Concept

Even harmonic cancellation — where it appears

The disappearance of the even harmonics from a transfer curve that is odd about its operating point, as a balanced pair's is. It is exact to the arithmetic rather than approximate, so what limits a differential pair is its third harmonic and not its second.

Named by 7 essays across 3 fields — each of them below, with the objects they name alongside it.

One exponential and one pair, both driven 20.0 mV. computed by solving, not by drawing. The pair's characteristic is odd, so its even harmonics vanish: the second comes out at 1.5e-16 of the fundamental against 18.88% for the single stage. It is not a small residue but the floor of the arithmetic. The price is the third harmonic, 1.202% against 2.404%, and total distortion of 1.202% against 19.03%.

What a pair cancels, and what it only halves

A differential pair's transfer characteristic is an odd function, and an odd function driven symmetrically produces no even harmonics at all. Measured, the second harmonic comes out at 10⁻¹⁶ of the fundamental — the arithmetic's own floor, not a small physical residue. The third harmonic is a different story, and it comes out at exactly half the single stage's, which is a reduction and not a cancellation.

semiconductors · Differential pair
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
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 a third-order account never measures.

The product that is not the third

Every distortion result in this field is odd-order, and every two-tone calculation behind those results computed the second-order product and threw 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
Degenerated to gₘRₑ = 1, a pair with 1 mV of offset peaks at 0.41% of second harmonic, against 0.74% bare. computed by solving, not by drawing, every harmonic read from a solved waveform. The second (solid) and third (dotted) harmonics, as fractions of the fundamental, of a differential pair with 1 mV of input offset and emitter degeneration gₘRₑ = 1, against the peak drive; the same pair undegenerated in grey (dashed: second; dot-dash: third). Degeneration lowers both and moves both to larger drives. At small drive the second harmonic follows (u/4)·g″/g′² at the offset's operating point, which for a small offset is (u/2)·(Vos/2Vₜ)/(1 + gₘRₑ)³. Its largest value over all drives is 0.406 per cent at 177 mV, against 0.736 per cent at 76 mV undegenerated.

Two decibels of matching for one of gain

An offset gives a differential pair back the second harmonic its symmetry removed. Put resistors in the emitters and that second harmonic falls as the cube of one plus gₘRₑ at a fixed input and as its square at a fixed output — measured slopes of −3.00 and −2.00 — so each decibel of gain given up buys two decibels of distortion, or two decibels more offset at the same distortion. A matching specification that allowed 0.535 mV for −60 dB at 10 mV of output allows 2.14 mV at gₘRₑ = 1. The worst case over all drives relaxes more slowly: the offset below which the second harmonic can never reach one per cent goes from 1.36 mV to 4.39 mV as gₘRₑ goes to three, because the peak it limits sits where the pair saturates, and degeneration only moves that further out.

limits · Small-signal
The second harmonic in each cycle of a transformer's inrush, as a share of the fundamental. computed by solving, not by drawing, the harmonics taken per cycle of the marched current. The first cycle carries 79% second harmonic and the second 93%. The share falls below the 15% a differential relay is commonly set to restrain at only in cycle 30, when the peak is 205 mA against a first peak of 88 A — the harmonic outlasts the inrush it identifies.

The harmonic that names an inrush

A transformer's differential relay sees an inrush as current going in and nothing coming out, which is what an internal fault looks like, and it tells them apart by the second harmonic. Marched cycle by cycle, an inrush starts at 79% second harmonic and stays above the usual 15% restraint for thirty cycles — long after its current has fallen from 88 A to 205 mA, because the current unwinds in a few cycles and the flux offset under it takes seconds. A fault's offset is a smooth exponential and peaks at 14.6% only when X/R is as low as three.

power · Inrush
At a fixed output a 0.5 mV junction mismatch's second harmonic falls as (1 + gₘRₑ)⁻² and a 1% resistor mismatch's as (1 + gₘRₑ)⁻¹: past gₘRₑ = 1.93 the resistors set it. computed by solving, not by drawing, every waveform solved arm by arm in currents. A pair on a 1 mA tail, its output held at what the bare, balanced pair gives for 10 mV of peak drive; the second harmonic over the fundamental from a 0.5 mV VBE mismatch alone (dashed), from a 1% mismatch between the emitter resistors alone (dotted), and from both in the same sense (solid), against one plus the degeneration. Between gₘRₑ = 5 and 20 the junction's falls with slope −2.00 and the resistors' with −0.89, still approaching −1, since the local slope of gₘRₑ/(1 + gₘRₑ)² is 1/gₘRₑ − 1. The resistors' mismatch is an input offset of Vₜ·gₘRₑ·δ, which grows with the degeneration that divides it, so the two cross where gₘRₑ = ΔVBE/(Vₜδ) = 1.93; beyond it each decibel of gain given up buys one decibel of distortion, not two.

The resistor mismatch that is only an offset

Two emitter resistors that differ by δ unbalance a degenerated pair, and with an ideal tail they do it in the simplest way possible: the arm currents always sum to the tail, so the mismatch adds the constant (Iₜ/2)·R·δ = Vₜ·gₘRₑ·δ to the input and nothing else. Solved arm by arm, a 1% mismatch at gₘRₑ = 3 is the balanced pair moved by 0.776 mV to 4 × 10⁻¹⁶ of the tail, and a trim of that size leaves no second harmonic at any drive. What separates it from a junction's offset is that it grows with the degeneration meant to divide it: at a fixed output its second harmonic falls as gₘRₑ/(1 + gₘRₑ)², one decibel per decibel of gain rather than two, and it takes over from 0.5 mV of base–emitter mismatch at gₘRₑ = 1.93. For −80 dB at 10 mV of output the resistors must match to 0.823% at gₘRₑ = 1 even with perfect transistors.

limits · Small-signal
An exponential stage at 20% output, with a loop gain of 1: second 2.5% from 4.97%, third 0.0415% from 0.165%, and fifth 0.68 ppm from 0.814 ppm. computed by solving, not by drawing: the harmonics of an exponential stage with emitter resistance, gm·Re = T, solved point by point on the implicit curve and transformed, at a fixed output fundamental of a fifth of the quiescent current, for T = 1 (dots a) against no feedback (dots b). 2: 2.5%, against 4.97%; 3: 0.0415%, against 0.165%; 4: 5.36 ppm, against 40.9 ppm; 5: 0.68 ppm, against 0.814 ppm; 6: 0.0202 ppm, against 0.0135 ppm; 7: 0.000245 ppm, against 0.000192 ppm; 8: 0.0000478 ppm, against 0.00000239 ppm; 9: 0.00000178 ppm, against 2.62e-8 ppm. The series gives the second and third at small drive: the implicit function's Taylor coefficients are 1/(1 + T), 1/2(1 + T)³ and (1 − 2T)/6(1 + T)⁵, and they agree. The third therefore has a null at T = ½, and past that it has the opposite sign, and feedback that reduces the low orders puts some of them back into the high ones.

The fifth a resistor adds

Local feedback's reputation is that it reduces distortion, and on the second order it does: 4.97 per cent falls to 2.5 at a loop gain of one and 0.46 at ten, with the output held fixed. The third has a null at a loop gain of exactly a half and comes back with its sign changed. And from the fifth upward, a harmonic rises above its open-loop value before it falls: the fifth 1.42 times, the sixth 5.23, the seventh 24.6, the ninth 960, peaking at a loop gain between six and twenty. The second order, carried back through the resistor, mixes with the signal and makes them. A total-harmonic reading shows only improvement throughout, because the second dominates it. A differential pair has no second order to carry back, and its fifth never passes its open-loop value.

semiconductors · Distortion

Named alongside it

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

Emitter degenerationDifferential pairModel rangeDesign tradeoffDevice mismatchGain compressionNonlinearityOffset voltageSmall-signal modelTotal harmonic distortionTransfer curveComponent tolerance

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