The resistor that buys the margin back

The load that gets inside the loop measured the damage. Hanging 2.2 nanofarads on the output of a unity-gain inverter changes nothing a reader can find in any expression for its gain and takes the phase margin from ninety degrees to 30.11. The mechanism is fifty ohms of the amplifier’s own output resistance, which no data sheet page puts next to the stability page: the load works against it, and the pole that results is in the forward path.

The repair everybody reaches for is a small resistor between the amplifier and the capacitor. It works, and the reason it works is worth an essay, because it is not the reason it looks like.

What it does, measured #

The figure cuts the loop at the inverting input, injects, and reads what comes back — the same measurement the loop-gain essay makes, with the load and the resistor in place — and bisects the value that restores each margin:

Forty-five degrees arrives at 9.90 Ω, sixty at 23.20 Ω and seventy-five at 45.35 Ω. Those are ordinary resistors, and the repair costs a component and a place to put it.

Why it works, which is not obvious #

The tempting explanation is that the resistor “damps” the resonance, in the way a resistor damps a tuned circuit. That explanation is wrong, and the figure’s second curve is what shows it wrong.

What the resistor actually does is move the load’s pole out of the loop. Without it, the capacitor works against the amplifier’s fifty ohms of output resistance, the pole is at the amplifier’s output node, and the feedback is taken from that node — so the pole is inside the loop and costs phase at crossover. Insert $R_{\text{iso}}$ and take the feedback from the amplifier’s side of it, and the node the feedback watches is no longer the node the capacitor is on. The capacitor now works against $r_{\text{out}} + R_{\text{iso}}$ in series with the impedance looking back — and, crucially, what the loop sees at its own node is a much smaller capacitance, because the resistor separates them.

The test that settles it is to put the resistor in and take the feedback from the load instead, which is what anybody who cares about the load’s voltage would do. Then the resistor’s pole is inside the loop rather than outside it, and:

Every value is worse than none. The same component, in the same place, with one wire moved, has the opposite sign. That is not a damping effect; a damping effect would not care where the feedback comes from.

What it costs, and the cost is not bandwidth #

The usual account of the isolation resistor stops at “it costs a little bandwidth at the load”, which is true and is the smaller of two costs. The larger one is a direct-current error.

Feedback removes an amplifier’s output impedance by dividing it by the loop gain: fifty ohms with a hundred thousand of open-loop gain becomes milliohms at the amplifier’s output node. But the loop is now watching the amplifier’s node, not the load’s, and between them is $R_{\text{iso}}$ — which the loop cannot see and therefore does not correct.

So the impedance the load sees is $R_{\text{iso}}$ itself, at every frequency down to direct current. A load that draws current develops a drop across it that the feedback will never remove. Solved with a kilohm of resistive load in parallel with the capacitor:

$$\frac{R_{\text{iso}}}{R_{\text{iso}} + R_{\text{load}}} = \frac{10}{1010} = 0.990\%$$

which the figure measures on the closed circuit rather than from the divider expression, and which grows to 4.3% at the forty-five ohms that would restore seventy-five degrees.

That is a large error by the standards of anything a precision amplifier is used for, and it is entirely invisible in a stability measurement. The repair converts a stability problem into an accuracy problem, and the two are usually owned by different parts of a design.

The band, because both edges are hard #

Put the two costs together and the answer is a band rather than a direction, which is the shape this collection keeps finding.

Below about ten ohms the margin is not restored: at 4.7 Ω it is 37.6°, which rings visibly and leaves 3.85 dB of peaking at the load.

Above a few tens of ohms the direct-current accuracy at the load is gone, and with it most of the reason for using a feedback amplifier at all.

The width of that band is set by the ratio of the load resistance to the resistor the margin needs, and both ends move. A heavier load narrows it from above; a larger capacitance widens the resistor the margin needs and narrows it from below. With a kilohm of load and 2.2 nanofarads there is a comfortable region around ten to twenty ohms. With a hundred ohms of load there is none: forty-five degrees needs 9.9 Ω and 9.9 Ω into 100 Ω is 9% of error.

That is the condition under which the simple repair stops being available, and it is computable before anything is built.

The repair that costs nothing at direct current #

There is a second arrangement that gets both, and it is worth naming because it is what a design that cannot afford the error actually uses.

Take two feedback paths: a resistor from the load back to the summing junction, which closes the loop at direct current and low frequency and therefore regulates the load’s node, and a small capacitor from the amplifier’s own output to the summing junction, which takes over above a frequency and closes the loop at the amplifier’s node where the phase is good.

The result is a loop that watches the load where accuracy matters and watches the amplifier where stability matters, with a crossover between the two set by the capacitor. Its cost is a pole and a zero in the loop transmission — which is to say a doublet — and the settling behaviour that comes with one.

So the three arrangements form a ladder with no free rung. Feedback from the amplifier: stable, and inaccurate at the load. Feedback from the load: accurate, and unstable. Both: stable and accurate, and carrying a doublet whose settling tail is set by how well the two paths’ time constants match.

What the load sees, which is a third response #

There are three voltages in this circuit and the essay has been talking about two of them. The third is the one a reader most often wants and it behaves differently from both.

At the amplifier’s output, the response is the closed-loop one and it is well behaved: flat to the loop’s bandwidth, with peaking that falls as the margin is restored.

At the load, the same response with an extra pole at $1/2\pi R_{\text{iso}}C_L$. At the ten ohms that restores forty-five degrees that pole is at 7.3 megahertz — far above the loop’s own bandwidth, so it costs nothing. At the forty-five ohms that restores seventy-five degrees it is at 1.6 megahertz, which is beginning to matter. At two hundred and twenty ohms it is at 329 kilohertz and the load is being fed through a first-order filter.

And the loop’s own transmission, which is neither of those and is what the margin is a property of.

The pattern in that list is the useful part. As the resistor grows, the margin improves, the direct-current error grows linearly, and the load’s bandwidth falls as its reciprocal. Three quantities, one knob, and only one of them improving — which is what makes the interior of the band the answer rather than either end of it.

It also explains a measurement that surprises people. Probing the amplifier’s output shows a clean step; probing the load shows a slower, rounder one. Nothing is wrong: those are two different nodes with two different transfer functions, and the resistor is the difference between them. A designer looking for peaking should probe the load, and a designer measuring the loop should probe neither and cut the loop instead.

Where the pole actually goes #

It is worth being precise about the amplifier’s node, because the informal account — “the resistor isolates the capacitor” — suggests the capacitor stops mattering, and it does not.

Looking out of the amplifier’s output, the load is $R_{\text{iso}}$ in series with the capacitor’s impedance. At low frequency that is a large impedance and the amplifier sees nothing. At high frequency the capacitor is a short and the amplifier sees $R_{\text{iso}}$, a resistive load, which costs no phase at all. In between there is a pole and a zero: the pole at $1/2\pi(r_{\text{out}}+R_{\text{iso}})C_L$ and the zero at $1/2\pi R_{\text{iso}}C_L$.

The zero is what does the work. The pole is still there and still costs ninety degrees eventually, but the zero at a higher frequency gives them back, and the ratio between the two frequencies is $(r_{\text{out}}+R_{\text{iso}})/R_{\text{iso}}$. So the phase dip has a floor set by that ratio: with $R_{\text{iso}} = r_{\text{out}}$ the two are a factor of two apart and the worst phase lag is about twenty degrees; with $R_{\text{iso}} \gg r_{\text{out}}$ they nearly coincide and the lag vanishes.

That is why 45.35 Ω — very nearly the amplifier’s own fifty ohms — restores seventy-five degrees, and why nothing above a few times $r_{\text{out}}$ buys much more. The natural scale of the isolation resistor is the amplifier’s own output resistance, which is a number the data sheet does have, somewhere.

When the capacitance is not a capacitance #

Everything above assumes the load is a capacitor. Two common loads are not, and both change the answer in the direction of making the repair easier.

A length of cable is a transmission line, and below the frequency at which its length matters it looks like a capacitance — but it also has loss, and its far end has a termination. A metre of coaxial cable is about a hundred picofarads and behaves as one to a few tens of megahertz; ten metres is a nanofarad and does not, because by the frequency the loop cares about it is an electrical length rather than a lump. Which model applies is the lumped-element boundary Kirchhoff’s own frequency computes — the current law assumes the signal crosses the circuit in no time, it crosses at about two-thirds the speed of light, and the law therefore has a frequency of its own set by nothing but physical size. On the wrong side of it the load’s impedance stops falling and settles at the characteristic impedance, which is resistive and costs no phase.

What it costs instead is measured in the lines field, and it is not nothing. The staircase in time is what an amplifier driving that cable actually sees: a source driving a metre of cable does not know what is on the far end of it for 4.83 nanoseconds, what it drives into during that interval is the characteristic impedance and nothing else, and when the far end answers it does so as a staircase whose limit is the resistive divider the circuit was going to be all along. A loop closed around that node is being asked to regulate something whose value changes in steps at intervals of a round trip, which is a different problem from a pole and is not what a phase margin describes.

A capacitor with real equivalent series resistance brings its own isolation resistor. An electrolytic of a few hundred milliohms in series with its capacitance puts a zero in exactly the place this essay has been buying one, which is why a circuit that is unstable into a ceramic capacitor is often perfectly stable into an electrolytic of ten times the value. Two requirements pulling one capacitor states the same fact the other way round, as a window that resistance has to be inside — with the lower edge bisected at 939 milliohms for 45 degrees, and with the upper edge turning out not to be a stability edge at all but a transient one, so that the resistance giving the smallest droop sits ten per cent inside the region the loop must not be built in.

And the capacitance in such a part is not the printed one either. The capacitance that is not one number measures a class II ceramic at its rated voltage as 2.000 µF read as a slope where the part is printed as ten, which moves a load capacitance across most of this essay’s horizontal axis without anybody having changed the component. The load a designer specified and the load the loop sees are different numbers, and the difference is larger than the margin being bought.

Both are worth knowing before reaching for the resistor, because in both cases some of it is already fitted.

There is a third load that is not a capacitance and it is the one this repair cannot help at all. A load that draws its own current — a second circuit, a bias network, anything that is not purely reactive — meets the isolation resistor as a source impedance rather than as a pole, and what the load sees looking back measures what that is worth: ten ohms with no loop gain in it whatever, so a load step leaves an error that never goes away. The repair here is specific to a load that takes no current at direct current, and the specificity is easy to miss because a capacitor is exactly such a load.

Where this repair sits in the ladder #

The trade this essay measures is the first of four made about the same load, and it is worth naming what each of the others does with the cost identified here.

The load that gets inside the loop is the rung below, and it establishes the thing being repaired: hanging a capacitor on an amplifier’s output changes nothing a reader can find in any expression for its gain and takes the phase margin of a unity-gain inverter from ninety degrees to thirty, with the mechanism being fifty ohms of output resistance that no data sheet page puts next to the stability page, and forty-five degrees arriving at 905 picofarads — a metre of coaxial cable.

The path that buys the error back takes the cost measured here — a loop that has stopped regulating the load’s node — and buys it back with a second feedback path. What that charges is not an error and not a margin but a range: at twelve picofarads it settles to a hundredth of a per cent in 0.745 microseconds, faster than the circuit it repairs, and at a hundred it takes 9.18, with the phase margin there being better. So the repair for this essay’s cost has a cost of its own that no stability measurement reports.

What the load sees looking back then asks the question from the other end, which none of the three rungs below it does. A load that draws its own current sees an impedance looking back, and with the feedback taken from the amplifier that impedance is the isolation resistor with no loop gain in it whatever: ten ohms, and a load step leaves an error that never goes away. That is this essay’s direct-current cost restated as an impedance, and it is the form in which it is largest.

What is checked #

Three assertions, and the second is the one that tells the mechanism from the story.

That the resistor restoring forty-five degrees is found on the solved loop — bisected on the cut and injected loop transmission rather than computed from the pole it makes — and that the margin there really is forty-five to a tenth of a degree.

That taking the feedback from the load makes every value worse, asserted at two hundred and twenty ohms against no resistor at all. That is what separates “the resistor damps something” from “the resistor moves a pole out of the loop”, and only the second survives it.

And that the loop stops regulating the load’s node — the direct-current error into a kilohm with the resistor in, against the error with none, on the closed circuit. It is the cost, it is invisible to every stability measurement, and it is the reason this repair is a trade rather than a fix.