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

Desensitivity — the series

5 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. How much of the amplifier reaches the answer, at 35° of phase margin. computed by solving, not by drawing by multiplying the forward path's gain by 1 + 10⁻⁵ and solving again. At low frequency 0.0999% of a fractional change in the device reaches the closed-loop gain, which is one over the loop gain of 995.04. It changes sign at 258 Hz — above there a better amplifier gives a smaller gain — and reaches 1.61 at 8.08 kHz, which is worse than having no feedback, and settles at exactly 1 above crossover (5.73 kHz), where there is no loop gain left to spend. The second route, the real part of 1/(1 + T) from the cut loop, agrees to 9.6e-6.

    How much of the amplifier gets through

    The reason to build an amplifier from a bad amplifier and two resistors is that a change in the device barely reaches the answer — a tenth of a per cent of it at low frequency, measured by making the device better and solving again. Near crossover the same fraction rises above one, so feedback makes the gain more device-dependent than no feedback at all, and below fifty degrees of margin it changes sign on the way.

    part 1 · feedback
  2. The gain's sensitivity to the device and to the divider add to one at every frequency, and at crossover each is exactly one half (34.9° of margin). computed by solving, not by drawing, three times at each of 241 frequencies: nominal, with the amplifier's gain raised by one part in 10⁵, and with the feedback fraction raised by one part in 10⁵. The fraction of the device's change reaching the closed-loop magnitude is 9.97 × 10⁻⁴ at 10.0 Hz and 1.60 at its peak near 7.94 kHz; the fraction of the divider's change is 0.99899 at 10.0 Hz and falls to −0.60 at 7.94 kHz. They add to one within 1.16 × 10⁻⁵ everywhere, and at the loop's crossover, 5.73 kHz, they are 0.5000 and 0.5000: the real parts of 1/(1 + T) and T/(1 + T), which are one half at |T| = 1 for any phase. The cut loop, a second route, agrees to 9.55 × 10⁻⁶.

    Half to the amplifier, half to the resistors

    A feedback amplifier is supposed to hand its accuracy to the resistors, and at low frequency it does — 99.9 per cent of a change in the divider reaches the gain and 0.1 per cent of a change in the device. The two fractions are measured here by nudging each and solving again, and they add to one at every frequency, to five parts in a hundred thousand. At the loop's crossover each is exactly one half, whatever the margin: the gain's magnitude answers the amplifier and the resistors equally. As complex numbers both are 1/(2 sin(φ/2)) there, above one for any margin under sixty degrees, and around the crossover lies a band where the gain is more sensitive to both than it would be with no feedback at all.

    part 2 · feedback
  3. On a linear frequency axis the loop's desensitisation is an area, and the area it gives back above crossover is the same: 6.74 and 6.74 kilohertz-nepers at 34.9° of margin. computed by solving, not by drawing. The logarithm of the device sensitivity |1/(1 + T)|, from the cut loop at 3601 frequencies between 0.01 Hz and 10 GHz, drawn in decibels against linear frequency to 60 kHz. It is −60.0 dB at low frequency, crosses zero at 4.47 kHz (the crossover is 5.73 kHz), peaks at +5.70 dB at 6.87 kHz, and returns towards zero as the loop gain goes. The area below zero is 6.74 kHz·Np (58.6 dB·kHz); the area above, integrated to 10 GHz, is 6.74, of which 6.11 lies inside the drawing. The net is −1.48 × 10⁻⁸ Hz·Np. Bode's sensitivity integral says it must be zero for a loop that falls two poles or more and is stable when cut.

    The area that feedback only moves

    A feedback amplifier divides its device's variation by the loop gain, sixty decibels of it at low frequency, and pays for it with a sensitivity peak near crossover. The payment is exact. Integrated over linear frequency, the logarithm of the sensitivity has as much area above zero as below — 6.74 kilohertz-nepers each side at 34.9° of margin, net zero to eight parts in a billion — at every margin and every gain, because the loop falls by more than one pole. Margin decides only the shape of the repayment: a 14-decibel spike at 11.7°, a one-decibel hump spread past half a megahertz at 80.4°. Only a loop that falls as a single pole keeps any of it, (π/2) times its crossover, and no amplifier is one.

    part 3 · feedback
  4. A 10 µs delay in the loop: the peak rises to 13.53 dB and the sensitivity above crossover becomes a comb, and the two areas are still equal — 7.14 kHz·Np each side. computed by solving, not by drawing, from the loop's closed form, held against the netlist's cut loop to a part in a billion. The three-pole amplifier of the area essay (second pole 4 kHz) with a pure delay of 10 µs in its loop. Its margin falls from 34.9° to 14.3° at an unchanged crossover of 5.73 kHz, and the peak of |1/(1 + T)| rises from +5.70 to +13.53 dB. Above crossover the delay keeps turning the loop's phase, so the sensitivity rises and falls about zero with a period of 100 kHz. The area below zero is 7.144 kHz·Np, the area above 7.144, net 9.00 × 10⁻⁵ Hz·Np — integrated to 50 MHz on a grid that takes sixty steps a tooth.

    The phase a delay never stops adding

    Put a pure delay in a feedback loop and Bode's sensitivity integral does not notice: with 10 µs in the three-pole amplifier's loop the logarithm of the sensitivity still has equal areas above and below zero, 7.14 kHz·Np each side. What changes is where they are. The delay keeps turning the loop's phase above crossover, so the sensitivity becomes a comb with a tooth every 100 kHz; the desensitisation shrinks from 6.74 to 6.48 kHz·Np, the main peak rises from 5.70 dB to 13.53 and overpays it by 7%, and the comb hands the excess back below zero. The margin reads the peak from below by about 1.4 dB at every delay. Past 16.9 µs the loop is unstable and the integral stops balancing — by exactly the growth rate of the pole pair that has crossed.

    part 4 · feedback
  5. Sampled at 100 kHz, the loop repays its whole debt below 50.0 kHz: 6.60 kHz·Np each side, with a peak of 8.72 dB — the continuous loop had repaid 89% by then. computed by solving, not by drawing, from the loop's exact zero-order-hold equivalent. The three-pole amplifier of the area essay (second pole 4 kHz), closed through a sampler and hold at 100 kHz. The device sensitivity, in decibels, against linear frequency from zero to half the sample rate, where the discrete response ends. Margin 24.7° at a crossover of 5.71 kHz, against 34.9° for the continuous loop; peak +8.72 dB at 6.45 kHz. The area below zero is 6.598 kHz·Np and the area above, all of it inside the band, 6.598; net 1.84 × 10⁻² Hz·Np. The continuous loop, by 50.0 kHz, had repaid 88.6% of its own.

    The band a sampled loop repays in

    Close the three-pole amplifier's loop through a sampler and hold at 100 kHz and Bode's sensitivity integral still balances — but only over the band below 50 kHz, because that is all the band a sampled loop has. The whole debt, 6.60 kHz·Np, is repaid there; the continuous loop had repaid 88.6% of its own by the same frequency and spends decades above on the rest. Below a tenth of the sample rate the sampled loop is the continuous loop with half a sample of delay, 24.7° of margin against 24.6°. And the finite band turns the equality into a floor: hold the sensitivity 20 dB down over a third of the band and no loop can peak below 10 dB. This loop's floor is low and its margin sets its peak; the sample rate at which the peak is 3 dB above the continuous loop's is 18 times the crossover.

    part 5 · feedback

All series