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

Differential pair — where it appears

Two matched devices sharing a tail current, whose output depends on the difference of their inputs and, to the extent they match, not on what is common to both. The symmetry removes every even harmonic exactly, so its leading distortion term is the third and its two amplitude edges come in the opposite order to a single device's.

Named by 10 essays across 4 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 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 The step at which the output impedance stops being a number. computed by solving, not by drawing. The excursion divided by the step, against the step. The flat line is the linear model, and it is flat to 0.0 parts per million across four decades — which is what an impedance is. The rising curve is the same netlist with the differential pair's tanh in the transconductor, and it leaves at 10.6 mA: the input error there is 3.63 thermal voltages, so the boundary is an amplitude in the pair's own units rather than a current with the amplifier's name on it. At 300 mA the ratio is 54.5 Ω against the linear 23.6 — 131 per cent, and it is no longer a property of the circuit at all. The slew rate that decides it is 3.25 V/µs, which is twice the thermal voltage times the gain-bandwidth in radians, and contains no design choice.

The step too large to have an impedance

The rung below drove the load node with a current step and reported an impedance: a voltage divided by a current, which is a number only if the ratio does not depend on the current. Give the amplifier the differential pair's own tanh in place of a linear transconductor and it is a number up to 10.6 milliamps and not above — where the input error is 3.63 thermal voltages, and where the slew rate that decides it is twice the thermal voltage times the gain-bandwidth in radians, containing no design choice at all.

feedback · Capacitive load 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 Two coupling capacitors 1% apart reject common mode by only 46.0 dB at their corner, 0.159 Hz — and perfectly at direct current and far above it. computed by solving, not by drawing at 281 frequencies: one common-mode source driving two coupling networks, each 1 µF into 1 MΩ, with the capacitors 1% apart. The difference between the two outputs, per volt of common mode, agrees with the difference of the two first-order high-passes to 6.9e-14. It is zero at direct current, where both block, and zero far above the corner, where both pass; between, it peaks at the corner, 0.159 Hz, at 5.000e-3 — half the mismatch — which is 46.0 dB of rejection. Behind a 90 dB amplifier (grey), the coupling network decides the rejection from 0.000503 Hz to 50.3 Hz.

The rejection two capacitors lose at their corner

AC-couple both inputs of a differential instrument and the two coupling networks become part of its common-mode path. Two 1 µF capacitors one per cent apart turn common mode into difference at exactly half their mismatch at their shared corner — 46.0 dB of rejection at 0.159 Hz, from a pair that is perfect at direct current and far above. Behind a 90 dB amplifier they decide the instrument's rejection over five decades around that corner. A 10 V common-mode step arrives as a 36.8 mV differential pulse one time constant later. Four hundred pairs of 20 per cent electrolytics reject a median 25 dB. Coupling the difference instead of each side removes the dip and hands the common mode to the amplifier.

instruments · AC coupling 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 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 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

Named alongside it

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

Thermal voltageSmall-signal modelDevice mismatchEmitter degenerationEven harmonic cancellationEven-order distortionInput offsetValidity regionComponent toleranceDevice matchingOdd symmetryOffset drift

All concepts