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1 MODULE TITLE : ELECTONICS TOPIC TITLE : AMPLIFIES LESSON : FEEDBACK EL - 3 -

2 INTODUCTION This lessn trduces the ideas f negative feedback, which we shw can vercme the disadvantages f wide parameter variat fund transistr amplifiers. Negative feedback is fund t stabilize ga, reduce nnlearities, and generally enhance the perfrmance f an amplifier. But there are caveats t the use f negative feedback because, under certa cndits, stability can ccur. YOU AIMS At the cnclus f this lessn, yu shuld be able t: deduce the ga f clsed lp systems appreciate the effects f negative feedback n amplifier perfrmance assess the cndit fr scillat simple systems. STUDY ADICE Many f the prciples established this lessn will have imprtant applicat the subsequent lessns. Make a list f the imprtant frmulae and make sure yu understand hw they are derived.

3 2 NEATIE FEEDBACK PINCIPLES I Input + E Frward O Output F Feedback H FI. FIUE shws the general negative feedback situat. An put I is fed t a frward path funct, typically an amplifier, with ga. A fract H f the utput O is subtracted frm the put t give a signal E. The equat f the system can be fund as fllws. O E ( I F) ( I HO)... () earrangement f Equat () gives an express fr the ga f the system. O I + H Frward path ga + Lp a... ( 2) Dividg bth numeratr and denmatr by gives an alternative frm f equat (2). O I + H

4 3 O Frm this, we see that if is large enugh then I ga is defed almst entirely by the feedback ga. ; H that is, the system This is very imprtant. In the amplifiers s far discussed, we have seen that it is nt difficult t design high ga circuits. Hwever, because f transistr parameter variats, it is very difficult t design amplifiers with fixed stable gas. Negative feedback prvides the answer. The required ga is we must make >>. It is als clear that, if this cndit hlds, it des nt B matter if is nn-lear; the feedback effectively learizes the verall system. H and EXAMPLE O If H 0., fd when 00 and when 000. I Insert values, takg 00. O I ( ) 909. Insert values, takg 000. O I ( ) S, a 0 : change leads t nly a 0% change the verall ga.

5 4 The thery s far has been develped a general way. Suppse we nw apply it t a vltage amplifier situat. OLTAE AMPLIFIE In FIUE 2, we represent the amplifier by a blck with put and utput defed by the fllwg equats.... ( 4) i H i I i i 2 H 0 FI. 2 Frm FIUE 2, we can als see that H is a cnstant fract f the utput. H + 2 ( ) i

6 5 Frm this, we bta the vltage ga.... ( 5) + H Therefre, frm equat (5), if is large, a simplified apprximat fr the vltage ga is btaed. H ( 6) The ga is therefre defed by the rati f tw resistrs plus! We nw vestigate Z, the put resistance f the system. i is the put resistance f the amplifier. Z I i i i H i H Substitute fr frm equat (5). Z + H... ( 7) i ( ) Hence, this circuit, the put resistance f the amplifier is greatly enhanced by the negative feedback, sce is usually very large.

7 6 We must als cnsider the effect f feedback n the utput impedance f the amplifier. See FIUE 3. i I i 2 H 0 FI. 3 Assumg that the feedback path current is small, the system utput resistance Z is given by: Z I i Substitute fr i, rememberg that 0. Z H + H... ( 8) Hence, the utput resistance f the amplifier is much reduced by the negative feedback.

8 7 WOKED EXAMPLE Fr the circuit f FIUE 2, let 50, i kω and kω. The feedback is defed by 0 kω and 2 00 kω. Determe the system vltage ga, the put impedance and the utput impedance. H + 2 ltage ga is calculated as fllws. v + H Input impedance is derived frm the fllwg relatship. Z + H i ( ) kω Output impedance is btaed frm the relatship belw. Z + H Ω

9 8 OLTAE AMPLIFIE 2 A secnd vers f the vltage feedback amplifier is shwn FIUE 4, where the put is shwn cnnected t the ther end f the feedback fract B, and the ther put is put equal t zer. 0 i 2 I i FI. 4 T fd i, we use the Superpsit Therem. i Nte that this can als be derived by cnsiderg the current flw frm t 2 and and neglectg any small current enterg the amplifier. ( + ) ( ) i i 2

10 9 Nw substitute fr i t give a relatship between and. i ( + H) H 2 H + ( ) 2 earrangement f this equat gives an express fr vltage ga. H 2 + H... (9) Nw, if H >>, a simplified apprximat is btaed (0) S the ga is defed merely by the rati f tw resistrs. What else d yu ntice abut the ga?

11 0 It is negative, s the signal will be verted. S this is an vertg amplifier; the previus circuit is therefre a nn-vertg amplifier. Equat (9) can be rewritten as fllws H ( ) It is nw much easier t see that, if is large, equat (0) is valid. It is als clear that the utput impedance f this circuit will be reduced the same way as that fr the first amplifier because, if 0, there is n difference between the circuits. Hwever, the put impedance is quite different! Z I ( ) i + i +... ( 2) Nw, if is very large, this simplifies t the express belw. Z... ( 3 )

12 Let us summarize the results btaed s far. Amplifier Nn-Invertg Invertg a (apprx.) Input Impedance i ( +H) Output Impedance + H + H Therefre, negative feedback a vltage amplifier: reduces ga stabilizes ga reduces the effect f nn-lear ga reduces utput impedance enhances put impedance the nn-vertg mde defes put impedance as the vertg mde.

13 2 STABILLITY In the precedg analysis, we have cnsidered essentially DC r lw frequency cndits. What happens at higher frequencies? Any real system cannt respnd stantaneusly. In practice, all systems have a frequency respnse (remember the lessn n cmplex numbers), perhaps well defed, but ften nt. S, bth and H the feedback analysis can be frequency dependent, i.e. (jω) and H(jω). The cmplex clsed-lp ga f a system CL (jω), such as that shwn FIUE, is btaed frm equat (2). ( ) CL j jω ( ω ) + ( jω) H( jω)... ( 4) Cnsider the denmatr f the ga. At certa frequencies it culd becme 0, i.e. (jω) H(jω). Physically, this means that the cmbed phase shift frm and H is 80 and the magnitude is. If the denmatr is zer, the ga apparently becmes fite, and the system scillates at this frequency. emval f the put has n effect; the system cntues t scillate. This is ne defit f stability, the Barkhausen Criter. It is usually an undesirable situat, but it is als ne which we can put t gd use if we want t design se-wave scillatrs, as we shall see the next tpic.

14 3 WOKED EXAMPLE 2 FIUE 5 shws a system with a number f C netwrks the feedback path, buffered by emitter fllwer amplifiers with a ga f +. Fd what the ga must be rder t cause scillat, and als what the frequency f scillat will be if kω and C μf. + C C C FI. 5 Each f the C netwrks is a lw-pass filter. Fr each, the ga is given by the fllwg express: ( jω ) jωc + jωc jωc + Sce there are three f these netwrks, the verall feedback fract, H(jω), is derived as fllws:

15 4 3 H( jω) ( jω) ( jωc + ) 3 2 ( 3[ ωc] ) + j 3ωC ωc ( [ ] 3 ) Prve fr yurself that this is true. emember the cndit fr scillat! It is when H. In this case: ( 3[ ωc] ) + 3ωC [ ωc] j( ) 2 3 Nw, is a real number, and we can assume is real. S, under what frequency cndit is the left hand side f this equat real? It must be when the denmatr is real, i.e. when the imagary part f the denmatr is zer. 3ωC [ ωc] 0 3 This gives tw sluts: either ωc 0 r [ ωc] 3 2 The first slut is f n terest, sce ω 0 crrespnds t DC, and yu cannt scillate at DC! The secnd slut is the ne which we want. At this

16 5 frequency, puttg [ωc] 2 3, and the imagary part f the denmatr t zer, we bta a value fr A. 3 3 ( ) S, fr scillat 8. Als, the frequency f scillat is btaed frm [ωc] 2 3. ω 3 C rad/s And, sce ω 2πf, 732 f 2π 276 Hz

17 6 FEEDBACK CONFIUATIONS FIUE 6 shws the fur pssible feedback arrangements. 2 I 2 H H (a) (b) 2 I 2 H H (c) (d) FI. 6 FIUE 6(a) is the type we are nw familiar with. A fract f the utput vltage is fed back series with the put. It is vltage derived, series-fed feedback. In FIUE 6(b) the utput current is sampled by the feedback element. The put half f the circuit is the same as fr FIUE 6(a). This is current derived, series-fed feedback. FIUE 7 shws hw the utput current might be sampled practice. The vltage develped acrss the lw value resistr f is prprtal t the utput current flwg thrugh it. It is this vltage that is fed back.

18 7 I 2 f H I 2 f FI. 7 In FIUE 6(c) the feedback is vltage derived. It is a current, thugh, that is fed back t the put. As the feedback element appears acrss the put, this arrangement is vltage derived, shunt-fed feedback. In FIUE 6(d) the utput is current derived and the put is shunt fed. This is current derived, shunt-fed feedback. Effects upn ga The derivat and methd f feedback will tgether determe the effect upn the amplifier's ga. We first nte that FIUE 6(a) it is the utput vltage that is sampled t prduce a feedback vltage. The feedback transfer funct is H and is a vltage rati. FIUE 6(d) is similar. It is the utput current that is sampled t prduce a feedback current. The feedback transfer funct is H I and is a current rati. In FIUE 6(b), thugh, it is the utput current that is sampled t prduce a feedback vltage. The feedback transfer funct is H Z and is nt a rati, it has units f vlts per ampere. It is represented by an impedance.

19 8 In FIUE 6(c) it is the utput vltage that is sampled t prduce a feedback current. The feedback transfer funct is H Y and nw has units f amperes per vlt. It is represented by an admittance. The effects f each cnfigurat upn vltage and current gas are summarized the table belw. The table shws that series-fed feedback affects the vltage ga whereas shunt-fed feedback affects the current ga. Fr this t be true we must apply tw cndits : (i) In the case f vltage derived, shunt-fed feedback the signal surce is assumed t be an ideal cnstant vltage generatr (zer surce resistance) (ii) In the case f current derived, series-fed feedback the signal surce is assumed t be an ideal cnstant current generatr (fite surce resistance). Type f feedback... Effect n... f If ltage derived, series-fed educed by Nt affected ( + H ) Current derived, series-fed educed by Nt affected ( + H ) ltage derived, shunt-fed Nt affected educed by ( + I H I ) Current derived, shunt-fed Nt affected educed by ( + I H I )

20 9 Each cnfigurat will affect the put and utput impedance f the amplifier but the analysis nw given we keep n familiar territry and cnsider vltage-derived, series-fed feedback nly. The effects f the ther cnfigurats can then be deduced. THE EFFECTS OF FEEDBACK UPON INPUT AND OUTPUT IMPEDANCE Input impedance FIUE 8(a) shws the put half f an amplifier with negative feedback applied. The amplifier has an put resistance f r and a current I f flws t it. The feedback element is represented by its Théven's equivalent f a vltage generatr H 2f and a series resistance r B. Up till nw we have mitted this resistance and we shall see shrtly that we were justified dg s. Fr the sake f accuracy, thugh, we shall nw clude it and then justify its miss. I f f r r B H 'lkg ' 2 (a) H 2f (b) FI. 8

21 20 Yu might fd it helpful t nw refer t FIUE 8(b) which shws hw H 2f might be generated practice. The Théven equivalent circuit is taken 'lkg t' the feedback element acrss the termals x-x The pen circuit vltage is that generated acrss 2 with the put f the amplifier remved. The equivalent resistance is that 'seen' lkg t the termals x-x. It can be shwn that this equivalent resistance, which we have dented r B, will equal 2( + ), where represents the parallel cmbat f the amplifier's utput resistance and its lad resistance. The vltage put equat fr the amplifier f FIUE 8(a) is I f r + I f r B + H 2f...() and usg the relat 2f f I f r I f r + I f r B + H I f r I f r ( + H ) + I f r B Dividg thrugh by I f gives I f ( ) + r + H r B Nw, I is the put resistance f the amplifier with feedback applied. We f will dente this resistance by r f. We can nw write an express fr the amplifier's put resistance with feedback: r f r ( + H ) + r B

22 2 If we make the assumpt that r ( + H ) >> r B then r f r ( + H )...(2) The put resistance has been creased by a factr ( + H ). This crease is nt simply due t the effect f the series resistance f the feedback element, the cmpnent f which, r B has been represented separately the equat. It is primarily due t the feedback vltage that ppses current flwg t the amplifier. Calculate a suitable value f 2 if the amplifier f FIUE 8(b) is t have a clsed lp ga f 00. The pen lp ga is 0 4 and kω

23 22 Fr such a large pen lp ga and mdest clsed lp ga we can use the apprximat H 00 f Als H and transpsg t make 2 the subject: 2 H H and ntg ( ) H then 2 H Ω 00 2 des deed have a lw value and if the lp ga is als high then the apprximat r ( + H ) >> r B is mre than justified.

24 23 The table belw summarizes the effects f negative feedback upn the different feedback cnfigurats. In all cases negative feedback will reduce distrt. Type f feedback Effect n f Effect n If Effect n Input resistance Effect n Output resistance ltage derived, series-fed educed by ( + H ) Nt affected Increased by ( + H ) educed by ( + H ) ltage derived, shunt-fed Nt affected educed by ( + I H I ) educed by ( + I H I ) educed by ( + I H I ) Current derived, series-fed educed by ( + H ) Nt affected Increased by ( + H ) Increased by ( + H ) Current derived, shunt-fed Nt affected educed by ( + I H I ) educed by ( + I H I ) Increased by ( + I H I )

25 24 EXAMPLES OF FEEDBACK CICUITS FIUE 9 gives sme cmmn examples f feedback as used transistr amplifier circuits. The first three examples are f lcal feedback: by this we mean that the feedback is cnfed t ne stage f amplificat. This is cntrast t the last tw examples where the feedback is ver tw stages. Fr each example we shall shw (by a much simplified analysis) that the vltage r current ga f the amplifier can be estimated simply frm the rati f tw resistrs. We shuld nte at this stage that an amplifier may have a.c. r d.c. feedback, r bth. In the first circuit, fr example, if the emitter by-pass capacitr C E is cluded the circuit then there will be n a.c. feedback. If C E is excluded bth a.c. and d.c. feedback are present. We shall assume an a.c. analysis, s all the vltages and currents FIUE 9 are r.m.s. Furthermre, we shall assume that the utput is nt laded (i.e. is pen circuit) and that the put is fed frm an ideal vltage surce (i.e. zer ternal resistance).

26 25 + CC + CC + CC C C I c I f 2 I e f I c 2 2 E C E E (a) (b) (c) + cc C C2 2 f I f E f E2 (d) + cc 2 f I f I e E (e) FI. 9

27 26 FIUE 9(a) is f a cmmn emitter amplifier. The feedback cmpnent is the resistr E. A current I e flws thrugh E t develp a vltage E I e that is fed back t the base f the transistr. The feedback cnfigurat is current derived series feedback. Analysis... Assume that the amplifier is pen circuit s that all the utput vltage is develped acrss the cllectr resistr C. The utput vltage is then C I c. The feedback vltage is E I e. Hence the vltage feedback rati is: I E I C e c E Ie I C e E C makg the assumpt that I e I c. E Thus H and if the h fe f the transistr is reasnably high, the vltage C ga f the stage is given by: H C E FIUE 9(b) shws a cmmn emitter circuit with an alternative frm f biasg. The base resistr f is taken t the cllectr f the transistr rather than directly t the supply rail. This trduces feedback because the current I f varies with variats 2. The feedback cnfigurat is vltage derived, shunt-fed.

28 27 Analysis... We assume that the bias/feedback resistr will be f a very high value. We are safe t assume that f is very much greater than the put resistance t the transistr. The current I c (fr which an arbitary direct has been assumed) splits at the cllectr termal t flw thrugh f and C. Thus I f I c C C + f I c C f as f >> C practice. The current feedback rati is therefre: I I f c C f and if the h fe f the transistr is high: I f C FIUE 9(c) is f an emitter fllwer. The utput is taken frm the emitter and it is this very vltage that is fed back t the put. The cnfigurat is vltage derived, series feedback. The circuit represents 00% feedback, ignrg the small vlt drp acrss the base-emitter junct.

29 28 Analysis... The vltage feedback rati is and therefre the vltage ga is. FIUE 9(d) is f a tw-stage amplifier emplyg vltage-derived, series feedback acrss the tw stages. A prt f the utput vltage appears acrss the put via the emitter resistr f the first stage. Analysis... In this example, each stage has current derived series-fed feedback (like that shwn FIUE 9(a)). Thus, the ga, say, f the tw, cascaded states (with f remved) is: C E C2 E2 Let the feedback ver the tw stages be H. Then if H >>, we can make the usual apprximat, as develped the previus lessn, that f + H H Nw H E C2 and therefre f C2 E

30 29 FIUE 9(e) is f a tw stage amplifier emplyg current derived, shunt-fed feedback. Analysis... We aga assume that the feedback resistr f is very much greater than the put resistance f the first stage. The emitter current f the secnd stage splits between E and f. Thus: I f I e E E + f I c E f makg the assumpt that f >> E Thus the feedback rati is: If H I I c E f and if the pen lp ga is high: I f E

31 30 POSITIE FEEDBACK Instead f ppsg put changes, which is what negative feedback des, it is f curse pssible t have psitive feedback. In essence, the stability situat utled the previus sect is an example. A 80 phase shift rund the lp means that negative feedback is transfrmed t psitive feedback. If the ga rund the lp is greater than unity, any put change is amplified rund the lp until sme system limit is reached; e.g. a transistr amplifier will reach cut-ff r saturat. Psitive feedback is nrmally an effect that lear systems shuld avid like the plague! But, nce understd, mst scientific cncepts can be develped t perfrm useful functs. Psitive feedback is used many switchg situats, as we shall see the next tpic. elays emply psitive feedback t ensure a gd psitive switchg act. Square-wave scillatrs and timers use psitive feedback. In the switchg act, there are nly tw stable states, e.g. ON-OFF, Lgic r Lgic 0. The transits between these states usually ccur very quickly, and it is impssible t dwell an termediate state. FIUE 0 shws a typical situat. A lear system with negative feedback has a well defed sgle-valued characteristic. A switchg system with psitive feedback will have tw stable levels, transits between them ccurrg at different put levels. This is a hysteresis effect. As examples f this, a thermstat will switch n and ff at tw different temperatures, and a relay energizes and de-energizes at tw different cil vltages.

32 3 Psitive feedback Lear Hysteresis FI. 0

33 32 SELF-ASSESSMENT QUESTIONS. In a negative feedback system, H 0.0. Fd the system ga when has values f 0, 00, 000, 0 4 and Suppse the required ga f a system is 00; what value f is needed t ensure the ga is with 2% f this value if H 0.0? 3. In a nn-vertg amplifier, FIUE, the ga is 50, the put impedance is 2 kω and the utput impedance is 200 Ω. A 00 m vltage surce has an impedance f 00 Ω, and a lad f 200 Ω is cnnected t the utput, as shwn belw. What will the utput vltage be? 00 Ω 00 m S 2 kω i 200 Ω 0 0 L 200 Ω FI A negative feedback fract H 0. is nw applied t the circuit abve. Fd the clsed-lp vltage ga, put impedance and utput impedance. What will the utput vltage be if the put is?

34 33 5. A feedback netwrk has the fllwg funct. H ( jω ) 3 ( jωc) + 5jωC + 6( jωc) + ( jωc) 2 3 What value f is required t make the circuit scillate? If 0 kω and C 47 nf, what will the scillat frequency be? 6. FIUE 2 shws the circuit f a cmmn-cllectr amplifier (r mre usually called the emitter fllwer). Sketch the amplifier's h-parameter equivalent circuit (usg just h fe and h ie ) and shw that the vltage ga is apprximately unity. + CC 2 E L 2 FI. 2

35 34 ANSWES TO SELF-ASSESSMENT QUESTIONS. Clsed lp ga is given by the frmula belw. a + H Usg H 0.0 gives the fllwg values fr ga. H a As creases, the ga appraches H The target clsed lp ga is 00; we have t fd t guarantee a ga f at least 98 with H 0.0. Substitut f these values t the ga frmula gives the required value f ( ) ( ) Frm the diagram, it is clear that the put resistance reduces the surce signal.

36 35 i S S i + i Similarly, the utput resistance reduces the utput vltage. 50 i + L L Calculate the utput vltage by substitutg these reduced values i S Calculate the clsed-lp vltage ga. + H ( ) 833.

37 36 Calculate the put impedance. Z + H i ( ) ( ( )) kω Calculate the utput impedance. Z + H ( ) 33. 3Ω Nw, use the same relatships as the last quest t calculate the reducts surce and utput signal, and fally the utput vltage. i S S Z + Z ( ) i Z + L L

38 i S 7.07 The ga is bviusly less, but the effects f put and utput impedance are reduced. 5. H ( jω ) 3 ( jωc) + 5jωC + 6( jωc) + ( jωc) j( ωc) 2 3 ( ) ( ωc) + j ωc [ ωc] This is real nly when the real part f the denmatr is zer, because then the j the numeratr will cancel with the secnd term the denmatr. Oscillat will therefre ccur at a frequency defed as fllws. 2 ( ) that is ω 6 ωc 0, 6C At this frequency, H 3 ( ωc) ( ωc) 3 5ωC ωc 5 ωc ( ) 2 ( ) 2

39 38 Substitute (ωc) 2 /6 t the abve equat. H S, rder t make H, must be 29! This is rather a surprisg answer. The frequency f scillat is then calculated as fllws. f ω 2π 2π 6 C 2π Hz 6. FIUE 3 shws the required equivalent circuit. This is perhaps best cnstructed by drawg the transistr mdel first and then addg the ther cmpnents arund it. I b h ie h fe I b B E L 2 FI. 3

40 39 Fr cnvenience, the tw base bias resistrs have been represented by B. Applyg Kirchhff's vltage law t the put circuit gives ( ) I h + I h + b ie b fe L where + L L E At the utput circuit, ( ) I h + 2 b fe L Dividg the put equat t the utput equat gives: 2 ( ) ( ) I h + b fe L b ie b fe L Ih + I h + ( hfe + ) L h + I h + ( ) ie b fe L ( ) >> But hfe + L hie 2.

41 40 Thus, the vltage ga f the emitter fllwer will be just under unity. This fact gives the circuit its name: the utput vltage fllws the put vltage. Nte als that the emitter fllwer des nt vert the utput vltage. The circuit des, hwever, have gd current ga and this, cmbed with its very high put impedance and lw utput impedance, gives it many applicats drivg lw impedance lads frm high impedance surces.

42 4 SUMMAY This lessn has shwn that negative feedback stabilizes and defes ga, enhances put impedance and reduces utput impedance. If the frward path ga is large, the ga is defed by feedback fract., where H is the H On the ther hand, a psitive feedback situat can lead t stability and scillat, a situat which we shall put t advantage subsequent lessns.

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