Scribd
Upload
56 views32 pages

Eca Unit-2

Uploaded by

karthiksai9315
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
56 views32 pages

Eca Unit-2

Uploaded by

karthiksai9315
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
You are on page 1/ 32
. oniT-IL u . FEEDBACK AMPLIFLERS @®. Explain the basic concept of feed back in amplifier with Suitable block diagram. mn.| Basic Concept Of feed back Feedback concept $ A portion of the ®utput signal is taken from the utput of the amplifier and 1S combined with the input Signal is called "feedback", Need for feedback + . x distortion should be avoided a8 far as possible, | * Gain must be inckpendent of external factors, lenceot of feploace Block diagram of feeclback amplifier corsist of Q basic amplifier, & Mixer (or)a Comparator, Qa gample and a feedback network. tt t o-tt | oo —_ — - i eee Signal comparator of Basic arnplifier Sampling + source| Ls] river oetwork|= forward tan fer VIF | [avo BF gain A network eft ; Li | = Feedbock network Vel Reverse transmission B Figs Block’ diagram Of amplifier with feedback where , A + Gain Of basic amplifier = a j p > Feedback ratige YE | | 0 | ACb) HOS Qa) g mplifier 2 {Op voltage amplifies VE > Feedback signal (current | Negative feedback + the advantages of -* tnchanced frequency Response. > High Input &mpedance. > lower output Dmpedarve, > Reduction in noise. > Increase in linearity. amplifiers. ->Figure Shows ‘ar thevenin’s equivalent art of WA. >>Rs then Vi =Vs. Rs > DF Ri >If RL>>Ro then Vo = AVVi = AyVs. AE —> Gain of feedback amplifier = “ ‘S | %$ — ac Signal in the input side Ceurvent or voltage) er voltage) Uist the characteristics of negative feedback amplifiers. using negative feedback amplifies is > Better Stabilized voltage gain [overall gain Reduced] explain the detail about basic amplifies used in feedback I> Ay Represent the Open circult voltage gain with R=, > guch Amp. called voltage amplifier Practically RF >>RS, facurvent Amplifier ¢ Ro<< Re RI>>Rs, Ro<< kl Vo AVVs and the proportionality doesn’t depend magnitude of Ror Re > An Gdeat voltage Amplifier chould R=, Ro =0. But — An tdeat current Amp defined o amplifier which - the Signal current and the prop. Alo | EProvides an @utput current Prop. to is tnolep. of RS,RL- Rdegl CA § Ri=0,RO=D . if practical f RIRE ea 6 RISrR = 2 (3) Fansconductance amplifiers Ri >>RS, Ro >>RL Po =Gervs j Q input is thevenins, output 1S Bi z + nortons - i RS gy ® Ro = to. ‘To v. | [ Gy) Gm = Vg Vi | nctep Ro Rs. "9 ‘ Fdeal Transcondcutance ampli $ Ri, Ry=00 ES (4) Trans Resistance Amplifier ¢ AR? BRRAREARCE IMPNPLE Goput —» nortons ot ko + a i Qutput > Thervinin’s y ] = 20 LR) 4 z = Fe | Brckp-0f Rokp FF i ‘ty Idd Snput_side Ro >rkt + Beds 1 ®utput Side R9<< RL V0 =Rmij = RFs - vo ; Rm = a ieee Rm {5 Oc transfer Resistance For prmactical Trans resistance Amps RIKRS, Ro << Rt. explain feedback amplifiers topologies with necessary diagram. Based on the type of Sampling at eutput side and type of mixing to the input Side, feedback ampifiers ® Classified into four topologies. (a) voltage- Series feadoack 6] Geries Shunt feedback (b) Current - Series feedback & Series - series feedback (0) Current - ghunt feedback a& Shunt- Series feedback (@ vokage- shunt feedback al shunt - shunt feedback. (a) voltage - Series feedback or series~ Shunt feedback (vottage amplifier with voltage Series feedback] 3 + ba B Ub) transconductance amplifier with Current — series feedback [Current-sevies feedback Or Sevies- Series feedback] + + +{ ‘Tronsconductance a Amplifier * Bo = vp B Kc) Current-shunt or Shunt - Series feedback (current amplifier with current - Shunt feedback J s- fot > ty — % rT | current Re ts g amplifier Ty =Bl0b | | 2 — B (da) voltage -Shunt feedback or shunt- shunt feedback (iranstesistance amplifier with voltage shunt feedback] 3- t= -——_—_______ Trans resistance [TZ at 95 . L Vo Penpli fier a . | Sg-Bx0 | 6 Note $ ~~ (voltage Series} - (Sertes~ shunt ] = voltage Auplifier (voltage spunt ) = { Shunt- shunt }> Svarns Resistance Arrplifior (curvent series] ~ [Series - series] = Trans conductance Arpliifie (current shunt) ~ [shunt - series ] 3 Current Amplifier. 30)| show that bandwidth of An amplifier cane improved by 3(b) USING Negative feedback, - The bandwidth of the amplifier is given as Bul = upper Cutoff frequency - Low Cutoff frequency #+ Bandwidth of the amplifier with feedback ts given as - = 7 ft Blip c fart = (+Amig B) ty — Fons PD it is very clear that Cig ~ fig) >(fy fi) ard hence bandusiath Of amplifier with feedback & grexter than bandwidth of amplifies igure, without feedback, a> Shown In f Gain Peoid 0-07 Bynid Ag raid 9-304 Ageaid fF te th k—— Bis ———4 Fi Tel fil requency 8w1¢ ———————_—__} fig: tffect of negative feedback on gain and bandwidth Key point 5 ve " Since bandwidth with” feedbak tncreases by factor (1+AB) and gain decreases by same factor, the gain- bandwidth product of an amplifier does. not altered, when negative feedback ts introduced. An amplifier hos vottage gain with feedback of 100 4 the © | : gata without feedback changes by aol, and the gain with Jjeal - | back Sould not vary mite than J. , determine the value of Opeo Loop gain A and feedback ratio BP Sot-| @iven Af = 100, OA Ag = 08}. 0 _ = 0:02. and DA. gor a = OQ we senow that DAE _ oa ! “Ap OA CCIE ABD 0-0a= Or Qx—t J+AR Gherefne , Ord + Ap= 2S 210 : p 0-0R -AlSo, we Know that the gain with feedback is A Aye = 5 t+ AB « re. hoe 1 i 100 =) Theresae » A = 1000 fy I+ AB= 105 i-e. AB =9 A = 0-009. Vherefore ¥ : , 1000 Peduce the expression of gain, Fnput and Output resfstang © AOS: [for a voltage Shunt feedback arpplifler, ‘ | Voltage shunt segdeack & A block diagram of voltage shunt feedback % shown iM figure. A fraction of ®utput voltage through the feedback network & applied in shunt with the Input voltage af the amplifier. It is also called aS shunt ckrived, shunt fed feedback. The feedback signal Zf iS Proportional to the @utput voltage. This type Of amplifier & called transresistance Amplifier. The feecback factor is p- + fe p ab {baste o—* 5 @ 7 transvesi stance to omeplifier - yf) [le — Feedback 9 network @ FI9$ Block diagyvom OF wltage Shunt feedback ‘The voltage Shunt feedback Provides a Stabilized veiall Bin and decreases both input and output ‘resistances by @ factor (1+AB). Input fesistance with feedback Rips The voltage shunt jfeedbiack topology a4 shown 0 figure. with amplifier input civcuit is Jepresented by the Norton's ‘fquivalent circuit and @utput etrcuit represented by fhevenin's equivalent. © 3s CT TL Dent? og I Ere | | . | Applying Kel at input node we get, Bo = Ppp \ = + pve The Output vottage Vo & gin os Rr Fj Ro Ro tRL Vo = = Rui —s a) where Rm = Ra Ro Ro+RL Substituting value of Vo for “eqca) into eq cr) we get, fg = +P Rm SF 224 (4B Rm) re input resistance with feedback Rip IS ghen as VI TC+B Rm) 0 vw, Rip = TE oe Re Vi wD T bla) Gutput Resistance with fealkack Rog § fo this topology, the Output resistance can be measure by Shorting the input Source Vg= 0 and looking into the Output terminals with RL disconnected 7% Shown in the figure. = whe y fg-0 Yee vi Ze Rott v Bu | Ly 4 %y Rep Applying KvL to the @utput Side we get» Rr if + 2Ro —V =0 a= v- Rm} tn The input current & given as Tis ~The -pv. ew Substituting 7 from aba) in eq ci) we gets v ae v+RmB Ro vCi+ Rmp) - Ko Rog a Deduce the expression OF Gain, Toput and utput resistance for A voltage Series feedback amplifies, | Any. *RLAG Geries femtback : A block diagram of voltage Series feedback ts shown in Figure, Here the put to the foodback Network 1 paratlel with the Sutput of Omplifier 8 Fraction of Output voltage through the feedback network iS applied in ¢ortes With the input voltage Of the amplifier. Basic amplifier with gain A ve f wt ld a Feedback vet network B Fig: Block diagiaM oF the Voltage -series foedback * THe Shunt oF parallel Connection at the ©utput reduces the output resistance 2. *# The series connection at the Input, increases the Output resistance Ri. This amplifier (Ss called voltage arplifier, The voltage feedback factor 1s pre. fo Input Resistance with feedback (Rip) | Dnput vesistance with feedback is “Apply KVt to ‘ioput toop, Wave VE , = GR+ Ve = PRE +BV0 = 2) Ri+B AVY " TRIB Av GRr £2; RE + BAVITR = Ti RI(4+ pav) Rit = “Se eilitpay) (+p av) | dnput resistance with fealback increases by a factor (AB). < ' Out put Resistance with feelback (Rog) For measuring output vesistance R iS disconnected and Vg is Set to yRr0- The external voltage V Applied across the ®utput and the current PF divicled by Vs ty 8Cb)) Sl. Figs valtoge-Series feedback civcult fox the cotulation of input and Output restStances Vv. Rop = Rue to feedback input voltage ve reduces the output voltage Avi which Opposes v. i- V-AVI_ VY TABY ~ Ro Ro @ - VCH+AB) R v. to G+AB) = Ro Ci+ap) Gutput vesistance with feedback reduces by a factor (I+AB). A voltage series negative feedback amplifier hag a voltage Gain without feedback of A=500, input resistance Ry =3kAy Sutput vesistance Ry = a0kr ANd feedback yatio B=O-ol- Calculate the voltage gain fy, input resistance Rip, And output resistance Rog OF the annplifier with feeltack. AV A = —— “§ I+ BAv 500 500 = a 0 63.33 10-01 x500 Rf = (I4BAVIR = (14 500 x00!) 3x10? = 6X3X 103 =1%,000 =I6k Ga) Rop = Ro I+ BAv | = 20x10 M4 (500x001) 80x13 6 3+ 33k " Show that the negative feedback reduces gain of an armpifier. Negative fealoack # Sf the feedback Signot-(vp) & Out Of phose with the Input signal (vs), then Vi=Vs- VF. SD applied input voltage is decreased and @utput also decreased, hence voltage gain reduced. This type of feedback called negative é& degenerative feedback. Qain, AE —% Vig = _ 1 4 te 0 Vo BB ! CYA} +6 Ap os rs Here lngl etal. 6(b) when Ap) >>! te. Jal >> (pp) - re. when A ts very High. Ag =p ; pis a feedback ratio. er Hence gain dlepends tess on epeiating fotentials, the chava- cteristics of the tsansistor o vaccum tube. > the Gain may te depend entirely OP feedback network. > $0, if the fealback network contains only Stable passive elements the gain of amplifier using negative feedback also stable + Analyse the effect Of negative foedback 0 Output resistance voitage Series and current Series feedback amplifier, generally, in feedback aroplifier, -> voltage sarnpling at Output + decrease lp tesistance at [Rog < J > current sampling @olp > increase op resistance | [Rog >I), co) voltage series fealback ¢ Rog [Resistance with feedback woking in the output] @btained by * PiSconnecting KR (+ R.=c0) x By making external Source voltage to 0’ (vs=0), > Replace Vo by v. v KOA Pop = V, pv Le —Ayvi7 DRotv =O Apply KvL to the Sutput Side, ~Ayvi- ERo tv =0 ee v-AvVi Ro The Input voltage is written as -V5-- Vi =0 Mi=-Mp = —BY [urvs=0] Substitute v; in the above qn R= v-+ BvBv Ro * Vv (I+ AvB) ~ K . - v=. & oe Kop © r+ By Rg = 2 HRAV a dap ce current Sevies fepdoack : Pn this topology, the output resistance can be measured ‘by Shorting the input Source ve=0 and looking into the l@utput terminals’ with Ri cigconnected, as Shown in Fig ff = Rog = a = Rol 1+Gmp) ae] @ |determine input and output vesistance of current shunt feedback amplifiers, Aas.) Current shunt feedback ¢ - The current shunt feedback is algo called series @® ckyivedy Shunt fed feedback «the shumt connection at the input reduces the input resistance and the Series comection at the @utput increases the Output resistance. This, 1S called cunent amplifier, The -feedback factor is B-t.. Basic amplifier [— ith gain . vol Is A = fF ——} Feedback Le network B Tf Fig: Block diagram Of Current Shunt feedback Input Resistance with feedback Rp : iz % v vi Ts i (LE The current shunt feedback topology ts shown in fig with aroplifier input and eutput circuit replaced by Noxston's equivalent circuit. Applying cL to the input Node we get Ig = DT; +Jp SD+B2 ery The @utput Current I) % given Ob Ai Ty Ro Ro + Re =A >) where Aj Ro r Ro+Re Substituting value of Ip from (a) into eq (1) we get, Rs = D+ PAT =2; (1+ pai) the Input resistance with feedback i¢ given w% Pee Rif ts z vi ~ B(1+B Ar) Rg = — Ri (+pal) ge Mow TT Quiput Resistance with Feedback Res dn this topology, the utput resistance can be measured by pen circuiting the Input source %&=0 and Woking into the Oltput terminals, with Rk cisconnected, shown in “figure. at , and the voltage gain without feedback & given by Ree hfe DpRe hfe Re vi vs Rs+hie where vg = fb (Rs + hie) The desensitivity % given by a he RE Det Pave Mt Rohe Rs + bie +hfe Re toe Rc + hie +Vec Rete figCay RS 8 aN . pW 7 a+ ay ~ ~ T ak € 7 © 7 t _ji- #49 C0) { ‘ ihe fIQ(OBIT emitter follower figlb) Amplifier without feedback ig cc) tow frequency equivalent Circuit where v. = sl B eat : qhen, the voltage gain with feedback can be written as Av hfe RE ee ra Ro+ hierh fee TE hpeRe >> Rgthie, then Avgq!. This sunit gain shows that it & an emitter follower Circuit. From fig ©), the Input resi¢tance without feadback we given by Ri =Ro the eee ee eee tence, for voltage Series feedback amplifier, the input resistance with fib tocreases due to Series mixing at the input and Wt & given by Ry = RD = (Rs thie) x Rethie hfe RE (Rs +hie) = Ro+hiet hfe Re | From Fig (ce), the @ueput resistance Of amplifier with feedback, without considering the external toad resistance (RL=0) & given by Bou = Ror = I+BAy where Ro=0 aod aes Ay =a. This indeterminacy can oe sesolved by frst eae Ro and then apply the at Ro. she Sutput resistance of amplifier with feelback by Considering the external toad, an be written as Roy = Ro, . Re (Rs+bied D Re thie-+hfere where Ro = RL=ReE Rop elt Rope _Rstoie_ Ri 00 be Hence, the teelback desemitiges voltage gain wr-to changes in hfe and it increases the input resistance and decreases the Output resistances fav vottage Series Feedback topology, eu) by) by} In the ByT emitter follower Circuit as Shown tb fig, the circuit -|component values are = bor, Rony AMG =a kA, bio = KA calculmee Aye, Rip, Rog Ord Ror. Sol.| Given Re = 600. Rea yak Re = 9k hie: = 5 KA : ; aye ticke— Ro +hie \ g0x 2x103 ~ BOX AXIO” _ 00+ XI0> = Bg 5F Abe cemensitivity & given by De 1+ BAY - Re-thier eRe Ro +hie ~ 600+ 5x103-+80xQ x03 600+5 x103 | = 89.53, where = ins pe vais ee wohen, the voltage gain with feedback can be @ritten as Av AE 5 = 0966 =1 Aye = Ye “$y 99,59 This unfty gain shows that it t& ao emitter follower Circuik, From fig, the foput resistance without featback ws given by RE= Rs thie = 6005 x18 =e SOM Hence, -for voltage Series feedback amplifier, -the joput resictana with $b increases due to Series at Input and it is gren & Rig = Ri D= 56x103 x 29,53 = 165959 KN From fig, the Gutput resistance of the amplifier with feedbex voithout Consideying the internat wad Yesistance (Ri=R) is given by eee Os ttBAy = where Reo and Ave lim Ay zoo, ‘this inckterminaey can be vevyplved . Rio by first evatuating Roy ancl when apply the Linit Rr the @utput resistance OF the amplifier wlth foedback by Considering the exteMMal toad, can be wasitten as 1 Rb. axio3 = = SEH OF bY Ro QU) t ee Way Gu jwhere R =R=Re. This Shows that the output resistance decreases due © voltage Sampling to .the output. 600+ 5 X03 Rpett- Qs f Row of €0 i = FON. os Rop =For. A RC coupled amplifier has a mid-frequency gain of 300 and Qa frequeny response from 100 Hx to 20 Ky. A regative feedback network fe with B=0-0a 18 Incorporated into the amplifier civcutt. Determine +the new System performance, Given o Mid- frequercy gpin of 200 ; Frequency responce fiom co Hx to 80 tix. P= o.0a eo ee a; +AB 1+Q00x0+0A OO ge FAB '+00 x 0103 tog = = fa x + AB) = QxI03x (14-Q00K 0-08) © 100 KHz, Bug “fag —fip 2100 x107— Go%0 kHy AXBw =&00 x 2on03 = Yooo KHz fb Shows that +h? gain- bandwidth product of the amplifier with negative feedback is same as that OF the gain- bandwidth product of amplifier without feedback. A Ab» Explain the effect of -ve feedback on toput resistarca for iculvent shunt and voltage shunt feedback amplifier, any loput to shunt with applied SIQDOL , Rig [SSE OF negative eedlbock go input resictance : when negative feedback - gignat fedback to the & decreaied, since Is = 2,495, then Pew > ond | Rig = a »becames smaller than &} (without feedback ) ts Hence for voltage, cusrent Shunt feedback Gp = — RP (rap) = BL D I> is) Ws = Tr +0 ‘ | Lo ‘tf =pvo Shunt current Shunt feedback ; man aoe SERIO to F ] , 1 | + : 7 Rs \ eg | gv vil 1 | 5) a ‘ | Pity a ' 2 Yo Te BS \ . le HTL. ' RIF = Vig For Q current shunt feedback amplifier , input and Qutput circurt veplaced by Norton's model tpply KCL to foput Side y Ws = D+ VY Is =) + pL0 The Gutput current written as = (Aili) Ro+Re To = AP Tj ag - fo, AlRo_ Ti | | R+R Substituting the value Of To in the Above KCL eqn, we get tg 2 Ti+ Bag. 2 2s =F) (4 par) : So, the peat ioput resistance, Rig = a i +s (+ par ay pad ey Aj = tt Ryo AL voltage Shunt feedback 3 ‘ (doput circult — Norton's model eurput Clicuit - Shevenin's rhecet - 1 i fe | | [| \. . a(t) OPM vy | ZS (2 A ud | l gee vi een itt Apply KCL +o the fnput Side, we get fg fp4+d5 = 27 +B Vo abe total Output voltage % written a3 Yo = (Rn f; ) Ro Ro+Rt Vo = Rw Sf where ve. _RmRo Re = 9 RL Substitute the value Of % iM above Kcl egn, Be = Tp +B RMS eli (I+B Rw) The Input fesistance with feedback gen ay Rig = ML ds vi (HB Ra) B Rm, Rm 7 0-c tramresistane Wo feedback. Rm oc tansvesistance wlo feedback taking R in account, R70 \oa)| Compare various tYyPes OF Pi. 1 Ree le Ree feedback amprr fiers, Types Of Feedback arnpliffes Charactenistigl a - = — current. cuts voltge ie vo 19 ol ee Rig TPncreases | @ncrease Decrease Decrease Rog doer oe “Decrease Decrease tncrease Ave Decrease Decrease Decrease Decrease r Bandwidth | Increase | Trerease Pocrease Lnerease ie shen gn | DEease | Decrease Decrease | decrease Noise Decrease Decsease Decsease Decrease 6 “¥ [80 F No % Ivo F190 \ab)| Compute Rn ANd Kmg USING feedback Principle for circutt Shown in Figure . Agsume and bye =50 and hie =1-1K* +Ve | | ; wx Given hfe =50 bie = blew Rashiea + Cithtea Re =I) x103+ 51x 50 = B65 EN we know that Ve Rin = Js Vo i fby . Ibo * Is Mo. che ibs (Re Ri) ae eee aa tbo 9, © -50( 0.5 x103 |]. x 103 ) = 1466 xjo3 2C1 2 hye =50 21 2 hype = Lor f bo. _ Rel ta Rey + Ria -3 x103 3BK103+ 3.65 x jp 3 = 0-45) Tr (Rs RD AS CRS TRE Rig a 162 x107} 8 x 104 © (jeax102 |) 14103) 1+1 x03 = 0.353 whe re Rta chie = hIkA Rin = ~ 19+ 6 X10 (9 -G61 ) X50 XO- 353 = BUAVKEA os 1x03 1 = 0833 X 107 Rng = Reo 1+ RoB 139.99 «103 14189-949x103x 0 833K 107 IQ KA W (ov) Rin Sa — ~ "0833 x10 = 1Qkr

You might also like