits.Also,the inSight gained b)thinking In terms of feedback can beextremely profi table. 10.1TheGeneralFeedbackStructure Figure10.1 shows thebaSIC structure of a feedback ampltfierRather than showing voltagesandcurrents.Fi g.10. 1 IS a signal-flowdiagram,where each of the quant itiesx can represent either a voltage or a current signal.The apen-loop ampltfier hasa gain A;thusits outputx, isrelatedto themput xby (10.11Source A ..t Load-0-\. E /l Figure 101GeneIIr h .d Ih quanlili';' ..ras ructure 0te feedback amplifi er. Thi s is a signal-now diagram. an e represent either voltage or current signals. 10.1 TheGeneralFeedbackStructure805 The output x, is fedtotheload aswellastoa feedback netwokh'hd T.,r,wIC prouces a sample of the output.hiSsample xf IS relatedto x, bythe feedbackfacto r /3, (10.2) The feedbacksignal XI issubtracted fromthe source signalrwhl'ch 'th.h .I.- J 'ise mput tote com-plete feedbackamphfier,toproduce the Signal x" whichis the input tothebasicamplifier. x,= x'-''f(10.3) Herewenotethatitisthis subtractionthat makesthefeedbacknegative. Inessence, nega-ttvefeedbackreduces thatappearsatthe input of thebasicamplifier. ImphcltInthedescnpttonaboveisthatthesource, theload, andthefeedbacknetwork donot load the basic amplifier. That is,thegain A doesnot depend onanyof these threenet-works. In practicethiswillnot be the .case, and weshallhave tofinda methodforcastinga realCIrcuitmtothe Idealstructure depicted InFig.10. 1. Fi gure10.1 also implies that thefor-wardtranstrusslOnoccursenttrelythroughthebasicamplifierandthereversetransmission occurs entirely through thefeedbacknetwork. Thegainof thefeedbackamplifiercanbeobtainedbycombiningEqs.( 10. 1) through (I OJ): A (10.4 ) I+A/3 ThequantityA/3 iscalledtheloopgain, anamethatfollowsfromFig.10. 1. Forthefeed-backtobenegative,the loopgain A/3must bepositive;that is,thefeedbacksignal xfshould have the same sign as x"thusresulting ina smaller difference signal x,. Equation (10.4) indi-catesthatforpositive A/3the gainwithfeedback Af willbesmaller thantheopen-l oop gain A bya factorequaltoI + A/3,whichis called theamount of feedback. If,asisthe case m many circuits, theloop gain A/3is large,A/3I,thenfromEq. (IDA) itfollowsthat I Af=7J(10.5) whichIS a verymterestmg result:The gam of the feedback amplifier is almost entirel)' deter-mined by the feedback network. Since thefeedbacknetworkusually consists of passive com-ponent whichusuallycanbechosentobeasaccurateasonewishes.theadvantageof negattvefeedbackIn obtaining accurate.predictable,andstablegainshouldbeapparent.In other words,theoverall gainwillhave veryItttle dependence on thegainof thebasic ampli-fier,A,a desirablepropertybecause thegain A isusuall y a functionof manymanufacturing andapplicationparameters, some of whichmight have wide tolerances. Wehaveseena dra-maticillustrationof allof theseeffectsinop-ampci rcuitsinChapter 2.wheretheclosed-loopgain(whichisanother nameforthegain-with-feedback) is almostentirely detemlined bythefeedbackelements. Equations(10.1)through (10.3) canbecombined to obtainthefollowingexpressionfor thefeedbacksignal Xi .'1= A(J. I+ A(J " ( 10.6) lin earlier chapters, we used the subscn pt "sig" for quantities associated with the signal source (e.g . I' and R",).WedId thaito avoidconfusionwithIhesubscnpl "s." which is usuall y usedWIth FETs 10 de-note quantitI es associated withthe sourcetermInal orthe transistor. At thi s point. however, it is expected that readershave become sufficiently familiar wi thIhe subJecllhat thepossibility of confusionis min-Imal. Therefore.we willreverttousingthe si mpler subscript sfor SIgnal-source quantities. o o o 806Chapter 10Feedback o Thusfor AfJ:!>I weseethat'r= X .' whIchimplies that theSignal"at theInputofthe bas,crtier isreducedtoalmostzeroThusif a largeamount of negatIvefeedback is employed, signal xJ becomes analmost Identicalreplica of theinput Signal x.AnOutcomeofthIS propertyisthetrackingofthetwoInputtermInalsof an amp.Thedifferencebetween xandx/,whichis x.'IS someumesreferredtoas .theer.rorAccordingly. theinput differencing circuit isoftenalso called ac.ompansonCirCUIt . (ItIS al so known asamixer.l An expresSIOnfor XicanbeeaSIlydetellllInedas I x,=I + A fJx,(10.1)fromwhichwecanvenfythatforAfJ :!> I. xbecomesverysmall .Observethatnegati,e feedback reducestheSIgnalthat appearsattheinputtellllInalsof thebasic amplifier b)the amountof feedback.(I+ A/lJ.Aswillbeseenlater.itisthisreductionof Inputsignalthatresults Inthe increased lInearityof thefeedbackamplIfier Thenomnvertingop-ampconfi gurationshownInFi g.10.2(a)pro, idesadirectImplementatIOn of thefeedback loopof Fig. 10. 1 R \+ R --R --(a) R. 0 + \ V+ --R --R. --rb) Figure 10.2(a)A nonlnvertlng op-ampCITCUII forExampteto. t withIts equivalentcircuit. \ R, --I' AIR, --hmp replaced(b)The ClfCUItIn(a)withteop-a ,....fCE amplifierWehave In fact already seen exampl esof Ihl s:adding a reSlSlanceR111 theemItter aafor( '..'.Ie . becauseor a reSIstance RIn Ihe SOurce of a CS amplifi er) increasesthe hneanty of these ampIIlers thesameInpUI SIgnalas before.v.and vare now smallar (bythe amountof feedback). 10.1 TheGeneralFeedbackStructure807 (al Assume that the op amphasinfimte InpUIresistance andzero output reSIstance. Findan expressionfor thefeedbackfactor fJ(blFindthe conditionunder whichthe closed-loop gainAJ is almost entirely determlOedby thefeedback network. (c)I f the open-loop gainA=10'VN. findR,IR,toobtain a closedloop gainAfof 10VN. (d) What isthe amount of feedbackin decibels? (e)IfV,=IV.findVo.Vf' andV,. (f) If A decreases by 20%, whatisthe corresponding decrease inAf? Solution (al Tobeabletoseemoreclearlythedlfect correspondencebetweenthecircuitinFig.10.2(a) andthe blockdiagramInFig.10. 1, wereplacetheop ampwithitsequivalentcircuitmodel. asshowninFig. 10.2(b).Sincetheopampisassumedtohaveinfiniteinputresistanceandzerooutputresi stance.its modelissimplyanidealvoltage-controlledvoltagesourceof gainA.FromFig.10.2(b)weobservethat thefeedbacknetwork,consisting of thevoltagedi vider(RI,R, ).isconnecteddirectlytotheOUlputand feedsa SIgnalVf tothe inverting inputtermi nal of theopamp.ltisimportant atthispointtonotethatthe zero outputresi stance of the op ampcausesthe outputvoltagetobe A V,irrespectiveof thevalues of R I andR,andof RL.Thatiswhatwemeantbythestatementthatintheblockdiagramof Fig.10. 1. the feedbacknetworkandthe load areassumednot to loadthebasicamplifier.Nowwecaneasily determine thefeedbackfactorfJ from p=!i =RI Vo RI+R, Let ' snextexaminehowVf IS subtractedfromV,atthetnputside. ThesubtractioniseffectIvelyper-formedbythedifferentialaCllonof theopamp;byits verynature.a differential -tnputamphfier takesthe dIfference betweenthe signalsatustwoinputterminal s.Observe alsothatbecausetheinputresistance of theopampIS assumedtobeinfinite, no currentfl ows tnR,. Thusthevalueof R,ha, nobearing onI ', ; orthesource"doesnotload"theamplifierinput.SimIlarly.becauseof thezerotnputcurrentof theop amp.Vf will depend only ontheralloRIR,andnot on theabsolutevalue, of RIandR, (b) The closed-loop gainAfisgIvenby TomakeAfnearlyIndependent of A,wemust ensurethattheloopgain AfJi' muchlarger thanunity. Sinceunder such a condillon. the condItIon canbe stated as AP :!> 1 A(RI):!>I R, +R. I Ai7J = R, +R, RI R, - 1+-= RI (c)For A=\04 VNandAf=10VN. weseethatA:!> AJ thuswecanseleciRI I P =- =0. 1 If andR,toobtain 808Chapter 10Feedback Example10.1continued Thu,. whJehYield, R, . R, =9 -A moreexact yaluefortheratioR1! R Icanbeobtamed from whichresultsin and. (d) The amount of feedback" whichis60dB. (e)ForI '=IV. 10= A I+AfJ 10' 1+10'fJ fJ =0.0999 R, 9.0 I- -= R, I+AfJ=:i-A, 10' 10=1000 J',=AJ J'=lO xI =10V JJ=fJl o=0.0999 x10=0.999V V=~=10=0.001V ,A10' NotethatifwehadusedtheapproximatevalueoffJ =0. 1, wewouldhaveobtamedJf =IV andJ',=OV, (f) If A decreases by20"k,thusbecoming A =0.8x10'VN thevalue of A!becomes 0.8 x10' AI=- - - - ' ~ ~ ~ - - =9.9975V' V I +0.8x10' x 0.0999 thatis,It decreasesbyO.025"k,whichIS lower thanthepercentage changeInAbyapproximately a factor (I+ A fJ). -10.2 Some Properties of Negative Feedback809 10.1RepeatExample10. 1,(c)to(f) forA=100VN. Ans.(c)10.11; (d)20dB;(e)IOV,0.9V,0.1V;(f)2.44% 10.2RepeatExampleIO.I,(c)to(f) forAf=10'VN. For(e)useV, =0.01V. Ans.(c)1110. 1; (d) 20dB;(e)10V, 0.009V,0.001 V; (f)2.44% 10.2SomeProperties of Negative Feedback Thepropertiesof negativefeedbackwerementionedintheIntroduction. Inthefollowing, weshallconsider some of theseproperties inmore detail. 10.2.1GainDesensitivity Theeffectof negativefeedbackondesensitizingtheclosed-loopgainwasdemonstratedin Example10. 1,wherewesawthata20%reductioninthegainof thebasicamplifiergave risetoonlya0.025%reductioninthegainof theclosed-loopamplifier.Thissensitivity-reductionproperty can be analytically established asfollows. AssumethatfJ isconstant. Taking differentials of bothsides ofEg.(10.4)results in d4=dA I(I+AfJ)2 ( 10.8) DlVldmgEg.(10.8) byEg.(lOA) yields dAr_IdA Af(I+AfJ)A (\ 0.9) whichsaysthat thepercentage changem AI (duetovariationsinsomecircuit parameter)ISsmaller than the percentage change inA by a factor equaltothe amount of feedback. For thiS reason,the amount of feedback,I+ AfJ,isalso knownasthe desensitivity factor o 10.3 Anamplifier witha nom malgam~=1000V'Vexhibitsa gamchangeof 10% astheoperating temperature changesfrom25'C to75CIf it IS reqUiredtoconstramIhechangeto0.1 %byap-plying negativefeedback,whatisthelargest closed-loop gampOSSible?If three o;lhese feedback amplifiers areplacedincascade, whato,erall gamandgamstabllt!}areacllle,ed Ans.10VN;1000VN,WithamaxmlUm,anability of 0.3%overIhespeCifiedtemperature range. 810Chapterl0Feedback 10.4 o o 10.2.2BandwidthExtension Consideranamplifierwhosehi gh-frequencyresponseis characteri zedby a single pole gainatmidandhighfrequencies canbe expressed as AM A (s)=-I-+-s"' l :-cv-'H (10.10)where Alf denotes the midband gainandcvH isthe upper 3-dBfrequency. Application of neg.ativefeedback.withafrequency-mdependentfactorP.aroundthisamplIfierresults in a closed-loop gam AJs) givenby A(s) A!.s)=1+ PAis) Substituting forA (s)fromEq. (10. I 0)results.after a littlemanipulatlon. in A!.s)=(10.11)Thus thefeedbackamplifierwillhavea midbandgam ofA All( I + A MfJ)andan upper J.dBfrequencymHIgiven by (10.12)Itfollowsthattheupper3-dBfrequencyIS increasedbyafactorequaltotheamountoffeedback. Similarly. itcan be shown thatif the open-loop gamis charactenzed by a dominant low frequencypolegivingri seto alower3-dBfrequencycv,.thenthefeedbackamplifierwillhave a lower 3-dB frequencym'l" (IO.IJ)Notethattheamplifier bandwidthis increasedbythesamefactor by whichits midbandgainis decreased. mailltallling the gainbandwidth product at aconstantvalue.This p01n1 is furtherillustratedbytheBodePlotinFig.10.3. Finally. notethattheactionof negativefeedbackinextendingtheamplifier bandwidthshouldnotbesurprising: Negativefeedbackworkstominimi zethe changeingammagnl' tude.includmg its change withfrequency. Consider thenon inverting op-amp CIrCUIt of Example10. 1. Lettheopen-loop gam A have a lowfre quencyvalue of 10' anda un,form6-dB/octaverolloffat highfrequenciesWitha 3-dBfrequency of 100Hz.Fmdthelow-frequencygainandtheupper3-dBfrequencyof a closed-loop amplifier WithR, = I kGandR, =9 kG. Ans.9.99VN;100.1 kHz ----10.2Some Propert,es of Negat,ve Feedback811 Gam (dB) I __________ _____--;;-____.!. )_ ,t',, ,I.20 dBldecade ,, ,20 log( ItI Ifll I,'0 dB/dc'.ldc ,, ,, ,, ,, ---....,; 20log(..1,,,1, ( ,, ,,, , I , ,, ,,I log(ItA"IlI,Ilog lltI ifill ' .'' .. ' ,,I , ,,I , I , ,,, , I , ,,, ,I ftfH fHI f(logscale) ("ItI ifill fl' -I+'\"IlI Figure 10.3 Applicallonof neg alivefeedback reduces Ihemidbandgam.mcreasesfll' andreduces/"allby Ihe samefactor. (I +A.,/J), whICh isequal to Ihe amounlof feedback. 10,2,3InterferenceReduction Negati vefeedback can beemployed to reduce themterferenceinanamplifier or.morepre-cisely.to increase the ratio of signal to interference. However,asweshallnowexplain.Ihls mterference-reducti on processis possibleonlyunder certam conditions.Consider Ihesitua-tion illustrated inFig.l OA.Fi gure10.4(a) showsanamplifierwithgainA" aninputsignal V,. andinterference,V,.ItISassumedthatforomereasonthisamplifiersuffersfrom IOlerferenceandthat theinterferencecanbeassumedtobeintroducedattheinput of the amplifier.The signal-to-i nterference ratioforthi s amplifier is S/1 = V,I v., (10.14) Consider next the ci rcuit inFi g.10.4(b).Hereweassumethatitispossibletobuildanother amplifier stagewith gainA,that does not sufferfromtheinterference problem. If thisis the case,then wemayprecede ourori ginalamplifierA,bythecleanamplifierA,andapply negative feedback aroundthe overallcascadeof such anamounlas tokeepthe overallgam constant.The output voltage of the circuitinFig.10.4(b) canbefoundbysuperposition: A,A,VA, v;,=V. 1 +A ,A, p+" I +A,A,p Thus the stgnal -to-interference ratioaltheoutput becomes SV - , IV. , (10.15) (10.16) whichisA,timeshigher thaninthe ongi nalcase..'. We emphasize once moreIhal the improvemenl10s,gnal-to-.mterferenceratIObyIhe applicati onof feedbackispossibleonl y if one canprecedetheIOlerference-pronestage 812Chapterl0Feedback v, + -+A, - -v, tI AI (a) -( {3(b) I AI + I Figure10.4Illustratingthe applicationof negativefeedbacktoimprovethesignal-la-interference ratio Inamplifiers. bya(relallvely)Interference-freestage.ThisSltuallOn,however,ISnotuncommon InpracticeThebestexampleisfoundintheoutputpower-amplifierstageofan audioamplifierSuchastageusuallysuffersfromaproblemknownaspower-suppl)humTheproblemarisesbecauseof thelargecurrentsthatthisstagedrawsfromthepowersupplyandthedifficultyofprovidingadequatepower-supplyfilteringinexpensivelyThepower-outputstageI,reqUiredtoprovidelargepowergainbutlillieornovoltagegain.Wemaythereforeprecedethepower-outputstagebya,mall-signal amplifier thatprovideslargevoltagegain,andapplyalargeamount of negativefeedback,thus restoringthevoltagegaintoIIsoriginalvalue.Sincethesmall-signalamplifiercanbe fedfromanother,lessheft y(andhencebellerregulated)power,upply,itwillnotsufferfrom the humproblem.Thehumattheoutput willthenbereducedbytheamount of the voltagegainof thisaddedpreamplifier. 10.5 Consider a power-output stage \\lIh voltage gainA I =I,anInputsignalI=IV,anda humI;of IVAssumethatthiS power stageIS precededby a smail-signalstagewithgam~ , =100VNandthat overall feedbackwith fJ = Iis applied.I fIand,"remamunchanged, findthesignaland mter-ferencevoltages at the output and hencethe ImprovementinSf! Ans,= IV;= 0.0 1 V;100140 dB) 10.2Some Properties ofNegativeFeedback813 Vo (V) - 0.08- 0.06- 0.04-0.02 4 3 I - 3 - 4 (a)(b) 0.020. 08 ,VIFigure10.5Illustratingtheapplication ofnegauve feedbackto reducethe nonlinear distortionInampli-fiers.Curve(a)showstheamplifiertransfercharacteristi c(voversus1'/) withoutfeedback.Curve(b) showsthe characteristic (vo versus vs) withnegativefeedback(P=0.01)appli ed. 10.2.4ReductioninNonlinear Distortion Curve(a)inFig.10.5showsthetransfer characteristic Voversus v, of an amplifier.As indi-cated,the characteristic ispiecewise linear, with thevoltagegain changing from 1000 to100 andthentoO. Thisnonlinear transfer characteristic willresultinthis amplifier generallng a largeamount of nonlinear distortion. Theamplifier transfer characteristic canbe consi derablylinearized (i.e ..made lesnon-linear)throughtheapplicationof negativefeedback.Thatthisispossible shouldnot be too surpOSlng,since wehave already seenthatnegativefeedbackreducesthe dependenceof the overallclosed-loopamplifier gainontheopen-loopgainof thebasic amplifier.Thus large changesinopen-loop gain (1000to100 inthis case) give risetomuch smaller corresponding changesIn the closed-loop gain. Toillustrate,let usapplynegative feedbackwithfJ = 0.0 I to the amplifier whoseopen-loopvoltage transfer characteristicisdepictedinFig.10.5. The resulting transfer characteris-ticofthe closed-loopamplifier,Voversus vs'is showninFig.10.5ascurve(b).Herethe slope of the steepest segment is givenby 1000 A/I=1+IOOO xO. OI= 90.9 andtheslope of thenext segmentIS givenby 100 Af2= -I+--:-:10""0c::.x--::0--:: .0:71 =5 0 Thus theorder-of-magnitude change inslope hasbeen considerably reduced. The pricepaid, of course,isareductioninvoltagegain.Thusiftheoverall gainhastobe.restored,a preamplifier shouldbeadded. This preamplifier shoul d not presenta severe nonhnear-d" tor-lionproblem,since itwillbe dealingwi thsmaller signals. 814Chapterl0Feedback Finally.It shouldbenotedthatnegativefeedbackcandonothingatallaboutam lifisaturation.sincem saturatIOnthegamIS verysmall(almostzero)andhencethPlet.e amOunt offeedbackisalmostUnIty. 10.3 TheFourBasicFeedback Topologies Basedonthequantitytobeamplified(voltage or current)andonthe desi redform of or current).amplifiers can beclassifiedintofour categories.These discussedm ChapterI.[nthefollOWIng.weshallreviewthiS amplifierclassificat' ...POInt outthefeedbacktopologyappropnate Ineach case. 10.3.1Voltage Amplifiers VoltageamplIfiersareIntendedtoamplifyaninputvoltageSignalandprovide anoutvoltagesignal.Thevoltageamplifierisessentiallyavoltage-controlledvoltage The input resistanceisreqUiredto be high.andthe output resistance isrequiredtobe I 0'. Sincethesignalsourceisessentiallyavoltagesource.ItisconvenienttorepresentIt intermsof a Thevenin equivalent circuit.[n avoltageamplifier.the output quantity ofinter estistheoutputvoltage.[tfollowsthatthefeedbacknetworkshouldsampletheoutputvoltage. Just asa voltmeter measures avoltage.Also.because of the Thevenin representatlOnof thesource.thefeedbacksignalXIshouldbeavoltagethatcanbemixed with thesource voltageinseries. Themostsuitablefeedbacktopologyforthevoltageamplifieristhevoltage-mixing. voltage-sampling one showninFig.10.6. Because of the series connectIOnatthemput andtheparallelorshuntconnectIOnattheoutput.thisfeedbacktopologyisalsoknown :II series-shunt feedback.Aswillbeshown.thistopologynotonlystabilizes thevoitagegain butalsoresultsIna higherInputresltance(IntUitively.aresultof theseriesconnection attheinput)andaloweroutputresistance(intuitively.areultof theparallelconnection atthe output).whichare deSIrableproperties fora voltage amplifier. R, + - voltageR, I ' Iamplifier J Iv, IIFeedback 1 , -) network Figure10.6BlockdI I5 .hlagramaafeedbackvoltageamplifierHereIheappropriatefeedbacktOpO 01) I senes- -5 unt.10.3 TheFour Basic Feedback Topologies815 Themcreasedinputresistance becauseVIsubtracts from V,.resulting in a smaller signalV,atthemputof the basIcamplifier.Thelower V,.in tum. causes the input currentto besmaller.Withtheresultthatthe resistance seen byV,willbe larger.We shallderive a for-mulaforthe inputresistance of the feedbackvoltage amplifier inthe next section. Thedecreasedoutpmresistanceresultsbecausethefeedbackworks tokeepV. ascon-stantas If thecurrent drawnfromthe amplifier outputchangesbyMa.the change6Va '?VawIllbelower thanItwouldhave been if feedback were not present.Thus theoutputresistance6V/Ma Willbelowerthan that of the open-loop amplifier.Inthe fol-lowingsectionwe shallderive anexpressionforthe outputresistanceof the feedbackvolt-ageamplifier. Three examplesof series-shunt feedbackamplifiers areshown inFi g.10.7.The ampli -fierinFig.10.7(a)isthefamiliarnoninvertingop-ampconfi gurati on.Thefeedbacknet-work.composedof thevoltage divider (R,.R,). developsa voltageVthat is appliedtothe negativeinput terminalof the opamp.TheubtractionofVIfromV: is achieved by utiliz-ingthedifferencing action of the op-amp differentialmput.For the feedback tobenegative. VImustbe of thesamepolarityasV, .thusresultingina smal ler signal atthe inputof the basicamplifier.ToascertainthatthisIS thecase.wefoll owthe signal aroundthe loop. as follows: AsV,increases,Vaincreasesandthevoltagedivi der causesVIto increase.Thus the change inIf is of the same polarity asthe changeinv,.andthe feedback is negati ve. Thesecondfeedbackvoltageamplifier.shownmFig.10.7(b).utili zestwoMOSFET amplifier stages incascade. The output vol tagev" issampled by the feedback networkcom-posed of thevoltage divider (R, R2).andthe feedbacksignalVIis fed to the sourcetermi -nalofQ"ThesubtractionisimplementedbyapplyingV,tothe gate ofQ,andVItoits source.withtheresultthatthesignalatthisamplifier inputV, =Vg,=V,- VI'To ascertain thatthefeedbackisnegative.letV,increase.ThedrainvoltageofQ,willdecrease. and sincethisisappliedtothegate ofQ2'itsdrainvoltageVawillincrease.The feedback net-workwillthencauseVItoincrease.whichisthesamechangeinpolarity initially assumed forV,. Thus thefeedbackisindeednegative. ++ Q, V tT I Rz Rz \' V, IR, IR, -- - -- - - -- -' whichistheresistanceseenbyRLmthecircuitof Fig.10.22(a),wesubtractRL fromRol ' Roo> ="02(I+ A Igm2RF) whichisanintuitivelyappealing result:The senes connection atthe output raisesthe outputresistance of Q2("02) bya factor equalto theamount of feedback.... Finally,wenote thatwehavedeliberatelysolvedthisproblemm greatdetailtoIllustrate the beauty 01013Forthe circuit analyzedinExample10.5, select a value forRF that willresultinAI=5 mAN. Now,forA,=200VN,gm2=2mAN,R,d=100kO," 0' =20kO ,andassumingthat R, , Forthecalculationof Roo>, assume that ,.oof Q3 IS 25kO . Solution (a)When AfJ~I, I =-fJ wherethefeedbackfactorfJcanbefoundfromthefeedbacknetwork.Thefeedbacknetworkishigh-li ghtedin Fi g.10.23(a), andthe determination of thevalue of fJis illustrated inFig.10.23(b), fromwhich wefi nd + \ + V, --Re i=9kn --(b) --100x100=I 1.90 100 +640 +100 1 R, --) ) Re) =600n i,=:::::;---- ---I, R R, ------R (a) --(b) - -- -(C) \ R, --Vo + \ - -- -Figure 10.27(a)A feedbackIransreSi stanceamplifier;(b)the !3CJrcuit; (e)theA ClfCUIt. t \ --10.6TheFeedback Transresistance Amplifier (Shunt-Shunt)851 (a)If theloopgamIS large.findanapproximateexpressionforthec10sdI00fth C dbkI' fi e- oopopen-clfculltransresls-tance 0e leeacamp1 ter. (b)Findthe A Clfcuit andexpressIOnsforA.R,. andRo'(c)Find expressions fortheloopgam.AIRRRand R .Ii'In'0/'out' (d)Findthevaluesof R, .Ro.A./3.AI'R'I'Rm.Rol'andRoO! forthecasef.1 =10'VNR ro =100Q,RF=10kQ,andR,=RL =IkQ.,,/=(e)If insteadof a current sourceIhavinga sourceresistanceR- I,,-,theaI' fi .fd f.l..f- mp1leT IS eroma voltage V,havmgasourcereSistanceR, =IkQ.findanexpressionforandthevalueof the voltage gamVol V, . Solution (a) If the loop gainA/3islarge. Vo I AI=- =-I,/3 where/3can be foundfromthe/3circuitinFig. 10.27(b)as Thus. I{I /3=-=--Vo R, Vo- =-RF I, (10.48) Notethat inthi s case thevoltage attheinputnode (lhe inverting inputterminalof fl)willbeveryclose to groundandthusverylittle. if any.currentflowsintotheinputtenninal of theamphfier.NearlyallofI , willflowthroughRF resultinginVo =0 -I,RF =-I,RF' Thisshouldbereminiscentof theinverting op-amp configuration studiedinSection 2.2. (b)SincethefeedbacknetworkconSists ofR,. theloadingeffectattheamphfierinputandoutputwill simplybeRF.ThisisIOdicated10theAcirCUit showninFig.10.27(c). Theopen-looptransresistance A canbe obtained asfollows: where R,=R'd IlRFIIR, (RF II RL) Vo =-flV,dro +(RFIIRL) CombinlOgEqs. (10.49)and(10.51) gives A = =-f.1R (RFIIRL) I,'ro+(RFIIRL) (10.49) (10.50) (10.51) (10.52) The open-loop outputresistance canbe obtamed byinspecuon of the A circuit withI,setto O. Weseethat V,d=O.and Ro =rollRFIIRL(c)Theloop gamA/3can be obtained bycombiningEqs.(10.48) and(10.52), (R, )(RFIIRL) A/3=flR"ro+( RFIIRL) ( 10.53) 852Chapter 10Feedbac k Example10.7continued ObservethatalthoughbothA andjJarenegative.AjJispositive,acomfortingfactconfirming thatthe feedbackisnegative.AlsonotethatAjJisdimensIOnless,asitmust always be. The closed-loop gamAJcannowbe foundas f'o A AJ= I, =1+ AjJ Thus (10.54) Notethatthecondition of A/l pIwhich resultsInAJ =-RF corresponds to (10.55) The input resistance withfeedback,R'J' isobtainedby diVidingR,by(I+ AfJ>withtheresult R_R, 'J - I+ AjJ or IIij}Ii!... -=-+=-+ R'JR,R,R,RF (RF II RL) 1'0+ (RF II RL) SubstitutingforR,fromEq.(10.50)andreplacing,II(RF II RL)[1' 0+ (RF II RL)]by,II',wherefl ' IS lower thanbut usually close tothevalue of ,II, resultsin. R,, = R,dRF II R. II (RF ,II' ) The twoterms containingRFcanbe combined, R,r=RII R,dlllRFI (,II'+I)](10.56) SinceR'J =RII R". wesee that UsuallyR,dISlargeand thus R=R,='!.!: Inp'+Ip' (10.57) fromwhichwe observe that for largeamplifier gain,II, theInput resistance Willbe low. The output resi stancewithfeedbackRoJcanbe foundby dividingRoby(I+ A fJ>R_Ro oJ- I+AjJ Thus, 10.6The Feedback Transresistance AmplIfier(Shunt-Shunt)853 Substi tutingforRofrom Eq. (10.53), IIIIR,I =-+-+-+,11--RoJRLRFroRFro III (R, =- + - + - I+ ,11-RLRF1'0R Thus, Si nce,moreover, weobtam for RoUifrom whi ch wesee that for large ,II, the output resistance willbe considerablyreduced. (d)For thenumeri cal values given: R,=R'd IlRFII R, =~1110 III=0.91kQ Ro =ro II RFIIR, =0. 1 1110III=90Q (RF II Rd A=- ,IIR' l'o+(RFIIRL) _- 10' x 0 91 x(10 III )=-8198kQ - .0.1+(10111) jJ=-i=-AjJ=819.8 I+AjJ =820.8 I - =-0.1mAN 10 =A=_ 8198=-9.99kQ AII+ AjJ820.8 whichisvery closetotheidealvalue of - RF =- 10kQ. R,_910_1.11Q R'J=I +A/l- 820.8-I I =I.IIQ R;o - --II II ----R'J R, 1.111000 854Chapter 10Feedback Example10.7 continued which is verylow, a highl y desirableproperty. We alsohave R ., - Ro=90=0.1\0 0,- I+AP820.8 I Ro", =II - -=_ ~ I ' - - - , - - =O. \I0 II 0. 1\1000 which as wellis verylow, another hi ghly desirable property. (el If the amplifier is fedwitha voltage sourceV,having a resistanceR,=IkO , the outputvoltage canbe foundfrom Thus, ~=~=_9.99 kO=-9.99VN V,R,IkO 10.15For the transresistanceampli fie r m Fig. E10.15,replacetheMOSFETwithits equlvalentClrcult modeland usefeedback analysIs to showthe followmg: t1 (ideal) _Q R(lU'R, R, - -- - --(a)Forlargeloop gam(which cannot be achi eved here)A ,f (b)A=-( R, II R,}gm("o II R,) FigureE10.15 "V.l I ,=- R,. fI+ (R,IIRf }gm("o II RrJI R, R (c)R=( m[I +gm( roIlR,) ] 10.7 TheFeedbackCurrent Ampli f ier (Shunt-Series)855 _IIR, (d)Ro", -"01+ gm(R,II R,} (e}Forgm=5rnAIV' ''o=20 kO , R,=10 kO , andR,=IkO , find A, p , Ap,A, R" Ro' R./, RIOIRo/'andR out-Ans.(e)-30.3kO;- 0.1rnAIV;3.03;-7.52kO(comparetotheIdeal valueof- 10 kO ); 9090; 6.67kO; 2260 ; 2910 ; 1.66 kO; 1.66 kQ ---- -- - -- - ----- -.-.--10,6.3AnImportant Note Thefeedbackanalysis methodispredicatedonthe assumption tbat all(or most) of thefeed-forwardtransmissionoccursinthebasicampli fierandall(or most) of tbefeedback trans-mIssionoccursinthefeedbacknetwork.TheClfCllltconsideredinExercise10.15above is SImpleandcanbe analyzed directly (I.e.,withoutmvokingthefeedbackapproach) to deter-mineA,.In thiswaywe cancheck thevalidity of our assumptions.This pointis illustrated toProblem10.58, where we findthatforthe ci rcuit m Fig. E I 0.15, allof thefeedback trans-miSSIOnoccursinthefeedbackcircuit.Also,aslongasgm is much greaterthan1/ R"the assumption thatmost of thefeed forwardtransmi ssion occurs inthebasic amplifiers is valid, andthusthe feedback analysisisreasonably accurate. 10.7 TheFeedback Current Amplifier (Shunt-Series) 10.7.1TheIdealCase AsmentIOnedm Section10.3, theshunt- serIesfeedback topologyis best suitedfor current amplifiers: The shunt connection at the mputreduces theinputresistance, makingiteasier to feedtheamplifier witha currentsignal;thesampling of outputcurrentstabilizes10 ,whi ch istheoutputsignalina current amplifier,andthe series connectionattheoutputincreases theoutputresistance,makingtheoutputcurrentvaluelesssusceptibletochangesinload resIstance. Figure10.28(a) shows the ideal structure for the shunt- series feedbackamplifier.Itcon-sists of a unilateral open-loop amplifier (the A circuit) andan idealfeedback network. The A circuithasan inputresistanceR"a short-circuit current gainA ,, 1/Ii ' andanoutputresi s-tanceRo. Thep ci rcuit sampl esthe short-circuit output current10andprovides a feedback currentIfthat issubtracted from the signal-source currentI,at theinput node. Notethat the f3 circuit present s a zeroresistancetothe outputloopandthus does not loadtheamphfier output.Al so,thefeedbacksignalIf=PIo is proVIdedasan idealcurrentsource,andthus the P ci rcuit doesnotloadtheamplifier input.Al so observethat bothA andp arecurrent gains andAPis a dimensionlessquantity. Finally,note that the source andloadresistances have been absorbedinsidethe Acircuit (more onthislater). Smce thestructureofFig.10.28(a}follows theIdeal feedback structure of Fi g.10. 1, we can obtainthe closed-loop current gamA,as A I+AP (10.59) Thefeedbackcurrent ampli fiercanberepresented by the equi valentcircuitinFig.10.28(b). 856Chapter 10Feedback o o I, S' S S' ------------ -II,) t1 J I_ R,AI,R" 4ClfCUlt /' I I I , -----------, I I I L-____I __-' I I I I L- _________ __f3 Ctrcult II _ _ Rf Al l.,R ol 0' (b) I.0 0' Figure 10.28(a)IdealstruclureforIheshunl-seriesfeedbackamphfier(b) Equlvalenl CIrculi oftheamplifier In (3). NotethatA,istheshort-circuitcurrentgain.TheInputresistanceR" IS foundbydividingRby(I +A/ll, "hlch IS a result of theshunt connectionattheinput.Thus, R_R, 'f- I+AjJ (10.60)TheoutputresistanceRoJ IS theresistanceobtainedbysettingI=0,breaking theshoncirCUitoutput loop,at say00' , andmeasuring theresistance between thetwotenmnals Ihuscreated.Since theseriesfeedbackconnectionalwaysraisesresistance, wecan obtainR,( bymultiplYingRoby (I+ A/ll, ( 10.61)10,7.2 ThePracticalCase .I fi To beFigure10.29showsa blockdiagramforapracttcalshuntsenesfeedbackampI ler abletoapplythefeedbackequattonstothisamplifier,wehavetorepresentit bytheIdealstructureof Fig.10.28(a).OurobjectivethereforeistodeViseaSimplemethodforfindmgtheAandjJCIrCUitS.Buildingontheinsightwehavegainedfromthestudyof thethfeeother topologies,wepresent themethodfortheshuntsenes casewithout derivation, In Fig10.30. AsInprevIOuscases,themethodof Fig. 10.30 assumes thatthebasiclateral (or almost so) andthat thefeed forwardtransmissioninthefeedbacknetworkIS n g gibly small 10.7 TheFeedback CurrentAmplifier(Shunt-Series)857 I, t R, RiJ RonBasic amplifier Feedback network 10 E YY'L I" RoJRoUiFigure 10.29Block dIagramfor a pracllcalshunlsenesfeedbackamphfier. (a)TheACIrculiis I, t R, R,whereR!Iisobtainedfrom CD R Feedback network andIhegamAISdefinedas I A - 2 I, ( b)f3 IS oblatnedfrom f3'L 1>0 Basic amplifierR" andR"ISobtainedfrom Feedback network CD Feedback network RLyY' I" Figure 10.30Finding the A circuit and f3 for the current-mixmg current-sampling {shuntseries} feedback amplilier of Fig.10.29. AsmdicatedInFig.10.30.theAcircuitIS obtainedbyIncludingR,acrossthetnputtermi-nals of theampltfier andRLInsenes withits output loop.Theloading effect of thefeedback networkon theampitfierinputIS representedbytheresistanceR II_ andItS loading effect at theamplifier outputisrepresentedbyresistanceRn.Thevalueof RIIisobtained by look-IngintoportIof thefeedbacknetworkwhtleitsport2 IS open-cirCUited(becauseIt IS con-nectedinseries).ThevalueofR 22 IS obtainedbylookingintoport2ofthefeedback R 858Chapter 10Feedback network whileIts portIIS short-Clfcuited (because ItIS connected In shunt ),Finally, observethatsincethefeedbacknetworksenses10 , ItIS fedbya currentl a'andSinceIt deliv currentI,thatIS mixedtnshuntattheinput,its portIis short-circUIted andf3isf o u : : ~ IIwhereIIS the current thatflowsthroughthe short cIrculI. ,"I'. Theopen-loopresistancesR,andRoaredeterrmnedfromt ~ eAcIrcuItasindicated. ObservethatRaIS foundbybreakingtheoutputloopatsay} Yandmeasunng the resis.tancebetweenYandY'ResistancesR,andRoarethenusedinEqs.( 10.60)and(10,61), respecttvely.todetermtneR'fandROf Finally,theresIstancesRoo andRoUi thatcharacter.izedthefeedbackampltfierareobtainedfromR'iandRafbyreferencetoFig,10,29,asfollows: ( 10.62)( 10,63)Figure10.31shows a feedbackcurrent amplifier formedby cascading aninverting voltage amplifier Jlwitha MOSFETQ.The output currentlaisthedraincurrent of Q.Thefeedbacknetwork,con sisting of resistorsR,andR2 sensesanexactly equalcurrent,namely,thesource current of Q.and provIdes a feedback current signalthat ismIxedwIthIat theinput nodeNote that thebIas arrange-ment isnot shown. --'f f RR --/l Q -I R, I + R, R, (a) Figure 10.31CirCUit for Example10.8. --I, \o I,0 I ,+ --I , ~ R, R, R, --(b) ---+ -/l - -R, --(e) R, -R idR, 10.7 TheFeedback Current Amplifier (Shunt-Selles)859 tI" --R, + V,-- ---R IIt-' R, IR. /lV, R. -R, 860Chapter 10Feedback Example 10.8 continued TheamplIfier Jicanbeimplemented10a \'anety01'\\ a)"s,includmgbymeansof anopamp,a dif-ferentialamplifier.ora smgle-endedm\erting amplIfier.Thesimplest approachIS toImplement 11\\ Itha CSMOSFETamplifierHowe\er.10 sucha casetheloopgam\\ "I beverylimited. Assume thattheamplIfierJihasaninputresistanceR,d'anopen-circuitvoltagegainJi,andanoutput resIstancerol . (a)If theloopgamIS large.findanapproximate e"presslOnfortheclosed-loop gamAt"loll,. (b)Find the A circuit andderive e"presslOnsfor R. andR, (c) Give e"presslOnsforAp.A,.R". R'" . Ro,' andRoul '(d)FmdnumericalvaluesforA.p,AP.A,.R,.R.I' R,o'Ro'R""andR OUI lorthefollowingcase' Ji=1000YV.R=00.R d =00,1"0 1=Ikn. R=10kn.R,=90kn. androrQ' g.5mA YandI"=20kn Solution (a)When theloopgamAP1.Aj Thus. I Ip. To determineP refertoFig.10.31 (b). If P - - - - -I , RI RI+ R, I(R,) Af- 7J =\ I + II ( 1064) ( 10.65) To seewhathappensm thIScase more dearly. refer toFig.Ill., I (c).Herewehave a"umed theloop gamtobelarge.sothatI,0andthusI {= I ,AlsonolethatbecauseI ,().' ,\\" I bedosetozero. Thus.wecan easilydetermmethe\oltage atthesourceof Q asI, R,- IR,ThecurrentthroughRI WillthenbeI, R, IR IThesource current of Q \\ III be'(/, + I ,R, IR I ) \\ hllhmeansthattheoutput cur rent10 w1I1 be lo=/ l I+t ) whichconfirms theexpre"ion forA} obtamedabove (Eq.10.65). (b)ToobtamtheA circuit weloadtheinputsideof thebaSiCamplifier \\ IthRandRIIThelatter inthiscaseissimplyR,+ R,(becauseport2 of thefeedbacknetworkIS opened).Wealsoloadtheoutput of the basic amplifier withR ,.whichInthiscaseisR III R,(because portI of thefeedback networkisshorted).Theresultmg4 CIrCUitIS shownm Fig . IO.3i(d).wherewehavereplacedtheamplifierJiwithitsequivalentCIrcuit.AnalySiSof theACirCUitIS stralghtlilf\\ ardandproceeds asfollows: R=RII R,d II (R I+R,) ,- =I R ,, I , .I1',2 ='Ji ,'1 / gm+(R11IR,II .. 0')I"u +(RIIIR2) Combining Eqs.(10.67) and(IO.6R)resultsin.4 I" A_i, =Ji l1'", + ( R III R, ) (10.661 (10.67) ( I 0.68) (10.69) 10.7 The Feedback Current Amplifier (Shunt-Series)861 For the caseI' gm""(R, II R, lIro') ' R A- Ji' RI IIR,lIro'Whi ch reduces to A=- JiR,(10.70) R III R, Notmg thatRois the output resistance of Q,which hasa resl'stance(RII R).' tIdI,10I S sourceea. we can wnte (c)The loop gam IS obtamedby combmmg Eqs. (10.64) and(10.69). The mputreSistanceR'fIS foundas r 0'1 R I "u,+(RIII R,)RI+R, R'f =R,.(I+ A/ll -'-- =.!.+ :iP R'fR,R, (10.71) (10.72) (10.73) We can substitute forAPfrom the fullexpression10 Eq.(10.72).Fortheapproximatecase.weuseAP from Eq. ( 10.73). ThatIS,I - = R, R'f=R,II...; Ji SubslltutingforR,from Eq. (10.66).wewnte R, R'f=R,IIR",II (RI +R,) II Ii' SincebydefiOilion. wecaneasilyfindRmas (10.74) 862Chapter 10Feedback Example10.8 continued Usually the thirdcomponent onthe nght-hand sideisthesmallest; thus, R, R= -= 00 11For theoutputresistance,wehave Rol=Ro(I +AjJ)=APRo (10.75) Subst,tuting forRoforEq.(10.71)andforAP from theapproximate express tontnEq. ( 10.73), we have R, Rol=I1RR(gmro2) R, , +, -Finally, wenote that Cd) For thenumericalvalues given, SinceI / gm= R, ==100kQ 0.2 kQ 3 thisCIrCUit becomes unsta-ble.Thismight appear tocontradict our earlier conclusionthat thefeedback amplifier witha second-order responseIS unconditionally stable.Note,however, thatthecircuitinthisexampleisquitedifTerentfrom thenegallve-feedbackamplifierthatwehavebeen studying. Herewehaveanamplifierwitha positive gamKandafeedbacknetworkwhosetransfer functIonT(s)isfrequencydependent.ThisfeedbackIS III fact positive.and the circuit will oscillate at the frequencyforwhichthephase of T(jlO)iszero. Example10.9illustrates the use of feedback(positi vefeedbackinthiscase)to movethe polesof anRCnetworkfromtheir negativereal-axislocationstocomplex-conjugateloca-tlons. One canaccomplish thesametaskusi ng negati vefeedback.astheroot-locusdiagram of Fig.10.37 demonstrates. The process of pole control is the essence of active-filter design. aswillbe discussed inChapter16. 10.11.5Amplifiers with Three or More Poles Figure10.41showstheroot-locusdiagramforafeedbackamplifierwhoseopen-loop responseIS characteri zedbythreepoles.Asindicated.increasingtheloopgainfromlero movesthehi ghest-frequencypoleoutwardwhilethetwOotherpolesarebroughtcloser together.AsA"fiisincreasedfurther,thetwOpolesbecomecoincidentandthenbecome complexandconjugate.Avalue of A,fiexlStsatwhichthiSpair of complex-conjugatepoles enters the ri ghthalf of the spl ane, thus causmgtheamplifier tobecomeunstable. 878Chapter 10Feedback 10.23 . JW()( splane oa Figure10.41Root-locusdiagram foranamplifierwiththreepoles. The arro\\ S mdlcatethepolemovementas AoP ISIncreased. ThtSresultisnotenttrelyunexpected.sinceanampitfierwiththreepoleshasa phaseshiftthatreaches-270asWapproaches ThusthereeXIStsaI'lOttefrequency. atwhichtheloop gainhas180'phaseshift. Fromtheroot-locusdiagramof Fig.10.41.weobservethatonecanalwaysmaintainamplifier stabilitybykeeptngtheloopgam AofJ small er thanthevaluecorresponding to thepoles entering therighthalf-plane. Interms of theNyquist diagram.thecnticalvalue of AoPisthatforwhichthediagrampassesthroughthe(- J.0)point.ReducingAo/l belowthisvalue causes theNyquist plottoshrink and thus tntersect thenegative realaxts tothenght ofthe(-I. 0)pOtnt.tndicattng stable amplifier performanceOntheother hand. increasingA.,pabovethecriticalvaluecausestheNyquistplottoexpand.thusenctrclingthe(-1.0) pointandtndtcattngunstableperformance Fora givenopen-loopgatnA"theconclusionsabovecanbestated10termsof thefeedbackfactor fJ.ThatIS. thereexistsa lIIa.mnulIIl'aluefor fJ abovewhichthefeedbackamplifierbecomesunstableAlternatively.wecanstatethatthereexistsa IIIl1lillltllll\'altle for theclosed-loop gatnAjU belowwhtchtheamplifier becomes unstableTo obtain lower values ofclosed-loop gatnoneneedsthereforeto alter thelooptransferfunctionL(s). ThISisthe processknownas !requellc\compensalion.Weshallstudythetheoryandtechniquesoffrequency compensationinSection10. 13. Beforeleavingthtssecttonwepointoutthat constructtonof theroot-locusdiagramforampitfiershavtng three or morepoles aswellasfinttezerostS antnvolvedprocess for whicha systematic procedure exists.However. such a procedure willnotbepresented here. and thetnterestedreadershouldconsultHaykm(1970).Althoughtheroot-locusdtagramprovidestheamplifierdesignerwithconSiderableinsight.other.'tmpler techniquesbasedonBodeplots canbe effecttvely employed. aswillbeexplatnedinSection10. 12. Consider a feedbackamplifierforwhichtheopen-loop transferfunctton4(5)isgtvenby I A( .I)(10)3 1+5110' 10.12Stability Study Using Bode Plots879 Letthefeedbackfactor pbe frequencytndependent.Find the closed-loop poles asfuncttons of P. and show thattheroot locus isthat of Fig.EI0.23. Alsofindthevalue of pat whichthe amplifier becomes unstable. (Nole.This IS thesame amplifier thatwasconsidered m Exercise10.20.) JW60' 60 Ans.SeeFig. EIO.23;Pm'". = 0.008 splane(normalized to10' rad/s) o 10.12Stability Study UsingBodePlots 10.12.1Gain andPhaseMargins Figure 10.E23 .- -FromSecttons10. 10and10. II weknowthatwhether a feedbackamplifier is orisnotsta-blecanbedeterminedbyexamimngttsloopgainA/l asafunctionof frequency. Oneof thesimplestandmost effective meansfordoing thisisthroughtheuse of a Bodeplotfor AfJ. suchasthe one showninFig.10.42. (Notethatbecausethephaseapproaches -360.thenet-workexaminedisafourth-orderone.)Thefeedbackamplifierwhoseloopgainisplottedin Fig.10.42willbestable. since atthefrequencyof 180 phaseshift. WIS.' themagnitude of the loopgamtS lessthanunity(negative dB).Thedifferencebetweenthevalue ofIAfJlatwlSOandunity.calledthegainmargin. isusuallyexpressedindecibels. Thegainmarginrepre-sentstheamountbywhichtheloopgaincanbemcreasedwhtlestabthtytS mmntamed. FeedbackampltfiersareusuallydeSignedtohavesufficientgainmargintoallowforthe inevttable changesinloop gainwithtemperature,time. and so on.. Anotherwaytoinvestigate thestabiltty andto expresits degreeistoexamttle theBode plotatthefrequencyforwhichIA fJl=I.whichis thePOttltatwhtchthemagnttudecrossestheO-dB line.If at thisfrequencythephae angleis less(ttlmagnttude)than180 thentheamphfierIS stable.Thisisthesi tuationillustratedinFig.10.42.The.dtfference betweenthephaseangleatthi sfrequencyand180istermedthephase. margttl . Onth: .tdethephase lagtS m excess of180. other hand. tf atthe frequencyof unttyloop-gammag"'u. the ampitfier willbeunstable. 880Chapter 10Feedback 10.24 IAPIdB Gainmargin Wl80 0-----w,I : 1\1 II II II w(logscale) w,II o - - ; ; ; , _ ~ : . c--tl-:- w:- ,,-o----w-(-Io-g-s-c-ale) ~ o "-00 " '" " ~ '" -90 -180 .c-2700 0..-360 I I t Phasemargin Figure 10.42Bodeplotfortheloopgam AfJ ,lIustratmg the defimttons of thegam andphasemargins Constder anopamphaving a Single-pole.open-loopresponsewith Ao =10and,t;,=10Hz. Letthe opamp beIdealotherwtse (tnfinttetnputimpedance, zero output impedance, etc.). If this am plIfier tS connected 10thenonlnverting configurationwith a nomtnallow-frequency, closed-loop gatn of 100,findthefrequency at whichIAPI =I. Also,findthephase margtn. Ans.10' Hz;90 10.12.2Effect of PhaseMargin on Closed-loop Response Feedback amplifiersarenormally designed witha phasemarginof atleat 45. Theamountof phasemarginhasaprofoundeffectontheshapeof theclosed-loopgainresponse. To see thIS relationshipconsidfdbk.... ,eraeeacamplIfIerwithalargelow-frequency loop gam, A,j3jl>I. It foll ows that the closed-loop gain atlow frequenciesis approximatelyliP Denot 109thefrequencyatwhIchth "F'e magnttude of loop gamisunitybyOJ wehave(refer toIg.10.42),. (10.102a) where ()=180 - phase margi n (IO.IOlb) AtOJ, theclosed-loop gainISAfj 0J, 1 =-;--A--,Ue.,.OJ -,-,I.!..,l---:I +AUOJ,l/l SubstitutingfromEq.(l0. 102a) gives Afj 0J, ) Thusthemagnitude of the gainatw,is = (1/{J)e-)8 I + e-)8 I l /l II+e')81Fora phase margin of 45, ()=135; andweobtain 10.12Stability Study USingBodePlots881 (10103) (10.104) (10105) (10.106) That is,the gain peaks by a factor of 1.3above thelow-frequencyvalue of 1/fl. Thispeaktng increasesasthephasemarginisreduced,eventuallyreaching~whenthephasemarginis zero.Zerophase margin, of course,impliesthatthe amplifier cansustainoscillations[poles onthejwaxis; Nyquist plot passing through(-1, 0)]. .-.-_.----EXERCISE 10.25 Findthe closed-loop gatn atW, relativetothelow-frequency gatnwhenthephasemargtnIS 30. 60, and90. Ans.1.93;1,0.707 10.12.3An Alternat ive Approach for Investigating Stability InvestigatingstabilitybyconstructingBodeplotsfortheloopgainAP canbeatediousand time-consuming process, especiallyif wehave toinvesti gate thestability of a givenamplifier for avarietyof feedbacknetworks.Analternativeapproach.whichismuchsimpler,isto construct aBode plotfortheopen-loop gainA(jW) only.AssumingforthelImebeing that P IS tndependentof frequency,wecan plot 20 log( 11/3) asa horizontal straight line onthesame planeusedfor20 10g1AI.The differencebetweenthetwOcurses willbe 2010g1AUOJl l- 2010g1=2010g1A/l1 (10.107) whichIStheloop gain(in dB). We maytherefore study5tabllit)by examIning the difference betweenthe twoplots. If wewI&htoevaluate stabihty fora different feedbackfactor.weSIm-plydrawanother hori zontalstraight lineatthelevel20log( 1//l). Toillustrate, consider anamplifier whose open-loop transfer functionischaracterited by threepoles.Forsimplicityletthethreepolesbewidelyseparated-say,at0.1MHz,1 MHz, and10MHz,asshowninFig.10,43.Notethatbecausethepolesarewidelyseparated.the 882Chapter 10Feedback 10 10 10' dB 100 90

20log 80 X, ,....:-----,- (a) I 85dB(stable) I25dBgainmargin 70 20logI 60----------+--forleromarginsII/X, -40dB/decade (b) 40 30 20 10 o 10' '" 20logI II 50dB(unstable)II 10'10' I III III III III I./,.. II I' II 10' I,II , I 10' I I 60dB/decade , 10I( Hl ) 10'010'10'II 10' II 10'I( Hl) - 45' -90" - 135' -180' - 270' II 108' ,---_11___ --------------I 72pha,eI Figure 10.43Stability analySisusingBodeplot orlAl. phaseis approxtmately -45'atthefirstpolefrequency,135 atthesecond,and-225' atthethird.Thefrequencyatwhichthephaseof A 60mY, the output saturates IftheamplifierIS connected10anegallve-feedbackloop, findthefeedbackfactorp thatreducesthefactor-of-IO changeingam(occurnngatIVII =10mY)toonlya10% change. What isthe transfer characteristic Vo versus Vs of the amplifier WIthfeedback? Section10,3: The Four Basic Feedback Topologies 0 10.24 Forthefeedback voltage amplifier of Fig.10.7(a) let the op amp have aninfinite input resistance. a zero output resIStance,andafimteopen-loopgamA=10'y,v.If R,=Ikn, findthevalueof R,thatresultsina closed-loopgainof 100YN.WhatdoesthegainbecomeifR,ISremoved? 10,25ConSIderthefeedbackvoltageamplifierofFig. 10.7(c).NeglectI' a andassume that(R,+ R,)RD' (a)FindexpressIonsforAandp andhencetheamountof feedback. (b)Notmg thatthefeedbackcanbeeliminated byremoving RIandR,andconnectingthegateof Q toaconstantdc voltage(signalground)give themputresistanceR,andthe output resIstanceRaof theopen-loop amplifier (c)Usmgstandardcircuit analysis (i.e, WIthoutinvokmg the feedbackapproach),findtheinputresistanceR'fandthe outputresistanceRajof theCIfCUIt 10Fig.10.7(b).How doesR"relatetoR,' andRaftoRa? 10.26Thefeedbackcurrentamplifier10 Fig.PI 0.26uti-li7esanop ampwithanmputdifferentialresistanceR"I' an open-loopgainJI.andanoutputresistanceroo Theoutput current10thatISdeliveredtotheloadresistanceRris sensed bythefeedbacknetworkcomposedof thetwO reSIS-tancesR'IandR,.,andafractionIfIS fedbacktoIhe amplifierinputnode.FindexpressionsforA = 1(/ I,. Problems893 P = 1/I a'andAf'" I a'I,assumingthatthefeedback causes thevoltageattheinput nodetobe near ground.Ifthe loopgamIS large,whatdoestheclosed-loopcurrentgam become?Statepreciselytheconditionunderwhichthisis obtained. Forf.1 = 10'YIVRI= IMnI'= 100n 'II'0' RL = 10kn ,R\{ = 100n, andRF= 10kn, findA, p,and AI' ,----+ --I ,t - -- -Figure Pl0.26 10,27FigurePIO.27showsafeedbacktransconductance amplifierutilizinganopampWIth open-loopgainf.1,very largeinputresistance.andaverysmalloutputresistance, andanMOStransi storQ. The amplifier deliversits output currenttoRLThefeedbacknetwork,composedof resistor R.sensestheequalcurrentinthesourcetenninalof Q and deliversa proportionalvoltage' f toIhenegativeinputter-mmalof the op amp. V,+ --Figure Pl0.27 + V,--+ J.A.\R --(a)Showthat thefeedbackisnegative. (b)Openthefeedbackloopbybreaking the connection of R tothenegativeinput of the opamp andgroundmg thenega-tiveinputtenninal.Find anexpressionforA == 101 V,VI ... .... CDo a:; no o .... 894Chapter 10Feedback (c)Find anexpressionfor/3=J'r' l o (d)FmdanexpressIonforAI = 10J', (e)WhatIS thecond,tIOnto obtam10 = J'R ', 10.28 FigurePIO.28showsafeedbacktransconductance amplifier ImplementedusmganopampWIth open-loopgain J.1. a verylargemputresIstance.andanoutputresistance1"0 ' Theoutputcurrent10 thatIS deliveredtotheloadresistance RLissensedbythefeedbacknetwork composed of thethree resistancesR".R,. andR"anda proportionalvoltageIf isfedbacktothe negative-inputterm malof theopamp. Find expressionsforA=1/ v./3=Vr10,andAJ ,,101', If theloopgainIS large.fmdanapproXImateexpressionfor AIand state precisely the condItIOnfor whichthis applies. + \ v,---R, \ ----FigurePl0.28 10.29 ForthefeedbacktransresistanceamphfierinFig. 10.11 (d),usesmall-SIgnalanalYSIStofindtheopen-loop gainA =Voll,.thefeedbackfactor/3= 1/ J'Q'andthe closed-loopgainAI=V.lI, .Neglect"0 of eachofQ, andQ2 and assume that Rc