Concept

Even-order distortion — where it appears

Harmonics at twice, four times and other even multiples of a signal's frequency, produced by any curve that is not symmetric about its operating point. A balanced stage cancels them by symmetry, so how much returns is a measure of imbalance — a mismatch, an offset or an unequal load.

Named by 4 essays across 2 fields — each of them below, with the objects they name alongside it.

What a pair does to the three boundaries: removes two, moves one. computed by solving, not by drawing, with every harmonic taken from a transform of the waveform rather than from a series. A differential pair's transfer is an odd function, so the mean and every even harmonic are zero — -3.2e-17 and 9.1e-17 at a drive of two thermal voltages, which is absent rather than small. Two of the three one-per-cent boundaries therefore do not exist for a pair at all. The third, the loss of gain, moves by √2 in the limit of a tight criterion and by 1.42610 at the one per cent quoted here — 10.42 mV against 7.304 mV at 27 °C, against √2 = 1.41421 and 1.41433 at a criterion a hundred times tighter. So a pair reached for as headroom has bought forty per cent of it and a pair reached for as linearity has bought something else entirely. Twenty millivolts of imbalance brings the even orders straight back, which is what says the cancellation belongs to the symmetry rather than to the topology.

Two boundaries removed, and one moved

A differential pair's transfer is odd, so its mean and every even harmonic are zero — −3.2×10⁻¹⁷ and 9.1×10⁻¹⁷ at a drive of two thermal voltages, absent rather than small. Two of the three one-per-cent boundaries therefore do not exist for a pair at all. The third, the loss of gain, moves by √2 in the limit and by 1.42610 at the one per cent usually quoted: 10.42 mV against 7.304. So a pair reached for as headroom has bought forty per cent of it, and twenty millivolts of imbalance brings the even orders straight back.

limits · Small-signal $At gₘRₑ = 1 a pair holds its gain to one per cent up to 29.2 mV, a single transistor to 29.9 mV. computed by solving, not by drawing, every point a waveform solved sample by sample and its fundamental read from its transform. The fractional departure of the fundamental from the small-signal gain against the peak drive, for a differential pair (solid) and a single transistor (dashed), each device with an emitter resistor of 1 times its own 1/gₘ at 27 °C. The pair reaches one per cent at 29.2 mV; the cubic law 0.4·Vₜ·(1 + gₘRₑ)^(3/2) says 29.3 mV, which holds while the drive leaves most of the tail unsteered and overstates the boundary beyond; the single transistor at 29.9 mV, a ratio of 0.974. Without degeneration the ratio was √2 in the limit; with heavy degeneration it approaches two; and near gₘRₑ = ½, where the single transistor's cubic vanishes, the single transistor briefly holds its gain longer.$

Symmetry and a resistor together

A pair linearises by symmetry and an emitter resistor by feedback, and the two are normally used together. Solved together, the pair's one-per-cent gain boundary is 0.4·Vₜ times (1 + gₘRₑ) to the three-halves power up to gₘRₑ = 3: 10.4 mV bare, 29.2 mV at gₘRₑ = 1, 80.9 at 3. The √2 by which it outlasted a single transistor does not survive: at gₘRₑ = 1 the two are within three per cent, near ½ the single transistor outlasts the pair fourfold, because its cubic vanishes there and a pair's never can, and with heavy degeneration the ratio climbs to 1.95 at gₘRₑ = 50 — two, for a reason that has nothing to do with junctions: a single transistor cuts off when the drive reaches Ic·Re, and a pair has steered its whole tail at 2·Ic·Re, its two emitter resistors in series. What symmetry alone buys is the even orders, at every degeneration.

limits · Small-signal

With 1 mV of offset the pair's second harmonic returns in proportion to the drive, and the third harmonic reaches one per cent first. computed by solving, not by drawing, every harmonic read from a solved waveform. A differential pair at 27 °C with 1 mV of input offset: its second harmonic (solid) and third (dashed) over its fundamental, against the peak drive, beside a single transistor's second harmonic (dotted). The third is the balanced pair's own, u²/12, and reaches one per cent at 17.9 mV whatever the offset. The second is (u/2)·tanh(Vos/2Vₜ) at small drive, the single transistor's second harmonic times tanh(Vos/2Vₜ) = 1.93e-2, and never reaches one per cent: it peaks at 0.74% as the pair begins to saturate and falls again.

The offset that brings the even orders back

A balanced pair has no second harmonic. An offset gives it one, and the closed form is exact: the second harmonic over the fundamental is (u/2)·tanh(Vos/2Vₜ), with u the drive over 2Vₜ — a single transistor's second harmonic times the fraction of the tail the offset steers aside, 1.93% of it at a millivolt, not the twentieth usually estimated. It is first order in the offset up to 9.0 mV. It does not grow with the drive for ever: it peaks as the pair saturates and falls again, so below 1.36 mV of offset it never reaches one per cent at any drive, and above 3.14 mV it reaches one per cent before the pair's own third harmonic does, at 17.9 mV. Read the other way, a matching specification is a distortion specification: −60 dB of second harmonic at 10 mV of drive allows 0.535 mV of offset.

limits · Small-signal

A 1 MHz sine of 1 V into an emitter follower with 1 nF on its emitter and a 2 mA sink: the input falls at up to 6.28 V/µs, the output can fall at only 2.0, and the transistor switches off. Marched, the base driven by a stiff source and the emitter current Is·exp((vᵢₙ − vₒᵤₜ)/Vₜ). The input (dashed) and the output (solid) over the last of four cycles. The input's steepest fall is 2πfA = 6.28 V/µs; the load can be discharged no faster than the sink allows, I/C = 2.00 V/µs. On every falling stretch the base–emitter junction reverse-biases, the emitter current drops below a hundredth of the sink's (dots), and the output falls as a straight line of slope −2.000 V/µs until the input comes back up to meet it. On the rising half the transistor pushes whatever current the capacitor needs and the output follows a Vbe below the input.

The follower that lets go on the way down

An emitter follower can push current into its load without limit and pull it out only through whatever sinks the emitter. Marched as a large-signal circuit — a 2 mA sink, 1 nF on the emitter — a 1 V, 1 MHz sine asks the output to fall at up to 6.28 V/µs, the sink allows 2.00, and on every falling stretch the junction reverse-biases, the emitter current drops below a hundredth of its quiescent value for 57% of the cycle, and the output falls as a straight line of exactly −I/C. The boundary is a slope, 2πfA ≤ I/C: 2.87 V at 100 kHz, 0.309 V at 1 MHz, creeping above the line only as the period approaches the junction's own 12.9 ns. Past it the follower rectifies — its output's average rises 415 mV at 1 V of drive and 1.30 V at 2 V — and with a resistor in place of the sink the fall slows as the output descends, from 2.41 V/µs to 1.99. None of it is visible to the small-signal model the earlier essays used.

semiconductors · Emitter follower

Named alongside it

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

Small-signal model Differential pair Thermal voltage Validity region Capacitive load Emitter degeneration Emitter follower Input offset Large-signal Marching Model range Odd symmetry

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