The resistor that moves the lag

The current that does not reach the input built a guard ring around an electrometer’s input, held it at the input’s own potential with a follower, and measured what it bought: a teraohm of board leakage multiplied into 10¹⁸ Ω at direct current, and a hundred picofarads of cable bootstrapped nearly out of the signal’s way. The sign of what the guard gives back then found the price nobody quotes. The follower’s copy of the input lags it, and a capacitance driven by a lagging copy of its own voltage takes current out of phase with that voltage. So the guarded input presents a negative conductance that rises as the square of frequency and flattens at $-\omega_t C$: −1.59 kΩ behind a one-megahertz follower and 100 pF of cable.

That essay ended on the fix everybody reaches for. A resistor between the follower’s output and the ring is standard practice, usually to keep a follower stable when it drives a long cable, and it puts a pole in the guard’s own path. The question left open was what that pole does to the negative conductance — its worst value against the resistor, and whether one value of resistor works across the range of cables a design might see. Both have answers, and they point in opposite directions.

The negative resistance, for reference #

The mechanism is worth restating in the form it will be needed in. The cable’s capacitance $C$ runs from the input to the ring. With the ring at a fraction $G(j\omega)$ of the input’s voltage, the current through the cable is $j\omega C(1 - G)$ times that voltage, and whatever part of $1 - G$ is imaginary becomes a real part of the admittance. A follower lags by an amount that grows with frequency, $1 - G$ acquires a positive imaginary part, and $j\omega C$ times a positive imaginary number is a negative real one: a conductance that delivers power instead of absorbing it.

The ring has a capacitance of its own #

A resistor in the drive does not act alone, and the first thing to put in the netlist is the capacitance it acts on. A guard ring, or the guard conductor of a triaxial cable, is a conductor with a large area, and it has capacitance to the grounded shield or chassis around it as well as to the input it surrounds. In a triaxial cable the two are of the same order; here the ring’s capacitance to ground, $C_s$, is taken equal to the cable’s $C$ from input to ring.

On a circuit board the same is true of a guard ring etched around an input pad. The ring is a track, usually wider than the pad it surrounds, and underneath it are the board’s ground planes, a few tenths of a millimetre away through the laminate. Its capacitance to ground is set by its area and that spacing, and it is rarely smaller than its capacitance to the pad inside it. Treating $C_s$ as equal to $C$ is therefore not a worst case chosen for effect; it is the ordinary proportion, and the bound below says what happens as the proportion moves.

Without a resistor, $C_s$ is simply a load on the follower and changes nothing at the input. With a resistor $R_g$ between the follower and the ring, the ring’s voltage is set by a divider: the follower through $R_g$ on one side, the input through $C$ and ground through $C_s$ on the other. The ring then follows

$$G = \frac{A + j\omega R_g C}{1 + j\omega R_g (C + C_s)}$$

of the input, where $A$ is the follower’s own gain. Even with a perfect follower, $A = 1$, the ring lags the input once $\omega$ passes $1/(R_g(C + C_s))$ — which means the resistor has not removed the lag. It has supplied one of its own.

Weaker, and set by the resistor #

Every resistor makes the worst negative resistance weaker. The bare guard’s −1.59 kΩ becomes −6.5 kΩ through 1 kΩ, −42.6 kΩ through 10 kΩ and −403 kΩ through 100 kΩ. The curves also change shape: without a resistor the negative resistance falls as the square of frequency and then flattens, and with one it falls to a minimum and then rises again, so the worst value sits at a frequency the resistor chooses — 681 kHz for 1 kΩ, falling as the resistor grows.

The largest resistor’s worst value is the one to look at. With a perfect follower the cable takes $j\omega C \cdot j\omega R_g C_s/(1 + j\omega R_g(C + C_s))$, whose real part settles at $-C C_s/(R_g(C + C_s)^2)$. The negative resistance that corresponds to is

$$R_{\text{worst}} \to -\frac{R_g (C + C_s)^2}{C\,C_s},$$

which for a ring as capacitive to ground as to the input is $-4R_g$: −400 kΩ at 100 kΩ, which the solved circuit reaches to within a per cent. The amplifier’s gain-bandwidth product is not in it. The resistor has taken over the job of making the lag, and the negative resistance is now the resistor’s.

The high-frequency end of the negative band has an explanation too, and it is where the resistor stops mattering. Above the follower’s own gain-bandwidth product the follower’s gain falls away, and so does the part of the ring’s voltage it supplies. The ring is then held only by the capacitive divider, at $C/(C + C_s)$ of the input, which is a real fraction with no lag in it. With no lag the cable’s current is in quadrature with the voltage again and the conductance returns to positive. So every curve in the figure ends near a megahertz whatever the resistor: the resistor decides how negative the band is and where its worst point sits, and the follower decides where the band stops.

Two lags, and which one wins #

Between the bare guard and the large resistor there are two lags in series: the follower’s own, with a time constant of $1/\omega_t$, and the resistor’s, with a time constant of $R_g(C + C_s)$. Whichever is longer sets the negative resistance, and the two are equal at a resistor of $1/(\omega_t(C + C_s))$ — about 800 Ω for this cable and follower. Well below it the follower’s lag dominates and the input behaves as the bare guard did, at $-1/(\omega_t C)$. Well above it the resistor’s lag dominates and the input approaches $-4R_g$. The figure’s three resistors sit either side of that boundary: 1 kΩ is on it, and its −6.5 kΩ is neither the bare guard’s −1.59 kΩ nor $4R_g$’s −4 kΩ but a mixture; 10 kΩ is a decade past it, and its −42.6 kΩ is within seven per cent of $4R_g$; 100 kΩ is two decades past, within one.

The boundary moves with the cable. A longer cable has a larger $C + C_s$ and hands control to the resistor at a smaller value, which is why the three lengths in the next figure converge on the $4R_g$ line at different places — the nanofarad cable by a hundred ohms, the ten-picofarad one only past ten kilohms.

The follower’s speed is the other half of that boundary, and it cuts the opposite way from what a faster amplifier is usually bought for.

Behind a follower ten times faster, the bare guard gets ten times worse: −159 Ω against −1.59 kΩ, since the negative resistance of the bare guard is $-1/(\omega_t C)$ and a faster amplifier has a larger $\omega_t$. The negative band also stretches ten times higher, to 9.77 MHz. Through 100 kΩ the worst value is −400 kΩ behind either follower, because once the resistor’s lag dominates the amplifier’s bandwidth has left the expression. A faster follower makes the resistor more necessary, not less, and moves the crossover between the two lags down to about 80 Ω. The resistor is the part of the design that makes the guard indifferent to which amplifier is fitted.

One value for every length #

This is the answer to whether one value works across cables, and for the negative resistance it is yes. Without a resistor, the worst value runs from −16.7 kΩ on 10 pF of cable to −254 Ω on a nanofarad: a factor of sixty-five, because $-1/(\omega_t C)$ has the cable’s capacitance in it, and a longer cable is a stronger negative resistance. With a large resistor all three lengths converge on the same line, $-4R_g$, and at 1 MΩ they agree to a per cent.

The reason is in the bound’s form. $-R_g(C + C_s)^2/(C C_s)$ depends on the two capacitances only through their ratio, and a longer piece of the same cable scales both. So one resistor gives every length of one cable the same worst negative resistance. A designer who knows the cable type but not its length can choose the resistor from the bound alone.

Where the length went instead #

The same divider that supplies the lag also ends the bootstrap. Well below $1/(2\pi R_g(C + C_s))$ the ring follows the follower and the cable carries almost no signal current; the source sees a small fraction of a picofarad. Well above it, the resistor is effectively open and the ring is left to float on the capacitive divider between the input and ground, following $C/(C + C_s)$ of the input. The source then sees the cable in series with the ring’s capacitance to ground — 50 pF here, half the cable — which is exactly what an unguarded input with the same conductors would present.

For 10 kΩ and 100 kΩ the capacitance crosses a quarter of the cable at 73.6 kHz and 7.94 kHz, against 79.6 and 7.96 from the expression. For 1 kΩ it crosses at 341 kHz rather than 796, because the follower’s own bandwidth arrives first and pulls the corner down: the resistor is not yet the thing setting the lag.

And this frequency has the length in it. A cable twice as long has twice $C$ and twice $C_s$, and gives up its bootstrap at half the frequency through the same resistor. So the resistor that gives every length the same worst negative resistance gives every length a different bandwidth, and the longest cable the least.

The trade, and what it costs a gigohm source #

Put the two results side by side and the resistor’s value is a single number trading two things. The worst negative resistance is about $-4R_g$; the bootstrap ends near $1/(2\pi R_g \cdot 2C)$. Raising $R_g$ tenfold weakens the negative resistance tenfold and brings the end of the bootstrap down tenfold. There is no value that improves one without costing the other, and the cable’s length enters only the second.

Whether the negative resistance matters depends on what else is at the node. A source with a resistance $R_s$ in parallel adds a positive conductance $1/R_s$, and the node’s net conductance is negative wherever $1/R_s$ is smaller than the guard’s negative conductance. For the net to stay positive at every frequency the worst negative resistance must exceed the source’s, $4R_g > R_s$. For a gigohm source that asks for a resistor of 250 MΩ in the drive — and with 200 pF of cable and ring behind it, the bootstrap would end at about three hertz.

So for the sources a guard exists for, keeping the node’s net conductance positive everywhere and keeping the bootstrap at all are incompatible. That does not mean the guarded input is unusable. A negative net conductance at a capacitive node is not an oscillation by itself; the node needs something to resonate with, and the sign of what the guard gives back noted that the failure it causes needs a source with an inductance. What it does mean is that the resistor in the drive is not a cure for the negative conductance. It is a way of choosing where it lives and how large it is.

Choosing the resistor from the bound #

The two expressions are enough to size the resistor for a given source, and the arithmetic is worth doing once because it shows where the approach works. For the node’s net conductance to stay positive at every frequency, the source’s resistance has to be smaller than the magnitude of the worst negative resistance, $R_s < R_g(C + C_s)^2/(C C_s)$, which for $C_s = C$ is $R_s < 4R_g$. The bootstrap then ends near $1/(2\pi R_g(C + C_s))$.

For a source of 100 kΩ, the resistor must exceed 25 kΩ, and a metre of the cable here — 100 pF of it and 100 pF of ring — keeps its bootstrap up to about 32 kHz. That is a usable guard: the input sees a small fraction of a picofarad through the audio band and a positive net conductance everywhere. For a 10 MΩ source the resistor must exceed 2.5 MΩ and the bootstrap ends near 320 Hz. For a gigohm source, as above, a few hertz.

So the resistor makes a guard safe for moderate sources at the price of its bandwidth, and the price rises in proportion to the source’s resistance. The sources for which a guard is indispensable — gigohms and teraohms, where the leakage it removes is the whole error — are exactly the ones for which this remedy leaves nothing of the bootstrap. For them the choice is between a guard with a negative conductance and a source that must not resonate with it, and a resistor so large that the guard only guards against leakage and not against the cable.

Why the standard fix is still standard #

The resistor is usually installed for a different reason, and on that reason it does exactly what it promises. A follower driving a large capacitance directly has a second pole in its loop — its own output resistance against the cable — and can ring or oscillate at its own crossover. The load that gets inside the loop measures what a capacitance at a follower’s output does to its margin, and the resistor that buys the margin back measures the isolation resistor’s effect on the amplifier; what the cure at the base costs measures the same kind of resistor in front of an emitter follower. The resistor in a guard drive is the same resistor for the same purpose: it keeps the follower stable with any length of cable.

What this page adds is that it also changes the input’s admittance, and in the direction the follower’s stability does not care about. The follower sees a resistor and is happy. The source sees a negative resistance that the resistor has made weaker and has made its own, and a bootstrap that ends sooner the longer the cable. A designer choosing the resistor for the follower’s sake has also chosen the input’s negative resistance and its bandwidth, whether that was the intent or not.

The probe is part of the circuit makes the same point about instruments generally: whatever is attached to measure a node becomes part of the node, and a guard is an instrument whose whole job is to be attached. Here the attachment has three parts — the follower, the resistor, the ring’s own capacitance — and the input’s behaviour is set by all three together.

How the numbers were obtained #

The input is a netlist: a follower modelled as a transconductance into a compensation capacitance followed by a buffer, with a one-megahertz gain-bandwidth product and a direct-current gain of a million; a cable capacitance from the input to the ring; the ring driven from the follower’s output through $R_g$ and loaded by $C_s$ to ground; the board leakage split between the rail and the ring. The admittance is found by removing the source, driving the input with a one-volt test source at each frequency and reading the current it delivers, over three hundred frequencies from 1 Hz to 10 GHz. Its real part gives the parallel resistance and its imaginary part the parallel capacitance. The worst value is the most negative conductance on that grid, and the closed forms are compared with it.

What it leaves out #

The source. Every figure here is the input’s own admittance, with the source removed; whether the node actually misbehaves depends on what the source is, which is the next question and a different calculation.

The follower’s output resistance and its own response to a capacitive load. The follower here is ideal at its output, so the resistor’s effect on the follower’s stability, which is why it is usually installed, is not in this netlist.

And a ring whose capacitance to ground differs from its capacitance to the input. The bound $-R_g(C + C_s)^2/(C C_s)$ is smallest when the two are equal, at $-4R_g$, and grows when either dominates: a ring with little capacitance to ground gives a weaker negative resistance and a bootstrap that ends later, which is an argument for guard geometry rather than for any value of resistor.

Still open: a source that rings, the geometry of the ring, and the probe’s own bootstrap #

A source with an inductance. A negative conductance needs a resonator to become an oscillation, and a source with inductance supplies one. Solving the input’s poles with a series inductance in the source, against the inductance, would give the value at which a given cable and follower oscillate — and then what the resistor in the drive does to that value.

The ring’s capacitance to ground as a design variable. The bound is weakest when the ring is much less capacitive to ground than to the input. A guard geometry that achieves that — a ring close to the input and far from the shield — would weaken the negative resistance with no resistor at all, and the ratio $C_s/C$ it needs could be measured against the bound directly.

The oscilloscope’s bootstrap. A probe’s front end bootstraps its own input capacitance with a much faster amplifier and a much smaller capacitance. The product $\omega_t C$ sets the bare negative conductance, and whether that product lands anywhere near the resistances an oscilloscope’s inputs see is the same calculation in a different place.