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
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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.
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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.
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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
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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
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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
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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. From this the current transfer of the feedback network: ßi = H/Ai = 4/25. The actual value of the resulting input resistance: * Rin = Rin/(1+H) = 2k/(1+4) = 0.4 k ( 0.5 k O.K.) The actual value of the resulting output resistance: * Ro = Ro(1+ainHsp) = 2k(1+0.2 50 [4/25]) = 5.2 k ( 5 k O.K.) In this calculation:
Rg 0.5k ain 0.2 Rg Rin 0.5k 2k
Applying the feedback theory to the analysis of the common emitter stage (a piece of reading) The a.c. equivalent circuit (observe the way the RE is VS+ drawn): g21vB i RC CC2 B iC CC1 vi vB 1/g11
Rg Rg RB R RB vo RL C RL vi vg RE RE vo vg VS-
S-C feedback Substitution of the feedback network: RE RE
The transfer parameter
of the feedback: Zß = RE REiC REiB 0 The transfer parameter of the amplifier: YA = YAsc = g21 (YA=YAsc because g22 = 0)
The loop-gain: H = Hsp = YAZß = g21RE The resulting transconductance of the feedback section: * * YA = YA/(1+H) = g21/(1+g21RE) = g21 (that is the transconductance of the reduced transistor). Since the output and input voltage of the feedback section are identical with those of the whole CE stage, the voltage gain:
CE * g21 (RC xRL ) Av = -Ya (RC x RL) = 1 g21RE (the same formula that we derived with the reduced transistor);
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CE * 1 Rin = RB x Rin = RB x 1 g21RE g11
(again the same formula we derived with the reduced transistor); Here, neglecting RE at the output of the feedback network is justified, if for RE, which is series connected in the input circuit, the following condition is fulfilled: 1 R 1 g R = h + h R (O.K.) E « 21 E 11 21 E g11 The output resistance: CE * Ro = RC x Ro = RC x 8 = RC (again the known result). The RE at the input of the feedback network has no influence on the voltage gain, because it is not included by the output points (of course it makes load for the g21vB current generator and therefore has an effect on the maximum output voltage).
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The negative feedback of one-pole amplifiers
One-pole amplifier: - DC amplifier - it has one high frequency break-point Its gain formula: 1 A = A0 s 1 0 Conditions: the feedback circuit is ideal and has a real, transfer factor . The closed-loop gain: A A 1 1 A* = = ... = 0 = A* , 1 A s 0 s 1 A 0 1 1 * 0 (1 A0 ) 0
* A0 * where A0 and 0 = 0 (1 A0 ) , 1 A0 or introducing the DC loop gain, H0 = A0:
* A0 * A0 and 0 = 0(1+H0). 1 H0 Conclusion: the mathematical structure of the open-loop and the closed-loop gain is identical. The DC gain decreases in the same manner as the break-point frequency increases. The so called gain band-width product is independent of the feedback: * * A00 = A0 0 Graphical representation of the feedback-effect: A, dB 3 dB
A0 A 1+A0 -20 dB/D 3 dB
A0* 1/ A* 0 0* (log) A, if A0 » 1 ° 0°
-45° A* A -90°
Example for a one-pole amplifier: the type 741 op-amp, where A0 = 106 dB and 0 = 2*5 r/s
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The negative feedback of two-pole amplifiers
The two-pole amplifier is a DC amplifier with two high-frequency break-points. The open-loop gain: 1 1 A = A0 = A0 s s 1 1 1 1 1 1 s s2 1 2 1 2 1 2 Conditions: the feedback circuit is ideal and has a real, transfer factor . The closed-loop gain (introducing the DC loop-gain, H0 = A0): A A 1 A* = = ... = 0 1 A 1 H 1 1 1 1 0 1 s s2 1 H0 1 2 1 H0 12 Comparing the identical mathematical structures of the open-loop and the closed-loop gain with that of the second order low-pass filter section: A, dB
A0 1 -40 dB/D A A0 1 1 2 1 s 2 s = = Qp p p (log) 0
1 p 2 where p is the pole-frequency and Qp is the pole quality number, we find that these parameters are influenced by the negative feedback as follows:
* p = 1 2 and p = p 1 H0 1 2 * Qp = and Qp = Qp 1 H0 1 2
The three cases of different quality:
2 real roots
* 1 Qp = double real roots monotonously decreasing 2 * 1 conjugate complex roots: Qp = maximally flat 2 peaking
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The Bode-plot: The asymptotic Bode-plots are very A, dB similar to those in case of the one- A0 pole amplifier, but here the actual -40 dB/D frequency response can be qualita- tively different (see the peaking). 1+H0 H0 H
0H * A0 1/ 0A * p p (log)
° 0°
A, H A*
-90°
-180°
* More detailed, Qp can be expressed like this:
12 1 Q* 1 H 1 H p 0 0 1 2 1 2 2 1 2 * 1 If » 1 and H0 » 1, then Qp H0 1 2
1 2 E.g. in case of maximally flat transfer: H0 = . 2 1 The corresponding Bode-plot:
³H³, dB
-20 dB/D
H0, dB 0.52
2 0 6 dB 1 (log) -40 dB/D
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STABILITY OF AMPLIFIERS WITH FEEDBACK
The one-pole and two-pole amplifiers with real feedback are structurally stable. Nyquist diagrams one-pole two-pole
ImH ImH
-1 ReH -1 ReH
These curves can not embrace the -1 point, therefore are these cases structurally stable ones. A three-pole loop-gain may embrace the -1 point depending on the that is the H0 value, this is the conditional stability: ImH -1 H01 H02 ReH
The notion of stability reserve (or margin): a a: amplitude margin Im[H] : phase margin In minimum phase systems (most feedback prohibited Re[H] amplifiers belong to this) only one of the two area margins can be used. With amplifiers usually -1 the phase margin is used. Problem: what is the required minimum phase margin, if the closed-loop gain must be always smaller than the open-loop gain? The answer is with certain approximation: 1 1 ImH A* A A H 1 H 1
* 1 -1 ReH A A H 1 * A = A, if |H-(-1)| = 1 H-(-1)=1 From the figure: min = 60°. H In the practice smaller phase margin is allo- wed (e.g. 45°).
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Frequency (phase) compensation
The purpose of frequency compensation is to alter the frequency response of the loop gain in order to get stability and great enough phase margin.
Frequency compensation methods on the example of a three-pole loop-gain. The prescribed phase margin is: = 45°.
The expression of the loop-gain: 1 H H 0 s s s 1 1 1 1 2 3
³H³, dB original, unstable
H0 -20 dB/D 4 -40 dB/D 1 2 and 3
** 1 3 (log) 0
* * 2 2 d 1 1 -60 dB/D
° 2 and 3 4 original (log) 0° 1 -90° -135° 4 -180° = 45° -270°
-360° below -180°, because of this: unstable
Compensation method #1: new dominant pole (d) d is positioned so that the -20 dB/D asymptote arrives the 0 dB line at 2, that is: d = 1/H0
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Compensation method #2: shifting the first break-point to the left by increasing the time- constant determining it (if it is possible physically).
Compensation method 3: shifting the first break-point to the left by neutralising it and * introducing a new first break-point 1 . This can be done inserting a new block in the feedback loop with the following transfer function: s 1 1 s 1 * 1 ** Compensation method #4: combining the method #3 (shifting the 1 less to the left to 1 ) with the shifting the second break-point to the right by neutralising it and introducing a * new second breakpoint at 2 . The necessary additional transfer function is: s 1 2 s 1 * 2 * ** * * (Conditions: 2 3 and 1 /1 = 2 /2.)
The main difference between the compensation methods: the achieved high-frequency limit of the closed-loop gains. This limit frequency is approximately equal to the frequency, where the loop- gain intersects the 0 dB axis. The easiest way to accept this is the case of the one-pole amplifier with real feedback, if the DC loop-gain A0 is great enough (see figure).
A, dB 3 dB
A0 A i.e. H 1+A0 A0 = H0 -20 dB/D 3 dB 0H
A0* 1/ 0 A A* 0 0* (log) A, if A0 » 1 °
The Bode-plot of A (the open-loop gain) gives approximately the Bode-plot of the H (the loop- gain) in the co-ordinate system with the shifted zero line, 0H, and we find the intersection with * this zero dB line really at the frequency 0 . In mathematical formulation: where the H intersects the 0H dB line, there H = 1 and the phase-angle is (being far from the break-point 0) -90°, that is H = -j, and therefore at this place: A A A* = 0 = 0 1 H 1 j A thus ³A*³ = 0 which means the -3 dB point. 2
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If the loop-gain has more than one break-points, but the intersection point with the 0H dB axis is on the -20 dB/D asymptote of the bode-plot of H and far enough from the second break- point, practically the same rule applies. If the second break-point is not far enough (or as in our 45° phase margin case, it is identical with the intersection point) the rule can be applied as a thumb-rule, knowing that the true high-frequency limit is somewhat smaller than (but close to) the frequency of the intersection point. According to this thumb-rule, compensation method No. 1 (new dominant pole) results poor high frequency limit, methods No. 2 and 3 (shifting the dominant break-point) give better high-frequency limit, and method No. 4 is the gives the best one.
Circuit realisation of the compensations.
Somewhere in the loop, e.g. at the junction of the amplifier stages the following intervention is made. The compensating components are indicated by bold lines.
Method No. 1: 1 R1 New factor in the loop: Cc s 1 R2 c where: c = 1/Cc(R1xR2)
Method No. 2: R1 C C c * 1 is changed to 1 , where R 2 * 1 = 1/[(C+Cc)(R1xR2)]
Method No. 3: the introduced new factor in H: R1 s Rc 1 1 R2 s 1 * 1 Cc where: 1 = 1/RcCc and * 1 = 1/Cc(Rc+R1xR2) Method No. 4: the introduced new factors: the first identical Cc2 with method No. 3, and the second s 1 2
R1 R2 s 1 * 2 where 2 = 1/R1Cc2 and
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* 2 = 1/(R1xR2)Cc2. Practical example (a piece of reading):
Type 741 op-amp: it has an internal 30 pF capacitance determining a dominant break-point (see the analysis of the internal circuitry of the 741 op-amp).
Type 748 op-amp: it has no internal capacitance but an external capacitance can be connected to the same internal point, between which the 30 pF is connected in case of the 741 (the internal structure of the two types of op-amp are basically identical, giving almost identical amplifier parameters).
Suppose that a closed-loop gain of A* = 10 is to be achieved with a minimum phase margin of 45°. Determine the available high-frequency limit using both 741 and 748 op-amps.
|Avd| H748
106 dB 3 pF
˜H0 H741 -20 dB/D
30 pF 10 dB * Av0 (log) f 0 dB 5 Hz 50 Hz -40 dB/D
fhl 741 100 kHz
fhl 748 1 MHz
In the case of the 748 we can apply smaller capacitance and still the zero transition point of the loop-gain is before the 2nd break-point on the -20 dB/D asymptote, that is the phase-margin is at least 45°. Due to the smaller capacitance the dominant break-point frequency and the high- frequency limit is higher with the 748 op-amp (the phase-margin of the 741 op-amp is greater, approaching 90°).
End of class 15.
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