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

Emitter degeneration — where it appears

A resistor in series with an emitter, which divides the stage's gain and linearises its transfer curve by the same factor. The distortion edge it buys grows as the square of the factor and the gain error's edge does not, so the two boundaries move apart as it is increased.

Named by 8 essays across 2 fields — each of them below, with the objects they name alongside it.

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 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.

Where the two exponents come from

What a resistor in the emitter buys measured two exponents and could explain neither: the distortion edge moves as the square of the degeneration factor and the gain edge as its 0.950 power, and at a factor of exactly three halves the gain error vanished. The degenerated transfer curve has no closed form forwards and an exact one backwards, and reverting that series gives all three. The distortion exponent is exactly two; the gain edge is proportional to D squared over the root of the absolute value of three minus twice D, which is infinite at three halves and tends to a three-halves power; and the two edges are 7.07 apart on the bare device and 1.62 at a factor of eleven, against 7.06 and 1.61 measured.

semiconductors · Emitter degeneration The fourth-order term explains 95% of what the third leaves out. computed by solving, not by drawing. Two closed forms for the amplitude at which a degenerated stage reaches 1% of second harmonic, each against the same measurement — a bisection on the harmonic content of a Newton-solved curve, which shares no arithmetic with either. The leading expression is 4Vₜ·t·D², derived at the rung below this one and exact in the limit of small drive; its error grows as the 1.96 power of the drive it is evaluated at. At a degeneration of eleven that drive is 4.53 thermal voltages and the expression is 6.89% optimistic. Carrying the reversion one order further takes it to 0.362%.

What the fourth order says about the third

A closed form derived by neglecting the fourth-order term is a claim with an error, and the error is a quantity the fourth-order term can be asked about. Carried one order further, the degenerated exponential's reversion predicts 7.33 per cent where the leading expression was measured to be 6.89 per cent optimistic — and takes the residue to 0.362 per cent. The correction has its own edge, in the same quantity, and it is measured too.

semiconductors · Emitter degeneration Every order buys less range than the one before, and above 9 Vₜ the sixth is worse than the fourth. computed by solving, not by drawing. The error of the same expression truncated at three orders, against the drive it is evaluated at, with the measurement it is chasing being a Newton-solved transfer curve that knows about no series at all. Each truncation's error grows as its own order in the drive — fitted at 2.00, 4.00, 5.88 against 2, 4 and 6 — so each buys a further range at a stated accuracy: inside 1% the leading expression is good to 1.12 thermal voltages, the fourth order to 3.88 and the sixth to 7.03, factors of 3.46 and 1.81. Beyond all of them the series stops helping: at 8.9 thermal voltages, where the second harmonic is 11.4%, the sixth-order expression is exactly as wrong as the fourth and is worse above it. What a designer does there is bisect the curve.

The order that stops helping

Three essays in this field have derived expressions for the amplitude at which a degenerated stage's distortion reaches a target, each one order longer than the last, and each one nearer the measurement. This is where that stops. The error of an expression truncated at order m grows as the m-th power of the drive — 2.00, 4.00 and 5.88 measured — so every added order buys a range that ends sooner than the last one bought, and above 8.9 thermal voltages the six-term expression is further from the device than the four-term one.

semiconductors · Emitter degeneration A junction and its resistor in one loop: equal shares at 12.50 mV, and quietest against both at 49.98 mV. computed by solving, not by drawing. A junction carrying 25 µA in series with a resistor, the loop closed into a short and solved as a netlist with each noise current injected across its own element. Against the drop across the resistor: the bare junction's 2qI, the bare resistor's 4kT/R, each one's share of what reaches the outside, and the total. The shares are equal at 12.50 mV (500 Ω), not at the 49.98 mV where the bare floors cross; there the resistor supplies 80.0 per cent and the total is 0.5556 of either floor. The total is below both floors at every drop.

The resistor in the same loop

A resistor's noise and a junction's are equal as bare densities at 49.98 millivolts of drop, and a junction in series with the resistor that carries its current is the arrangement every current source is built from. In one loop each noise current has to cross the other element, so the two supply equal shares at 12.50 millivolts, a quarter of the crossing; at the crossing itself the resistor supplies 80 per cent and the loop is 2.553 decibels below both floors, which is further than it gets anywhere else. The same resistor multiplies the stage's input-referred noise by five.

noise · Shot noise 300 mirrors built to one design, with 2% device mismatch. computed by solving, not by drawing. Every pair in the population is a full Newton solve of the same netlist with two saturation currents drawn from a normal distribution, the Early conductances iterated to self-consistency for each. The mean is 3.836 per cent, which is the systematic error the rung below computed with identical devices (3.937 per cent) — the mismatch does not move it. The spread about it is 1.955 per cent, which is the device mismatch arriving with nothing dividing it, and the worst pair of the 300 is 8.43 per cent out. A design whose specification is the mean has specified the one mirror nobody has.

The error that is a distribution

The rung below solved a mirror and separated two errors — one that falls with beta and one that does not. Neither is what limits a real mirror. Two transistors on the same die differ, a fractional difference in saturation current is a fractional difference in collector current with nothing dividing it, and the honest object is a spread rather than a number: mean 3.84 per cent, standard deviation 1.96, worst of three hundred 8.43. Degeneration divides it by one plus gm·R and stops at the resistors' own tolerance, and where it stops is a voltage — a hundred millivolts, containing nothing but the ratio of two tolerances.

semiconductors · Current mirror Degeneration removes the second-order product 9 times less well than the third. computed by solving, not by drawing. Both intermodulation products of a degenerated stage at 5 mV a tone, against the degeneration factor. The second-order product falls as D⁻² — the straight reference is exactly that law, anchored at D = 1 — and the fitted exponent is -2.000. The third-order product falls faster, as D⁴/|3 − 2D|, which is one more power of D at large factors, and it collapses altogether at D = 1.5 where its coefficient changes sign. So the ratio between the two goes from 20.7 at D = 1 to 183 at D = 16: the more linear the stage is made, the more completely its distortion is the product this collection had never measured.

The product that is not the third

Every distortion result in this field is odd-order, and the two-tone machinery has computed the second-order product on every call since the day it was written and thrown it away. On a bare exponential it is the drive over twice the thermal voltage — 1.934 × 10⁻² of the fundamental at a millivolt, against 1.870 × 10⁻⁴ for the third-order product, a ratio of 4Vₜ/a and a hundred and three to one. A differential pair puts it at 6.2 × 10⁻¹⁶. And degeneration removes it as D⁻² where it removes the third order as D⁴/|3 − 2D|, so a stage linearised until its third-order product is negligible is a stage whose distortion is almost entirely the one nobody measured.

semiconductors · Distortion 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 rangeTotal harmonic distortionCurrent mirrorDesign tradeoffLocal feedbackClosed formConvergence orderDevice matchingDevice mismatchEarly effectEven harmonic cancellationGain compression

All concepts