The resistor that pays the amplifiers back

The boundary that improves when the part gets worse took apart the most unsettling property of a synthetic inductor. Built from four 10 kΩ resistors, a 10 nF capacitor and two 1 MHz amplifiers, Antoniou’s arrangement is one henry to within a per cent over three and a half decades, and its series resistance — the quantity that is copper in a wound coil and therefore positive at every frequency — changes sign at 63.0 Hz. The essay found that frequency to be a coincidence between two unrelated imperfections: a small positive resistance from the amplifiers’ finite gain, and a negative one from their finite bandwidth that grows as the square of frequency. It moves when either does, and it moves down when the amplifiers get better at direct current.

That essay explained the boundary and stopped. A circuit designer wants it gone, and the obvious move — faster amplifiers — only scales the negative term down; it does not change its sign. The less obvious move is to look at what the negative term is. A real part that grows as $\omega^2$, in an impedance of the form $Z_1Z_3Z_5/(Z_2Z_4)$, is exactly what a resistor in series with the capacitor $Z_4$ produces. So the amplifiers’ contribution can be read as a negative resistor hidden in series with the gyrator’s capacitor — and it can be paid back with a positive one.

Why a resistor in that place has the right law #

The arrangement’s input impedance is $Z_1Z_3Z_5/(Z_2Z_4)$, and with $Z_4$ the capacitor and the rest resistors it is $sCR_1R_3R_5/R_2 = sL$. Put a resistor $r_4$ in series with the capacitor and $Z_4$ becomes $r_4 + 1/sC$, so

$$Z_{in} = \frac{sL}{1 + sCr_4} \approx sL - s^2LCr_4,$$

and at $s = j\omega$ the second term is $+\omega^2 LCr_4$: a positive series resistance that grows as the square of frequency, with the inductance untouched to first order. That is the frequency law of the term the amplifiers subtract. The first-order expansion is good while $\omega Cr_4 \ll 1$, which for 10 nF and 64 Ω is four parts in a thousand at 1 kHz and four in a hundred at 10 kHz.

It is worth being exact about what this does and does not claim. It does not claim that the amplifiers’ negative resistance is a resistor in series with the capacitor; it is a residue of two amplifier loops that fail to cancel, as the earlier essay found by taking the residue apart term by term. It claims that the residue has the same frequency dependence as that resistor over the band where the amplifiers are a single pole, and therefore that one element can match it everywhere in that band at once — not at one frequency, which any resistor anywhere could do.

The value of the hidden resistor #

If the amplifiers’ term is equivalent to a resistor, that resistor has a value, and the value can be measured without any algebra about the amplifiers. Solve the bare inductor’s series resistance at 10 Hz, where the $\omega^2$ term is negligible and only the small positive part is left, and at 100, 200 and 300 Hz; the difference at each, divided by $\omega^2LC$, is the resistance that would have to be added in series with the capacitor to make it up.

For the one-henry design the fitted value is 63.48 Ω. The number has a closed form, and it is a simple one: $4/(\omega_tC)$, with $\omega_t$ the amplifiers’ gain–bandwidth in radians a second. For 1 MHz and 10 nF that is 63.66 Ω. Across six designs — resistance levels of 3.3, 10 and 30 kΩ with the capacitor scaled to keep one henry, gain–bandwidths of 300 kHz, 1 MHz and 10 MHz, and open-loop gains of $10^5$ and $10^6$ — the fitted value lies on $4/(\omega_tC)$ within 1.8 per cent.

What the law leaves out is as informative as what it contains. The resistance level is not in it: a 3.3 kΩ design and a 30 kΩ design with the same capacitor-per-ohm need resistors that differ only through the capacitor. The open-loop gain is not in it: a gain of $10^6$ gives 63.56 Ω against $10^5$’s 63.48. The open-loop gain decides the small positive part, which is why it moved the sign change in the earlier essay, and it has nothing to do with the negative part. Only the amplifiers’ gain–bandwidth and the capacitor decide that.

One more thing is not in it, and it was the hypothesis this measurement started from. The two amplifiers in the arrangement play different roles — one holds the capacitor’s top, the other its bottom — and it is natural to expect the negative resistance to come from their disagreement, so that a matched pair would cancel it and a mismatched one expose it. Solved with the two amplifiers at 1.1 and 1 MHz, and then at 1 and 1.1 MHz, the resistance at 1 kHz is −24.08 Ω and −24.09 Ω: swapping them changes nothing, and both are what a matched pair at the harmonic mean of the two would give. The negative resistance is a sum over the two amplifiers, each contributing $2/(\omega_tC)$ of hidden resistor, not a difference between them. Matching the amplifiers buys nothing here — the same structure the mismatch that cancels itself found in a current mirror, whose two transistors’ current gains enter only as a sum of reciprocals and so cannot be traded against each other.

What the resistor moves #

Put 63.66 Ω in series with the capacitor and the hero figure’s black curve is the result. The small positive resistance at low frequency is untouched — 0.099 Ω at 100 Hz, because $r_4$’s contribution there is $\omega^2LCr_4 = 0.025$ Ω — and the $\omega^2$ term is gone: −0.075 Ω at 1 kHz against −25.2 bare. The sign change has moved from 63 Hz to 754 Hz, by a factor of twelve.

It has not gone, because the amplifiers are not exactly single poles cancelled exactly by a series resistor. What remains is a term of higher order in $\omega/\omega_t$, and above about a kilohertz it shows: with exactly the cancelling value the resistance is −1.2 Ω at 3 kHz and then swings positive, to +23 Ω at 10 kHz, where the inductor is close to its upper edge anyway. A tenth more resistance — 70.0 Ω — over-compensates the $\omega^2$ term enough to keep the resistance positive across the whole band below the amplifiers’ edge: 2.44 Ω at 1 kHz, a loss, but a loss of the right sign.

The quality factor makes the point more starkly. Bare, the inductor has a respectable positive Q at low frequency — 2,429 at 30 Hz — and a negative one from 63 Hz up, −250 at a kilohertz, which is a component that adds energy to any tank built around it. Exactly cancelled, the Q at 1 kHz is −84,197: the residual is under a tenth of an ohm, of a sign that depends on the higher-order terms, and a Q that large and that uncertain is not a design value. At 1.1 times the cancelling resistance it is +2,579, and at 1.3 times +843. A little over-compensation trades an enormous, unreliable Q for a large, positive, predictable one.

The Q the amplifier decides found the same thing in an active filter section: a finite gain–bandwidth raises a section’s Q above its design value, and the fix there was also a deliberate compensating error. Here the compensating error is one resistor, placed where its frequency law matches.

A residue of a residue #

There is a numerical question hiding in that −84,197, and it is worth answering before trusting any of the small numbers above. At a kilohertz the synthetic inductor’s impedance is 6,283 Ω of reactance and −25.2 Ω of resistance, so the real part is already a part in two hundred and fifty of the number it is read from — the situation the digits the arithmetic did not have studied, where a small answer is the difference of large ones. With the cancelling resistor in, the real part is −0.075 Ω, a part in eighty thousand. A solve that keeps sixteen significant figures still holds that to about eleven, so the residual and its sign are real properties of the netlist rather than of the arithmetic; and the fitted resistor values agree with the closed form to a per cent or two, which a numerical artefact of that size could not arrange.

What the precision cannot do is make the residual meaningful for a real circuit. A twentieth of an ohm is smaller than the difference between two 10 kΩ resistors of 0.1 per cent tolerance would make in the low-frequency positive part, and smaller than the resistance of the copper on any board. The quality factor of −84,197 is a correct statement about an idealised netlist and a useless one about a built inductor, which is the second reason the design figure below over-compensates rather than aiming for zero.

A tank built on the repaired inductor #

The reason for wanting a positive, predictable series resistance is that a synthetic inductor is built to resonate. The inductor that is an amplifier put a capacitor across the bare inductor above 63 Hz and found not a high-Q tank but an oscillator, starting on its own noise, because the negative resistance supplied the energy a tank is supposed to lose. With 70 Ω in series with the gyrator’s capacitor, the same tank at 1 kHz has an inductor Q of about 2,600, and the tank’s Q is then set by whichever of its two parts is worse — the rule the Q the components allow established for passive resonators, that the reciprocals of the parts’ quality factors add.

That makes the repaired synthetic inductor an unusually good component. A wound one-henry coil at a kilohertz has a Q of a few tens, limited by its copper and its core; where the Q comes from found an active section’s Q error set by the ratio of its amplifier’s gain–bandwidth to the section’s own frequency, an error that a designer can size but not remove. Here the Q of a few thousand comes from a deliberate, positive loss of a couple of ohms, and the instability is a known distance away — exactly the distance between the slowest amplifier the specification allows and the one the resistor was chosen for.

The catch: a cancellation is a tolerance #

A resistor of 63.66 Ω cancels the term of a 1 MHz amplifier. An amplifier’s gain–bandwidth is not a precision quantity: data sheets quote a typical value, and production spread and temperature move it by tens of per cent.

The law for what is left is the difference of two hidden resistors: the one the amplifiers have, $4/(\omega_tC)$, and the one the designer put in, $4/(\omega_{design}C)$. At frequency ω the residual series resistance is about

$$R_0 + 4\omega^2L\left(\frac{1}{\omega_{design}} - \frac{1}{\omega_t}\right),$$

and the solves agree with it to a quarter of an ohm. Faster amplifiers than designed for leave a positive residual, which is loss; slower ones leave a negative one, which is the original problem back. Ten per cent slow gives −2.90 Ω at 1 kHz, a Q of −2,170 — a tank built on it would oscillate. Thirty per cent slow gives −10.97 Ω.

So the cancelling resistor converts the inductor’s sign of loss from a property of the circuit into a property of a tolerance. Without it the sign was certain and wrong. With it the sign depends on which side of the design value the amplifiers fall — and the earlier essay found that the boundary improves when the amplifiers get worse at direct current, which is a reminder that the amplifier parameter on which a gyrator’s loss depends is rarely the one its data sheet guarantees.

Designing for the slowest amplifier #

The way out is the usual one for a tolerance: design for the worst case. Choose $r_4$ for the slowest amplifier the specification allows, with the tenth of margin the hero figure showed the residual needs, and every amplifier in the spread is then at least as fast as the design value, leaving a residual that is positive — loss — everywhere.

The price of a wider tolerance is paid in the Q at the fast end, where the designed-in resistor most exceeds the amplifiers’ own. At ±10 per cent the Q at 1 kHz runs from 805 to 2,338; at ±30 per cent from 312 to 1,847; at ±50 per cent from 163 to 1,344. Every one of those is positive and every one is predictable, against an uncompensated inductor whose Q is −250 at every amplifier in the range. A designer who needs a Q of at least 300 at a kilohertz from 1 MHz amplifiers can now read off the gain–bandwidth tolerance that allows it — about ±30 per cent — which is a specification on a part rather than a hope about one.

This is the general shape of every cancellation in circuit design. What the second path costs at the floor found a compensation for a capacitively loaded amplifier that was suspected of paying in noise and paid instead in settling; the payment here is in tolerance. A cancellation converts an error into a difference, and a difference is only as small as the two things subtracted are known.

What the resistor costs, and where else it could go #

The resistor itself costs almost nothing. It adds Johnson noise — 64 Ω in series with a capacitor whose voltage the amplifiers force, so its noise reaches the input through the same impedance conversion as the capacitor’s — and against the four 10 kΩ resistors whose noise eight amplifiers, and what they add measured, sixty-four ohms is negligible. It does not change the inductance to first order: the 1 per cent band, from 1.0 Hz to 3.65 kHz, is the same with and without it to the third figure. And it does not move the upper edge, which the amplifiers’ running out of gain decides.

The position is the one place where the resistor’s real part becomes $\omega^2$ in the input impedance. A resistor in series with any of the other four elements adds a real part with a different law — a constant, or one that falls with frequency — and could cancel the amplifiers’ term at one frequency only. That is the distinction between this repair and an ordinary trim: a trim is right at the frequency it was set at, and this is right at every frequency where the amplifiers are a single pole, because its law is the error’s law.

The numbers, collected #

A one-henry Antoniou inductor from 10 kΩ, 10 nF and 1 MHz amplifiers has a series resistance of −25.2 Ω at 1 kHz and changes sign at 63 Hz. The amplifiers’ negative term is equivalent to a resistor of $-4/(\omega_t C)$ in series with the capacitor, fitted at 63.48 Ω against the formula’s 63.66 and within 1.8 per cent across six designs; it contains neither the resistance level nor the open-loop gain, and it is a sum over the two amplifiers rather than a mismatch between them. A real 63.66 Ω there moves the sign change to 754 Hz; 70.0 Ω keeps the resistance positive below the edge and gives a Q of 2,579 at 1 kHz. Amplifiers 10 per cent slower than designed for leave −2.90 Ω; designed a tenth over the slowest amplifier a ±30 per cent spread allows, the Q at 1 kHz is 312 to 1,847 and never negative.

Still open: the higher-order residue, a floating inductor, and the resistor’s temperature #

The residue after cancellation. With the $\omega^2$ term removed, what is left is a term of higher order that changes sign near 3 kHz. Its law, and whether a second element — a small capacitor across one of the resistors, say — could cancel it as $r_4$ cancels the first, would say how far up the band the technique can be pushed before the amplifiers’ edge itself is the limit.

The floating inductor. One inductor, and ten components built a floating inductor from four amplifiers. Whether its negative resistance is likewise a hidden resistor in series with each of its capacitors, and whether one cancelling resistor per capacitor restores it, is the measurement that would carry this repair into the ladder filters the floating gyrator was built for.

A cancellation that tracks. A gain–bandwidth falls with temperature, so a fixed $r_4$ is a cancellation at one temperature. A resistor whose value tracks the amplifiers’ — or an amplifier whose bandwidth is set by an external capacitor of the same type as the gyrator’s — could make the two hidden resistors track each other, and the residual would then be set by matching rather than by tolerance.