Devices, and the amplitude they stop being linear at

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.

Two devices sharing a tail current, driven in opposition, produce an output current proportional to tanh(v/2V_T). That characteristic is the whole subject of this page, and one property of it does all the work: it is an odd function.

An odd function has no even terms in its power series. Drive it with a sinusoid and no even harmonic can appear in the output, because there is nothing in the series that could produce one. The cancellation is structural — a consequence of the shape of the curve rather than of matching, trimming or biasing — and that is a much stronger kind of statement than any number.

One exponential and one pair, both driven 20.0 mVcomputed 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%.-101-0.100-0.05000.0500.100differential drive (volts)output, normalisedthe pair: odd, and it saturatesone exponentialharmonics, as a fraction of the fundamentalsingle, h218.876%pair, h21e-16single, h32.404%pair, h31.202%single, h40.231%pair, h46e-17single, h50.018%pair, h50.017%solved, then checked — evenness measured, not assumedthe pair's second harmonic is 1e-16
Fig. 1 An exponential and a differential pair, driven twenty millivolts either side of their bias. The pair saturates gracefully in both directions; the single stage runs away in one and flattens in the other. The bars are the harmonic content of each, and the pair’s even harmonics are at 10⁻¹⁶ — which is not a small number but the floor of double-precision arithmetic. The slider is the drive.

Zero, and how it is different from small

The measured second harmonic of the pair is 1.5×10⁻¹⁶ of its fundamental at twenty millivolts of drive, and 4.1×10⁻¹⁷ at fifty. The single stage’s is 9.6% and 42.2%.

The pair’s numbers are not measurements of a physical quantity. They are the residue of summing five hundred and twelve floating-point samples, and they move about with the drive in a way that has no pattern because there is nothing underneath them to have a pattern. That is worth saying plainly, because a figure that reported “second harmonic: 1.5×10⁻¹⁶” without comment would invite the question of what physical effect produces 10⁻¹⁶ of second harmonic, and the answer is none.

The distinction between small and zero is the reason this topology is used in almost every amplifier input stage ever built. A small second harmonic could be made smaller by driving the stage less hard, and would have to be re-checked every time anything changed. A zero one is zero at every drive, every bias current and every temperature, because it follows from the symmetry of the circuit rather than from the values in it.

What is not guaranteed is that a real pair achieves it. The cancellation requires the two halves to be identical, and two transistors on a real die are not. A mismatch in saturation current appears as an offset in the characteristic, the characteristic is then no longer odd about the drive’s zero, and the even harmonics reappear in proportion to the mismatch. That is a genuine and well-known effect, and it is worth being clear that this page does not measure it: the model here has two identical halves, so what is being shown is the ceiling that matching approaches, not what a real pair delivers.

The third harmonic, and the factor of exactly two

The interesting result is not the even harmonics — they are zero and that is the end of it — but what happens to the odd ones.

Measured at the same drive, the pair’s third harmonic is exactly half the single stage’s. Not approximately: at two, five, ten and twenty millivolts the ratio is 0.500000 to six figures. At forty millivolts it is 0.499920, at eighty 0.497686, and at a hundred and fifty 0.480333.

An exponential driven 10.0 mV either side of its biascomputed by solving, not by drawing. A sinusoid in, and out comes a waveform whose peaks are taller than its troughs are deep. The second harmonic is 9.61% of the fundamental, measured by transforming 512 samples and predicted independently as I₂(0.387)/I₁(0.387) = 9.61%. The two routes agree to 5e-10 over the 5 harmonics that stand above the arithmetic's own floor, and share nothing but the amplitude.the drive: a sinusoidthe current out, and a symmetric one for comparisonone cyclemeasured, against the Bessel ratioharmonic 29.61%harmonic 30.62%harmonic 40.03%harmonic 51.2e-5harmonic 63.7e-7agreement: 5e-10 relativesolved, then checked — a transform against a seriessecond harmonic 9.6% at 10.0 mV
Fig. 2 The single stage’s harmonics, measured and predicted. The third harmonic of an exponential is I₃(x)/I₁(x) exactly, and it is that number the pair halves. Both routes to it agree to better than 10⁻⁷ relative wherever the harmonic stands above the arithmetic’s floor.

Where the half comes from is visible in the two series. The exponential’s cubic coefficient relative to its linear one is 1/6V_T², since every derivative of exp(v/V_T) is the function over a power of V_T. The hyperbolic tangent’s expansion is tanh u = uu³/3 + …, and with u = v/2V_T the cubic coefficient relative to the linear one is 1/12V_T². Half.

The departure at large drive is equally explicable: the halving is a statement about the leading cubic terms, and once the fifth-order terms contribute, the two curves stop being related that simply. The measured ratio holds to six figures up to about a thermal voltage of drive and then drifts, which is exactly the range in which a third-order truncation is a good description of either curve.

So the pair’s benefit is asymmetric in a way the usual summary — “a differential pair cancels even harmonics” — states correctly and incompletely. Even harmonics: gone. Odd harmonics: halved, until the drive is large, and then less than halved.

Where the tail current goes

The hyperbolic tangent is bounded, and that boundedness is a physical statement rather than a mathematical convenience: the two devices share a fixed tail current, so the most either of them can carry is all of it.

The measured split is worth setting out because it is much steeper than most readers expect. At ten millivolts of differential drive one side carries 59.6% of the tail and the other 40.4%, a difference of 19.1%. At twenty-six millivolts — one thermal voltage — the difference is 46.4%. At fifty it is 74.7%, and by a hundred and thirty-seven millivolts one device carries 99.4% and the other has effectively switched off.

A hundred and thirty-seven millivolts is 5.29 thermal voltages, and it is the whole dynamic range of the pair as an amplifier. Below about a tenth of that it is linear enough to be useful; above it, the pair is a switch. That dual character is not a defect — the same circuit is the input stage of every operational amplifier and the core of every emitter-coupled logic gate, and which of the two it is depends only on how hard it is driven.

The saturation also explains why the pair’s gain boundary arrives so early relative to its distortion. A bounded odd function must flatten, and flattening is gain compression; the compression begins as soon as the curvature does, while the third harmonic it produces stays small because the curve is symmetric about the origin. Hence 10.4 mV for one per cent of gain error and 18.2 mV for one per cent of distortion.

The fifth harmonic, and how far down the series goes

The pair’s odd harmonics fall away quickly, and how quickly is a useful measure of how well behaved the characteristic is.

At five millivolts of drive the third harmonic is 7.775×10⁻⁴ of the fundamental and the fifth is 7.257×10⁻⁷ — three decades further down. At twenty millivolts the third is 1.202×10⁻² and the fifth 1.745×10⁻⁴, now only two decades down. The gap narrows as the drive grows, which is the series becoming a worse and worse approximation of itself, and it narrows fast enough that by a hundred millivolts the higher harmonics are no longer negligible in any sense.

The useful consequence is that a total-harmonic-distortion figure for this device is very nearly its third-harmonic figure over the whole range where the device is being used as an amplifier: 1.2022×10⁻² of total against 1.20204×10⁻² of third harmonic at twenty millivolts. Anything that reduces the third harmonic reduces the total, and anything that claims to have reduced the total without touching the third has measured something else.

A diode fed from 5 V through 1.0 kΩcomputed by solving, not by drawing. The operating point is where the exponential meets the load line: 0.692544 V and 4.3075 mA, reached in 13 damped Newton steps from a cold start. The one-line Newton on Vs = v + R·i(v), which touches no matrix, gives 0.692544 V. The "drop" is not a constant: it moves about 60 mV per decade of current.02.5057.5000.2000.4000.6000.800voltage across the diode (volts)current (milliamperes)load line: (5 V − v)/1.0 kΩ123450.6925 V, 4.307 mAthe circuit5 V1.0ksolved, then checked — two Newtons, no shared arithmeticthe drop moves 60 mV per decade
Fig. 3 An operating point found by Newton on the netlist. A differential pair has one too — the tail current sets it, and everything on this page is a property of the curve at it. The pair’s special property is that its curve is odd about that point by construction rather than by choice, which is what the whole essay turns on.

The boundaries change places

The previous essay measured two amplitude boundaries for a single exponential and found them seven times apart, with distortion arriving first: one per cent of harmonic content at 1.03 mV and one per cent of gain error at 7.30 mV.

Run the same measurement on the pair and the order reverses.

A differential pair: where its distortion and its gain give waycomputed by solving, not by drawing at 61 amplitudes. Total harmonic distortion reaches one per cent at 18.2 mV and the gain falls one per cent short of its small-signal value at 10.4 mV. They are 0.6 times apart, and the reason is that a second harmonic is first order in the drive while a gain error is second order.-5-4-3-2-101e-41m10mdrive amplitude (volts)log₁₀ of the errorharmonic distortiongain errorone per cent1% distortion at 18.2 mV1% gain error at 10.4 mVsolved, then checked — both edges bisected0.6× apart
Fig. 4 The differential pair’s two boundaries. One per cent of distortion arrives at 18.2 mV — eighteen times further up than the single stage — but one per cent of gain error arrives at 10.4 mV, which is earlier. For this topology the useful range ends with gain compression rather than with harmonic content, which is the opposite of the single stage and follows from the same fact that removed the even harmonics.

One per cent of total harmonic distortion at 18.18 mV. One per cent of gain error at 10.41 mV. The ratio is 0.572, where the single stage’s was 7.06.

The reason is the missing second harmonic. For the single stage, the distortion at small drive is almost entirely second harmonic, which is first order in the drive and therefore arrives very early. Remove it and the leading distortion term is the third harmonic, which is second order in the drive — the same order as the gain error. Two quantities of the same order arrive at comparable amplitudes, and which of them arrives first is then a matter of the coefficients rather than of the orders.

That is a satisfying place for the argument to land. The factor of seven in the previous essay was presented as a consequence of one quantity being first order and the other second; here is a device where the first-order quantity is structurally absent, and the factor of seven duly disappears.

One exponential and one pair, both driven 80.0 mVcomputed by solving, not by drawing. The pair's characteristic is odd, so its even harmonics vanish: the second comes out at 1.1e-16 of the fundamental against 57.81% for the single stage. It is not a small residue but the floor of the arithmetic. The price is the third harmonic, 12.580% against 25.277%, and total distortion of 12.747% against 63.76%.-101-0.100-0.05000.0500.100differential drive (volts)output, normalisedthe pair: odd, and it saturatesone exponentialharmonics, as a fraction of the fundamentalsingle, h257.808%pair, h21e-16single, h325.277%pair, h312.580%single, h48.798%pair, h41e-16single, h52.531%pair, h52.056%solved, then checked — evenness measured, not assumedthe pair's second harmonic is 1e-16
Fig. 5 The same comparison at eighty millivolts, which is well outside anything either device should be asked to do. The pair has flattened at both ends — it is a hyperbolic tangent, and it saturates at the tail current — while the exponential has run away upward. The pair’s even harmonics are still at the arithmetic’s floor, which is the point: the cancellation does not degrade with drive, it simply becomes less relevant beside the odd harmonics that do.

What breaks the symmetry, and how gently

The cancellation is exact for an odd characteristic driven symmetrically, and both halves of that sentence are conditions that a real circuit meets approximately.

The characteristic must be odd about the drive’s zero. Two devices with different saturation currents have a characteristic that is odd about some other point, so a drive centred on zero sees a curve with a quadratic term. The second harmonic that reappears is proportional to the offset for small offsets, which is the gentlest possible failure: a one-millivolt input offset on a stage driven at twenty millivolts restores roughly a twentieth of the second harmonic the single stage would have had, not all of it.

The drive must be symmetric. A pair driven on one input with the other held fixed is a single-ended drive, and the tail node then moves — which is a common-mode excursion, and the characteristic seen from one input alone is not odd. In practice the tail’s own impedance decides how much it moves, and a good current source keeps it small enough that the pair behaves as though both inputs were driven. That is another reason the tail source’s output resistance appears in the list of costs below.

Neither failure is catastrophic and both are proportional, which is a large part of why the topology is robust enough to be universal. The symmetry does not have to be perfect to be worth most of what it promises, and the amount it delivers is proportional to how well it is kept.

Two devices, and what the second one costs

Nothing in this collection is free, and the pair’s price is worth stating.

Half the transconductance. The tail current splits between two devices, so each carries half of it, and the differential transconductance of the pair is g_m/2 where g_m is what one device carrying the whole tail current would have had. A pair therefore has half the gain of a single stage at the same total current, and the eighteen-fold improvement in linearity is bought partly with that.

A tail current source that has to be good. The cancellation assumes the tail is constant. A tail that moves with the common-mode input turns common-mode into differential and destroys the symmetry, which is why the output resistance of the tail source is one of the numbers that decides a real pair’s common-mode rejection.

Twice the input-referred noise power. Two devices contribute noise and only their difference is signal, so the noise voltage referred to the input rises by √2. This is the boundary the noise field is about, and it is worth flagging here because it is the one cost that gets worse rather than better as everything else is improved: every technique on this page raises the amplitude at which the stage stops being linear, and none of them lowers the amplitude at which it stops being able to see anything.

Those three together are the reason a differential pair is a choice rather than an obvious improvement. Half the gain, a component that has to be good, and √2 more noise, in exchange for the exact removal of one harmonic and the halving of another. For an amplifier that will be wrapped in feedback, the trade is overwhelmingly worth it — the loop gain lost is recovered by adding a stage, and the linearity gained is not recoverable any other way. For a single stage that has to do everything itself, it is a genuine decision.

A common-emitter stage with 2.0 pF from collector to basecomputed by solving, not by drawing. The stage's midband gain is 144.7 and its −3 dB point is at 503 kHz. The Miller approximation lumps 311 pF at the input and predicts 643 kHz — 21.7% high. The network's second pole is at 336 MHz and its right-half-plane zero at 3.08 GHz, both of which the approximation has no room for.-20020401001k10k100k1M10M100M1G10Gfrequency (hertz)gain (decibels)midband 43.2 dBsolved: 503 kHzMiller says 643 kHzsecond pole 336 MHzzero at g_m/C_μsolved, then checked — two networks, one measurementMiller is 22% optimistic
Fig. 6 What a device’s own capacitance does to a stage’s bandwidth, from the next essay. A pair does not escape any of it: each half has the same internal capacitances and the same Miller multiplication of the collector-base one, so the linearity argument on this page and the bandwidth argument on the next are independent and both apply.

The general shape

The pattern this essay is an instance of is worth naming, because it recurs well beyond electronics.

A symmetry in a system removes a whole class of terms exactly, rather than making them small. That is qualitatively different from a numerical improvement and it behaves differently: it does not degrade with amplitude, it does not need re-checking when a parameter moves, and it fails only when the symmetry itself is broken — at which point it fails in proportion to the breaking rather than gradually.

The measured consequence here is a second harmonic of 10⁻¹⁶ where there was one of 10⁻¹, achieved by adding a device rather than by reducing a drive. What it does not achieve is anything at all for the odd harmonics beyond a factor of two, and a design that treats “differential” as a synonym for “linear” has assumed the second half of a result whose first half is exact.

The discipline it asks for is specific and is worth carrying into any claim of this shape. Establish what the symmetry is, in the form of a property of a function rather than a description of a circuit. Then check that what it removes is removed to the arithmetic’s floor rather than merely made small, because a symmetry argument that produces 10⁻³ has an error in it somewhere and a numerical argument that produces 10⁻¹⁶ is being flattered by a coincidence. And then say what the symmetry does not remove, which in this case is everything of odd order — half of it, and only while the drive is small.

There is a version of this essay that stops after the first of those three, and it is the version most often written. It is not wrong about anything it says.

Where four of this site's models stop being trueIn order: the ideal operational amplifier at 1.42 kHz, a 10 V output at full amplitude at 7.96 kHz, Kirchhoff's laws on 10.0 cm at 3.97 MHz, the ideal 100 nF capacitor at 4.69 MHz. The fifth boundary is an amplitude rather than a frequency and cannot share this axis: a small-signal model is 1% wrong above 7.3 mV, at every frequency there is.101001k10k100k1M10M100M1G10Gfrequency (hertz)the ideal operational amplifier1.42 kHz — a gain of 100 from a 1 MHz part is 1% low herea 10 V output at full amplitude7.96 kHz — above this the output cannot move fast enoughthe ideal 100 nF capacitor4.69 MHz — 1.2 nH of lead makes it 10% wrong hereKirchhoff's laws on 10.0 cm3.97 MHz — the board is one degree long hereeach bar is where the model may be used; the rule at its end is the numbersolved, then checked — each boundary from its own modeland one that is not a frequency: 7.3 mV
Fig. 7 The frequency boundaries the collection has measured. The pair’s boundaries are not on it, because they are amplitudes: 18.2 mV for one per cent of distortion, 10.4 mV for one per cent of gain error, and 137 mV for the point at which one device has taken 99.4% of the tail and the circuit has stopped being an amplifier at all. Three numbers, all in millivolts, for a device usually described as “large signal” above a volt.