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 11 essays across 3 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 Parts sharing 90% of their error: the divider's ratio spreads 0.091% and the chain's total resistance 0.559% — one from the same four parts, cancelling, and one adding. computed by solving, not by drawing, over 20,000 seeded builds: four 1% resistors whose errors share a fraction ρ = 0.9 of their variance — as parts from one reel, one wafer or one temperature do — built into a divider tapped at the middle. The ratio's error (solid bars) has a standard deviation of 0.0912%, half the parts' 0.577% times √(1 − ρ); the total resistance of the same four in series (outline bars) has 0.5587%, the parts' spread times √(ρ + (1 − ρ)/4). Correlation that makes a ratio better makes a sum worse, by the same arithmetic.

What a ratio does not see

Parts from one reel, one wafer or one temperature share part of their error, and a circuit that uses them as a ratio cancels exactly the part they share. Four 1% resistors sharing ninety per cent of their error make a divider whose ratio spreads 0.091 per cent and a series chain whose total spreads 0.559 — the same four parts, one answer six times better than independent parts would give and one nearly twice as bad. The ratio's spread falls as √(1 − ρ), the total's rises as √(ρ + (1 − ρ)/4). From −40 to 85 °C the chain moves by up to 0.76 per cent and the ratio by 0.135, inside the ±25 ppm/K the coefficients differ by. And a resistor array whose parts share 99.5 per cent of their error puts a divider inside ±0.1 per cent every time that four discrete parts manage it one time in four.

networks · Component tolerance Two 1% parts sharing 90% of their error: the divider's output spreads 0.129% at a half, 0.236% at ten to one and 0.258% at a hundred to one. computed by solving, not by drawing, over 20,000 seeded builds at each of seven ratios, every ratio computed exactly: a divider of two 1% parts, each with a uniform part's 0.577% spread and sharing a fraction ρ = 0.9 of it, with the top resistor N times the bottom. The output ratio's standard deviation (dots) against N, beside N/(N + 1)·√2·σ·√(1 − ρ) (line) and the same law for the other shared fractions (faint). N 1: 0.1286%; N 2: 0.1712%; N 5: 0.2145%; N 10: 0.2356%; N 20: 0.2477%; N 50: 0.2536%; N 100: 0.2581%. What is shared cancels at every ratio; what is not reaches the output with a leverage of N/(N + 1), which is a half for a halving divider and nearly one for a large ratio — so the same two parts make a ten-to-one divider 1.83 times worse.

The small leg of a large ratio

A divider of two 1% parts sharing ninety per cent of their error spreads 0.129 per cent at a half and 0.236 at ten to one — the same two parts, nearly twice as bad, because what they do not share reaches the output with a leverage of N/(N + 1) rather than one half. Correlation cancels at every ratio; the leverage does not. Built from equal units of one array, a divider of N to one spreads as σᵤ·√(N/M), so at any one area a ten-to-one ratio costs √10 against a half and needs ten times the units to match it, and the units it needs are the bottom leg's. And a ratio far from a half heats its own parts unequally: at 50 V across 11 kΩ the top of a two-part ten-to-one divider runs 46 K hotter than the bottom and the ratio moves 1,055 ppm with perfectly matched coefficients, where a chain of eleven equal units moves at most 29.

networks · Component tolerance A pair with 2.0% saturation-current mismatch, nulled at 300 K: trimmed by a constant it drifts 1.706 µV/K, trimmed by its load ratio 0.000 µV/K. computed by solving, not by drawing, at 31 temperatures — each a Newton solve of both junctions, a bisection for the tail current and a bisection for the input that nulls the output — a bipolar pair at 10 µA a side whose second transistor has 1.02 times the first's saturation current. Untrimmed (grey) it needs 511.9 µV at 300 K and 682.6 µV at 400 K. Trimmed at 300 K by subtracting a constant (dashed) it is zero there and 170.6 µV at 400 K, drifting 1.706 µV/K — the untrimmed offset's drift, untouched. Trimmed at 300 K by making the collector loads 1.02000 to one (solid) it is 0.0 µV at 400 K, drifting 0.000 µV/K. The load ratio acts inside the same logarithm the mismatch does, so it cancels the mismatch at every temperature, not at one.

The trim that takes the drift with it

A bipolar pair's offset from a saturation-current mismatch is proportional to absolute temperature, so trimming it to zero at one temperature can trim its drift to zero too — but only if the trim acts inside the same logarithm the mismatch does. Trimmed by its collector-load ratio, a 2% mismatched pair is zero from 250 to 400 K; trimmed by subtracting a constant after it, it drifts the same 1.706 µV/K it drifted untrimmed. Add 20 kΩ of source resistance and a 10% current-gain mismatch, and a fifth of the offset follows the current gain instead: the load trim now leaves −0.574 µV/K, the constant trim 1.256, and above 34 kΩ the constant trim is the better one. A trim cancels the drift of the share whose temperature law it shares and converts the rest; two knobs, set for value and slope, leave 11 µV of curvature.

semiconductors · Differential pair Two junctions at 8 to 1 in area, at four currents from 10 µA to 10 mA. computed by solving, not by drawing: each drop is the diode law solved at its own temperature, from 200 to 400 K, with a saturation current that rises 15.0 orders of magnitude over the range. The two drops themselves sit 179 mV apart across the four currents; their difference does not move at all. It is Vₜ·ln 8 to 0.40 parts in a million at 10 mA and to 0.040% at the worst point, 10 µA at 400 K, where the larger junction's saturation current is no longer negligible against the current — 53.758 mV at 300 K, rising 179.19 µV/K — and the straight line through it reaches 0.001 µV at absolute zero (dashed).

The voltage at which the slope is zero

Two junctions carrying one current, one of them eight times the other's area, differ by exactly the thermal voltage times ln 8 — 53.758 mV at 300 K, rising 179.19 µV/K along a line that reaches zero at absolute zero to a thousandth of a microvolt — while each drop alone moves 1.9 mV/K. Weighted by 10.72 and added to one drop, the difference cancels it, and the sum is then 1.171704 V. That voltage is not a coincidence of this design: every weighting of the sum leaves the band gap at absolute zero, so a reference's slope at 300 K is its distance from 1.1717 V divided by 300 K, and two hundred references built wrong three different ways sit on that one line to 0.16 nV/K. A reference is trimmed by its voltage, at one temperature, and the voltage it is trimmed to is a statement about how its current is made.

semiconductors · Temperature What balancing the source resistances is worth, against how correlated the two input currents are, for a bias-cancelled input. computed by solving, not by drawing; 400 seeded parts at each correlation, each solved unbalanced and balanced at a 1 MΩ source, input currents of zero mean and 1 nA spread. The ratio of rms errors (dots) follows 1/√(2(1 − ρ)) (line): 0.72 for independent currents — balancing makes it 38% worse — one at ρ = ½, and 6.5 at 0.99. The cure removes what the two currents share, and is a cure only if they share more than half.

The cure that needs one current

Balancing the resistances an amplifier's inputs look into removes whatever its two input currents have in common and leaves their difference. On a plain bipolar input the two are one current at two pins, correlated at 0.991, and the cure is worth ten. On an input whose bias current has been cancelled they are two residues of either sign, correlated at 0.10, and over 400 seeded parts the cure makes the error 38 per cent worse. It is worth 1/√(2(1 − ρ)) for a zero-mean current, nothing at ρ = ½, and in general exactly the input current over the offset current.

instruments · Input bias current

Named alongside it

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

Model rangeThermal voltageComponent toleranceMonte carloTemperature coefficientDesign tradeoffInput offsetCommon-mode rejectionComponent sensitivityCurrent mirrorDevice mismatchDifference amplifier

All concepts