Circuit Response
every model has an edge, and the figure carries it
Series

Capacitive load — the series

10 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. An inverting unity gain driving 2.2 nF, and the pole that is inside the loop. computed by solving, not by drawing. Two ten-kilohm resistors around a 10 MHz amplifier make a gain of 1.00, and a loop that closes against 2.01 — one plus the ratio, not the ratio. Hanging 2.2 nF on the output leaves the closed-loop gain at a kilohertz unchanged — 0.99998002 against 0.99997988 — and takes the phase margin from 90.0° to 30.1°. The mechanism is at the other end of the amplifier from the summing-junction case and the arithmetic is the same: the load works against the amplifier's own fifty ohms of output resistance, which puts a second pole in the forward path — inside the loop, where the feedback has to live with it — while the gain the loop closes against does not move at all. Forty-five degrees is reached at 905 pF, bisected on the netlist. The capacitance is not part of the signal path and does not appear in any expression for the gain.

    The load that gets inside the loop

    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. The mechanism is fifty ohms of output resistance that no data sheet page puts next to the stability page: the load works against it, the pole that results is in the forward path, and forty-five degrees arrives at 905 picofarads — which is a metre of coaxial cable.

    part 1 · feedback
  2. 9.9 Ω restores 45°, 23 Ω restores 60°, and the load pays for it in ohms. computed by solving, not by drawing. Phase margin against the resistor placed between a unity-gain inverter's output and 2.2 nF of load capacitance, with the feedback taken from the amplifier's own side of it. With no resistor the margin is 30.11°; 9.90 Ω restores 45° and 23.20 Ω restores 60°. The lower curve is the same resistor with the feedback taken from the load instead, where it makes every value worse — the pole is then inside the loop rather than outside it, and at 220 Ω the margin is 13.4°. The rising curve is what it costs: the loop no longer regulates the load's node, so 1 kΩ of resistive load pulls the output down by 0.990% at 10 Ω, uncorrected, at direct current.

    The resistor that buys the margin back

    Two point two nanofarads takes a unity-gain inverter's phase margin from ninety degrees to thirty. Ten ohms between the amplifier and the load restores forty-five, twenty-three restores sixty, and it works for a reason that reads as a cheat: the feedback is taken from the wrong side of the resistor, so its pole is outside the loop. Take the feedback from the load instead — which is what anyone controlling the load would do — and the same resistor makes every value worse. What it costs is that the loop no longer regulates the load's node at all: ten ohms is one per cent of error into a kilohm, at direct current, uncorrected.

    part 2 · feedback
  3. The second path costs nothing at 12 pF and an order at 100 pF. computed by solving, not by drawing. Settling time to 0.01% of final value, marched on the closed loop, against the value of the second feedback path's capacitor — with the phase margin of the same circuit divided by ten drawn on the same axis so the two can be compared. The direct-current error the previous rung recorded as the isolation resistor's cost, 0.99% into 1 kΩ, falls to 1.20e-4% with the second path in. What the second path costs instead is a range: at 12 pF the circuit settles in 0.745 µs against the isolation resistor's 1.419 µs — faster than the thing it repairs — and at 100 pF it takes 9.18 µs, 12 times longer, at a phase margin of 47.9° that reports nothing whatever about it. What it is settling by there is one exponential of time constant 0.99 µs, which is the feedback network's own RC and contains no amplifier.

    The path that buys the error back

    An isolation resistor restores a capacitively loaded amplifier's phase margin and costs it the thing feedback was for: the loop stops regulating the node the load is on, and a kilohm of load pulls the output down by a per cent. The standard repair is a second feedback path, and its cost is not an error or a margin — it is a range. At twelve picofarads it settles to a hundredth of a per cent in 0.745 microseconds, faster than the circuit it repairs; at a hundred it takes 9.18, and the phase margin there is better.

    part 3 · feedback
  4. The quietest capacitor is 18× the fastest one, and the margin prefers neither. computed by solving, not by drawing. The total noise at the load of a capacitively loaded stage against its compensation capacitor, with the settling time on the same axis at ten microseconds to the microvolt. Three independent sources are put in the netlist and solved separately — the amplifier's own 4 nV/√Hz at its input, and √(4kTR) in series with each of the two feedback resistors — and added in power. The noise falls monotonically with the capacitor, from 50.7 µV at 1 pF to 12.1 µV at 220 pF. The peak in the noise gain falls with every larger capacitor and is gone entirely from 12 pF upward, where the uncompensated stage's peaks at 3.85 times its own low-frequency value. What the capacitor costs is settling: the fastest is 12 pF at 0.74 µs — the same capacitor that flattens the noise gain, because one handover decides both — and the quietest takes 20.3 µs, at a margin above 40° everywhere in that range.

    What the second path costs at the floor

    The arrangement that repaired a capacitively loaded amplifier was suspected of paying for itself in noise, because that is how compensations usually pay. It does not: it has no peak in its noise gain at all, and the total at the load falls from 54.9 microvolts to 28.9 as the capacitor is added. What it costs is settling, and the capacitor that is quietest is eighteen times the capacitor that settles fastest — a trade the phase margin says nothing about, because the margin is comfortable at both.

    part 4 · feedback
  5. The load sees 10.0 Ω, 1.2e-3 Ω or 1.2e-3 Ω at direct current, and the peak is lowest for the arrangement with both paths. computed by solving, not by drawing. The impedance at the load node of all three arrangements, measured by grounding the input and driving a unit current into the load. Feedback from the amplifier leaves the load looking at the isolation resistor — 10.0 Ω, with no loop gain in it at all. Feedback from the load gives 1.2e-3 Ω, and the two-path arrangement has the same, which is what its direct-current path is for. All three resonate with the load capacitance near 3.2 MHz, and the two-path arrangement's peak is the lowest — 27.6 Ω against 37.4 and 59.2. What it gives up is between: above the 159 kHz handover it has let go of the load node.

    What the load sees looking back

    Four rungs of this argument have measured what the amplifier does to the signal — the margin, the settling, the error, the noise. None has asked the question from the other end. 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. The two-path arrangement recovers to a thousandth of it and charges for that in a quantity none of the four rungs below measured.

    part 5 · feedback
  6. The step at which the output impedance stops being a number. computed by solving, not by drawing. The excursion divided by the step, against the step. The flat line is the linear model, and it is flat to 0.0 parts per million across four decades — which is what an impedance is. The rising curve is the same netlist with the differential pair's tanh in the transconductor, and it leaves at 10.6 mA: the input error there is 3.63 thermal voltages, so the boundary is an amplitude in the pair's own units rather than a current with the amplifier's name on it. At 300 mA the ratio is 54.5 Ω against the linear 23.6 — 131 per cent, and it is no longer a property of the circuit at all. The slew rate that decides it is 3.25 V/µs, which is twice the thermal voltage times the gain-bandwidth in radians, and contains no design choice.

    The step too large to have an impedance

    The rung below drove the load node with a current step and reported an impedance: a voltage divided by a current, which is a number only if the ratio does not depend on the current. Give the amplifier the differential pair's own tanh in place of a linear transconductor and it is a number up to 10.6 milliamps and not above — where the input error is 3.63 thermal voltages, and where the slew rate that decides it is twice the thermal voltage times the gain-bandwidth in radians, containing no design choice at all.

    part 6 · feedback
  7. Which limit binds is a property of the load, and they change places near 22 nF. computed by solving, not by drawing. Each limit measured on its own, as the departure of its march from the linear one, at a load step of half the output stage's rating. The input pair's departure falls with load capacitance — a bigger reservoir holds the node while the loop responds, which is the sixth rung's own result — and the output stage's does not fall nearly as fast, because what it has to supply is the charge the capacitor wants. Below about 22 nanofarads the thermal voltage decides the answer and above it the output stage does, and nothing about the amplifier changed.

    The current above which there is no impedance

    The sixth rung found the impedance leaving at 10.6 mA, where the input pair's own tanh takes over and the slew rate is twice the thermal voltage times the gain-bandwidth in radians, with no design choice in it. A real output stage has a second limit that is nothing but design choice, and the two do not bind at the same load: at 0.47 nF the input pair's departure is 19.4 per cent against the output stage's 4.2, at 22 nF it is 0.9 against 2.3, and above the output stage's rating the excursion does not come back at all — 2,254 Ω for a quantity that was 37.

    part 7 · feedback
  8. One channel's load step reaches another through the supply, and the compensation decides by 4497×. computed by solving, not by drawing. Two identical amplifiers on one rail — sharing no signal node — with an ampere of load step pulled from the first and the second's output read. With the wiring left out the coupling is exactly zero, which is what the seven rungs below this one computed. With 30 nanohenries and fifty milliohms of rail and 10 microfarads of decoupling it is not: 3.84 microvolts per ampere at 271 kHz if the compensation capacitor returns to ground, and 17.29 millivolts per ampere at 2.33 MHz if it returns to the rail. That is a factor of 4497 decided by a modelling choice, which is why both are drawn. The channel that caused the step is unaffected: its own loop corrects the disturbance along with everything else, and the crosstalk is entirely a problem for the channel that did not.

    The rail the load moves

    Seven rungs of this ladder end by saying the same thing: the supply is an ideal voltage source, so a load step is drawn from a node that cannot be disturbed. Giving the rail an impedance turns out to change the disturbing channel's own output impedance by three parts in ten million — its loop corrects the supply along with everything else — and to open a path to a second amplifier that shares nothing with it but a wire. How large that path is is a modelling choice: 3.84 microvolts per ampere with the compensation capacitor returned to ground, 17.3 millivolts with it returned to the rail, a factor of four and a half thousand.

    part 8 · feedback
  9. The resistors own the floor between 2.91k Ω and 85.9k Ω, and the part owns it outside. computed by solving, not by drawing. The noise at the load of the two-path compensation against the impedance of its own feedback network, with the resistors scaled together and the compensation capacitor taken down in proportion so that Rf·Cf — the handover between the two feedback paths — does not move. Four contributions are integrated over 10 Hz to 100 MHz: the amplifier's 4 nV/√Hz, fitted as the 0.005 power of the impedance and so flat; the two resistors' √(4kTR), the 0.501 power; and the amplifier's 0.60 pA/√Hz flowing in the feedback resistor, the 0.997 power. Two different powers of one quantity cross twice. The resistors carry more than half the power only between 2.91k Ω and 85.9k Ω; outside that window, in both directions, the part does. The part's share is least at 15.8k Ω, which is not eₙ/iₙ — it is that ratio multiplied by the noise gain of 2.000 and again by 1.187, the square root of the ratio of the bandwidths the two generators actually see; there it carries 26.26 per cent. The model stops where the amplifier's output current does: at 100 Ω the feedback resistor alone draws 10 mA a volt.

    The window the resistors own

    Eight essays have priced one compensated stage, and the fourth of them left two of the amplifier's own generators named and uncounted. With the current generator put in the netlist the floor at ten kilohms goes from 28.88 microvolts to 30.13, and the resistors carry more than half the noise power only between 2.91 kΩ and 85.9 kΩ — outside that window, in both directions, the part does. The flicker corner turns out to be worth 1.00009 in this stage's own band, and 3.474 one band away.

    part 9 · feedback
  10. Two large-signal limits, each alone and then both, at 20 mA and half of it. computed by solving, not by drawing. Four marches of one netlist at each load: neither limit, the input pair's tanh alone, the output stage's 20 mA alone, and both, driven by a 10 mA step. The three curves are each limit's departure from the linear march and the departure with both present; the faint line is the two singles added. Both lies on the sum and a little above it — 1.112 times it at 0.47 nF and 1.022 at 47 nF — so the limits are present together rather than taking turns. Where the two singles cross, near 10 nanofarads, the pair costs 1.88 times what the worse of them costs alone.

    The load that neither limit owns

    Nine rungs of this argument asked which of an amplifier's two large-signal limits binds, and drew the load capacitance where the answer changes hands. Both are present at every load: the excursion with both in the netlist is the two departures added and between 2 and 12 per cent more, never the larger of them. So the crossing is not a handover but a maximum — at 12 nanofarads the pair costs 2.084 times what the worse of them costs alone, against 1.35 at 2.2 nanofarads and 1.07 at 47 — and the same peak sits on the resistance axis at 20 ohms and the gain-bandwidth axis at 50 megahertz.

    part 10 · feedback

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