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

Closed-loop response — where it appears

How an amplifier with its feedback connected behaves against frequency, following the feedback network until the loop gain runs out. Its peaking near crossover is the same quantity as the step response's overshoot, measured in the frequency domain instead of in time.

Named by 7 essays across one field — each of them below, with the objects they name alongside it.

One loop, two measurements of the same margin. The loop gain crosses unity at 5.73 kHz with 34.9° of phase left. The closed-loop step overshoots by 35.2%, which the second-order relation says corresponds to 34.8°. They differ by 0.1°, and the difference is the third pole.

Two measurements of one margin

A phase margin is computed from the loop gain in the frequency domain, without ever looking at a step. An overshoot is measured from the closed-loop step response in the time domain, without ever looking at a Bode plot. Inverting the standard relation on the second returns 34.9° against the first's 34.9°, and the residue is the third pole.

feedback · Margin overshoot What the summing junction of an inverting amplifier actually is, at 1.00 MHz of gain–bandwidth. computed by solving, not by drawing by driving a current into the node and reading the voltage. It is 100 mΩ at direct current, rises 1.000 decades per decade of frequency, and settles at 909.5 Ω — which is the 1 kΩ and 10 kΩ in parallel, with the amplifier contributing nothing. It passes one per cent of the input resistor at 995 Hz, a factor of 1,005 below the gain–bandwidth. The second route — the open-loop impedance over one plus the return ratio from the cut loop — agrees to 0.045%.

The node that is at ground for a while

An inverting amplifier's summing junction is held at ground by the loop, so it is at ground exactly as well as the loop is strong. Driven with a current and measured, it is a tenth of an ohm at direct current, ten ohms at a kilohertz, and 909 ohms above a megahertz — which is the two feedback resistors in parallel, with the amplifier contributing nothing.

feedback · Virtual earth A summing junction with 8 inputs on it, and the ceiling that fell. computed by solving, not by drawing by driving a current into the node and reading the voltage, with the input source zeroed, at each of 6 channel counts. With one input the node's ceiling is 909.5 Ω — the input and feedback resistors in parallel — and with 32 it is 31.15 Ω, a factor of 29.19 stiffer for 32 times the inputs — less than the count, because the feedback resistor is in the parallel combination too. Only the ceiling moves: the direct-current value and the whole rise are the same to 5.4 per cent at every count, because the open-loop impedance and the loop gain both scale as 1/N and cancel. So what the channel count does is bring the ceiling down to meet the rise earlier, and the per-pair leakage saturates rather than reaching unity — -0.8 dB at one input against -30.1 dB at 32. The total into every other channel together does not saturate: it is (N − 1)·Z/Rin, and with Z ≈ Rin/N that tends to unity — 0.966 at 32 inputs, 0.476 at two. A summing junction above its loop hands over essentially all of one input's signal to the others collectively, and the count decides only how it is divided.

The node that does not care how many

Put thirty-two inputs on a summing junction instead of one and its ceiling falls from 909 ohms to 31.2 — the input resistors in parallel with the feedback resistor, twenty-nine times stiffer. Every other part of the curve is unchanged to five per cent, because the open-loop impedance and the loop gain both scale as one over the count and cancel. So the stiffer node buys nothing: the per-channel leakage saturates instead of reaching unity, and the leakage into every other channel together tends to exactly one, whatever the count.

feedback · Virtual earth An integrator's summing junction is flat at 15.7 Ω, and worst at direct current. computed by solving, not by drawing. A current is driven into the node and the voltage read. The dashed curve is the same amplifier with a 10 kΩ resistor as its feedback element: a tenth of an ohm at direct current, rising a decade per decade, 909.5 Ω at the top. With a 10 nF capacitor instead, the loop gain has no frequency in it — the amplifier's gain falls as 1/f while the feedback factor rises as f — so it is flat at 62.83 and the node is flat with it, at Rin/(1 + 2π·GBW·C·Rin) = 15.666 Ω, measured 15.668 Ω and holding to 0.50 per cent from 100 Hz to 5.31 kHz — between the amplifier's own pole and the capacitor's corner, which is where a loop gain with no frequency in it lives. Both ends are the other way round from the resistive case: at direct current the capacitor is an open circuit, there is no loop at all, and the node is 843.0 Ω — a factor of 8388 worse than the resistor's 100 mΩ. Above the amplifier's crossover the capacitor is a short and the node is 47.6 Ω against 909.5. The two cross at 1.56 kHz, which is the frequency above which an integrator is the better virtual earth of the two.

The shelf a capacitor makes

An inverting amplifier's summing node rises a decade per decade to the two resistors in parallel. Replace the feedback resistor with a capacitor and it does neither: an integrator's loop gain has no frequency in it, so the node is flat at Rin/(1 + 2π·GBW·C·Rin) — 15.67 ohms over four decades, to half a per cent. Both ends invert. At direct current the capacitor is an open circuit, there is no loop at all, and the node is 843 ohms against the resistor's tenth of one; above the amplifier's crossover the capacitor is a short and the node is nineteen times better.

feedback · Virtual earth A 100 kΩ resistor across the capacitor: the node rises a decade per decade and stops at the shelf. computed by solving, not by drawing. The summing-junction impedance of the integrator (1 kΩ in, 10 nF feedback, 1 MHz amplifier) with 100 kΩ across its capacitor (solid), beside the ideal integrator (dashed) and the 10 kΩ inverting stage (dotted). At the lowest frequencies the resistor closes a direct-current loop and the node is 0.999 Ω; it rises a decade per decade, as a resistive stage's does, and stops on the integrator's shelf of 15.9 Ω at the feedback's own corner, 1/(2π·Rp·Cf) = 159 Hz. Its highest point below a tenth of the amplifier's bandwidth is 16.3 Ω.

The resistor that holds the bottom of the shelf

An integrator's summing junction sits on a flat shelf, 1/(2π·GBW·Cf), over its working band, and at direct current it is nearly its whole input resistor, because a capacitor closes no loop there. A resistor across the capacitor closes one. Below the feedback's own corner the node is then an inverting stage's, rising a decade per decade from Rp/A₀, and the rise meets the shelf exactly at that corner, 1/(2π·Rp·Cf), for any resistor and any amplifier — so the node never rises above its shelf at all. That holds for every resistor below 1/(2π·f₁·Cf), the one that puts the feedback's corner on the amplifier's own pole: 1.59 MΩ here. Above it the bottom climbs past the shelf. The same 1.59 MΩ is where a millivolt of offset becomes 1.58 V at the output, and the price below it is a flat gain of Rp/Rin and arctan(1/(2π·f·Rp·Cf)) of phase.

feedback · Virtual earth Three loops with exactly 45° of phase margin, and their step responses. computed by solving, not by drawing; three lags spaced by a ratio r, the amplifier's gain bisected in each for a 45.000° margin, the step from the closed loop's residues. Lags a hundredfold apart need a loop gain of 141 and overshoot 23.6%. Spaced by three they need 4.7 and overshoot 34.5%; coincident, 2.83 and 41.0%. The same margin, and overshoots 1.73 times apart.

One margin, and the gain under it

A phase margin of 45° is quoted as an overshoot of 23 per cent, and on a loop with plenty of gain below its crossover it is: three lags a hundredfold apart, closed at 45°, overshoot 23.6%. Bring the same three lags together and the loop reaches 45° at a loop gain of 2.83 instead of 141, and overshoots 41.0% — the same margin, 1.73 times the ringing. The closed-loop peak follows the overshoot where the margin does not: predicted from the peak the worst error over 27 loops is 3.1 points, from the margin 22.2. The gain margin moves with it, from 37.2 dB to 9.0.

feedback · Margin overshoot Two loops with exactly 45° of margin and the same crossover: with 100% of the extra lag a delay, the step overshoots 28.7%, against 23.3% for two poles. computed by solving, not by drawing: the closed-loop step of a loop K·exp(−sD)/(s(1 + s/p)) with its crossover at one, its phase margin exactly 45°, and 100% of the 45° beyond the integrator supplied by the delay (solid), against the same margin from a lag alone (dashed). Delay 0.785, lag's pole absent. Overshoot 28.7% against 23.3%; settled within 2% at 7.03 against 7.06; gain margin 6.0 dB against no limit. Marched with a history buffer, and at δ = 1 checked against the exact series of the method of steps.

The margin a delay spends

A phase margin can be made of a lag or of a delay, and the number is the same either way. Forty-five degrees from a lag overshoots 23.3 per cent, and forty-five degrees from a delay overshoots 28.7; at sixty degrees the sign reverses, 8.77 against 5.57, and the two cross near 54 degrees. The overshoot is the smaller difference. A two-pole loop's gain margin has no limit at all, and five per cent of its extra lag as delay brings that to 25.5 dB; with all of it delay the gain margin is 90°/(90° − margin), which is 6.0 dB at 45°, and nothing else is in the expression. Across twenty-one such loops the margin alone misses the overshoot by 13.1 points. The closed-loop peak, which already reads the whole band, misses by 1.2.

feedback · Margin overshoot

Named alongside it

The objects these essays reach for when they reach for this one.

Loop gainModel rangeSumming junctionDamping ratioLoadingOvershootPhase marginSecond-order approximationGain marginIntegratorBisectionCrossover frequency

All concepts