ELECTRONICS-1. ZOLTAI CHAPTER 6 AMPLIFIERS WITH FEEDBACK Purpose of feedback: changing the parameters of the amplifier according to needs. Principle of feedback: (S is the symbol for Signal: in the reality either voltage or current.) So = AS1 S1 = Si - Sf = Si - ßSo Si S1 So = A(Si - S ) So ß o + A from where: - Sf=ßSo * So A A ß A = Si 1 A 1 H where: H = Aß Denominations: loop-gain H = Aß open-loop gain A = So/S1 * closed-loop gain A = So/Si Qualitatively different cases of feedback: oscillation with const. amplitude oscillation with narrower growing amplitude pos. f.b. H -1 0 positive feedback negative feedback Why do we use almost always negative feedback? * The total change of the A as a function of A and ß: 1(1 A) A A 2 A* = A (1 A) 2 (1 A) 2 Introducing the relative errors: A * 1 A A A * 1 A A 1 A According to this, the relative error of the open loop gain will be decreased by (1+Aß), and the error of the feedback circuit goes 1:1 into the relative error of the closed loop gain. This is the Elec1-fb.doc 52 ELECTRONICS-1. ZOLTAI reason for using almost always the negative feedback. (We can not take advantage of the negative sign, because the sign of the relative errors is random.) Speaking of negative or positive feedback is only justified if A and ß are real quantities. Generally they are complex quantities, and the type of the feedback is a function of the frequency: E.g. if the loop-gain has three time-constants: Nyquist-diagram of Im[H] the loop-gain on the complex plain f = Re[H] -1 f = 0 unstable case stable case The four basic feedback arrangements of asymmetrical amplifiers (everywhere, the resistances R can be substituted by the corresponding impedances Z.) Voltage Current Av ocv1 YA ocv1 io R Rin Ro Rin o v1 v1 Rg Rg Series RL vo RL v g vg vin vin vvo Zßio ZA sci1 ß Ai sci1 Ro Ro i Rin io i1 Rin 1 iin iin ig RL arallel ig Rg P Rg vo RL iio Yßvo ß The features of the ideal feedback network: it does not load the amplifier; it provides the feedback signal like an ideal generator; it transfers the signal only in one direction. Elec1-fb.doc 53 ELECTRONICS-1. ZOLTAI In the case of the series feedback the summing of voltages is made (v1=vin-vß where for series voltage feedback v = v and for series current feedback v = Z i ). This type of feedback has ß ßv o ß ß o an effect on those transfer quantities, which contain the input voltage in their definitions. Thus for series feedback: A Y A* v ; Y * A ; but: A* A and Z * Z . v 1 H A 1 H i i A A RL - for series voltage feedback: H =Avßv where Av=Av oc Ro RL Ro - for series current feedback: H= YAZß where YA=YA sc Ro RL The input resistance for series feedback connections: * Rin = vin/iin = (v1 + vß)/iin = (v1 + Hv1)/iin = (1 + H)v1/iin = (1 + H)Rin . End of class 12. Class 13: 1st mid-term-test. In the case of the parallel feedback the summing of currents is made (i1=iin-iß where for parallel voltage feedback i = Y v and for parallel current feedback i = i ). This type of feedback has ß ß o ß ßi o an effect on those transfer quantities, which contain the input current in their definitions. Thus for parallel feedback: A Z A* i ; Z * A ; but: A* A and Y * Y . i 1 H A 1 H v v A A Ro - for parallel current feedback: H =Aißi where Ai = Ai sc Ro RL RL - for parallel voltage feedback: H= ZAYß where ZA=ZA oc Ro RL The input resistance for parallel feedback connections: * Rin = vin/iin = vin/(i1 + iß) = vin/(i1 + Hi1) = vin/i1(1 + H) =Rin/(1 + H). Output resistance: * vooc The application of the general definition: Ro is difficult, because in case of current iosc feedback the vooc and in case of voltage feedback the iosc will be calculated without the operation of the feedback. E.g. in the case of series voltage feedback: Simplification: Rg = 0. Elec1-fb.doc 54 ELECTRONICS-1. ZOLTAI Avoc Avoc vooc = vg where Hsp=Avoc v is the special loop-gain vg ß 1 Avoc v 1 H sp belonging to a special load condition (here open circuit load); Avoc iosc = - vg (at this calculation the feedback is inactive); Ro substituting into the definition formula, we get: * Ro Ro 1 Hsp If Rg 0, then shifting the Rg into the box of the amplifier, we can retrace the calculation to the derived formula with the only difference that instead of Avoc the modified gain ainAvoc must be Rin used everywhere (ain is the voltage division: ain = ). Ro Rin Rg With this, the general formula: R Rin Rin R g R * o o 1 ain Hsp vg vg Taking the parallel current feedback: Here, the simplification is: Rg = For open circuit load the feedback is inactive: vooc = -igAiscRo Aisc For short circuit load: iosc ig 1 Aisc i R * R (1 A ) R (1 H ) With these the output resistance: o o isc i o sp where Hsp = Aiscßi is the actual special loop-gain. If Rg , then shifting the Rg into the box of the amplifier, we can retrace again the calculation to the derived formula with the only difference that instead of Aisc the modified gain ainAisc must be R g used everywhere (ain is the current division: ain = ). R o R in With this the general formula: Rg Rin Rg Rin R * R (1 a H ) ig ig o o in sp Elec1-fb.doc 55 ELECTRONICS-1. ZOLTAI Without derivation, the formulas for the remaining two cases: - for series current feedback: R R * R (1 a H ) in o o in sp where ain = and Hsp = YAscZß Rg Rin - for parallel voltage feedback * Ro Rg Ro where a and H = Z Y 1 a H in sp Aoc ß in sp Rg Rin Numeric problems Problem 1 The input resistance of an asymmetrical amplifier is 12 k, its load resistance is five times the output resistance. The amplifier is provided with an ideal parallel voltage feedback. The resulting output resistance is given for two different generator resistances: * at Rg1 = 12 k, Ro1 = 250 and * at Rg2 = 24 k, Ro2 = 200 . Tasks: calculate the output resistance of the amplifier without feedback and its input resistance * with the feedback (Ro, Rin = ?) Solution: * Ro In case of the parallel voltage feedback: Ro 1 ain Hsp First, we can calculate the input current divisions in both cases: Rg1 12k Rg2 24k 2 ain1 0.5 and ain2 Rg1 Rin 12k 12k Rg2 Rin 24k 12k 3 * Applying the Ro formula for both cases: 250 = Ro/(1+0.5Hsp) and 200 = Ro/(1+[2/3]Hsp), we get a system of equations with two unknown quantities. The solutions are: Hsp = 6 and Ro = 1 k. The loop-gain belonging to the actual load: H = Hsp([RL/(Ro+RL)] = 6[5Ro/(Ro+5Ro)] = 5 And so the resulting input resistance: * Rin = Rin(1+H) = 12(1+5) = 72 k. Problem 2 Elec1-fb.doc 56 ELECTRONICS-1. ZOLTAI The parameters of a given asymmetrical amplifier:: Avoc = 160, Rin = 10 k, Rout = 1 k; input and output closing resistances: Rg = 1 k, RL = 15 k. * * * Prescribed parameters: Av = 10, Rin 100 k, Rout 0,15 k. (Purpose: no great input and output attenuation.) Tasks: a.) the type of the necessary feed-back; b.) the transfer factor of an ideal feed-back; c.) checking the prescribed parameters. Solution: * Rin Rin series feed-back is necessary; * Rout Rout voltage feed-back is necessary, So the applied arrangement: series voltage feed-back (see page 47). R L 15 A v A voc 160 150 R out R L 115 A* 1 10 H 14 v H 14 and A v 1 H 150 A v 150 * R in R in (1 H) 10(114) 150k O.K. R 1 R * out ... 0,073k O.K. out 1 a A 10 14 in voc 1 150 11 150 R in 10 10 where a in R g R in 110 11 Problem 3 (for self-contained understanding) The following data of an asymmetrical amplifier are given: Avoc = -50; Rg = 0.5 k; Rin = Ro = RL = 2 k. The next resulting parameters are to be realised by the use of negative feedback: * * * Ai = 5; Rin 0.5 k; Ro 5 k. Tasks: determine the type of the feedback, the transfer of the feedback network and the realised * * values of Rin and Ro . Solution: * * From Rin follows parallel, from Ro follows current feedback. Thus, the arrangement: From the given data, first the short circuit current Ai sci1 gain must be determined: Ro i i1 Rin o Ro iin A v o c R in i 1 R Avocv1 R o i RL o g Rg The searched short circuit current gain: i ßi o A v o c 5 0 A is c R in 2 k 5 0 R o 2 k Elec1-fb.doc 57 ELECTRONICS-1. ZOLTAI The current gain belonging to the actual load: Ro 2k Ai Aisc 50 25 Ro RL 2k 2k * From the prescribed resulting current gain: Ai = Ai/(1+H) that is 5 = 25/(1+H), we get H = 4.