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

Device matching — where it appears

How nearly two devices intended to be identical actually are, which sets the offset and the drift of every circuit built on their symmetry. Two devices with the same imperfection cost a symmetric circuit nothing, because their common error is rejected; only the difference between them survives.

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

What a 5% mismatched pair leaves behind, across 150 K. computed by solving, not by drawing. A saturation-current mismatch of 5.0% appears as an input offset of Vₜ·ln(m) — 1.2613 mV at 300 K — which is proportional to absolute temperature and therefore drifts at 4.2044 µV/K, exactly the offset divided by the temperature. That is 3333 ppm per kelvin at every mismatch on the slider, because the ratio is 1/T and contains nothing about the device. One junction on its own drifts 1.828 mV/K, 435 times harder.

What matching does about temperature

A pair cancels the 1.8 mV/K that broke the previous essay, and what it leaves behind is exact: a saturation-current mismatch of m shows up as an offset of the thermal voltage times ln(m), which is proportional to absolute temperature and therefore drifts in proportion to itself. 3 333 parts per million per kelvin, at every mismatch on the slider, because the ratio is 1/T and contains nothing about the device at all.

semiconductors · Differential pair A first stage of 100 buys 40.0 dB of rejection, and gives it back above 10.0 kHz. computed by solving, not by drawing. The rejection of a three-amplifier instrumentation amplifier against frequency, beside the one-amplifier difference stage it is built around. At low frequency the two differ by 39.99 dB against 20 log 100 = 40.00 dB, and the reason is that the input stage passes a common-mode voltage at exactly unity: the common-mode gain of the whole instrument is 1.998 mV/V, which is the difference stage's own. So the four resistors around the last amplifier decide the rejection and the two that set the gain do not — ten per cent between them moves it by less than a hundredth of a decibel. What ends it is bandwidth: above 10.0 kHz, which is the amplifier's gain–bandwidth divided by the gain that bought the rejection, the differential gain falls and the rejection falls with it at twenty decibels a decade.

The four resistors that decide, and the two that do not

A difference amplifier's rejection is decided by four resistors and one-tenth-per-cent parts give 54 decibels. Putting a two-amplifier stage in front adds exactly twenty times the log of its gain — 94 decibels at a gain of a hundred — and the reason is not that the input stage rejects anything. It passes common mode at exactly unity, so the common-mode gain of the whole instrument is 1.998 millivolts per volt at every gain tried, and the improvement is entirely the differential signal arriving larger. The two resistors that set that gain may be ten per cent apart without moving the answer a hundredth of a decibel.

instruments · Common-mode rejection An exponential with 6× of degeneration: both edges move, and not together. computed by solving, not by drawing at 61 amplitudes. Total harmonic distortion reaches one per cent at 36.6 mV and the gain falls one per cent short of its small-signal value at 85.0 mV. An emitter resistor dividing the gain by 6 moves the distortion edge by 35.4 times — the square of the factor, because the resistor both divides the drive reaching the junction and linearises what the junction does with it — while the gain edge moves by only 11.6 times. So the two edges close up: 2.32 times apart here against 7.06 bare, and a well-degenerated stage stops being limited by its linearity and starts being limited by how accurately its gain is known.

What a resistor in the emitter buys

Degeneration is described as a trade: give up gain, get linearity. Measured on the transfer curve, the two sides of that trade are not the same size. Dividing the gain by six moves the amplitude at which distortion reaches one per cent by thirty-five times — the square of the factor — and the amplitude at which the gain is one per cent out by only twelve. So the two edges close up, and a well-degenerated stage stops being limited by its linearity and starts being limited by how accurately its gain is known.

semiconductors · Emitter degeneration Matched parts cost nothing; a 2 dB difference between them sets a 112 dB ceiling. computed by solving, not by drawing. The common-mode rejection of a three-amplifier instrumentation amplifier against the gain of its input stage, with each amplifier's own rejection in the netlist as an input-referred error of the common-mode voltage over the rejection. The architecture's own figure rises decibel for decibel with the gain, because the difference stage sees a larger differential signal beside the same common-mode one. The parts' contribution does not rise with anything, and the part of it that matters is not their rejection but the difference between their rejections: two amplifiers of 98 dB that are identical cost 0.000 dB, while 100 dB against 98 dB leaves a ceiling of 111.7 dB with no gain in it. The two mechanisms cross: below a gain of 1903 the four resistors decide everything, and above it more gain buys no more rejection at all — 111.9 dB at a gain of 100000, where the arrangement alone would have been worth 148. The one place the instrument beats its own floor is a gain of 1000, where the two errors cancel; that is a coincidence of signs and not something a design can hold.

The rejection the parts have

Two essays measured the architecture: four resistors decide an instrumentation amplifier's rejection, two do not, and the answer is the one-amplifier figure plus twenty log of the first stage's gain — exactly, with amplifiers of infinite rejection. Give each amplifier its own and something unobvious happens: two matched but individually mediocre parts cost nothing at all, because their error is a common-mode signal at the difference stage and is rejected there. What costs is the difference between them, and it sets a ceiling with no gain in it.

instruments · Common-mode rejection One output in saturation takes 6.5% off another that is at five volts. computed by solving, not by drawing. A three-transistor mirror: a reference, an output taken down into saturation, and a third output held at five volts throughout. The upper curve is the saturating output's own loss and the lower one is the sibling's. At 50 mV the saturating output is 27.3 per cent down and the sibling, which is nowhere near saturation, is 6.54 per cent down. The reference current moves by -0.0188 per cent, which is nothing: a base current is a hundred and fiftieth of a collector current and the reference is set by a resistor from the supply. What does move is the base-emitter voltage every output shares — by -1.75 millivolts — and every output is an exponential of it.

The refusal, and what it was protecting

A current mirror's model has declined to answer below two hundred millivolts of collector-emitter voltage since it was written, because it has no base-collector junction and would return a forward-active current for a saturated transistor. Put the junction in and the refusal turns out to have been placed where the model it protects is still right to seven parts in ten thousand — and the mirror's real failure is somewhere else entirely: a saturated output takes six and a half per cent off an output that is sitting at five volts.

semiconductors · Current mirror Three mismatches, and only one of them reaches the output. computed by solving, not by drawing. Each of the three quantities that can differ between the two transistors is given a spread of its own, one at a time, and 200 pairs are solved at each. The saturation currents produce a spread that follows them exactly — exponent 0.998, so 1.910 per cent of copy error for two per cent of mismatch. The current gains produce a line of slope 2.003, which is second order rather than first, and land at 2.20e-4 per cent for the same two. There is no third line because there is no third component: with no emitter resistors there is nothing for a resistor tolerance to be a tolerance of, and matching a mirror is a statement about emitter area and about nothing else.

The mismatch that cancels itself

A current mirror's copy error is spread by three things the two transistors can differ in, and the population that measures it has always drawn all three at once. Turned on one at a time, a two per cent spread of saturation currents gives 1.910 per cent of copy error and a two per cent spread of current gains gives 0.00022 — because the gains enter only as a sum of reciprocals, which has no first derivative where they are equal. Then the standard cure un-cancels it, by a factor of 86.

semiconductors · Current mirror

Named alongside it

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

Model rangeThermal voltageCommon-mode rejectionComponent toleranceCurrent mirrorDesign tradeoffDifference amplifierEmitter degenerationGain–bandwidth productInput offsetClosed-loop gainCompliance range

All concepts