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

Compensation network — where it appears

A deliberate network added to make a response flat or a loop stable, by cancelling something the circuit already has. It is a value rather than a direction: a probe's capacitor, a shunt's parallel network and an amplifier's feedback capacitor are all wrong on both sides of one setting, which is why each of them is trimmed.

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

The corner a 2 nH shunt has, against the current it is sized for. computed by solving, not by drawing. A shunt held at the best burden voltage of 7.75 mV has R = u⁄I, so its own 2 nH of series inductance puts a corner at u⁄(2πLI) — 616 kHz at an ampere and 6.16 kHz at a hundred, for the same piece of metal. The optimum that contains no current at all therefore hands the bandwidth a current dependence: the corner falls in exact proportion. At 1 A the shunt is 7.75 mΩ with a time constant of 258.2 ns, so a 10 ns edge is read 2.58e+3% high and a 1 ns edge 259.20 times too large. A resistor and a 258.2 pF capacitor across it — the value found by search on the solved response, agreeing with L/(R·Rc) to 7.9e-5% — flatten the reading to 7.8e-5% across six decades, and a fifth too much makes it ten times worse. The dots are the corner bisected on the solved impedance rather than taken from R/2πL.

The optimum that hands back a bandwidth

The best burden voltage across a shunt is 7.75 mV and contains neither the current nor the resistance, which is what made it worth having. A shunt has two nanohenries whatever it is made of, so holding the burden fixed fixes the resistance at u*/I — and the corner R/2πL then falls in exact proportion to the current: 6.16 MHz at a tenth of an ampere, 616 kHz at one, 6.16 kHz at a hundred. A resistor and a 258.2 pF capacitor across it, found by search on the solved response, flatten the reading to 8×10⁻⁵ per cent across six decades.

instruments · Current sensing What a feedback tee buys, and the single thing it charges. computed by solving, not by drawing. Two 50 kΩ resistors with a 6.250 kΩ tap give 500.0 kΩ of transimpedance, which is R₁(1 + R₂/R₃) + R₂ solved on the netlist. That expression has no term for the noise gain, and the tap sets it to 1 + R₁/R₃ = ×9.00 — against exactly one for a single resistor of any value, because at direct current the source is a capacitor. One quantity then does all three things at once. The signal and R₁'s own noise are multiplied together, so the tee buys no signal-to-noise ratio at all: 1846 against 5582 in a hertz at a nanoamp for one resistor of the same transimpedance, a factor of 3.02 which is the square root of the noise gain. The amplifier's own voltage noise and offset are multiplied where they were not before. And the phase margin RISES, from 1.8° to 15.9°, because starting the noise gain high shortens its climb to the peak — so the tee is a compensation that charges for itself in noise, and a capacitor across the feedback is the same compensation for nothing. With no tap the three effects are absent rather than small.

The tee that charges for its own compensation

A feedback tee makes a large transimpedance out of small resistors, and R₁(1 + R₂/R₃) + R₂ is the whole of what is usually said about it. The expression has no term for the noise gain, and the tap sets that to 1 + R₁/R₃ — nine, where a single feedback resistor of any value gives exactly one, because at direct current the source is a capacitor. One quantity then does everything: the signal and R₁'s own noise are multiplied together so the tee buys no signal-to-noise ratio at all, and the phase margin RISES from 1.8° to 15.9°.

feedback · Transimpedance

Named alongside it

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

Design tradeoffBandwidthBurden resistanceCurrent sensingJohnson noiseModel rangeNoise gainParasiticsPhase marginSignal-to-noise ratioTransimpedance

All concepts