Half to the amplifier, half to the resistors

How much of the amplifier gets through measured one half of a bargain. It multiplied an amplifier’s gain by one part in a hundred thousand, solved the closed loop again, and found that a tenth of a per cent of the change reached the answer at low frequency, that the fraction turned negative below fifty degrees of margin, and that it rose to 1.61 near the crossover of a loop with 34.9° of it. It also said, in its list of what it did not measure, that there was nothing to draw for the resistors: the gain is one over the feedback fraction to a part in a thousand, so its sensitivity to the divider is essentially one.

That is true at direct current and it is the only place it is true. The point of feedback is that the accuracy moves from the device to the divider, and a move has two ends. This essay measures the other end — the fraction of a change in the feedback fraction that reaches the closed-loop gain — on the same netlist, by the same method, at the same seven settings of the second pole, and finds that the two fractions are not independent quantities at all. They add to one at every frequency. Where one falls the other rises by exactly as much, and at the crossover they meet at one half.

The resistors’ share, measured #

The loop is the one the earlier essay used: an amplifier with a low-frequency gain of $10^5$ and three poles, at 10 Hz, at the slider’s setting and at 2 MHz, closed through a divider for a gain of a hundred. Two perturbed copies are solved beside the nominal one at each frequency. In the first the amplifier’s gain is multiplied by $1 + 10^{-5}$; in the second the feedback fraction $\beta$ is. The fractional change in the closed-loop magnitude, divided by the fractional change that caused it, is the sensitivity, and the sign of the second is taken off, because a larger feedback fraction means a smaller gain and the interesting number is how much of the change arrives rather than which way it points.

At 10 Hz the divider’s sensitivity is 0.99899 and the amplifier’s is $9.98\times10^{-4}$. That is the textbook arrangement, measured: nine hundred and ninety-nine parts in a thousand of the answer’s variation belong to the resistors, which are the parts that can be bought at a tenth of a per cent, and one part belongs to the device, which is specified to a factor of three. A resistor that drifts by 0.1% moves the gain by 0.0999%. The two numbers add to one to five decimal places, which is suggestive and is not yet an argument.

Above a few hundred hertz both curves move. The amplifier’s share rises through the region the earlier essay mapped, turns negative at the setting drawn, and climbs to 1.60 near 7.9 kHz. The divider’s share falls in mirror image: to 0.5 at the crossover, to −0.60 at the same 7.9 kHz — a larger feedback fraction there makes the gain larger — and back to zero far above, where the loop has gone and the resistors are a load on a bare amplifier. The dashed line through the middle of the figure is the sum of the two measured curves, and it is one everywhere: the worst departure over 241 frequencies is $1.2\times10^{-5}$, which is the size of a difference quotient taken with a $10^{-5}$ nudge.

The second route shares only the netlist. The loop is cut at the amplifier’s input and the return ratio $T$ is measured by injection — the construction what is left at crossover bisects its margin on — and the two sensitivities are predicted as the real parts of $1/(1+T)$ and $T/(1+T)$. The measured curves agree with those to a hundred-thousandth of the larger. The first expression is the one the earlier essay confirmed; the second is new here, and a control engineer would recognise it as the complementary sensitivity, whose name says what the figure shows.

Why the two must add to one #

The sum is not a coincidence of this circuit and it does not need the return ratio to explain it. Scale the amplifier’s gain up by some factor $k$ and scale the feedback fraction down by the same factor. The loop gain $A\beta$ is unchanged — the product of the two is exactly what it was — so the loop behaves identically at every frequency: the same crossover, the same margin, the same peak. And the closed-loop gain is $A/(1 + A\beta)$, which has changed only through its numerator. It is multiplied by exactly $k$.

The figure makes that change with $k = 1.1$ at four solves a frequency, and the dashed curve — both changes at once — is 1.1 to sixteen figures in magnitude and in phase, from 10 Hz to 10 MHz. The two single changes are nothing like it. Ten per cent more amplifier moves the gain by 0.009% at 10 Hz, dips it by 1.6% at 3.98 kHz, and raises it by 16.7% at 8.41 kHz. Ten per cent less feedback fraction raises it by 9.99% at 10 Hz and lowers it by 5.4% at 7.94 kHz. Each is a frequency-dependent mess, and their combination is a flat line.

A change of one per cent in both at once is therefore a change of one per cent in the gain, and a change of one per cent in both is the sum of the two changes made separately, to first order. So the fraction of the amplifier’s change that arrives, plus the fraction of the divider’s change that arrives, is one. That holds for the complex gain, where the sensitivities are $1/(1+T)$ and $T/(1+T)$ and add to one as complex numbers, and it holds for the magnitude, whose sensitivities are their real parts. It holds at every frequency and every margin, for any loop, because it is a statement about where $A$ and $\beta$ sit in the expression and not about what either of them does with frequency.

The consequence is the whole content of this essay: no frequency makes the gain insensitive to both. Wherever the feedback succeeds in hiding the amplifier, it has handed the variation to the resistors, and wherever it stops hiding the amplifier, it has stopped handing it on. The earlier essay’s sensitivity curve is half of a single object, and the other half is not “essentially one” except in the band where the first half is essentially zero.

At crossover, exactly one half #

The point where the two measured curves cross is the loop’s crossover frequency, 5.73 kHz at the default setting, and they cross at 0.5000 each. The ideal amplifier, and where it stops being one found the closed-loop gain itself already one per cent low by 1.35 kHz on a loop of this kind; the crossover is several times further up, where the gain has stopped being $1/\beta$ in any sense a specification would accept. Move the slider and the crossover moves — to 1.98 kHz at 11.7° of margin, to 9.10 kHz at 65.3° — and the two curves still cross there, at one half.

That follows in a line from the expressions. At the crossover $|T| = 1$, so $T = e^{j\theta}$ for some angle, and

$$\frac{1}{1 + e^{j\theta}} = \frac{e^{-j\theta/2}}{e^{j\theta/2} + e^{-j\theta/2}} = \frac{e^{-j\theta/2}}{2\cos(\theta/2)},$$

whose real part is $\cos(\theta/2)/(2\cos(\theta/2)) = \tfrac12$ for any $\theta$. The phase of the loop gain, which is the margin, cancels. So a statement can be made about every feedback amplifier ever built, at the one frequency its designer most certainly knows: at the loop’s crossover, a fractional change in the amplifier reaches the magnitude of the gain at exactly half its size, and so does a fractional change in the resistors. A margin of eighty degrees and a margin of twelve give the same answer there. What they change is everything around it.

The geometry is cleaner than the algebra. The complex sensitivity to the device, $S = 1/(1+T)$, is a point in a plane; the sensitivity to the divider is $1 - S$, the arrow from that point to the number one.

As frequency rises the point leaves the origin — where the divider carries everything and the device nothing — swings out into the plane and settles at one, where the roles are exchanged. At the crossover it lies on the vertical line through one half, equidistant from zero and from one, so the two arrows form an isosceles triangle. Their real parts are one half each. Their lengths are equal to each other and to nothing in particular: short when the locus stays close to the real axis, long when it swings wide.

The rotation that the magnitude does not see #

The lengths of those arrows are the sensitivities of the complex gain — the magnitude and the phase together. A magnitude responds only to the part of a change that lies along it, which is the real part; the rest turns the gain without lengthening it. For a circuit whose output is read on a level meter, the real part is the whole story. For one whose output is compared with another signal — a loop inside a loop, a channel matched to a second channel, a filter whose phase is part of its specification — the arrow’s length is.

At the crossover that length has a closed form. With $T = e^{j\theta}$ and a phase margin $\varphi = 180° + \theta$, $|1 + T| = 2\sin(\varphi/2)$, so

$$|S| = |T/(1+T)| = \frac{1}{2\sin(\varphi/2)}.$$

The seven loops land on the curve to ten figures: 4.925 at 11.7°, 3.185 at 18.1°, 2.293 at 25.2°, 1.669 at 34.9°, 1.248 at 47.3°, 0.926 at 65.3° and 0.774 at 80.4°. The curve crosses one at $2\sin(\varphi/2) = 1$, which is sixty degrees exactly. Below that margin a change in the amplifier moves the complex gain at the crossover by more than itself, and so, by the same amount, does a change in the resistors. The rest of each change beyond the magnitude’s half is a rotation: at 34.9°, a one per cent change in either element moves the gain’s magnitude by half a per cent and its phase by about 1.6 hundredths of a radian, 0.9°.

Sixty degrees is a familiar number to anyone who has specified a loop, and it usually arrives as a rule of thumb about overshoot. Two measurements of one margin showed that a phase margin and an overshoot are one quantity read in two domains, 34.9° against 34.9°. Here the same margin appears a third time with a threshold attached that is not a convention: sixty degrees is the margin at which the triangle’s sides are exactly as long as its base.

A band that belongs to neither #

Both arrows can be longer than one at once, and near the crossover of a loop with less than sixty degrees they are.

The amplifier’s band opens a little below the crossover and runs far above it, because above crossover $|1/(1+T)|$ approaches one from above and takes decades to get within a per cent. The divider’s band opens far below the crossover, where the loop’s phase passes ninety degrees while its gain is still large, and closes a little above. Where they overlap, the gain is more sensitive to the amplifier than an amplifier with no feedback would be, and more sensitive to the resistors than a gain of $1/\beta$ with a perfect amplifier would be.

At 11.7° of margin that overlap runs from 1.42 to 2.78 kHz, at 34.9° from 4.50 to 7.94 kHz, and at 47.3° from 6.41 to 9.70 kHz: roughly the octave around the crossover, narrowing as the margin grows. At 65.3° it has gone — the device’s band starts above the crossover and the divider’s ends below it — and it does not come back at any larger margin, because at the crossover both sensitivities are then below one and the two bands separate there first.

The shape is the one the node that is at ground for a while found on a summing junction’s impedance, and for the same reason: every quantity a loop holds is divided by $1 + T$, and $|1 + T|$ is smallest near the crossover, so every one of them peaks there. The band is not a curiosity of an exotic loop. At 47.3° of margin, a comfortable design, the gain responds to its own resistors 25% more strongly at the crossover than a resistor-only circuit would, and to its amplifier 25% more strongly than an amplifier on its own. Neither the device nor the divider owns that octave. The feedback is working hard there and achieving the opposite of its purpose for both.

What a trim is worth, and where #

The practical reading concerns anything adjusted. A circuit trimmed by one of its resistors is trimmed through the divider’s sensitivity, and at direct current that is one: a 0.1% turn of the trimmer is 0.1% of gain, which is why trimming works. At the crossover the same trim moves the magnitude by half as much and turns the phase as well; in the octave around it, a trim of the magnitude at one frequency is a trim of the phase at the same frequency, in a direction set by the margin.

And a device that ages is trimmed out through the same bargain in reverse. The divider’s share of the variation is what the resistors’ accuracy is spent on; the amplifier’s share is what is left over for the device to disturb. A specification of “0.1% gain accuracy” is a statement that the divider’s share is near one and the device’s near zero, and the measurement above says how far up in frequency it stays true, and it says something less expected on the way. At 34.9° of margin the divider’s share does not fall from one: it rises, to 1.014 at 1 kHz and 1.121 at 3 kHz, because the device’s share is negative there by exactly the excess — −0.014 and −0.121. In that band a 0.1% change in a resistor moves the gain’s magnitude by 0.112%, more than the resistor-only arithmetic allows, and the extra is paid for by a device whose improvement makes the gain smaller. Only above about 4 kHz does the divider’s share begin to fall, to one half at the crossover and −0.60 at 7.9 kHz. They are not two facts about the circuit. They are one.

The resistors’ own share also says something about how resistors should be chosen. Below the band, where their share is one, everything about them arrives — tolerance, temperature coefficient, ageing, voltage coefficient — and the tolerance that is not on any part is what that costs in a divider built from several of them, where the printed worst case is one per cent and the measured spread 0.29. Above the crossover their share falls towards zero, which sounds like good news and is the opposite: it means the amplifier’s gain, the thing nobody specified, has taken over. There is no frequency at which both the parts that were paid for and the part that was not are kept out of the answer.

How the numbers were obtained #

Each loop is a netlist: a controlled source with a gain of $10^5$, three RC sections isolated by unity buffers for the three poles, and a resistive divider for a closed-loop gain of one hundred. The device perturbation multiplies the controlled source’s gain by $1 + 10^{-5}$; the divider perturbation multiplies the feedback fraction by the same factor by recomputing the upper resistor, so that the lower one and the amplifier are untouched. Each closed loop is solved at 241 frequencies from 10 Hz to 10 MHz, and each sensitivity is the difference quotient of the solved magnitudes. The second route cuts the loop at the amplifier’s input, injects a unit source there, and reads the return ratio from what arrives at the divider’s output; the predicted sensitivities are the real parts of $1/(1+T)$ and $T/(1+T)$ from that ratio, and the crossover is bisected on its magnitude. The scaling test solves four loops at each frequency — nominal, each change alone, both — with $k = 1.1$, and compares complex gains rather than magnitudes. The bands use 300 points a decade and a threshold of one per cent above one.

Where it stops #

One perturbation per side. The device’s gain is scaled at every frequency together, and the divider’s fraction is scaled as a real number. A resistor that carries capacitance changes the feedback fraction’s phase as well as its size, and a device’s poles move independently of its gain; each of those is a different sensitivity with a different curve. The sum rule holds for the two perturbations drawn and for any pair that leaves the loop gain unchanged, and for nothing else.

A loop with more than one crossover. The one-half result is a statement about any frequency where $|T| = 1$, so a loop whose gain passes through one several times has it several times. Stable, and unstable with less gain drew a loop whose phase crosses −180° three times; one whose magnitude crosses unity three times would have three frequencies at which the resistors and the device share the gain equally, and a margin at each.

A loop whose feedback is not a single fraction. A circuit whose output is taken from inside the loop rather than at the amplifier, or whose input enters partway round, has a gain that is not $A/(1 + A\beta)$, and the scaling argument has to be rewritten for it. The pair of sensitivities still exists — any loop has a return ratio — but the pair that sums to one is then a pair of other elements. Every derivative, and the one that is zero computes the derivative with respect to every element at once, and on such a circuit it would say which ones.

First order only. The sum rule is exact for the scaling and first-order for the separate changes. Ten per cent changes are not small near the crossover — the device’s 16.7% at 8.41 kHz is not ten times its small-change sensitivity there, which is 1.596 and would predict 16.0% — and a large change has to be solved, not added.

Still open: the price, the second divider, and the resistors that are not a fraction #

Whether the peak can be flattened without being moved. The two shares add to one at every frequency, so neither can be pushed below zero everywhere; but a sum rule says nothing about how large the excursions have to be. Whether the amplifier’s share can be kept below one at every frequency, for any loop that is desensitised at all, is a stronger constraint on the loop’s shape than this one, and it is the question how much of the amplifier gets through left when it found the peak monotone in the margin. The answer turns out to be a conservation law about area rather than a bound on height, and it is the next thing to measure.

Two dividers in cascade. An instrument built from two feedback stages has two pairs, and whether the overall gain’s sensitivity to the first stage’s resistors is simply that stage’s complementary sensitivity, or picks up the second stage’s too, is a solve on the cascade that nothing here has done.

Resistors with capacitance in them. A feedback resistor’s parallel capacitance turns $\beta$ into a complex number, and the divider’s sensitivity then includes a rotation of its own. The gain the loop closes against found nine picofarads at a summing junction taking a margin apart; the same picofarads are a perturbation of $\beta$, and their sensitivity is the divider’s arrow in the plane above, turned by their phase.