The area that feedback only moves

How much of the amplifier gets through found that a feedback amplifier’s sensitivity to its own device is a thousandth at low frequency and rises above one near the crossover: 1.61 at 34.9° of margin, 3.23 at 11.7°, and still 1.10 at 80.4°. Its table made the peak look like a function of the margin, which it is, and like a defect that enough margin would remove, which it is not — there was no margin on the slider at which the peak reached one. Half to the amplifier, half to the resistors then showed that the device’s share and the resistors’ share add to one at every frequency, so neither can be driven down without the other rising; but a sum rule is silent about how large the excursions must be.

There is a stronger constraint, and it is about area. Draw the logarithm of the sensitivity against frequency on a linear axis and the region below zero — the desensitisation, the reason for the feedback — has exactly the same area as the region above it. Nothing in the loop’s design changes the total, which is zero. What the design changes is only where the positive area is put: a tall spike just past crossover, or a low hump stretched over a decade. The result is Bode’s, from 1945, and this essay measures it on the netlist rather than quoting it.

The debt, measured as an area #

The loop is the same three-pole amplifier: a gain of $10^5$, poles at 10 Hz, at the slider’s setting and at 2 MHz, closed for a gain of one hundred. The return ratio $T$ is measured by cutting the loop and injecting, at 3,601 frequencies from a hundredth of a hertz to ten gigahertz, and the quantity integrated is $\ln|1/(1+T)|$ — the logarithm of the device sensitivity’s magnitude, in nepers, so that a sensitivity of a thousandth is −6.9 and one of two is +0.69.

At the default setting the curve sits at −60.0 dB from direct current to about 10 Hz, rises at twenty decibels a decade as the loop gain falls, crosses zero at 4.47 kHz — a little below the crossover at 5.73 kHz, because $|1+T|$ falls to one while $|T|$ is still above it, the phase having already turned $T$ back towards −1 — peaks at +5.70 dB at 6.87 kHz, and then falls back towards zero as the loop gain goes. The area between the curve and zero, below zero, is 6.74 kHz·Np, which is 58.6 decibel-kilohertz. The area above zero, integrated to ten gigahertz, is 6.74 as well. Their sum is $-1.5\times10^{-8}$ hertz-nepers, which is the error of a trapezoid rule on a logarithmic grid and nothing else.

It is worth being clear about what that number means physically. The desensitisation is not free and was never claimed to be; the claim is that it has a price that can be stated exactly, and the price is paid in the same currency — decibels of sensitivity multiplied by hertz — at the other end of the band. At 60 kHz the drawing stops with 6.11 of the 6.74 repaid; the rest is spread thinly over the next decade and a half.

Why it is conserved #

The integral is a theorem about functions of a complex variable. The sensitivity $S(s) = 1/(1 + T(s))$ has no poles in the right half of the $s$-plane if the closed loop is stable, and no zeros there if the loop gain itself has no right-half-plane poles — which is true of every loop on this page, since its three poles are all at negative frequencies. So $\ln S(s)$ is analytic in the right half plane, and the integral of its real part along the imaginary axis is fixed by its behaviour at infinity. If $T(s)$ falls at least as $1/s^2$ there — two poles more than zeros — then $\ln S \approx -T$ falls as $1/s^2$ too, the contour at infinity contributes nothing, and

$$\int_0^\infty \ln\left|\frac{1}{1 + T(j2\pi f)}\right|\, df = 0 .$$

Nothing in the statement refers to the margin, the gain, the pole positions or the crossover. They decide the curve; the curve’s area is decided before any of them are chosen. The measurement is the check that the netlist is a function of the kind the theorem is about, and it agrees to eight parts in a billion of the area on either side.

The same argument, applied to the phase of a function rather than the logarithm of its magnitude, is what the phase the magnitude already knows drew for minimum-phase networks: a network with nothing in the right half plane cannot choose its phase independently of its magnitude. This is the feedback version of the same constraint. A loop with nothing in the right half plane cannot choose where its sensitivity is small without choosing, in the same act, how much of it is large.

The axis the conservation hides on #

Every sensitivity curve in these essays, and nearly every one anywhere, is drawn against a logarithmic frequency axis, and on that axis the balance is invisible.

Against log frequency the area below zero is 16.74 neper-decades and the area above it is 0.254: sixty-six to one. The hollow fills the left of the picture and the peak is a small bump past crossover, and a reader would conclude that feedback buys a great deal and costs a little. On a log axis, a stretch of frequency is drawn with a width proportional to its ratio rather than its extent, so the decade from 1 Hz to 10 Hz is as wide as the decade from 100 kHz to 1 MHz, which is a hundred thousand times longer. The desensitisation lives in the first kind of decade and the repayment in the second.

That is not an argument against logarithmic axes, which are the only way to see six decades of anything. It is a warning about what their areas mean. The conserved quantity is the area against linear frequency, and a sensitivity curve read by eye on the usual axis systematically understates what the loop has spent.

Where the debt is paid #

The running integral — the area from zero up to a frequency, divided by the whole area below zero — shows the repayment as it happens.

Each curve falls to −1 at the frequency where the sensitivity first reaches one and climbs back to zero from there. At 34.9° of margin, half of the debt has been repaid by 11.9 kHz, twice the crossover; ninety per cent by 57.1 kHz, ten times it; and all but a hundredth by 423 kHz. The repayment is front-loaded — the peak does the first half — and then long-tailed, because far above crossover $\ln|S| \approx -\mathrm{Re}\,T$, and a loop gain falling as the square of frequency leaves a debt whose remainder falls only as one over the frequency — each further factor of ten in what is owed costs most of another decade.

The margin changes how the repayment is shaped. At 18.1° half is repaid by 5.01 kHz and ninety per cent by 23.3 kHz — a fast, concentrated payment, the tall spike. At 65.3°, half by 37.4 kHz and ninety per cent by 179 kHz, four times the crossover and twenty times it: a low, broad payment that is still going on two decades above the band the loop was built for. Neither is cheaper. The shape of the repayment is the only thing a designer chooses.

A spike or a hump, and never nothing #

Across the seven settings of the second pole the two areas are equal every time, and they are not the same area from one setting to the next.

The areas run from 2.95 kHz·Np at 11.7° of margin to 11.60 at 80.4°. They grow with the margin here because moving the second pole up moves the crossover up — 1.98 kHz at the lowest setting, 9.87 kHz at the highest — and the area below zero is roughly the crossover frequency times a neper: $\int_0^{f_c}\ln(f/f_c)\,df = -f_c$ for a loop gain falling as one over frequency, which gives 5.73 against the measured 6.74 at the default, the difference being the stretch where $|1+T|$ departs from $|T|$ near the crossover. A wider band of desensitisation is a larger debt, and it has to be repaid at the same rate.

What the margin does is set the height at which the repayment is made. At 11.7° it is a 14.02 dB spike, the sensitivity reaching five times its no-feedback value in a band a few hundred hertz wide. At 80.4° it is a 0.95 dB hump, a sensitivity of 1.12, stretched so far that at 60 kHz — six times the crossover — only 3.93 of its 11.60 kHz·Np have been paid. The earlier essay’s observation that no margin makes the peak reach one is this, read the other way: a peak of exactly one would repay nothing, and there is always something to repay.

More gain, deeper and higher #

The same bargain applies to the thing a designer most often wants more of, the low-frequency loop gain.

Tripling the amplifier with the poles held deepens the hollow by 9.5 dB and moves the crossover from 5.73 to 10.6 kHz, which widens it too. The area below zero goes from 6.74 to 14.18 kHz·Np, and the area above has to follow — so the peak rises from +5.70 to +9.52 dB, and the margin, which is how the peak is usually read, falls from 34.9° to 20.4°. Cutting the amplifier to 30,000 does the reverse: 2.72 kHz·Np each side, a +2.76 dB peak, 57.8° of margin.

This is the reason a loop cannot be improved by turning up its gain alone, stated as an accounting identity rather than as a stability rule. The extra desensitisation is real, and it must come out somewhere above crossover, and the only way to make the somewhere a place where it does no harm is to spread it wider — which, with the poles fixed, the loop cannot do. Compensation is the design of where the repayment goes: a lead network or a dominant pole reshapes the curve so that the positive area lands thinly over a band where nothing needs the accuracy, and it never makes the area smaller.

The one loop that keeps its savings #

The theorem has a condition, and the condition is the interesting case. It needs the loop gain to fall at least as the square of frequency far above crossover.

A loop that falls as a single pole is the exception. Its sensitivity rises monotonically to one and never exceeds it — the curve in the figure approaches 0 dB from below and stays there — and its integral is not zero but $-(\pi/2)\,\kappa$, where $\kappa$ is the limit of $s\,T(s)$ at high frequency. In hertz that is $-(\pi/2)f_u$, with $f_u$ the unity-gain crossover: 10 kHz here, so −15.71 kHz·Np, and the netlist gives −15.708 against a closed form of −15.708. That loop really does buy its desensitisation for nothing, and the saving is proportional to its bandwidth.

Add a second pole, anywhere, and the saving vanishes. The two-pole loop repays exactly, with a peak of +5.66 dB; the third pole at 2 MHz adds only 0.04 dB to that, because it sits two decades above the crossover and changes the curve only where the sensitivity is already nearly one. What matters is not how many poles a loop has but whether its gain eventually falls faster than one pole — and every physical amplifier’s does, since every one has a second pole somewhere, even if it is a gigahertz up. The single-pole loop is the idealisation the textbook argument for feedback is made on, and it is the only loop for which that argument is complete.

The same area in every quantity a loop holds #

The integral is about $1/(1 + T)$, and $1/(1 + T)$ is not only a sensitivity. It is the factor by which a loop divides everything it holds against: an output impedance, a summing junction’s impedance, a supply’s rejection, a distortion product generated inside the loop. Each of those is its open-loop value multiplied by the sensitivity, so the logarithm of the ratio closed-to-open is the same curve drawn on this page, and it obeys the same accounting.

A source below a frequency measured a regulator whose output impedance is 0.43 milliohms at direct current and 1.95 ohms at ten kilohertz, a third higher there than the same circuit with its loop cut. The first number is the desensitisation and the second is the repayment, and they are not two properties of that regulator: the area of the logarithm of its closed-to-open impedance ratio, over linear frequency, is zero, and the peak exists because the hollow does. The node that is at ground for a while is the same statement about a summing junction: a tenth of an ohm at direct current, ten ohms at a kilohertz, and the passive network’s 909 ohms once the loop has gone — a hollow whose area is owed somewhere near the crossover.

So a designer who has bought sixty decibels of desensitisation has bought sixty decibels less output impedance, less junction impedance and less distortion in the same band, and has agreed to pay for all of them at once, in the same octave, at the same rate. The resistor that buys the margin back is a repayment moved rather than cancelled: ten ohms outside a loop driving a capacitive load restore forty-five degrees of margin, which lowers the peak and, by this integral, spreads the same repayment further up. And two measurements of one margin is the peak read in the time domain, as the overshoot that the margin predicts to a tenth of a degree.

What the conservation does not say #

It says nothing about how high the peak must be. A fixed area can be spread arbitrarily thin, in principle; the integral bounds the average height over a band, not the maximum. What prevents indefinite spreading is that the loop gain falls, so there is a limit to how far above crossover the repayment can be pushed before $T$ is too small to carry it. Bounds on the peak itself exist, and they need the bandwidth over which $T$ can be made to do anything as an input.

It needs the loop to be stable when cut. A loop gain with a pole in the right half plane — an amplifier that oscillates on its own before feedback is applied — makes the integral positive, by an amount set by how far those poles sit into the right half plane. The debt is then larger than the desensitisation. None of the loops here have one, and a loop that does is a different kind of design.

It is a statement about the magnitude. The phase of $S$ has its own relation, the Bode gain–phase one, and the rotation half to the amplifier, half to the resistors found at crossover is not constrained by this integral at all.

How the numbers were obtained #

The return ratio is measured on the netlist by cutting the loop at the amplifier’s input and injecting a unit source there; nothing in the integral uses an expression for $T$. The grid is 300 points a decade from 0.01 Hz to 10 GHz, 3,601 frequencies, and the integral is a trapezoid in linear frequency on it, with any interval that crosses zero split at the crossing so that each part is counted on its own side; the interval from zero to 0.01 Hz is added in closed form, since the sensitivity is flat there. The single-pole loop’s closed form, $-(\pi/2)\cdot A_0\beta f_1$, is computed from the element values and compared, not used. The running integrals and the peak are read from the same grid. The integral is truncated at 10 GHz, where the three-pole loop’s integrand is below $10^{-12}$ and falling as the cube of frequency.

Still open: the bound on the peak, the sampled loop, and the loop with a delay #

The peak’s lower bound. With the loop gain required to be below some value above a stated frequency — which is what a real amplifier’s finite bandwidth enforces — the area above zero has to be repaid in a finite band, and that turns the integral into a lower bound on the peak’s height. Computing it for the three-pole loop here, and comparing with the peaks this essay measured, would say how close an ordinary compensation comes to the best possible.

The same integral for a sampled loop. A loop closed through a sampler has a sensitivity defined only up to half the sample rate, and the corresponding integral is over that finite band. The debt then has nowhere far away to go, and the peak is correspondingly harder to flatten. Half a clock against a pole found the phase a sampled loop loses to its hold; the area it loses is the same fact in this currency.

A delay in the loop. A pure delay adds no poles and no right-half-plane zeros in the usual sense, and leaves the integral at zero, but it makes $T$ fall no faster at high frequency while adding phase without limit. How the repayment is shaped when the phase keeps turning — whether it breaks into a comb of small peaks — is a measurement nobody has made on this loop.