1.1 Introduction Feedback plays an important role in almott all electronic chews, It is almost lnvatiably used in the amplifier to improve its performace and to make it more ideal. In the process of feedback, a part of output is sampled and fed back to the input of the amplifier. Therefore, at input we have two signals : Input signal, and part of the output ‘whieh i fed back 10 the sypyt. oth these signals may te in phase or out of phase. When Input cignal and part of output signal are im phase, the feedback ie clled pos feedback On the other hard, when they are i out of phase, the ferdtiack is ealed negative feedback. Use of positive feedback results in oseilations and ence not used in amphfies, In this chapter, we introduce the concept of feedback and show how to modify the characteristics of an amplifier by combining a portion or part of the output signal with the ‘put signal. We ao stucy the analysis of various feedback amplites. 12 Classification of Amplifiers. Before proceeding with the concepts of feedback, it 1s usehil to understand the chaofication of amplifers bated on the magnitides of the input and output impedances of an amplifier relative to the source and loed impedances, respectively. The amplifiers can be dassified into. four broad categories voltage, current, transconductince and anseesistance amplifiers 12.4 Voltage Amplifier Fg. 11 shows a Thevenin’s equivalent crclt of an ample 1 eve amptiner Input resistance Kj 18 large compared with the source resstance Ky thon Wj =\j. Ifthe extemal lead resistance Ry 6 large compared with the output resistance R. of the amplifier, then V, = A, Wj = Av V. Such amplifier cicuit provides a voltage output proportional te the voltage input, and the proportenality factor does not cepend fon the magnitudes of the source and load resistances. Hence, this ampliier is called voltage amplifier, An ideal voltage ampliier must have infin input resisiance &, ond (1-4)Fig. 1.1 Theverin’s equivtet circuits of a voltage amplifier 20 output resistance Ry. For practcal volage ampliier we must have Rj >> Ry and mR, 1.22 Current Amplifier Fg. 12 shows Norto’s equivalent circuit of a curent amplifer. If amplifier input resistance R, 0, then I, = Hf amplifier output resistance Ry ->% then = Ay. Sich ftnplfler proves « curent output propovtons! Whe sgial cent, ant he [Propotiontiy facor ie weeperdont af xtc ae ad wstancon Ti amie cao ‘curent amplifier An ideal current amplifier must hare zero input resistance Ry and Infinite ouput resistance Ry. For practi current amplifier we must have Ry << Ry and RoR RawoRe, RecRork, 220 Fig. 12 Norton's equivalent circuits of 2 curert amplifier 41.23 Transconductance Amplifier Fg 13 shows 8 eansondvctinceamplier witha Thevenns equivalent it input Cru and Nator's equivalent in its output cru. i this amplifier, an oxiput caren! is [preporional t the pt ignal voltage aad the peoporicnaity factors inept of {he mognitdes ofthe source and led restances eal, this amplier mat have an fnfnte inptrstnce Rand nate cpt rentance Ry For praccl eamcendectance amplifier we must have K, >> and Ry >>ask Fig. 14 13 Block Diagram 1h the previous secon we have sen four hase amplier types and thes Mea Garacertics. In each one ofthese cits we can sample he ouput eotage or caren ty meses of» sabe snpling sctreck sod apply te sgoal fo the pet Hoang = ‘wedtack two port network. as shown io the Fg'15.At the input dhe Kwacha ‘omtined wit the isp signal ough « mixer network andi fed io the ampli | ‘As shown inthe Fig 1.5 feedback connection has thee netwock + Sampling Network |Feedback Network + Mie NetworkTs i Fig 15 Block dagram of smplifier with feedback 13.1 Sampling Network ‘There ae two eye 0 simple the ouput according to the sampling parameter soltage ce curect The output role sumed by conectng fe kedoack Renee ‘shunt acs the output, as shown inthe Fg 16 (a This ype of connecton is retered 1 (2 von, anode, sapling, The autpl cured sampled by cmaecing the Kd, ‘network in sie with the cotpat at shown i the Fig 16) The tp of connection ‘elred 19 as curent of loop. sampling, 1s) Sota nace mein Fig 18 "43:2 Feedback Network Itmay consis of resto, capacon, and inductors. Mot often itis simpy a restive cenfigwaion. It provides reduced portion of fhe cup «© feedback signal wo te pet Imowr network Iie piven a8 yw =ayletronic Sirus 45 Feedback Ampitiors wher Pls «feedback tater o fndbeck rafla The embol fused in fendbach croute represents eck factor which aways ls between 0 ae 1k ttally diferent fom B symbol used ty reprecent cunent gan in common emer ampli, which greater ‘hae 7 1.3.3 Mixer Network ike sampling. there ae to ways of ming feedback sgn withthe input igs Thee ae = seres put conection and shunt input conmecton The Fig. 17 (and ©) show the simple and very common sees Soop) input and shunt inode) input Connections respecte ae bel (0) see mine (0) shar ing oad 1.34 Transter Ratio or Gain Ing 15, the rato 0f the Owpus signal fo the input sia of the Deak amples represered bythe symbol A. The Sly A given next. represents the different tanh o‘The four quai Ay, A, Gu and Re ane ere teas & ane gin’ TO acspifer without fudback and use cf only symbol A represent any cme of dese “The tranfer gun ith frac repro by the smb Ay. H i dfn w te ‘00 of the outpat sna to the input signal of the amplier confquntion shown Fig 15 Hee Ay i ed eprset ay on he fling four aos = pec = Vota gn with onda 6 = Ay = Comet gan with eta 6 Xe = Ou + Taman wi ck o 1K Rag + Tremere ence ° Fig. 1 shows the schematic representation of 1 feedback connection arcund a base snp. Recall tat, when part of tpt signal and put signal are in cut of pase the eeltack i ale negative Feedback The schematic diagram shown in Pg. 18 Yepresnts rogue feedback Decne the fetch sgl fad back othe input ofthe apa ot of phase with input signal ofthe amplifier. Fig. 1.8 Schematic reororentation of negatve feedback ampifier 1.4 Advantages of Negative Feedback 1 posable to improve inpertat darsctertcs of four bashe amplifier types lacs Insect 1.2 bythe pooper use of nga feedback ‘Normally high input reste ofa olageampier canbe made higher. 4 Normlly low output restanee of rita amplifier canbe lowered+ The wamte pin Ao te mpi: wae ea SSE CICS ‘artim ofthe hor hen paar fd aaa the puraen f (Se chr ecve device ted in apie + The piper ine of egatvefedbuck empore fgueney reponse of the pte +The gcant improvement inthe Inet of operation of the eck ptt compel wi te of sri woe leds May Peet 5 Ae abana met sh indo the pt of ein by {sh fateh hh Imre i emparn aT nar gan A of ir ‘out abc 415 The Four Basic Feedback Topologies ‘The hase amplifier shown in Fg. may be wltage. crt, tanaconducane, oe transesitance amplifier These canbe connects i 8 Feeback cafigurason a shown = the Rg 19, Fig. 18 (b)Transconductance ampifiar with current sures eadbackug sie soe Fig. 1.9 () Curent ampiter wit curert-shunt feedbock wn [Ee Fo $9 ld Tranerasitancs anpliter with veltaga shunt fudback 1.6 Gain with Feeback We have sen, the symbol Ais wed to represent trnaer yin ofthe base amphier stat feedbck and symbol Ay wed fe mpreent tater pln of the bak ampier With feedback. These are gien a8 a Sent a oe % os where X= Outpt voltage o output curert X, = put vokage or input current X= Saute vege ot source coment ‘9 tin «negative frac the reltion betwee X, and Kyi ven a8 LMF EXD ere 1X; = Feedback voluge of feedback cuert x Me Boyt:1.84 Loop Gain ‘The dferoce signal, Xin Fig. 18 muliplind by A in passing through the ampir ‘% mutptied by pin tansmisson through the fetabck network ae Is mule by ~ ‘im the mining o difermce network A path of» niga fom sap termine tough base “ple, rough the edback network and back wo the lap ermal forms 4 oop. The _i of this loop in the provdct~AB This gin koown a lop gain or ret rai. 1.62 Desenstivity of Gain The transfer gan the ampli isnot constant a it depends on the fates such 15 ‘opertog point emperture, ec. This lack of ably Ia ampli can be reduced by tnt negative feedback, We know that,‘Becwonie Crewe 1-0 Feecbeck Ampitiers ‘ierntiting bath sides with respect w A we get, aay, (1sBANI=BA a: Tear ecb TF a tA apa Dividing both sides by A we et ae apa) 4 ROA Ta = * eliesean where = Frsconal chang in ampficain with oodback as aE = rachnal chang: in amplification witout fedack oki teqtin () we am ay hat carga wih bck iin she cane in gin thou sean by fer (Is BA) The fection change axpicaton with fedock sve yf Sacer cnge wont edocs alle the sertviy ofthe tamer gin Hence the seein 6 rr—fzy The opel of the seville the desenaity ven De tem Terre tity fhe pier ners wi creme a debt x‘Sree A reprosnts ther Ay, Gye A) OF Ry and Ay mpreanis the conreponding teaser gains with fac ether Ay, Gy, Ay Gt Rg the equation sigrifies that Fo vg ete etch © © ° he = Cot pi std ® 1.7 Cut Of Fraquencios With Feedback none Ug a epuon cn tee 9 rh ® dew = o wt en zt)Soong aloe of Aue uation () wa,Ane o Stenting vate of Am equation 1) weg Aitigh » Aetna Team ‘she opp ct frauen th nts in inom = (Am Be ® Fram equation (0), we at say tat upper cal eguecy with Keedbick pear than upper cio mquecy wthost teach by far (Ve Ans R_ ThereBandi ‘he bandwidth ofthe ampli i given as [BW = Upper cutoff frequency - lower cu equency Bandwidth of he amplifier ith eaback ie gen a BY, = ts ne Be o 1 every le at Ue) > lig fo) and hance bandwidth of amplier with feedback grater tun bandwith of ample without feedback, as shown in Bg 11. ig, 1.10 Effect of negative fvcback on gain and bandwidth Key Point Sine hadith with nai fac racy by fcr (1 + A nd aie “tes by sao, roc ot ET aos ote et epee fin robe 4.8 Distortion with Feedback 1.41 Frequency Distortion From auton (of previous san we ca say ttf the fonback rato doce ot contain renctve lms the overall gain isnot a function of frequency. Under such ‘edition frequency and phase citron lo rtntialy reed ups made up of reactive components, the ractnces of thee components wl change swith fogemcy. changing the A a result gin wil ako change with froquency. The fact [tse in tured amplifier: Ta ined ampliers feedback network's designed sch at at tuned irequeny fr» 0 and at oer frequents +=. Ata reat, apie proves igh fm for algal a ed guy and rately a at ctr fearonPo. 112 eoiack signal ade 10 the ‘pat volage (ogardenof whether the Idback is stained by sampling the sutpt voltage oe ore), decreases the input resistance. Since =I, 1, the comet 1, raw rot the Signal source tainted overeat woul ‘Soe inte Pg 112account. snce throughout the dicen of feedback ampifer we wil cosiber Ry 10 be [pet of the mpl ard we will drop the subscript on the tarfer gain and inet Frobtance (A, letend of Awa! Ry inated Of Rg re 8 te dr 128 a dt mee o -Arpiying KVL tothe input ade we getKy =m (eBay) « ‘Current series foodback ‘The current seis feedback topology is shown in Fig. 14 eh amplifier Input drat ‘a resented by Thevert’s equals cat and outpat chou by Norton's equivalent wn nny, = ut ° ‘The uate sven as be SM aay, ® oe = (gate m= Key Pott: Gu repress the on cra rescondacincr thnk fot xd Gu the enonucionet thot flac tang he oa Ro tect ‘Substawing value of L tom eqaticn (6) io equaton (5) we gt, Vez hho Gu SURE GELR = MeEe nassew Soren T fm | | wt te ‘The current shant feedback topology is shown in Fig. 1.15 wth amplifier irput and ‘utp cc replaced by Nertn'sequitalen excat ‘Appling KCL to the input rode we get heken = hobh o ‘The output cuvent Lis given ascoin f as r oe th du | Al r em | i th a — ~ reuse seg NC pl ae beten 2h so% an “he cpt wotage Vo gen 8 Ral, a “ mo Be veto ce no te sen see geet reso SR en ran lpi canta wy Ben 1 KOs DD™* phe os 19.2 Output resistance ‘The negative feedback which samples the output vllage, egal of how this ‘utp signals vtumed to te pu, ends to decease the Opt rnstance, a Sn @ (onthe ober hand, the repave feedback wich samples the ouput cert. parle rosounce, of how ths output sgn 8 retures & chown te Fg 18 (2 iG i Fig 118 Nine, wt the flat of meguive ft: on sutpt vntence tn dient topologies (ways) of nteducing negative feedback and aban Ry, quanitatvely Voltage sores feedback 1h the tpotegy, the tpt setnaee com be snerane by sharing tee input oven \, = 0 and looking int the output terminals wih Ry disconnected a8 shown in the Fe us,Subang the V; fom equtn (32) eqution 1) we Rt, Ree TBAT cy © aon, , (bi )+B a oe ad BR ak FRLGSBAG) ” RFR BAR (ng meme tnd enema by a) we Rake RE Re = a ae a et) ay Poet Here Avs te op op vl aking Reo acteee arses In this topology. the omput reiatncy can be mens! by sharting the inp scarce V. = 0 and looking into the output terminals with Ry dscomeced, 2s shown in the “Pol f eT Fg 120 Appiving KVL. t te opt mde we get Ra riRe-V = 0 1. Sh as) “The npet corte given oe Ln tee. 0% etna wm epntion (9) etn 8) we gt pe MiB, VOsRaB) eee ne} Ree ns pot ite "a tg uniter me“The npet caret i given at Hosa me ny Yeap) 2 Kay Poi: He A; the open ep crrent gui witoting it acum R= melita = REEL a BePAO Re. RoR OBA)ee aa: etary aca coe In this topology the output resistance cn be measured by shorting the input \ =O and boking into the oufpu tominals with R, siscnnertes at shown nets Pace rosa Aepv HCL te out ode we te Leow ‘The input voluge ls given as -% Me sit Fe Sci ve fm auton 9 atin wt. oe te Zeon v rasan) =X tee} -Rasose)ay Point: Hee. Go the oes lap tosondcince wo king Rn acu ee = nat Bh hig enter Ke = BGP) a BAM a no Ge an ‘Ky Point: Nt tht he, Gi the open op caren i taking, tant weet Re = 1.10 Summary of Effect of Negative Feedback on Ampitier ‘ible 1.1 sumaizs the fet of negate feedback on ampli Parameter | Wotge series | Curent sarin | Curt shunt | Votage shant on a wh ‘eee BR a = oon — eon — Fae oe ces owe od neo] Race ren ‘ops sane vem) | Ramo owawomece| nye [Me MinGn | Ree READ | ae Tablet4.11 Method of Identifying Feedback Topology and Analysis of a Feedback Amplifier To analyse the Seach amplier tis necessary to go through the following steps. Step 1: Lenfy Topology (Type of fedback) 2) To nd the type of sampling naork 1. By song the output La V =, fend sgl (4) beeen te thn we s9y dat 6 "Volage Sapling” 2 By eperng the ouput loop 12.1 = Of frdback signal (2) becowes zero then we cam ay that i “Curent Sampling” 1b) To find the type of mixing network 1. the fetch sgpal esubleated from the externally applied signal as vlage 1 the imp lop, we cin ay a "eres aN 2 the feck signal subtracted fram the externally applied signal ‘in the input oop ee can sy hat “shunt mining” ‘Thus by deermning pe of sampling network and mixing network type of feedback amplifier can be dtemnine’ For ample, ampiter uses a vollge sampling and teres ‘motng then we can say that ts a vole seis amiptter. ‘Step 2: Find the input dec 1. For volage sampling make V = 0 by shorting the eutpet 2. For cunt sampling make |= 0 bs opening the otput loop Step 3: Find the output cet 1. or seis mining make I 0 by opening the put op. 2. For shunt ming make V =0 by shorting the np Suep 2 and step 3 ensure that the feedback is reduced to ze without altting the tonting on he bate ampier ‘Step 4: Optional Replace each sctve device by its parameter model st low frequency. ‘Step 5: Find the open lop gain gain without feedback, A ofthe ampli Step 6: Indicate X an on he cca and este oe Step 7 From and Rind, Ac eR Riarc rnsiogy ‘erage sree | Curent ares | Curent stunt | Voltage shunt sero or x | vote on oon = aes wet Vota Comet Cnet veer Yo tad not wee uso eo wo. bono Tota oan ery ee vee wea er Soe ee Tren | Then Neen wenn pense ws we we eRe enya | wet | ae w/e Datena tom | tee tba 150m 7 ne oo ~~ ~e a » ~~ 6 ee Roser) Roe |e Reem ges 3A) — Table 12 1.42 Analysis of Feedback Amplifiers 1421 Voltage Series Feedback a this setion, we wil ee two examples ofthe volge setesamglier, Fest we will aly tans emo ower cat an then asus tower sang FET {2:4 Tanita Emir Follower Fg. 123 shows the tansstr emiter ‘ollower reat Het ferback voltage is the ‘tage across R. and sample signa is, arom R.To find the input iru wat = & and hence \, in series wath Re appeass etween 8 and E To fil the output cca, st = fy 0, and here Re opens oni? fF the ouput lop. With tae cormecons ‘we chtsiy fw crit as shown in the Fe 2 Replace transtorby ts Irparameter equivalent eit‘Applying KVL to inpat lop we get Xe haehe) Substting valve of we get he = BaRe. 08100 yy > RRS" TSI ‘Step 6: Indist Vand Vand calcite p x We here = ye ni ling poa ‘Sip 7: CakuliteD, Ay Res Ra and Ra D = lay eo Tay De R= Rohe = ks 1aK =21K m= RD = kee = romK fier oh ‘ i f F1122 Current Series Feedback 1 thi sete, we wil se bro examples ofthe curr series oedhuck ample. Fit ‘we will analyse tranister common emer circit with onbypase emiter reste and Sen commen sour wie webypuned ure rawr 19221 Common Enter Contgoraion wih Unypssset Ry, Tig 134 shows the common enter ‘cll wits unbypaed Re The comme fonts cut wh wibypatied Lie ae ‘Sample of coment srs feedback. the ‘configuration resbtor K, is common ‘vet enlter put cect an wells clear 19 miter output cit and Input carent bas well os output cute [Cboth flow through it. The volage drop oe eG ALR ALR STR Th soltage drop snes thet the output current Li beng sampled fant tis converted to votage by feedback retwork. At input aide eltage Ve fe tsttraed fom. t0 produce | V, ‘Therefore, the feedback applied in sess. Sep 4: ently tpology By opening the cutput lop exp caren s = 0 ferdbick sgl becomes er a Inence it's carrent sampling Looking at Fg we can soe Ht feedback signal Vi sabtced from the extemal applied signal V, and ence iia series mixing, Combining ‘vo concsions we cn ay that i a crrentseres febick plier. ‘Step 2 and Stop 3: Fd input and eutput crew To fin inpt crcl etl «0, then Re appears at the inpt side To find ont rout at 1, = Otay appa in the ‘tpt “rst ‘The seahing ct i ‘hon inthe Fg 15.° > pats * RTE TESTER = -0ns ‘ep 6: Indicate Ay and Vy and calle f Hep 7: Cael D.C Any a Rae and Ri an cy ‘evolu pan Ag gn be Ye eaate one = 0) = -armao 22k ssi Lenina 136, cae om R= Robert oo = WrliRsi2Ke39K Ry = R.De33Ke9I8 = om looking at ig. 1368, given as Ry = mDe Ry = RoR =m =uK1.13 Nyquist Criterion for Stability of Feedback Amplifiers ‘A negative feedback amplifier designed for 2 patil frequency range may beak out {nto cscilition at sone hgh or low frequency. Ths sublty problem arses in fact ‘amples when the Inop gam hs tore tha to al pole. The estnce of pole with # [ontive real pact roit ha dhturtance nening exporently sit te Wher Bich Transient cstrborce pests indeinvely oF merease, the stem becomes unstable ec, the conditon which mist be sai, 5 be stb, tha the pols of the rarer fenton mast all le in the lethind hall ofthe complx-teuency lane Ifthe system witout feedback io tbl, the pokes of A doe tn th hha al line Therefore, from squat A, = A/I + AB we cin say thatthe sbi condition Feguires Mat Be eres of 1+ AB oll be isthe lethand hal of the compler-tequency pine ‘The Nqust cteron forms fe Bs of 9 stndy-sae metho of determining whether ‘x not an taper le eta. The Hyg rere expres condition for sai tn ore of the stds oF fequency response, characters. Let ws se the Nyguit citron. Since the product AB 64 complex number, it may’ be represented 36a pot in the wt Irequncy Consent FOI {be compen plane ae cine fr . fe vats of A cerning to y Mi clan from =a oS = Powe ecick he lous al the poi os Rp) hae carve The rein of Ngati tht the cnpifir Se salble thin carve encase the , ein’ 1+ [0 and the amplifier g Baable if the carve doce nat Cie Sten ETA Fig 144 Locur of 1+ A= “The Fi. 14 shows the Jonas of + AB| = 1. cece of ait raion, it ite contr at 1 6} Oil lor any fequeey, AB ees oube the cee the focdbak eguive since |! + AB > 2 however, ABs win thi cic hen [1+ AB] < 1, ad the Fednck poste, Am example ofthe Ngati tae in Pg. 148. The hans in Fg. 148) ls sable since t does not enclose the ~ 1 + j) point wheres the locus shown in Fig 1450) & unstable sie the cure des ence the ~ 1 + point DO punts Fig, 1.45 Stabity conditon using Myguist entrion Examples with Solutions mb eam 12 A fick amp Is op ly gain of 66 ak fos tor, ‘B00. Find theca op gis ath mate fleck. (oniDe3000 | ‘Sotuon: A N= Tea TRO = me