The second pole the ringing guard sees

The source that rings against its guard found why a guarded input can oscillate when nothing else in the circuit would. The guard drives a cable’s screen from the input’s own follower, so the cable’s capacitance is bootstrapped: seen from the input it is a capacitance in series with a negative resistance, $-1/(\omega^2 C/\omega_t)$, falling as the square of frequency. A source with resistance R and inductance L across it forms a resonator whose resistances cancel where $R = L\omega_t$, so the input oscillates once the source’s own time constant L/R passes the follower’s $1/\omega_t$. For a 100 Ω source behind a 1 MHz follower that is 15.9 µH by the expression and 17.6 µH by the poles of the whole netlist.

That essay’s follower had one pole, and it said so at the end: a second pole adds phase near the follower’s crossover and makes $1 - A$ larger there, which should lower the critical inductance for sources whose resonance falls near that frequency. Solving it would say whether the one-pole condition is safe or optimistic for a real amplifier. The solve says it is both, depending on the source, and it says where the line between them falls.

The follower with a second pole #

The follower is the same macro-model: a transconductance charging a compensation capacitor, set so that its gain crosses unity at 1 MHz, followed by a unity buffer. The second pole is an RC after the buffer and inside the loop, at a stated frequency — 30 MHz, 3 MHz or 1.5 MHz — so the amplifier’s open-loop gain acquires a second corner above its crossover, as a real amplifier’s does. Everything else is the earlier essay’s: a 100 pF cable whose screen is driven from the follower’s output, 10 pF of the amplifier’s own input capacitance that the guard does not bootstrap, and a source of resistance R and inductance L.

For each source resistance the critical inductance is found by bisection on the sign of the real part of the rightmost natural frequency of the whole netlist. Nothing about the oscillation is assumed: the bisection asks only whether a pole has crossed into the right half-plane.

Safer on one side, worse on the other #

With the second pole at 3 MHz — three times the crossover, an ordinary figure for an amplifier with a good phase margin — the critical inductance of a 9.49 Ω source rises from 1.66 µH to 6.85: the input can now carry four times the inductance before it oscillates. A 949 Ω source goes the other way, from 177 µH to 141. At 1.5 MHz, half an octave above the crossover, the low source becomes 7.7 times safer and the high one 23 per cent worse; at 30 MHz, a very fast amplifier, the changes shrink to ten per cent safer and three per cent worse.

So the one-pole rule $L/R > 1/\omega_t$ is neither a safe bound nor an optimistic one. For low-resistance sources it is conservative, sometimes by a large factor; for high-resistance ones it is optimistic, by up to 44 per cent among the sources and poles drawn here. A designer who used it as a guarantee would be right for a thermocouple and wrong for a high-impedance sensor.

Where the line falls #

The reason for the two signs is the frequency the input rings at. A low-resistance source reaches its critical point at a small inductance, and a small inductance against 100 pF resonates high: 20.7 MHz for the 3 Ω source, far above the follower’s 1 MHz crossover. A high-resistance source needs a large inductance and rings low, 0.65 MHz for the 3 kΩ source, below the crossover.

A second pole does two things to the follower’s gain $A$. It reduces its magnitude, by $1/\sqrt{1 + (f/f_2)^2}$, and it adds phase lag, by $\arctan(f/f_2)$. The bootstrap’s negative resistance is made of $A$: take away its magnitude and the bootstrap weakens and the negative resistance with it, which helps; add lag and the bootstrapped impedance turns further towards a negative resistance, which hurts. Far above the second pole the magnitude loss is large and wins. Well below it the magnitude barely moves while the phase is already appreciable — the lag grows as $f/f_2$ while the magnitude falls only as $(f/f_2)^2$ — so the phase wins.

Plotted against the ring frequency divided by $\sqrt{f_tf_2}$, the three second poles change sign at 1.014, 1.024 and 1.024 — within two and a half per cent of one, for poles spread over more than a decade. The geometric mean of the follower’s crossover and its second pole is where the input’s ringing stops being helped by the pole and starts being hurt. That placement is a measurement on three poles rather than a derivation, and it is worth stating as such; the reason it should be the geometric mean is that the bootstrap’s own negative resistance falls as $f_t/f$ while the pole’s phase grows as $f/f_2$, and the two are equal in size at $\sqrt{f_tf_2}$.

Which sources ring where #

The ring frequency at the critical point is set by the source’s resistance and the cable, since the critical inductance is about $R/\omega_t$ and the resonance is $1/\sqrt{LC}$: it is $\sqrt{\omega_t/(RC)}/2\pi$, falling as the square root of the source resistance. So the dividing line in frequency is a dividing line in resistance. For a 100 pF cable, a 1 MHz follower and a second pole at 3 MHz, $\sqrt{f_tf_2}$ is 1.73 MHz, and the ring frequency passes it between 300 and 500 Ω: sources below that are safer than the one-pole rule says and sources above it are worse.

That split lands in an awkward place for guarding. Guards exist for high-impedance sources — the current that does not reach the input is the leakage a guard is built against, and it matters only when the source’s own resistance is large enough for picoamperes to be millivolts — and high-resistance sources ring low, below $\sqrt{f_tf_2}$, where every second pole makes the one-pole rule optimistic.

The saving grace is size. The critical inductance grows with the source resistance, so a megohm source behind a 1 MHz follower needs a sixth of a henry before it can ring at all, and no lead has that. The sources for which the result is a practical problem are the ones in the middle: hundreds of ohms to tens of kilohms, with inductance of tens of microhenries to millihenries — a coil, a transformer’s secondary, a long twisted lead into a moderate impedance. Those sit near and below the dividing line, their critical inductance is within reach of real parts, and there the one-pole rule overstates the margin by a fifth to nearly a half.

Why the follower’s poles are in the input at all #

It is worth being clear why an amplifier’s second pole belongs in a question about an input’s stability, because without the guard it would not. The same source into the same input without its guard is stable at every inductance: the cable is then a capacitance to ground, the source and cable form a passive resonator, and a passive network cannot oscillate.

The guard changes that by putting the follower’s transfer function into the input’s impedance. The cable’s capacitance now sits between the input and a node driven at $A$ times the input’s voltage, so the input sees $1 - A$ of it — and $1 - A$ is complex, with a real part whose sign is set by the follower’s phase. The sign of what the guard gives back is the measurement of that sign; every pole the follower has is a term in it.

The same mechanism turns up without a guard wherever a buffer drives a capacitance and its input is watched. The buffer that is not a buffer finds an emitter follower’s output inductive, and the resistance that is below zero finds its input resistance negative over a band when a capacitance loads it — the same $1 - A$ read from the other terminal. The emitter follower has a second pole of its own, from its base resistance and input capacitance, and the geometric-mean rule would apply there too: which side of $\sqrt{f_tf_2}$ the ringing falls on decides whether that second pole helps or hurts — which is a prediction this page makes and does not test.

The one-pole result, for comparison #

The comparison needs the one-pole behaviour in view, and two of the earlier essay’s figures give it.

Each source crosses into instability close to its own $R/\omega_t$, and the crossing is sharp: a pole moves from the left half-plane to the right over a small change in inductance. The second pole does not change that shape; it moves the crossing, and in which direction depends on where the crossing’s frequency sits.

The one-pole rule was already only good to a factor of one and a half, and for reasons the rule does not contain — the amplifier’s input capacitance, which the guard does not bootstrap. The second pole is a larger departure for low-resistance sources and a comparable one for high-resistance sources, and unlike the input capacitance it has a sign that changes.

Two loops, and which one is unstable #

The oscillation is easy to misattribute, and the second pole makes the misattribution tempting. A follower with a second pole at 1.5 MHz and a crossover at 1 MHz has lost thirty-four degrees of phase at its crossover to that pole, leaving a margin in the fifties — adequate, not generous, and what is left at crossover is the measure of exactly that. It is natural to suspect that the guarded input rings because the follower has lost margin.

It does not. The follower’s own loop — its output fed back to its inverting input — is stable at every second pole drawn, and the guarded input with no source inductance is stable too: its rightmost pole sits at −4.4×10⁶ per second with the pole at 1.5 MHz. The loop that goes unstable is a second one, through the source: the source’s inductance and resistance, the cable, and the bootstrapped screen driven by the follower. Its gain crosses one at the resonance of the source’s inductance with the cable, not at the follower’s crossover, and its margin is set by the source’s L/R against the follower’s time constant.

That is why the second pole can help. A follower with less margin in its own loop is a less effective bootstrap at high frequency, and at the frequencies a low-resistance source rings at, a less effective bootstrap is a smaller negative resistance. Stable, and unstable with less gain is the general version of the surprise: a loop’s stability belongs to the loop, and a property that worsens one loop’s margin can improve another’s.

Three amplifiers, side by side #

The three second poles bracket what real amplifiers offer. A pole at thirty times the crossover is a fast, well-compensated part, and it changes the critical inductance by ten per cent for the lowest source and three for the highest: the one-pole rule is nearly exact. A pole at three times the crossover is the common case of a part compensated for a comfortable margin, and it changes the lowest source’s limit fourfold and the highest’s by a fifth. A pole at one and a half times the crossover is a part run close to its edge, and it changes them sevenfold and by a quarter.

The asymmetry between the two directions is the useful part of that summary. The helpful change can be large, because far above the pole the magnitude loss grows without limit; the harmful change is bounded, because below the pole its phase can only grow to its value at the ring frequency, and the ring frequency cannot fall far below the crossover before the source’s resistance is enormous. So the one-pole rule is never badly optimistic — by less than half, on everything drawn — and can be very conservative.

What a designer should do with it #

Read the rule as an estimate of the ring frequency, then check the side. $L/R > 1/\omega_t$ gives the critical inductance to within the factors above, and with it the ring frequency. If that frequency is above $\sqrt{f_tf_2}$ the rule is conservative and the input has margin the rule does not credit; if it is below, the rule overstates the margin, by up to half on the amplifiers drawn here.

Add loss where the ringing happens, not where the amplifier is slow. The earlier essay’s remedy — a resistor in the guard drive, which the resistor that moves the lag measured in its other role — raises the critical inductance by giving up bootstrap at high frequency, and it helps most exactly where the second pole hurts: below the geometric mean, where the bootstrap is strong and the pole’s phase is the problem.

And do not choose a slower amplifier to fix it. A lower second pole makes low-resistance sources safer, which looks like a cure, and makes high-resistance sources worse, which is where guarding is used. The amplifier’s second pole is not a knob for this problem, because it acts on both sides of a line the source decides.

How the boundary was found #

Every critical inductance is a bisection on the sign of the real part of the rightmost pole of the whole netlist — source, cable, screen, input capacitance, the follower’s transconductance, compensation and second pole — over twelve decades of inductance to sixty halvings. The natural frequencies are the eigenvalues of the netlist’s own matrices; nothing about the oscillation’s frequency or form is assumed.

The claims are checked in the figures: with any second pole the lowest source resistance drawn must have a higher critical inductance than with one pole, and the highest a lower one; and each pole’s ratio must cross one within twelve per cent of $\sqrt{f_tf_2}$ in ring frequency. They cross within 2.4 per cent.

Poles above the second, slewing, and leads that are lines #

The second pole is a clean RC. A real amplifier’s second pole comes with further poles and zeros above it, and its phase near the crossover is the sum of all of them; the geometric-mean rule would then be read against an effective second pole that matches the phase at the ring frequency.

The follower does not slew or saturate. Every result here is about whether a small disturbance grows. How large the oscillation becomes, and whether it shows at a slow instrument’s output as noise or as an offset, needs a march with the amplifier’s limits in it.

And the source is a lumped R and L. A lead that is a transmission line has many resonances rather than one, and the guard’s negative resistance, which grows towards low frequency, reaches the lowest of them first.

Still open: the amplitude it settles to, a lead that is a line, and the phase at the ring #

The amplitude of the oscillation. Marching the guarded input with a follower that slews and saturates would say what amplitude the oscillation settles to and whether it rectifies into a direct-current error at the input — the question that decides whether a guard that rings is merely noisy or is actually wrong.

A lead that is a transmission line. Replacing the lumped inductance with a line of stated length would give the input a series of resonances, the lowest at a quarter wavelength. For a lead of a few metres that lies in the megahertz range, near $\sqrt{f_tf_2}$ for ordinary amplifiers, which is exactly where the sign of the second pole’s effect changes.

The phase at the ring frequency, measured directly. The geometric-mean rule is inferred from where the ratio crosses one; a direct statement would compare the amplifier’s phase and magnitude at the ring frequency with and without the pole and say which of the two the negative resistance follows. That would turn the rule from a measured placement into a derived one.