The magic of negative feedback
I remember browsing in the engineering library when I was a student and coming across books devoted entirely to amplifiers. Surely amplifiers play an important role in electronics. There are plenty of tiny signals in need of substantial amplification, such as those generated by antennas, microphones, cartridges, thermocouples, strain gauges, even the human brain and heart. For some reason, there is a tendency to believe that the higher the gain of an amplifier, the better. As an instructor, I have observed students eagerly compare the gains of their amplifiers in the lab. Once, a student even lied, though after confronting the data for his lab report, he had to settle for a much lower gain than that bragged about to his peers.
So, paraphrasing Bob Pease, I am tempted to ask What’s all this high-gain stuff, anyhow? There are many applications where too much gain just won’t do (think of a unity-gain inverting amplifier to perform just plain signal inversion.). Besides, the very word amplifier seems somewhat restrictive (and even a bit nerdy to folks from the humanities) to warrant entire books on the subject. It so happened that somebody, too lazy to reshelf a book on Shakespeare’s Sonnets, had left it next to one of the amplifier books I was looking at, making the latter appear even more prosaic by comparison.I came to appreciate the importance of high gain only after getting exposed to negative feedback. Just a bit over 86 years ago, Harold Black conceived this milestone concept while trying to reduce amplifier distortion [1, 2]. In today’s parlance and notation, he was striving to implement a circuit that would accept an input vI and yield an output vO such that
where Aideal is the desired circuit’s gain (not necessarily huge - just as large as required by the application at hand). In the real world, ideality is unattainable, but we can strive to get as close to it as necessary. To quantify our effort, we need to define an error signal, say
and then we must devise a way to adjust vO so as to drive vE as close as possible to zero. Harold Black’s genial idea was to generate vO by merely amplifying vE with a high-gain amplifier, according to the arrangement of Figure 1. Aptly called an error amplifier, it gives
where is the amplifier’s anticipated high gain.What? If vE is an error, isn’t vO itself going to be an error, and an outlandishly magnified one at that? Have you ever heard of anything good coming out of glorifying an error? Evidently, this is not the
Figure 1 - Basic negative-feedback block diagram.Best way to look at it. A better alternative is to shift the attention from vO to vE, rewritten as
and to realize that an amplifier with a high will require a fairly small vE to sustain vO. (If you liken the amplifier to a binocular, this is like looking thorough the binocular backwards.) Rewriting Eq. (2) as
indicates that if vE is sufficiently small, vO will come very close to AidealvI. What’s just as important, should any disturbance attempt to increase/decrease vE, the amplifier will react in the opposite direction by decreasing/increasing vO. It is precisely that little “minus” sign that prevents vO from blowing up uncontrollably (think of it as the secret of negative feedback!)
Figure 2 - A circuit (a) before and (b) after the inclusion of an amplifier, showing the new voltages and currents in the limit → ∞, in which case we have vE → 0.To illustrate, consider the circuit of Fig 2a, with the voltages and currents shown. Next, connect a high-gain amplifier as in Fig. 2b, and observe how it alters the voltages and currents in order to make vE small. Indeed, in the limit → ∞, the amplifier will drive vE to zero, thus establishing a virtual short between nodes C and A. This will cause the 2-V source and 2-kΩ resistance to establish a current of (2 V)/(2 kΩ) = 1 mA. This current is drawn from the 1-kΩ resistance, making vA = –(1 kΩ)x(1 mA) = –1 V. By KVL, vB = vA + 2 V = +1 V, and vC = vB – (2 kΩ)x(1 mA) = –1 V = vA, thus confirming that vE = vA – vC → 0. The current proceeds through the 3-kΩ resistor into the amplifier’s output node, and finally to the negative power supply, not shown. So, vO = vC – (3 kΩ)x(1 mA) = –4 V.
How does the amplifier “know” to adjust vO to –4 V exactly? Well, let it “dare” raise vO by, say, 1 V, from –4 V to –3 V. Using simple voltage-divider reasoning, you find that vC will rise by 0.5 V and vA will rise by 1/6 V, causing vE (= vA – vC) to change from 0 V to –1/3 V. This, in turn, will cause the amplifier to swing vO in the negative direction, thus opposing the initial attempt. So, let it “dare” swing vO to –5 V instead. This will change vE from 0 V to +1/3 V, in turn redirecting the amplifier to swing vO in the opposing, positive direction. Evidently, any attempt to change vO from –4 V is met by a counteraction that tends to restore vO back to –4 V, the only value for which the amplifier finds itself “at peace.” As we know, this is negative feedback. Try swapping the amplifier’s input terminals so as to make feedback positive, and you’ll see that any attempt to swing vO away from –4 V (assuming vO ever got there), will cause vO to diverge, till the amplifier eventually saturates.
What if is not infinite, but, say, = 1,000 V/V? Still, vE will be small enough to cause very little change in the loop current, and hence, in the various voltages. Assuming vO is still in the vicinity of –4 V, Eq. (4) predicts vE ≈ –4/1,000 = –4 mV, so the loop current decreases from 1.0 mA to (2 – 0.004)/2 = 0.998 mA. Repeating the above step-by-step calculations with this new current value, you’ll find that vO changes from –4 V to –3.988 V, a negligible change!Summarizing, negative feedback uses a high-gain amplifier not to make vO uncontrollably large, but to make vE vanishingly small, or, ideally, to null vE. Call it a nullifier instead of an amplifier? If the word amplifier is too nerdy for an entire book on the subject, the word nullifier (network-theory books actually call it a nullor) is even less of a public-relations winner (like manufacturing binoculars to be used backwards). Would my students compare data in the lab and brag that “my amplifier nulls more than yours”? Would an entire book on nullors stand a better chance next to Shakespeare’s Sonnets?
An instructive exampleLet us put the above ramblings in a more practical framework by recreating a distortion situation of the type that might have inspired the genius of Harold Black. In Figure 3a we are trying to drive a 100-Ω load by means of a unity-gain (Aideal = 1 V/V) push-pull buffer. The push-pull has close-to-unity gain so long as vI > VBE1 or vI < VEB2, but zero gain for VEB2 < vI < VBE1, resulting in the highly distorted output of Figure 3b, top. Also shown in Figure 3b, bottom, is the error vE = vI – vO.
Figure 3 - (a) Push-pull buffer; (b) input/output waveforms (top), and the error waveform (bottom).Would you ever consider reducing vO’s distortion by amplifying the error vE, let alone with a high-gain amplifier? Harold Black did, and the results are depicted in Figures 4 and 5. The circuit of Figure 4 uses a preamplifier with = 100 V/V, along with a plain wire to feed back vO and establish the error vE = vI – vO at the input. The benefits are evident from Fig. 5, top, which reveals that vO is now a much closer replica of vI. Were we to jazz up by another decade, to 1,000 V/V, vO would change very little because it’s already very close to vI: the extra decade of gain increase would simply go into reducing vE further by a factor of ten (remember the analogy of the binocular being used backwards).
Figure 4 - Raising aε to 100 V/V, and using a wire to implement negative feedback for Aideal = 1 V/V.
Where has the distortion gone? A look at the amplifier’s output vA of Figure 5, middle, reveals the kind of contortions required of the amplifier in order to make vO closely track vI. And where does the amplifier get the instructions for these contortions? From the error signal of Fig. 5, bottom, which in this case is vE = vA/100, and is expressed in tens of millivolts. And how does the amplifier manage to pre-distort its own input? “That’s pure magic, the magic of negative feedback,” is the answer given in class by an ex-student of mine. What is the price we are paying for this magic? We are essentially throwing away 40 dB of error gain to approach an overall gain of only 1 V/V, or 0 dB – a price well worth paying, given the benefits.
Figure 5 - Waveforms for the negative-feedback circuit of Figure 4.Negative feedback is full of fascinating nuances that some students, rushing through homework and tests, don’t get the chance to absorb in full. Many will master them on the job after graduation, but others may not get the opportunity to go deeper. To honor the genius of Harold Black, I intend to post a series of tutorial blogs specifically for these engineers. My “analog bytes” will progress through levels of increasing complexity, from the very basics of the present byte all the way to a byte on the oft perceived as intimidating topic of frequency compensation in the presence of a right-half-plane zero.
And now, a quizThe circuits of Figure 6 are somewhat similar of those of Figure 2, except that once you connect the amplifier it will give vE = 0 regardless of its gain ( can be large, medium, small, or even 0!) Can you explain why? No math, no SPICE allowed – use just plain intuitive reasoning.
Figure 6 - A circuit (a) before and (b) after the inclusion of an amplifier.