Circuit Response
every model has an edge, and the figure carries it
Concept

Guarding — where it appears

A conductor around a high-impedance input, driven to the input's own potential so that the leakage across the board has no voltage across it. It multiplies the input resistance by the amplifier's loop gain, and it leaves behind whatever the copy of the input fails to be.

Named by 6 essays across 2 fields — each of them below, with the objects they name alongside it.

A clean board is 14.0 mV of error, and a guard makes it 1.0 nV. computed by solving, not by drawing. Two boards differing by one wire, solved with the leakage present and again with it removed, so the number plotted is the leakage's own contribution and nothing else. Unguarded, a teraohm across the laminate from a 15 V rail through a 1 GΩ source is 13.99 millivolts — and a humid morning takes that resistance down two decades, which is the left-hand end of this axis. Guarded, the ring is held at the input's own potential by the amplifier, so what is across the leakage is the amplifier's own error and the result is 1.00 nanovolts. The guarded line is flat in the rail and proportional to the signal: the offset has become a gain error of 1.00 parts per billion, and the reading is low by it rather than high: the ring sits a little below the input, so the last of the leakage pulls the input down.

The current that does not reach the input

A teraohm across a board from a fifteen-volt rail is fourteen millivolts of error through a gigohm source, and a humid morning takes that resistance down two decades. A ring held at the input's own potential leaves a nanovolt — proportional to the signal rather than to the rail, so an offset has become a gain error of one part in a billion — and the same wire multiplies the input resistance by the loop gain, which makes it 10¹⁸ Ω at direct current and 10¹² Ω at a megahertz.

instruments · Guarding The guard leaves a negative resistance, and it reaches −1.59 kΩ. computed by solving, not by drawing. The magnitude of the conductance a source sees looking into the input, guarded and not, with 100 pF of cable and a 1.00 MHz amplifier. The unguarded input's conductance is positive everywhere — a capacitance to ground and a leakage to a rail are both losses. The guarded one is negative above 0.0404 Hz, and its magnitude rises as the square of frequency: −15.9 MΩ at 10 kHz, −161 kΩ at 100 kHz, −3.18 kΩ at a megahertz. Above the amplifier's gain-bandwidth product it flattens at ωₜ·C, which is −1.59 kΩ. That is the same input the guard raises to 10¹⁸ Ω at direct current, and nothing about the leakage the guard was installed for appears in it: the negative resistance is a product of the amplifier's bandwidth and the cable it is driving.

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.

instruments · Guarding A resistor in the guard drive raises the worst negative resistance from −1.59 kΩ to −403 kΩ, and sets it itself. computed by solving, not by drawing. The parallel resistance a guarded input presents wherever it is negative, against frequency, for a 100 pF cable whose guard has 100 pF of its own to ground, driven by a 1.00 MHz follower through no resistor, 1 kΩ, 10 kΩ, 100 kΩ. Without a resistor the worst is −1.59 kΩ, −1/(ωₜC). With one it is −6.5 kΩ, −42.6 kΩ, −403 kΩ respectively: weaker, and for the largest resistor close to −R(C + Cₛ)²/(C·Cₛ) = −400 kΩ, a figure with the amplifier's bandwidth nowhere in it. The negative band still runs to about 970 kHz.

The resistor that moves the lag

A guarded input driven by a one-megahertz follower presents a negative resistance of −1/(ωₜC), −1.59 kΩ behind 100 pF of cable. The standard remedy is a resistor between the follower and the ring. It weakens the negative resistance, to −6.5 kΩ at 1 kΩ and −403 kΩ at 100 kΩ, but it does not remove it: the resistor and the ring's own capacitance to ground make a lag of their own. With a large enough resistor the worst value is −R(C + Cₛ)²/(C·Cₛ) for a drive resistor R — minus four times the resistor when the ring is as capacitive to ground as to the input — and the same at every length of one cable. The length moves somewhere else instead. The frequency where the bootstrap is given up, 1/(2πR(C + Cₛ)) for a drive resistor R, falls with every metre.

instruments · Guarding A guarded input driven through an inductance oscillates once the source's L/R passes 1/ωₜ. computed by solving, not by drawing. The real part of the rightmost natural frequency of a guarded input — a 100 pF cable bootstrapped by a 1.00 MHz follower, 10 pF of input capacitance — driven from sources of 10 Ω, 100 Ω, 1 kΩ with an inductance in series, against the inductance. Above zero the input rings with growing amplitude. Each source crosses at 1.75 µH, 17.6 µH, 187 µH, close to R/ωₜ — 1.59, 15.9, 159 µH — and oscillates at 11.4 MHz, 3.59 MHz, 1.14 MHz. The same source without the guard is stable.

The source that rings against its guard

A bootstrapped cable is a capacitance in series with a negative resistance that falls as the square of frequency, −1/(ω²·C/ωₜ). Drive it from a source with resistance R and inductance L and the loop's reactances cancel at 1/√(LC). Its resistances cancel where R = L·ωₜ, so the input oscillates once the source's own time constant L/R passes the follower's 1/ωₜ. The poles of the whole netlist say 17.6 µH for a 100 Ω source behind a 1 MHz follower, against 15.9 from the expression. The cable's length moves that by at most a factor of two and sets only the frequency it rings at. The same source without the guard is stable. A resistor in the guard drive raises the limit, to 164 µH through 1 kΩ, and pays for it in bootstrap bandwidth.

instruments · Guarding A 1% mismatch in the cable reads as 3.0% of full-scale displacement, at every frequency. computed by solving, not by drawing. A capacitive displacement sensor — two 10 pF plates driven at ±5 V in antiphase facing a common electrode — read by a charge amplifier through two metres of three-conductor cable with 30 pF a metre between each excitation conductor and the sense conductor. With the two cross-capacitances mismatched by 1 per cent and the sensor centred, the reading's in-phase part (solid) is 3.00 per cent of full scale at every carrier from 100 Hz to 1 MHz: the cable is a sensor capacitance, and its mismatch a displacement. The quadrature part (dashed), which the cable's resistance makes, rises with frequency and a synchronous detector rejects it. With each conductor in its own screen the cross-capacitance is gone and so is the error.

The sensor that is the capacitance

Read a capacitive displacement sensor through the same kind of cable that limited a strain-gauge bridge's carrier, and the roles exchange. On the bridge, a 1% mismatch in the cable's capacitance was a quadrature leak that grew as the square of the carrier and could be escaped by a lower one. On a sensor of two 10 pF plates it is 3.0% of full-scale displacement, in phase, at every carrier from 100 Hz to 1 MHz — because a capacitance between an excitation conductor and the sense conductor is in parallel with the measurand and is a sensor capacitance. No detector phase separates it and no carrier escapes it; only geometry does. Screen each conductor separately and there is nothing left to mismatch.

applied · Bridge A second pole at 3.00 MHz makes a 9.49 Ω source 4.1 times safer and a 949 Ω one 20% less safe. computed by solving, not by drawing. The source inductance above which a guarded input oscillates — a 100 pF cable bootstrapped by a 1.00 MHz follower with 10 pF of input capacitance — bisected on the sign of the rightmost pole of the whole netlist, against the source resistance, for a follower with no second pole and with one at 30 MHz, 3 MHz and 1.5 MHz; the thin line is R/ωₜ. With a second pole at 3.00 MHz the critical inductance is 6.85 µH at 9.49 Ω against 1.66 with one pole, and 141 µH at 949 Ω against 177: a low-resistance source rings high above the follower's crossover, where the second pole takes gain away, and a high-resistance one rings near it, where the pole adds phase.

The second pole the ringing guard sees

A guarded input driven through an inductive source oscillates once L/R passes 1/ωₜ — for a follower with one pole. Give the follower the second pole every real amplifier has and the rule stops being a bound in either direction. A 9.49 Ω source that rang above 1.66 µH now needs 6.85 µH with a second pole at 3 MHz, four times safer; a 949 Ω source that rang above 177 µH now rings above 141, a fifth worse. The dividing line is the frequency the input rings at: above the geometric mean of the follower's crossover and its second pole, the pole takes gain from the bootstrap and helps; below it the pole adds phase and hurts. On three poles a decade apart the change of sign falls within 2.4 per cent of √(fₜ·f₂).

instruments · Guarding

Named alongside it

The objects these essays reach for when they reach for this one.

BootstrappingNegative resistanceSource impedanceInput impedanceLoop gainMeasurement errorOscillationPolesCapacitive loadCapacitive sensorCharge amplifierDevice mismatch

All concepts