Measurement, which is a circuit on a circuit

The sign of what the guard gives back

A guard ring is sold on two numbers and they are both about magnitudes: a teraohm of leakage multiplied to 10¹⁸ ohms, and a hundred picofarads of cable bootstrapped out of the way. The guard is also driving that capacitance with a copy of the input that lags it, and a capacitance driven by a lagging copy of its own voltage takes current out of phase with the voltage across it. What the guarded input presents is a negative conductance rising as the square of frequency — −15.9 megohms at ten kilohertz, −3.18 kilohms at a megahertz, flattening at the gain-bandwidth product times the capacitance — and a faster amplifier makes it worse.

17 min read 6 figures Every model has an edge The parasitic is the component Measured, not tabulated

Assumes: The current that does not reach the input · The current the instrument draws

The current that does not reach the input measures what one wire buys. A ring of copper around a high-impedance input, held at the input’s own potential, turns fourteen millivolts of leakage error into a nanovolt, multiplies the input resistance by the amplifier’s loop gain to 10¹⁸ ohms at direct current, and drives the cable’s capacitance so that the source does not have to — which is worth 1,230 times the bandwidth.

Every one of those numbers is a magnitude, and every one of them survives this essay. What none of them says is what the guarded input looks like as an impedance: a complex number, with a sign on its real part.

The guard is driving a capacitance with a copy of the input’s own voltage, and the copy lags. A capacitance driven by a lagging copy of its own voltage takes current that is not in phase with the voltage across it, and the part that is in phase has the wrong sign. What the source sees is a negative conductance — small at low frequency, rising as the square of it, and reaching $\omega_t$ ·C.

Fig. 1 The magnitude of the conductance a source sees looking into the input, guarded and not, with 100 pF of cable behind a 1 MHz amplifier. Unguarded it is positive everywhere. Guarded it is negative — −15.9 MΩ at 10 kHz, −161 kΩ at 100 kHz, −3.18 kΩ at a megahertz — flattening at

\omega_t

·C = −1.59 kΩ. The slider is the amplifier’s bandwidth.

Where the sign comes from

The arithmetic is three lines and it is worth doing before anything is solved, because the measurement below is checked against it.

A capacitance C between the input and the guard carries a current jωC·( $v_\mathrm{in}$ − $v_\mathrm{guard}$ ). The guard is a copy of the input, so $v_\mathrm{guard}$ = A(jω)· $v_\mathrm{in}$ , and the admittance the source sees through that capacitance is jωC·(1 − A(jω)).

If A were exactly one the admittance would be zero, which is the whole point of a driven guard: the capacitance is still there and no current flows in it. A real follower’s gain is not one. For a unity-gain follower closed around an amplifier of gain-bandwidth $\omega_t$ , it is 1/(1 + jω/ $\omega_t$ ), so the quantity 1 − A(jω) is (jω/ $\omega_t$ )/(1 + jω/ $\omega_t$ ), and well below $\omega_t$ that is jω/ $\omega_t$ — a positive imaginary number. Multiply by jωC and the result is $-\omega^2 C/\omega_t$ , which is real and negative.

So the sign has nothing to do with the guard being badly made. It follows from the follower lagging, and every follower lags. The magnitude is the interesting part: it goes as the square of frequency, because one factor of ω comes from the capacitance and one from the follower’s own phase error.

Above $\omega_t$ the follower’s gain has fallen to nothing and 1 − A is one, so the admittance is just jωC — the cable is back, in full, with no negative real part. Between the two the conductance flattens at $\omega_t$ ·C, which is the largest it gets.

What it reaches, in ohms

Measured on the netlist rather than from the expression, with the input node driven by a test source and the current it delivers read back as an admittance, the guarded input’s conductance is negative from a fraction of a hertz upwards, and its equivalent parallel resistance is:

−15.9 megohms at ten kilohertz. −161 kilohms at a hundred. −3.18 kilohms at a megahertz. Above that it flattens at −1.59 kilohms, which is 1/( $\omega_t$ ·C) for a megahertz amplifier and a hundred picofarads.

The unguarded input, for comparison, has a positive conductance at every frequency — a capacitance to ground and a leakage to a rail are both losses, and nothing about an unguarded input can be anything else.

That is a large change to make to an input whose whole purpose is to be high-impedance. At direct current the guard raised it to 10¹⁸ ohms; at a megahertz it is −3.18 kilohms. The two statements are about the same wire and the same amplifier, and the second is fifteen orders of magnitude below the first.

What sets it, and the direction is wrong

The plateau is $\omega_t$ ·C, so it is a product of two numbers a data sheet prints and contains nothing at all about the board — not the leakage the guard was installed to defeat, not the laminate, not the humidity.

Sweeping the cable: ten picofarads gives −15.9 kilohms, forty-seven gives −3.39 kilohms, 220 gives −723 ohms, 470 gives −339. Exactly in inverse proportion, to a part in ten thousand. The cable is the thing to shorten, and shortening it helps by exactly the factor it is shortened by.

Sweeping the amplifier is the uncomfortable one. A hundred-kilohertz amplifier leaves −15.9 kilohms; a megahertz one −1.59 kilohms; ten megahertz −159 ohms. A faster amplifier makes the negative resistance ten times worse for every decade of bandwidth.

That is the opposite of the direction every other number about a guard points. A faster amplifier makes the guard a better guard at direct current, holds the ring closer to the input’s potential over a wider band, and buys more bandwidth from the cable. It also makes the one thing that can destabilise the input an order of magnitude worse per decade. The two considerations a designer would naturally weigh pull in opposite directions, and only one of them is usually written down.

A faster amplifier makes the negative resistance worse, not better. computed by solving, not by drawing. The magnitude of the negative resistance a guarded input presents above the amplifier's own corner, against the amplifier's gain-bandwidth product. It is 1/(ωₜ·C) at every point, to 1.0 parts in ten thousand — 100 kHz: −15.9 kΩ, 1.00 MHz: −1.59 kΩ, 10.0 MHz: −159 Ω. That is the awkward direction: a faster amplifier makes the guard a better guard at direct current, and makes the negative resistance it leaves ten times worse for every decade of bandwidth. The two things a designer would reach for pull opposite ways. — Fig. 2 The negative resistance the guard leaves against the amplifier’s gain-bandwidth product. It is 1/( $\omega_t$ ·C) at every point, to a part in ten thousand: −15.9 kΩ at 100 kHz and −159 Ω at 10 MHz. The faster amplifier is the worse one here.

When the node as a whole goes negative

A negative conductance in parallel with a positive one is only negative if it wins, so the question a design has to answer is where the total at the input node changes sign.

The source contributes +1/ $R_\mathrm{s}$ . The leakage, multiplied by the loop gain, contributes almost nothing. So the node is net-negative above the frequency where $\omega^2 C/\omega_t$ overtakes $1/R_\mathrm{s}$ , and that frequency is $\sqrt{\omega_t/(C R_\mathrm{s})}/2\pi$ — which the solve returns to half a per cent by bisecting the measured admittance rather than by evaluating the expression.

For a hundred picofarads behind a megahertz amplifier: 39.9 kilohertz from a megohm source, 3.99 kilohertz from a hundred megohms, 1.26 kilohertz from a gigohm, 399 hertz from ten gigohms.

Two things about that list matter. The dependence is a square root, so ten times the source resistance buys only a factor of 3.2 — a design cannot get away from it by being a little more careful about the source. And the direction is again the awkward one: the sources a guard exists for are the ones that go negative first. A gigohm source, which is the case the current that does not reach the input is written about, has a net-negative input node above 1.26 kilohertz.

The higher the source impedance, the sooner the guarded node goes negative. computed by solving, not by drawing. The frequency above which the total conductance at the input node is negative — the guard's −ω²C/ωₜ against the source's own +1/Rₛ — plotted against the source's resistance, for 100 pF of cable behind a 1.00 MHz amplifier. It is √(ωₜ/(C·Rₛ))/2π, which the solve returns to half a per cent: 39.9 kHz from a 1 MΩ source and 399 Hz from 10 GΩ. The slope is a half on this axis, so ten times the source resistance buys a factor of 3.2 and not ten. And the direction is the awkward one: the sources a guard exists for are the ones that go negative first. — Fig. 3 The frequency above which the total conductance at the input node is negative, against the source’s resistance. $\sqrt{\omega_t/(C R_\mathrm{s})}/2\pi$ , to half a per cent: 39.9 kHz from a megohm and 399 Hz from ten gigohms. The slope is a half, so ten times the source buys a factor of 3.2.

How the admittance is measured

The measurement is worth setting out, because “the impedance looking into a node” is a phrase with more than one thing behind it and the wrong one gives the wrong answer here.

The input node is driven by a one-volt source with the signal and the rail both zeroed, and the current that source delivers is read back. For one volt, that current is the admittance — real part and imaginary part together, with signs. The signal source and its own series resistance are removed first, because the question is what the source sees looking in, and a source does not see itself. It is the same measurement the instrument’s own rise time makes of a front end, asked as an admittance rather than as a step.

That is not how the current that does not reach the input measures the leakage path’s impedance, and the difference is deliberate. There the leakage resistances are twelve orders of magnitude above everything else in the netlist, and a one-amp probe into that node puts the matrix past the pivot tolerance the conditioning essays exist to talk about — so that figure reads a ratio of two quantities within one ordinary solve instead. Here the quantity of interest is the capacitance path, whose admittance is comparable with the rest of the netlist, and the direct measurement is well conditioned.

Two checks say the measurement is the right one. Driving a plain resistor and capacitor to ground returns exactly 1/R + jωC, sign and all, which fixes the current convention. And the plateau the guarded input reaches is $\omega_t$ ·C to a part in ten thousand across six cable capacitances and five amplifier bandwidths — a product of two numbers that were never given to the solver as a product.

What a negative conductance does and does not mean

A negative conductance at a node is not, by itself, an oscillator, and it is worth being precise because the alarming statement is easy to over-read.

What it does mean is that energy comes out of the input node into whatever is connected, at the frequencies where it is negative. If the thing connected is a pure resistance larger than the negative resistance’s magnitude, the pair is a net source and any disturbance at those frequencies grows.

What it does not mean is that the amplifier is unstable on its own. The measurement here is of the admittance looking into the input, with nothing in front of it; the amplifier’s own loop is perfectly well behaved, and its poles sit in the left half-plane at −8.15 ± j7,930. What has been built is a two-terminal negative resistance, which is a component rather than a fault — it is what the resistor that is not made of the resistors measures as a Thévenin resistance with the wrong sign, arriving here as a parasitic instead of a design.

And what makes it a fault rather than a component is the thing in front of it. A source with any series inductance — a long cable to a sensor, a wound resistor, a coil — makes a resonator with the input’s own capacitance, and a resonator whose loss is negative at its own resonant frequency is an oscillator. That is the failure a guarded electrometer is known for and this is where it comes from.

The sign error that hid it

Finding this required repairing the model, and the repair is worth recording because of what it did and did not change.

The amplifier in this site’s guarded-input model is a transconductance into a compensation capacitance followed by an inverting buffer, which is the usual macro-model. Its transconductance was written to produce gm·( $v_\mathrm{fb}$ − $v_\mathrm{in}$ ) rather than gm·( $v_\mathrm{in}$ − $v_\mathrm{fb}$ ) — the control terminals the other way round.

With that sign the loop closes as $v_\mathrm{out}$ = A· $v_\mathrm{in}$ /(A − 1) instead of A· $v_\mathrm{in}$ /(1 + A). The modelled follower then leads its input by ω/ $\omega_t$ instead of lagging it by the same amount, its closed-loop gain has a magnitude slightly greater than one at low frequency, and the network has a pole at +7.92 × 10³ — in the right half-plane. It is positive feedback that converges at direct current because the excess gain is one part in a million.

Nothing the earlier figures printed was wrong, and that is the part worth carrying. Every one of them reads a magnitude: |1 + A| and |A − 1| differ by two parts in a million at direct current and by nothing at all above the amplifier’s corner, so the 10¹⁸ ohms, the fourteen millivolts, the nanovolt and the twenty-decibels-a-decade rolloff are identical to every printed digit either way. What was wrong was every statement about sign — and with the terminals the wrong way round the input’s conductance came out positive $\omega^2 C/\omega_t$ , which is a plausible-looking loss that no argument would have questioned.

That is a familiar shape of defect arriving in a new place. A check that asks whether a number is right cannot see a sign that no figure prints, and the only thing that found this was asking a question the model had never been asked.

The input resistance is 10¹⁸ Ω or 10¹² Ω, depending on when you ask. computed by solving, not by drawing. The leakage current the signal itself drives, divided into the signal, which is the resistance the source sees. Guarded, it starts at 1.00e+18 ohms — the teraohm of laminate multiplied by the amplifier's open-loop gain — and falls twenty decibels a decade above the amplifier's own corner, arriving back at the bare leakage by 1.00 MHz. Half of what the guard bought is gone by 1.73 Hz. The dashed line is the unguarded board, which has no loop gain in it and is therefore flat. A data sheet's input resistance is a direct-current statement about a circuit nobody uses at direct current. — Fig. 4 The other half of what the guard does, and the half that was always right: the resistance the signal meets through the leakage path, 10¹⁸ Ω at direct current, falling twenty decibels a decade above the amplifier’s own corner and back at the bare teraohm by a megahertz. Every number here is a magnitude and none of them moved when the sign was repaired.

The cable, and the two things it is doing

Set beside the bandwidth result, the capacitance turns out to be doing two jobs at once and they part company at different frequencies.

Below about a hundred kilohertz the bootstrap works as advertised: the capacitance the source sees is a hundredth of a picofarad at ten kilohertz and one picofarad at a hundred, against the hundred picofarads that is physically there. That is the 1,230 times of bandwidth the current that does not reach the input measures, and it is real.

At a megahertz the source sees fifty picofarads — half the cable is back — and above $\omega_t$ it sees all hundred. So the guard’s bootstrapping of the capacitance and its production of a negative conductance give out at the same frequency and for the same reason, which is that the follower has stopped following.

The practical summary is a band rather than a number. Below $\sqrt{\omega_t/(C R_\mathrm{s})}/2\pi$ the guard is doing everything claimed for it. Between there and $\omega_t$ it is presenting a negative conductance at a node whose capacitance is climbing back. Above $\omega_t$ it is an ordinary unguarded input with an amplifier in front of it. A design that lives in the first of those three is fine; one whose signal or whose source’s resonance reaches the second is the one that oscillates.

The same wire, and the same loop gain, is 1230× the bandwidth. computed by solving, not by drawing. The −3 dB frequency from the source to the output, with 100 pF of cable on a 1 GΩ source. Unguarded the cable and the source are a low-pass filter at 1.59 Hz and there is nothing to be done about it. Guarded, the shield follows the input, so there is no voltage across the capacitance and no current into it — the amplifier drives it from its output, where it is a load rather than a pole, and the corner moves to 1960 Hz. It is not the full loop gain, because the guard's own accuracy runs out with frequency: the answer is where those two curves cross, and the netlist finds it. — Fig. 5 The bandwidth the guard buys from the same capacitance: 1.59 Hz unguarded against 1.96 kHz guarded, a factor of 1,230. It is the same capacitance and the same follower that produce the negative conductance, and they give out together.

What to do about it

Three remedies follow from the three quantities in $\omega^2 C/\omega_t$ , and they are not equally available.

Shorten the cable. The negative resistance is inversely proportional to the capacitance, exactly, so halving the cable halves it. This is the only one of the three that costs nothing and it is the one a layout can usually deliver.

Do not use a faster amplifier than the measurement needs, and weigh the input current the current the instrument draws prices against the bandwidth before choosing one. A decade of bandwidth is a decade of negative resistance, and the direct-current guarding it buys — a larger loop gain multiplying the leakage — is already far past what any board needs. The ten-to-the-eighteenth ohms at direct current is nine orders beyond the teraohm of laminate; the −159 ohms a ten-megahertz amplifier leaves is not nine orders beyond anything.

Put a resistor in series with the guard drive. This is the remedy that is actually used and this essay does not measure it: a few hundred ohms between the amplifier’s output and the ring, with the ring’s own capacitance to the input, makes a compensated divider of exactly the kind a divider with two ratios is about — the guard then drives a lagging copy of a lagging copy, and the second lag can be arranged to cancel part of the first. What resistance is right, and whether it can be chosen without knowing the cable, is the measurement this essay hands on.

Ten times the cable is a tenth of the negative resistance. computed by solving, not by drawing. The magnitude of the negative resistance a guarded input presents above the amplifier's own corner, against the capacitance the guard is driving. It is 1/(ωₜ·C) at every point, to 0.0 parts in ten thousand — 10 pF: −15.9 kΩ, 47 pF: −3.39 kΩ, 220 pF: −723 Ω. So the cable is the thing to shorten, and shortening it helps in exact proportion. — Fig. 6 The negative resistance against the capacitance the guard is driving: 1/( $\omega_t$ ·C) at every point, from −15.9 kΩ at ten picofarads to −339 Ω at 470. The cable is the thing to shorten and it helps in exact proportion.

Three numbers about one input, and none of them contradicts the others

It is worth putting the three descriptions of this input side by side, because the temptation is to decide that one of them is the real one.

At direct current the input is 10¹⁸ ohms. That is the teraohm of laminate multiplied by the amplifier’s open-loop gain, it is what the guard was installed for, and it is correct.

At a kilohertz the input is about −1.6 gigohms. The leakage path has begun to lose its multiplication and the capacitance path has begun to produce its negative conductance, and the second is already the larger. Nothing about the first has stopped being true; it has stopped being the dominant term.

At a megahertz the input is −3.18 kilohms. The follower is more than half a radian behind and the capacitance it is driving is most of the way back into circuit.

All three are readings of the same admittance at different frequencies, and an input impedance was never a number. What makes the middle one surprising is only that the first is what a data sheet quotes and the third is what a bench measurement at any convenient frequency would find — and between them the quantity changed sign, which no interpolation between the two end points would suggest.

Still open: the series resistor, a source with an inductance, and the other bootstrap

The resistor in the guard drive, and what it is worth. Adding a resistance between the amplifier’s output and the ring puts a pole in the guard’s own path, which adds phase and changes the sign of what the capacitance presents over some band. It is the standard fix and nothing here prices it: the measurement wanted is the negative conductance’s worst value against that resistance, and whether a single value works across the range of cables a design might see. A remedy that has to be tuned to a cable is a different kind of remedy from one that does not.

A source that resonates. Everything here is a resistive source. The failure this negative conductance actually causes needs a source with an inductance, and then the question is not the sign of a conductance but whether the loop’s poles cross into the right half-plane — a second-order resonator with a loss that is negative in a band. Sweeping the source’s inductance and watching the poles would give the inductance at which a given cable and amplifier oscillate, which is the number a designer would actually use.

And the other bootstrap, where the same arithmetic applies with a different sign. The probe is part of the circuit measures a probe loading what it is measuring, and the usual fix inside an oscilloscope’s front end is the same bootstrap applied to the input’s own capacitance. Whether that one leaves a negative conductance too, and whether its bandwidth requirement pulls the same way, is the same calculation in a place where the capacitance is smaller and the amplifier is very much faster — so the product $\omega_t$ ·C, which is all that matters, may land anywhere.