The Concept of Feedback

The chapter on feedback generally occurs toward the end of an electronics textbook, because it is considered rather difficult. Of course, feedback has appeared often before, above all in connection with operational amplifiers, where all the circuits use feedback. Only the name has been mentioned, however, not the theory. It is necessary to understand feedback, the theory as well as the practice, to be a useful electronics engineer. Nearly all practical circuits now make use of it, and certainly all the best and most interesting of them.

Indeed, feedback is heard in common language, where it refers to information from the object of some activity, and may help to improve that activity, making it more effective or more pleasant. It occurs in mechanism, in things like servos, where the actual output is compared with the desired output, so that steps can be taken to make them agree. Biology presents further examples, so there is some general utility to the feedback concept, whether it is applied qualitatively or quantitatively.

In electronics, feedback is used quantitatively, and has a definite mathematical expression. The figure at the right is a signal flow diagram, where the signal is represented by x, the input xs is at the left, and the output xo is at the right. The circle with the Σ in it is an abstract summing junction with the sign of each input beside the input arrow. The amplifier has a gain A, the ratio of its output to its input. The feedback network multiplies the output by the factor B, and sends its feedback signal to the summing junction. The algebra beneath the diagram shows that the closed-loop gain, Af, is the open-loop gain A of the amplifier divided by the feedback factor 1 + AB. An input fed into the amplifier input is multiplied by A, and then by B, as you follow the feedback loop, so that AB is the loop gain. Although x is an abstract signal here, representing any such quantity, there are corresponding quantities in any concrete feedback system.

Note that the feedback signal is subtracted from the input signal, and tends to reduce the output of the amplifier. This is called negative feedback, which is almost always the desired type. If the feedback signal were added to the input signal, the output would be greater than in the absence of feedback. This principle of positive feedback was used in regenerative receivers to amplify weak radio signals, but was always very tricky and unstable, requiring constant adjustment. Positive feedback can be seen with a microphone and an loudspeaker, usually appearing as a howl at some high frequency. Positive feedback does have its uses, but they are different from those of negative feedback, and not as beneficial.

If you differentiate the expression for the closed-loop gain Af with respect to the open-loop gain A, you will find that dAf/Af = (dA/A)[1/(1 + AB)], or that the fractional change in the closed-loop gain is reduced by a factor 1 + AB with respect to a fractional change in the open-loop gain. This means, among other things, that the output of the feedback amplifier will be much more linear and constant than that of the raw amplifier. In fact, when Howard Black of Western Electric introduced feedback in 1928, it was this factor that was so important in the design of telephone repeaters. If the open-loop gain is very large, so that AB is large compared to 1, then Af = 1/B, as constant as B can be made constant. The gain then depends only on the amount of feedback, and not at all on the characteristics (which may be wild) of the amplifier.

One way that the open-loop gain may vary is with frequency. We have just seen that the closed-loop gain can be made much more independent of frequency. The gain of an operational amplifier is purposely made frequency-dependent, and is of the form A(ω) = A / (1 + s/ωo. Here, s is the imaginary angular frequency jω, and ωo is the angular frequency at which the gain is down by 3dB (the corner frequency). At higher frequencies, the gain is inversely proportional to the frequency. This is called a single-pole response. If you find the closed-loop gain for this case, the result is exactly the same single-pole responsed, but for the corner frequency is now (1 + AB)ωo, larger by the feedback factor. This means that the range of constant response is greatly extended towards higher frequencies.

At high frequencies, the gain of an amplifier not only decreases, but the phase of the output may change as well. If the phase shift is 180°, it is the same as a negative A. B may also change the phase. If the loop gain becomes -1, then the closed-loop gain goes to infinity: that is, the system oscillates at this frequency. This will always happen unless the open-loop gain is less than 1 at the frequency where the phase shift is 180°, and that is why operational amplifier gain is made to decrease 20dB/decade, a single-pole response.

All of these considerations apply to negative feedback systems in general, not just to electronic ones. In particular, if the gain A is high, then the input to the amplifier is x = xs - B (xs/B) = 0. That is, the feedback acts to reduce the amplifier input to 0. For an operational amplifier, this means that the two input terminals are brought to the same potential, as we have seen.

Electronic Feedback

We now consider the actual voltages and currents in a feedback system. The input to the amplifier can be either a voltage source or a current source, and there is a similar choice at the output. There are, correspondingly, four cases to consider. If we are to get circuits that are as easy to analyze as the abstract signal-flow diagram, it is necessary to be very careful in the choice of basic circuits. We will be interested in the closed-loop properties of the circuits: the gain, input impedance and output impedance. It is always possible to find these by circuit analysis, but the complications can hide the general properties that interest us. We want to understand the circuit, not just get numbers.

The four ideal configurations are shown above. Note that the sources have no internal impedances, and are either ideal voltage or current sources. The outputs of the amplifiers are drawn with output resistances, but a voltage output is left open (no output current) and a current output is shorted (no output voltage), so that the output resistances have no effect. They will, however, be useful in finding the output resistances. There are no load resistances at all, since they would affect the output voltage or current. If you make any other assumptions, your results will be complicated and full of voltage and current dividers, and the beauty will be hidden. Source resistance and load resistance will be taken into consideration separately later.

An output circuit with a voltage source is always used for shunt sampling, where the feedback voltage or current is proportional to the output voltage vo. An output circuit with a current source is always used for series sampling, where the feedback voltage or current is proportional to the output current io. An input circuit with a voltage source is always used for series mixing, where the feedback is represented by a voltage source. An input circuit with a current source is always used for shunt mixing, where the feedback is represented by a current source. The feedback sources have no internal resistances, and do not load the outputs (draw current from a voltage output, or cause a voltage drop with a current output).

The case of series mixing, shunt sampling is specially appropriate for an amplifier with a high-impedance input, and a low-impedance output, that is meant to amplify a voltage appearing at its input. An operational amplifier is a very good example. The gain A of the amplifier is the ratio of the output voltage to the input voltage, as shown, and is a pure number. The feedback ratio B is also a pure number. The output voltage vo = A(vs - vf) = A(vs - Bvo), from which vo = [A/(1 + AB)] vs, exactly as in the signal flow diagram.

To find the input resistance, one imagines vs removed, and a test current i injected into the input terminal. The input resistance will be the resulting increase in voltage v divided by i. Here, the injected current will cause a voltage iRi across the input resistance, which will cause an output voltage of iARi and a feedback voltage of iABRi. Hence, v = iRi + iABRi (note the +!), and Rin = (1 + AB) Ri. This is a general result: with series mixing, the input resistance is increased by the feedback factor.

To find the output resistance, one imagines vs reduced to zero (we are dealing with a linear circuit, and it is no use keeping this extra source in the problem), and a voltage v applied to the output terminal. The output resistance will be the voltage v divided by the current that results. The feedback voltage is Bv, which is the negative of the voltage applied to the amplifier, so the voltage source in the output supplies a voltage -ABv. This makes a total voltage v - (-ABv) = (1 + AB)v across Ro, which gives a current i = (1 + AB)v/Ro. Then, Rout = Ro/(1 + AB). Again, we have a general result that in shunt sampling, the output resistance is decreased by the feedback factor.

This is an ideal case, but the results we have discovered will hold approximately in a real case that is not too far different. Negative feedback, applied to a voltage amplifier, divides the gain by the feedback factor (stabilizing it), increases the input resistance by the feedback factor, and decreases the output resistance by the same factor. Any voltage amplifier greatly appreciates these gifts.

In the case of shunt mixing, we represent the source by its Norton equivalent, which is just a current generator here; its internal resistance is considered infinite. The feedback is also a current generator, connected so that the feedback will be negative. For series sampling the current output is also appropriate, the output current being that through the short-circuit at the output. This is appropriate to a current amplifier, which receives a current at the input and amplifies it. The gain A is again a pure number, as is the feedback ratio B. The closed-loop current gain, io/is = A/(1 + AB), by algebra that you can easily supply by this time. Again, it is just like the signal flow diagram.

To find the input resistance, we apply a voltage v to the input. This causes an input current v/Ri to the current amplifier, an output current of Av/Ri, a feedback current of ABv/Ri, and so an input current of v/Ri + ABv/Ri (note the +!). The input resistance Rin = Ri / (1 + AB). This is a general result. For shunt mixing, the input resistance is decreased by the factor (1 + AB).

To find the output resistance, we reduce is to zero (as in the previous case), and inject a current i into the output. This produces a feedback current of Bi, and an output current of -ABi. The output voltage is the voltage drop of these two currents across Ro, or v = (1 + AB)Roi, and so the output resistance is Rout = (1 + AB) Ro. Again, as a general result, the output impedance is multiplied by the feedback factor with series sampling.

The other two cases are now easily handled. With series mixing and series sampling, the input is voltage and the output is current, so the amplifier is a transconductance amplifier, and the gain A is now a conductance G, while the feedback ratio B is a resistance R. The loop gain AB is GR, again a pure number, although the factor are not. With shunt mixing and shunt sampling, the input is current and the output is voltage, so the amplifier is characterized by a transresistance R, and the feedback by a conductance G. Again, the loop gain is a pure number, RG.

The input resistance is multiplied by (1 + GR) for series mixing, but divided by (1 + RG) for shunt mixing. The output resistance is multiplied by (1 + GR) for series sampling, but divided by (1 + RG) for shunt sampling. These results are easily derived by the same methods used above.

We now know the properties of each of the four ideal configurations. Where an actual amplifier is approximated by them, the results can be used directly. However, the same amplifier may use more than one feedback loop, and inputs and outputs may not be ideal--in fact, they are usually not. Many actual cases can be massaged to look like the ideal cases, and this makes the results of much more extensive validity.

Adding Source and Load Resistances

The source is represented by its Thévenin or Norton equivalent, depending on the type of amplifier input in the ideal case. Then the source resistance is combined with the input resistance of the ideal amplifier, and the input terminals moved. The source now appears just as in the ideal case. This process is shown in the figure below.

The load resistance is combined with the output resistance of the amplifier, and again the terminals are moved so the circuit looks like one of the ideal ones. It is usually necessary to apply a factor to the output controlled source in this step. In the figure below, the intermediate step is shown so that the final result can be understood.

The feedback network can also be included in the ideal equivalent. Usually, the feedback network is not a great load on the output, and the input is not a great load on the feedback network. Instead of making general rules, the examples will show how this is done in special cases. Feedback networks are various enough to make general procedures unprofitable. One thing that is always neglected is the effect of changes at the input communicated by the feedback network to the output, since these effects are much less than those amplified by the amplifier gain. One should always be on the lookout for conditions different than those assumed, of course. When any great accuracy is required, a circuit analysis will always give the necessary numbers, if not the necessary understanding. If you worry about the 47k feedback resistor in this circuit, you should also worry about the approximately 100k output resistance of the transistor collector at the same time. Neither is really worth worrying about, but you should be aware of their presence.

In some cases, load resistances, networks at the input, and such are not actually within the feedback loop, and do not affect the feedback. They should be ignored in the feedback analysis, and taken up later. The feedback network should be as simple as possible, for clarity of understanding. It does little good to worry about small effects.


The circuit at the right is intended to be a voltage amplifier, with input vs and output vo. It looks very much like the simple--and impractical--common-emitter transistor amplifier we saw earlier, but here the base resistor is returned to the collector, not to +V, which makes a great difference in the circuit. We want to find the voltage gain, the input resistance and the output resistance of the circuit. Let's first make a feedback analysis, and then test the circuit on the breadboard to see how accurate our analysis is. We are, as always, looking for understanding. Numbers can always be obtained from a SPICE analysis or similar.

The first thing to do is to determine the bias conditions, in particular the voltage at the output node "a" and the collector current of the transistor, which will be necessary for determining the circuit parameters. Let us assume that the base node is at 0.65V, and that β = 100. Write node equations for nodes "a" and "b" in terms of Va and IB. It is easy to eliminate IB and solve for Va = 4.425V. Then IB = 0.0153 mA, so IC = 1.53 mA. A quick bias calculation can be made assuming β infinite. Then, VC = 0.65 + (47)(0.65/10) = 3.70V, from which IC = 1.77 mA. This approximate calculation is obviously good enough for determining the size of the feedback resistor, here 47k. We also see that the value of beta has little effect on the bias, changing IC only from 1.53 mA to 1.77 mA as beta goes from 100 to infinity. The circuit behavior is approximately independent of beta, which makes it a good circuit.

From the collector current, we find that the conductance gm = 1.53/25 = 0.0612 S, re = 1/gm = 16.34 Ω, and rπ = 1.65k. We can now write down the equivalent circuit for the basic open-loop amplifier. The feedback, by inspection, is shunt mixing, shunt sampling, so the basic amplifier is a transresistance amplifier that converts current into voltage. Therefore, the input circuit is written as a current input (Norton source) and the output circuit as a voltage output. The 10k input resistance becomes part of the Norton source, and is combined with rπ in parallel, to get Ri = 1.42k. The voltage source in the output is the voltage across the collector resistor produced by the current gmv = gmRiis = 86.9is, which is -408k is. This gives the transresistance of the basic amplifier, -408k. Ro = 4.7k (we have made a Norton-Thévenin transformation in the output).

The equivalent circuit of the open-loop amplifier is shown at the right. There is always the question of how the feedback, here the 47k resistor, affects the basic amplifier. In this case, the 47k is large enough that it does not significantly load either the input or the output. Note also that if it is physically removed, the circuit no longer works. The equivalent open-loop amplifier is a kind of fiction, in general. If you must take the feedback resistor into consideration, it can be included returned to ground at both the input and output nodes, where it slightly reduces the open-loop input and output resistances. What exactly is the open-loop amplifier can be a source of dispute, but in most cases, as here, the details make little difference. If the feedback network is a significant load, taking it into consideration becomes a matter of importance, and perhaps wizardry. To keep things simple, let's just neglect the effect of the 47k on the open-loop amplifier here.

The loop gain is 408k / 47k = 8.7, so the feedback factor is 9.7. The closed-loop gain is, therefore, -408k / 9.7 = 42.1k, current in to voltage out. The voltage gain from the input node at the end of the 10k input resistor is, then, -42.1k / 10k = -4.21. The input resistance is 1.42k / 9.7 = 146Ω. Taking out the 10k in parallel, the input resistance looking into the base is 149Ω, and the input resistance looking into the 10k is 10,149&Omega, which is 10k for all practical purposes. The output resistance is 4.7k / 9.7 = 485Ω. Feedback has beneficially reduced the output resistance of this circuit. We have done what we proposed, and now know the voltage gain, input resistance and output resistance of the amplifier, which is a quite practical one.

Now breadboard the circuit, and determine its properties by experiment. I measured Va = 4.36V and Vb = 0.66V, so IB = 0.0127 mA, and IC = 1.54 mA (be sure to subtract the current that goes down the feedback resistor). This gives gm = 0.0616 and β = 121, so rπ = 1.98k. The open-loop gain is then -(0.0616)(1980)(4.7k) = -573k, and the feedback factor is 12.2. We expect a voltage gain of (-573k)/(12.2 x 10k) = - 4.69, an input resistance of 1.65k / 12.2 = 135Ω, and an output resistance of 4.7k / 12.2 = 385Ω. These figures are a little different from those of our initial analysis, but are obviously roughly equal.

To measure the voltage gain, apply a small voltage at the input (say 0.4V or so) with a low-impedance source (I used an op-amp voltage source) and measure the change in the output voltage. I found a voltage gain of -4.45, satisfyingly close to the predicted value. The output resistance can be determined by connecting a load (I used 10k) and noting the drop in output voltage. The measured output resistance was 381Ω. The closeness to the predicted value is only luck, but we must be doing something right. The input resistance can be found by measuring the voltage across the 10k input resistor, and finding the change in current (note that this resistor also carries a bias current!). Taking some care to be precise, I found 10180Ω, again very close to the predicted value (but the precision is doubtful, of course, because of the subtraction). All in all, the feedback analysis gave very good answers for this circuit.

The input resistance at the base can be found directly, using an interesting trick that is often useful. Note that the 47k resistor is connected so that if the voltage of the base end is raised by, say, v, then the voltage at the collector end is simultaneously raised by Av, where A is the gain from base to collector (negative, of course). So far as current at the base node is concerned, the 47k resistor acts like a 47k/(1 - A) resistor. In the present case, the gain from base to collector is approximately - 4700 / 16.34 = -288, so the feedback resistor looks like 47000 / 289 = 163Ω to the base. The input resistance at the base is then this resistance in parallel with 1980Ω, or 151Ω. This shows the source of the low input resistance, and that it is a direct result of the feedback. Be careful that you do not use the amplifier voltage gain of -4.45 in this calculation! The gain must be between the two ends of the feedback resistor, not just any gain that happens to be around. This analysis does not give anything about the output resistance--its low value comes from a different source.

There is an easy rule to estimate the voltage gain of this circuit. Since the input resistance looking into the base becomes small, the base voltage does not change much when the input and output change. Therefore, the feedback resistor and the input resistor form a voltage divider whose "center" remains at the same voltage, so they act like a lever with fulcrum at the base, and the voltage gain is accordingly about G = -Rf/Ri = -47k / 10k = -4.7.

Exactly the same feedback configuration is found in the most familiar op-amp circuits, where a feedback resistor Rf is connected from the output to the inverting input. The current gain of this circuit is -Rf, the output resistance is very low, the input resistance at the inverting input is practically zero, and the two inputs are forced to the same potential by the feedback. All of these effects are instantly plausible from our knowledge of the shunt-shunt feedback configuration. The voltage gain of the inverting amplifier is then -Rf/Rs, given by the same rule as in the preceding paragraph. If the input is to the noninverting input, then the output voltage will follow it, since the feedback current will be practically zero, and there will be no voltage drop in the feedback resistor (which can be made zero).

In the noninverting configuration, a voltage vf = voR2/(R1 + R2) is fed back by a voltage divider to the inverting input. The voltage divider is hardly loaded at all by the input, and the current through the divider is a small load on the output. The op-amp is itself the open-loop amplifier, a voltage amplifier of high gain, so that the closed-loop gain is determined by the feedback network alone. This is the series-shunt (series mixing, shunt sampling) feedback configuration, appropriate to voltage amplifiers. The input resistance at the noninverting input is very high (a multiple of the open-loop resistance of a megohm or so), and the output resistance is very low (a few ohms at most), while the gain is stable and equal to 1 + R1/R2. The two inputs of the op-amp are provided to make feedback convenient; they are essentially the + and - inputs of the summing junction in the abstract feedback network.

Suppose the feedback resistor in the shunt-shunt configuration is replaced by a capacitor. The same analysis holds, but now the reactance of the capacitor, 1/jωC appears in place of Rf. This makes the feedback factor 1 + jωRmC, where Rm is the transresistance of the open-loop amplifier. Now the gain becomes Rm/(1 + jωRmC), which is a single-pole, low-pass response, as we mentioned in connection with stability at the top of the page. This is an easy way to use a small capacitor to compensate an amplifier for high-frequency stability, and is used in op-amp design. It also shows that a capacitor across a feedback resistor can increase stability markedly.

At the right is another feedback voltage amplifier, and we want the voltage gain and the input and output resistances. Note that you cannot simply connect a voltage source to the input! Normally, the input would be an AC voltage applied through a capacitor to avoid disturbing the bias conditions. This is a familiar circuit, and may be analyzed without feedback formalism. However, feedback gives us new insight. We observe that the output current is sampled, by the 1k emitter resistor, and the voltage across this resistor is fed back to the input (the emitter is the "inverting input" of the transistor) with the proper phase to give negative feedback. Therefore, we have series-series feedback in this case. The emitter current is slightly different from the collector current, but since most transistors have a beta over 100, the difference is negligible in practical work. The transistor is biased by the voltage divider feeding the base. We want a collector current of around 1.5 mA, so the necessary base voltage will be about 0.65 + (1.5)(1) = 2.15 V. The base current will be about 0.015 mA (or less), so the divider current should be about ten times larger, to swamp it, or about 0.15 mA. Now it is easy to design the (lightly loaded) divider, since we must have R1 + R2 = 12.0 / 0.15 = 80k, and R1/R2 = 9.85 / 2.15 = 4.581. The solution is R1 = 14.3k and R2 = 65.7k. Use the standard values 15k and 68k, which will give about 2.17V, close enough for government work.

The resulting collector current will be 1.52 mA, giving gm = 1.52 / 25 = 0.0608, re = 16.45Ω, and rπ = 1645Ω. Before proceeding with the feedback analysis, it is very useful to notice that neither the input bias voltage divider, nor the 4.7k load resistor, have anything to do with the feedback. We might as well eliminate them from the start, and be left only with the transistor and the 1k emitter resistor. We want a voltage input and a current output for the series-series case, and we now have that quite naturally. The equivalent circuit is shown in the figure, and it is simple indeed. The output resistance of the transistor collector has been shown, and given a typical value. Feedback can now be applied; the feedback ratio is simply 1000Ω, so AB = (0.0608)(1000) = 60.8, and the feedback ratio is 1 + AB = 62. The closed-loop input and output resistances are (62)(1.645k) = 102k and (62)(100k) = 6.2 M (!). Feedback has very nicely raised the input impedance to a respectable value, and the output is almost an ideal current source.

The closed-loop gain is 0.0608 / 62 = 0.0009806 S. The output is into the 4.7k collector resistor, giving a voltage gain of -(4700)(0.0009806) = -4.61, and, incidentally, an output resistance of 4.7k for the complete amplifier. If gmRe is large compared to unity (it is 60.8 here), then the voltage gain is given by G = -Rc/Re. Here, this is -4700 / 1016 = -4.63. Here, re has been included in Rf where it really belongs, but it is a small correction. We also note that the input resistance at the base of the transistor is β(re + Re). A resistance added at the emitter is multiplied by beta when reflected into the base.

The input resistance of the complete amplifier is dominated by the voltage divider. In fact, it is only 12.3k in this case, in spite of the fact that the input resistance at the base has been raised to over 100k. This is a good argument against this circuit design, but it can be tolerated occasionally. Such amplifiers appear mainly in electronics courses. It is a good example of series-series feedback, however, which is why it is here.

The circuit can be tested, and the results compared with our calculations. Remember not to use a low source resistance voltage source at the input! The input voltage can be changed by connecting a resistor in parallel with one of the divider resistors for the purpose of measuring the gain. The best way, however, would be to use an AC input to find both gain and input impedance. I used a function generator to supply a 1 kHz sine wave, and a dual-channel oscilloscope to measure the amplitudes of the input and output signals, with an 0.1 μF input coupling capacitor. The gain was -4.7, the input impedance 11.6k, and the output impedance 4.7k. The output impedance was measured by connecting a 10k resistor across the output, and noting the fall in output voltage (from 5.47V to 3.72V). The input impedance was measured by finding the input voltage with and without a 10k resistor in series with the function generator, remembering that the function generator has an internal resistance of 600Ω, and using the voltage divider rule. The agreement is pretty good.

The fourth feedback configuration, shunt-series, appears a little more rarely than the others, but is handled in exactly the same way. An example is given in Sedra and Smith, p. 704 (the first example above is also treated in S&S). Feedback is of great use in studying the frequency response and stability of amplifiers, but I shall leave that topic for elsewhere.

There is a remarkable feedback system, called a phase-locked loop that holds a voltage-controlled oscillator in exact phase relation to an external signal. It is not appropriate to treat them here, but they should be mentioned as another example of the wide application of feedback in electronics.

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Composed by J. B. Calvert
Created 11 July 2001
Last revised 14 July 2001