The boundary that improves when the part gets worse

The inductor that is an amplifier built one henry out of four ten-kilohm resistors and ten nanofarads, measured it, and found three boundaries where a wound coil has one. It stops being an inductor at a frequency the amplifiers decide. One end of it is soldered to ground. And its series resistance — the quantity that is wire in a real coil and is therefore positive at every frequency there has ever been — changes sign, at 63.0 Hz, well inside the band over which the same component is still the inductance the five parts predict.

That third boundary was reported the way a data sheet reports a corner: one frequency, for one set of parts. It is not that kind of quantity. Sixty-three hertz is an arithmetic coincidence between two imperfections that have nothing in common, and the useful statement is what happens when the parts change — which turns out to include a direction nobody would guess.

What is being solved #

The whole of it is one complex number at each frequency. A current source of one ampere drives the input node, the netlist is solved, and the voltage that comes back is the impedance. Its imaginary part divided by $\omega$ is the inductance, and its real part is the series resistance.

The real part is the interesting half and it is a small residue of two large quantities. At a kilohertz the imaginary part is 6,297 Ω and the real part is −25.2, so the real part is a part in two hundred and fifty of the number it is subtracted out of. That is the arithmetic situation the digits the arithmetic did not have is about, except that here the cancellation is not a numerical accident but the design: the arrangement works by two amplifier loops cancelling each other’s contribution to the real part, and what is left over is whatever the loops fail to cancel.

With ideal amplifiers there is no residue at all, and that is worth stating as a measurement rather than as a definition. Replacing both with nullors in the same netlist returns an inductance of 1.000000000 H at ten kilohertz — five parts in $10^{16}$ — and a real part of $-5.6\times10^{-11}$ Ω, which is the solver’s own floor rather than a resistance. So everything below belongs to the parts, not to the arrangement, and the arrangement itself is exact.

Two separate failures leave something behind, and they have nothing to do with each other.

The amplifiers have finite gain. Each of the two holds a node at the input’s voltage, and holds it short by the reciprocal of its open-loop gain. What that leaves in the real part is $r/A_0$ — a resistance, positive, with no frequency anywhere in it.

The amplifiers have finite bandwidth. The same node is also late, by a phase that grows in proportion to frequency, and a phase error appearing in a large imaginary part shows up in the small real part with the sign of a source rather than of a load. What that leaves is $-4\omega^2 L/\omega_t$ — negative, and growing as the square of frequency.

The reproduction, before anything is claimed #

Separating a measurement into two asymptotes is the kind of move that works on any curve if the asymptotes are fitted, so neither of these is fitted. Both are written down from the parts and then checked against the solve.

$r/A_0$ is checked by changing $r$ over two decades with the inductance held at one henry, which means changing the capacitor to match. The residue at one hertz comes out 0.0105, 0.0305, 0.1005, 0.3005 and 1.0005 Ω for resistance scales of 1, 3, 10, 30 and 100 kilohms, against a predicted 0.01, 0.03, 0.1, 0.3 and 1. The largest departure is at the smallest scale and is 4.7 per cent, which is the second residue already showing at the bottom of the range rather than an error in the first.

$4\omega^2L/\omega_t$ is checked by its power. Over 300 Hz to 3 kHz the measured magnitude fits the 2.0212 power of frequency, and at a kilohertz it is 25.24 Ω against the expression’s 25.13 — a part in two hundred, on a number that is the leftover of a cancellation two hundred and fifty times its own size.

That is the calibration this whole essay rests on, and it is the same move the assumption that is a geometry makes with a full window: find the case where the simple statement must be exactly right, agree there, and only then quote a disagreement.

A quarter of a degree #

There is one more way to look at the second residue and it is the one that says why the sign is negative rather than merely small.

An ideal inductor’s impedance has a phase of exactly ninety degrees. At the crossing this one’s is $90.0000°$ — the imaginary part is 396.563 Ω and the real part is zero to four decimals, which is what a crossing is. At a kilohertz the imaginary part is 6297.4 Ω and the real part is −25.24, and the phase is $90.2296°$.

Twenty-three hundredths of a degree past ninety is the whole of it. A phase of a hair more than ninety is a component whose current lags its voltage by more than a quarter cycle, which over a full period means it returns more energy than it took, which is what a source does. The magnitude of the impedance has barely changed; only its argument has, by less than a quarter of a degree, and that is enough to move a resistance of tens of ohms from one side of zero to the other. The same sensitivity is why the inductance reads 0.23 per cent out at that frequency while the real part has gone from +0.1 Ω to −25: a small angular error is a small fractional change in the large component of the impedance and a complete reversal in the small one.

That asymmetry is the general form of the trap, and it is the same one the cancellation that leaves a tail measures in the time domain: a pole and a zero a per cent apart move the magnitude by 0.078 dB and the settling time by a factor of thirty-six. Here a quarter of a degree moves the magnitude not at all and the sign of the loss completely. No amount of extra precision helps in either case, because the small quantity is genuinely where it is measured to be.

The law #

Equating the two residues gives the crossing, and every quantity in the answer belongs either to the design or to the amplifier:

$$f_0 = \tfrac{1}{2}\sqrt{f_c\,f_p}$$

where $f_c = r/2\pi L$ is the arrangement’s own corner — the frequency at which the resistor chain would be comparable with the synthesised reactance — and $f_p = f_t/A_0$ is the amplifier’s open-loop pole, the frequency at which its own gain starts falling. The inductance appears in $f_c$ and nowhere else, and it appears there divided into a resistance.

The exponent is the part worth dwelling on. Half a power is what a geometric mean produces, and the same half appears in all four variables when each is swept alone: the crossing goes as $r^{0.5004}$ with the inductance held fixed, as $L^{-0.4973}$ with the resistance scale held fixed, as $f_t^{0.5126}$, and as $A_0^{-0.5011}$ — 200.7, 63.0, 19.9 and 6.30 hertz for open-loop gains of $10^4$ through $10^7$, which is a factor of $\sqrt{10}$ per decade to three figures each time. Four independent quantities, four fits, four halves; and the one gain that list leaves out is $10^3$, where the crossing is 685.6 Hz against the 631 half a geometric mean puts it at, because there the positive residue is ten ohms and is no longer the small quantity in anything.

The second reading from that figure is the one a designer needs. The crossing and the inductance edge move together. At every gain–bandwidth drawn the crossing sits between 1.61 and 2.30 per cent of the frequency at which the inductance leaves one per cent, so there is no amplifier that removes the active region from the useful band. A faster part moves the whole picture up and keeps its shape, which is the same conclusion every model has an edge reaches about edges in general: buying a better component moves a boundary and rarely deletes one.

Where the law stops #

The two-term picture is an asymptotic one and it fails in the ordinary way, from both ends.

At the top of the gain–bandwidth sweep — 100 MHz parts, an open-loop pole of a kilohertz against a corner of 1.59 — the prediction is 8.7 per cent low. Nothing has gone wrong with either residue; they have simply stopped being separable, because the frequency at which they cross is no longer far below the frequency at which the arrangement stops behaving as an inductance at all. Half a geometric mean of two frequencies is only meaningful while there are two frequencies.

At the bottom, below an open-loop gain of about $10^4$, the positive residue is large enough to still matter at a kilohertz: with a gain of a thousand the real part at a kilohertz is −11.38 Ω rather than the −25.24 the bandwidth term alone would give, because 10 Ω of the finite-gain residue is still sitting on top of it.

Both failures are boundaries of the decomposition rather than of the component, which is a distinction the edge that is a region collects examples of. The component behaves identically either way; what stops is the ability to say the answer in two terms.

The direction that is wrong #

Now the reading that decides what the crossing is worth. $A_0$ is in the denominator, so a better amplifier — one with more open-loop gain at the same gain–bandwidth, which is exactly what a precision part is — puts the sign change lower.

A boundary that improves when the part gets worse is not measuring anything anybody is buying. What it locates is the frequency at which a positive term that was already a tenth of an ohm finally becomes smaller than a negative one — and a better amplifier reaches that point sooner precisely because its positive term was smaller to begin with. The negative term is untouched: at a kilohertz the real part settles at −25.24 Ω and stays there for every open-loop gain above $10^5$.

So the component with the higher gain is worse by the crossing and identical by every quantity that decides a circuit. That is the failure mode a boundary is a model and a tolerance is about, arriving from an unusual side: the boundary is correctly computed and correctly reported, and it is about the wrong quantity.

What the crossing was standing in for #

The question the sign change is a proxy for is whether a resonator built round this component will start on its own noise, and that question has an answer with no proxy in it at all. Above the crossing the real part is $-4\omega^2L/\omega_t$ and the imaginary part is $\omega L$, so the ratio is

$$Q = -\frac{\omega_t}{4\omega} = -\frac{f_t}{4f}$$

and the inductance cancels, the resistance scale cancels, and the open-loop gain cancels. What is left is the amplifier and the frequency.

That is a design rule the sign change cannot state. A tank whose other losses give it a quality factor of $Q$ has net negative loss anywhere above $f_t/4Q$ — 2.50 kHz for a hundred with one-megahertz parts, and the same 2.50 kHz whatever inductance the arrangement was built to make. Resonance and its bandwidth measures what a quality factor is for; this is the frequency above which the component supplies one of the wrong sign, and it is set by the part number alone.

It also explains why the crossing looked like a boundary in the first place. Between the crossing and $f_t/4Q$ the arrangement is active and harmless — it is supplying gain, but less gain than the rest of the circuit is losing. A boundary marked where a quantity reaches zero, when what matters is where it reaches a stated size, is a boundary drawn at the wrong contour.

The arrangement that has none of this #

The obvious generalisation from all of the above is that an inductor made of amplifiers has a negative resistance because amplifiers are late. That generalisation is wrong, and the counter-example is the four-amplifier version this ladder already built: one inductor, and ten components needed a floating inductance for a low-pass ladder’s series arms and made one out of two halves back to back.

The reason is in the netlist rather than in the parts. Each half takes its reference from the other terminal instead of from ground, so its resistor chain stands directly between the two terminals: the floating inductor is an inductance in parallel with the resistance scale. A real ten-kilohm resistor across the terminals produces 39 Ω of positive series resistance at a hundred hertz and buries a residue of 0.153 Ω underneath it, which is why the sign never changes.

What that costs is the quality factor. The floating arrangement reads 15.96 at a hundred hertz — the $r/\omega L$ of a resistor in parallel with an inductance, falling to nothing at the corner of 1.59 kHz where the inductance itself has halved — against the grounded arrangement’s −4123. Passivity here is not a property the extra two amplifiers bought. It is a resistor, and the resistor is a factor of 258 in the magnitude of $Q$.

What a wound coil does instead #

The comparison worth ending on is the component all of this is a substitute for, which has its own edge and a completely different one.

A coil’s boundary is a resonance, it is set by the geometry of the winding, and past it the component becomes a different passive element — which is what the inductor that is a capacitor is about. It never supplies energy, at any frequency, because there is nowhere for the energy to come from. The synthetic version’s boundaries are all consequences of a supply rail, and one of them is a failure mode that has no analogue in the passive part at all: the same shape appears in the resistance that is below zero, where an emitter follower with wire between its source and its base has a negative real part over a band, and a load capacitance resonating inside that band oscillates.

What it does not say #

None of this says the arrangement should not be used. Below its crossing it is a passive inductor to within a residue of a tenth of an ohm, it scales in impedance without limit — which is what the same filter a thousand times larger spends its time on — and one henry of wire is several kilograms of copper.

It says the boundary in the data sheet is the wrong one to design against, and that the right one is cheaper to state. The negative quality factor is a two-parameter statement: the amplifier’s gain–bandwidth and the frequency. There is no need to know the resistance scale, the capacitance, the open-loop gain or even the inductance being synthesised, and the measurement above shows all four genuinely cancelling rather than merely being absent from the algebra.

What the crossing does answer #

The crossing is still worth computing, for a reason that has nothing to do with a circuit starting.

It is the frequency above which the component stops adding loss to whatever it is in and begins subtracting it. A passive ladder is designed with every element lossless and its damping supplied by the two terminations; a realisation whose inductors have a small positive resistance comes out slightly more damped than the design, which rounds the passband corners and is the ordinary direction for a realisation error to go. Above the crossing the sign of that correction reverses, and the realisation is less damped than the design — a passband that peaks where the design has no peak anywhere, which is a defect no component tolerance produces.

The size of that correction is a design number and it is entirely the amplifier’s. At a kilohertz one-megahertz parts contribute −25.24 Ω of series resistance to every synthesised inductor and thirty-megahertz parts contribute −0.70, a factor of 35.8 for a factor of thirty in gain–bandwidth. So the question “how much does the realisation move the passband” has the same answer as “how fast are the amplifiers”, and neither the resistance scale nor the inductance being synthesised enters it.

That is a question about a passband rather than about a resonator, and it is the one the crossing genuinely answers — the one it was doing all along while being quoted for the other.

The number worth carrying #

Half the geometric mean of the arrangement’s corner and the amplifier’s pole for where the sign changes; the gain–bandwidth over four times the frequency for how much gain is on offer past it; and the second of those is the one to design with, because the first improves when the amplifier gets worse.

The habit that goes with it is more general than this component. A boundary defined as the zero of a quantity is only as useful as the quantity’s size near it. Where a residue crosses zero is decided by whatever else happens to be the same size there, which may be an unrelated imperfection moving for unrelated reasons; where it reaches a stated fraction of the thing it is a residue of is decided by the mechanism. The first is easier to bisect and easier to print, and eight amplifiers, and what they add found the same asymmetry in a different quantity — the noise floor of an active ladder rises as the square root of the resistance scale while its ceiling falls as the scale, so the ratio moves as the three-halves power and neither number alone says so.