Devices, and the amplitude they stop being linear at

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.

Assumes: The distortion a linear model cannot have · How small is small signal

The point the device is never at drove three transfer curves with two equal tones, fitted the one-for-one and three-for-one asymptotes, and found the third-order intercept sitting 7.84 decibels above the largest output a differential pair can produce. It is the sharpest linearity result in this field and it is entirely about the third order.

So is everything else here. The rung below it measures the third harmonic; the degeneration ladder reverts a series and quotes the third-order coefficient; the distortion edge is a third-order statement with a second-harmonic target attached. Odd order, every one.

The machinery that produced them has been computing the second-order product on every call since the day it was written, and throwing it away. Two tones at $f_1$ and $f_2$ through a curve with a squared term make components at $f_2 - f_1$ and $f_2 + f_1$ , and those have never been read.

A pair's third-order intercept is 7.8 dB above anything it can produce. computed by solving, not by drawing. Two equal tones through a differential pair, transformed coherently so every product lands in a bin of its own. The fundamental rises with slope 1.000 and the third-order product with slope 3.000, both fitted over the decade marked, and the dashed extensions are the extrapolation a specification quotes. They meet at a drive of 4.00 thermal voltages and an output of 2.00 — against a largest output of 0.8108, which is 8/π² and is what two equal tones give through a limiter. The intercept is 7.84 dB above it, which is π²/4 exactly. — Fig. 1 The rung below, recalled: two equal tones through a differential pair, the fundamental fitted at slope 1.000 and the third-order product at slope 3.000, and the extrapolated intercept at 4.00 thermal voltages against a largest possible output of 0.8108. Every number in it is odd-order.

Where the second-order product lands, and why it was skipped

The reason the third order gets the attention is a good one and it is worth stating before taking it apart. Third-order products sit at $2f_1 - f_2$ and $2f_2 - f_1$ , which for two closely spaced tones are beside them — inside whatever band the signal occupies, at a spacing no filter can reach. Second-order products sit at the sum and the difference, which for two tones near $f$ are near $0$ and near $2f$ : an octave away in one direction and at baseband in the other.

For a narrowband receiver that is the end of the argument, and it is why the specification everyone quotes is a third-order one. It stops being the end of the argument the moment the signal path is more than an octave wide. A baseband amplifier, a video chain, a direct-conversion mixer’s output, an instrumentation front end and a photodiode transimpedance stage all have the difference product landing squarely in the middle of what they are meant to pass, and none of them has anywhere to put a filter that would remove it.

What it actually is, on the field’s own device

The second-order product of a bare exponential, which nothing here has ever read. computed by solving, not by drawing. Three products of two equal tones through a bare exponential, against the drive. The tones sit at bins 60 and 65 of one record, so the difference product at bin 55, the sum product at 70 and the second-order pair at 5 and 125 each land in a bin of their own. The second-order product is the largest of the three at every drive drawn: at 1 mV it is 1.93e-2 of the fundamental against 1.87e-4 for the third. The two are equal only at 103.4 mV, which is 4Vₜ — above every drive a small-signal model of this device survives. — Fig. 2 Three products of two equal tones through a bare exponential, against the drive. The tones sit at bins 60 and 65 of one record, so the difference product at bin 5, the sum at 125 and the third-order pair at 55 and 70 each land in a bin of their own and no window is needed. At a millivolt the second-order product is 1.934 × 10⁻² of the fundamental, the third 1.870 × 10⁻⁴ and the fifth 1.166 × 10⁻⁸. The two lower orders cross the second only at 103.4 millivolts, which is 4Vₜ.

The second-order product of an exponential driven by two equal tones is $a/2V_T$ , exactly, and the third is $(a/V_T)^2/8$ . Both are asserted against the transform in the figure and both hold to three parts in a thousand at a millivolt of drive. Their ratio is therefore $4V_T/a$ — no device parameter, no bias current, no fabrication detail, just the drive against the thermal voltage.

At a millivolt that ratio is 103. At a quarter of a millivolt it is 414. The second-order product of a single-ended exponential stage is two orders of magnitude larger than the third at any drive a small-signal model survives, and it becomes the smaller of the two only at 103.4 millivolts, where the third-order product is already 6.8 per cent of the fundamental and nothing about the stage is linear any more.

The same arithmetic gives the second-order intercept directly, because $a/2V_T$ reaches one at $a = 2V_T$ : 51.70 millivolts, against the third-order intercept’s $2\sqrt2 V_T$ of 73.12. The second-order intercept of a bare exponential is exactly $\sqrt2$ — 3.01 decibels — below its third-order intercept, at every bias current and every temperature, and that is the whole of what those two specification numbers say about each other.

The third thing that was computed and discarded

The fifth-order product was thrown away too, and it is the term that connects the two halves of this essay. At a millivolt of drive on a bare exponential it is 1.166 × 10⁻⁸ of the fundamental, which is a sixteen-thousandth of the third-order product and nothing a measurement would look for.

It is not negligible where it matters. The extrapolated intercept below is a straight line fitted to the third-order product, and what bends that line — the only thing that can bend it — is the fifth-order term underneath. On a differential pair it is 1.247 × 10⁻⁴ of the third-order product at a millivolt, 4.46 per cent of it at twenty, 14.5 per cent at forty-one and 31.3 per cent at eighty-three — and those last two are exactly the windows at which the extrapolation goes ten per cent and eighty-nine per cent wrong. So the term nobody measures is the term that decides how far the specification everyone does measure can be trusted, and both were on the same page of output the whole time.

And a one-tone measurement understates it, by a different factor

The distortion a linear model cannot have measures harmonics, one tone in. That measurement is related to this one and the relation is a counting argument rather than a scaling.

A cube of a sum of two cosines has three ways of producing $2\omega_1 - \omega_2$ where a cube of one cosine has one way of producing $3\omega$ , so a third-order intermodulation product is three times the third harmonic the same drive makes — a factor the rung below asserts, and 9.54 decibels. A square of a sum of two cosines has two ways of producing $\omega_2 - \omega_1$ where a square of one has one way of producing $2\omega$ . So the second-order product is twice the second harmonic, which is 6.02 decibels, and it is measured here at 1.99998 of it.

Two different understatements from one substitution is the useful form of that. A designer who measures harmonics on a bench and reasons about intermodulation is out by 9.5 decibels in one product and 6.0 in the other, in the same direction, and the two errors do not cancel anywhere.

What a pair does to it, and what it does not

A pair's second-order product is at the arithmetic's floor and its fifth is untouched. computed by solving, not by drawing. Three products of two equal tones through a differential pair, against the drive. The tones sit at bins 60 and 65 of one record, so the difference product at bin 55, the sum product at 70 and the second-order pair at 5 and 125 each land in a bin of their own. The second-order curve is not drawn low, it is drawn at the floor: an odd characteristic has no even terms, so the product is under 10⁻¹² of the fundamental at every drive below 20 mV. The third-order product is exactly half a single device's and the fifth is exactly equal to it, so the arrangement removes an order and halves the next and does nothing above that. — Fig. 3 The same three products through a differential pair. The second-order curve is not drawn low, it is drawn at the floor: 6.159 × 10⁻¹⁶ of the fundamental at a millivolt and 2.307 × 10⁻¹⁵ at twenty, which is the transform’s own arithmetic and not a residue. The third-order product is 9.349 × 10⁻⁵ against the single device’s 1.870 × 10⁻⁴, and the fifth is 1.165 × 10⁻⁸ against 1.166 × 10⁻⁸.

The differential pair’s characteristic is odd, so every even term in it is zero, so the second-order product is not reduced — it is absent. That is the arrangement’s whole reason for existing, and what a pair cancels, and what it only halves establishes it for harmonics of one tone. Arriving from the two-tone direction it says something that essay could not: the product a wideband stage cannot filter is gone at the arithmetic’s floor, and the product it cannot filter either — the third order — is merely halved.

One exponential and one pair, both driven 50.0 mV. computed by solving, not by drawing. The pair's characteristic is odd, so its even harmonics vanish: the second comes out at 4.1e-17 of the fundamental against 42.23% for the single stage. It is not a small residue but the floor of the arithmetic. The price is the third harmonic, 6.330% against 12.667%, and total distortion of 6.350% against 44.19%. — Fig. 4 The harmonic version of the same cancellation, from the essay that measured it: at fifty millivolts of drive the pair’s second harmonic is 4.1 × 10⁻¹⁷ of the fundamental against 42.23 per cent for the single stage, and the price is a third harmonic of 6.330 per cent against 12.667.

Halved, and no better than that. And the fifth-order product is not halved at all: it comes out equal to the single device’s to five figures, which is the honest limit of what the arrangement buys. A differential pair removes one order completely, divides the next by two, and does nothing whatever above that — a sentence with three different verbs in it, where the received summary has one.

The reason is one line of series algebra. Writing $x = a/V_T$ , an exponential’s odd part has coefficients $x/1! + x^3/3! + x^5/5!$ and a hyperbolic tangent of $x/2$ has $x/2 - x^3/24 + x^5/240$ . The linear terms are in the ratio two to one and the cubic terms in the ratio four to one, so the normalised third-order product is halved; the fifth-order terms are in the ratio two to one, the same as the linear terms, so the normalised fifth-order product is unchanged. Measured: 0.500 and 1.000.

What degeneration does to it, which is the finding

Local feedback is this field’s other linearisation, and the whole of what has been measured about it is third-order.

Fig. 5 Both intermodulation products of a degenerated stage 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 as D⁴/|3 − 2D| and collapses altogether at D = 1.5 where its coefficient changes sign. The ratio between them goes from 20.7 to 183 across the sweep.

The second-order coefficient of a degenerated exponential falls as $D^{-2}$ and nothing else — fitted over every factor above the null rather than asserted at one point, the exponent is −2.000. The third-order coefficient is $(3 - 2D)/D^4$ , which for large $D$ is $-2/D^3$ : one more power of the degeneration factor.

So the two products separate. At five millivolts a tone the second order is 20.7 times the third on a bare device and 183 times on one degenerated by sixteen. At one millivolt, where the third order is smaller and the second is not, the same numbers are 103.4 and 913.

That is the result this rung exists for, and it inverts what the anchor has been saying. Degeneration is presented — here and everywhere — as the thing that makes a stage linear, and every measurement that supports it is third-order. Where the two exponents come from reverts the series and quotes the third-order coefficient. What the fourth order says about the third carries it one term further. The order that stops helping asks where the truncation fails. All of them are about the term degeneration attacks hardest, and none of them is about the term it attacks least. A stage degenerated until its third-order product is negligible has a distortion spectrum that is almost entirely second-order, and on any signal path wider than an octave that is the product that matters.

The second-order product of a degenerated stage, which nothing here has ever read. computed by solving, not by drawing. Three products of two equal tones through a degenerated stage degenerated by 6, against the drive. The tones sit at bins 60 and 65 of one record, so the difference product at bin 55, the sum product at 70 and the second-order pair at 5 and 125 each land in a bin of their own. The second-order product is the largest of the three at every drive drawn: at 1 mV it is 5.37e-4 of the fundamental against 1.30e-6 for the third. Degeneration removes one power of the factor fewer from the second order than from the third, so the gap between them widens as the stage is linearised. — Fig. 6 A stage degenerated by six, on the same axes as the bare exponential above. At a millivolt the second-order product is 5.373 × 10⁻⁴ and the third 1.299 × 10⁻⁶ — a ratio of 414 rather than the bare device’s 103, because the factor of six has taken 36 off the first and about 1,440 off the second.

The intercept the same coefficient names

A second-order intercept is a real specification — it is what a direct-conversion receiver is bought on — and this collection can now compute it, without a fit and without an extrapolation, because the product is a straight line of slope one all the way down. It reaches the fundamental at $a = 2V_T D^2$ exactly: 51.7 millivolts on a bare device, 206.8 at a degeneration of two, 1,861 at six and 6,256 at eleven, and the figure asserts the measurement against that closed form to five parts in a thousand.

Put beside the third-order intercept’s $2V_T\sqrt{2D^4/|3 - 2D|}$ , the ratio between the two is $\sqrt{|2D - 3|/2}$ — a pure number in the degeneration factor with no thermal voltage and no bias in it. It is 0.7071 on a bare device and 0.7071 at a degeneration of two, it passes through one at $D = 2.5$ where both intercepts are 323.1 millivolts, and it is 2.12 at six and 3.08 at eleven.

So a data sheet for a heavily degenerated stage reports a second-order intercept above its third-order one, which reads as the second order being the better-behaved of the two — and at every drive a small-signal model survives, the second-order product is the larger. Both statements are correct. They differ because an intercept is where a curve of slope one and a curve of slope three each reach unity, and the ordering at unity says nothing about the ordering at a millivolt.

The null at $D = 1.5$ is the sharpest illustration available, because it is a case where the two coefficients are visibly unrelated. At three halves the third-order coefficient passes through zero and the third-order product collapses by more than a factor of twenty; the second-order product sits exactly where the $D^{-2}$ law puts it and does not notice. Two coefficients that share no term, and a linearisation that was only ever measured against one of them.

That null is a small-signal feature, and the figure’s slider shows where it stops being one: above about half a thermal voltage of drive the fifth-order term fills it in, because what vanishes at three halves is the cubic coefficient and nothing else.

What the intercept rests on

The other thing the machinery computes and nobody reads is the straightness of the two fits the extrapolated intercept is built from. The intercept is a crossing of two straight lines, so the only evidence for it is that the lines were straight — and the fits’ own $r^2$ has been returned on every call and never assembled into anything.

The extrapolation is 89% wrong at an r² of 0.9812. computed by solving, not by drawing. The third-order intercept of a differential pair, fitted over a decade of drive and extrapolated, against where that decade was placed — with the two fits' own straightness beside it. Fitted where the product is tiny the answer is the closed-form 4.0000 thermal voltages to 2.2e-4 per cent. Fitted over a decade ending at 82.7 mV, where an analyser can actually see the product, it is 7.566 — 89 per cent, or 5.54 dB, optimistic. The lower curves are one minus r² for the two fits, and they are the reason the error is invisible: at the window where the answer is ten per cent wrong the third-order fit still has r² = 0.996882. The departure is curvature rather than scatter, and a statistic about residuals cannot see a gentle bend. — Fig. 7 The pair’s third-order intercept, fitted over a decade of drive and extrapolated, against where that decade was placed. Fitted at the transform’s floor it returns the closed-form 4.0000 thermal voltages. Fitted over a decade ending at 82.7 millivolts it returns 7.5655 — 89.1 per cent, or 5.54 decibels, optimistic — and the third-order fit’s r² there is 0.98121. At the window where the answer is ten per cent wrong, r² is 0.996882.

The straightness statistic is nearly blind to the only error this fit can make. At a window ending at 41.4 millivolts the extrapolated intercept is ten per cent high and $r^2$ is 0.9969, which every convention treats as an excellent fit. At 82.7 millivolts the answer is out by a factor of 1.89 and $r^2$ is still 0.981.

The reason is that the departure is curvature rather than scatter. A fifth-order term bends the product’s line gently across the window; a straight line through nine points on a gentle curve has residuals in the fifth decimal place and a slope that is displaced; and the displaced slope is then multiplied by the length of the extrapolation. $r^2$ measures residuals. It has almost nothing to say about a slope error, which is the only kind of error there is here.

The extrapolation is 11% wrong at an r² of 0.9997. computed by solving, not by drawing. The third-order intercept of a bare exponential, fitted over a decade of drive and extrapolated, against where that decade was placed — with the two fits' own straightness beside it. Fitted where the product is tiny the answer is the closed-form 2.8284 thermal voltages to 2.4e-2 per cent. Fitted over a decade ending at 29.2 mV, where an analyser can actually see the product, it is 3.134 — 11 per cent, or 0.89 dB, optimistic. The lower curves are one minus r² for the two fits, and they are the reason the error is invisible: at the window where the answer is ten per cent wrong the third-order fit still has r² = 0.999674. The departure is curvature rather than scatter, and a statistic about residuals cannot see a gentle bend. — Fig. 8 The same sweep on a bare exponential, whose exact intercept is 2√2 = 2.8284 thermal voltages. Its window has to stop sooner, because the third-order product reaches a quarter of the fundamental at a lower drive: at a decade ending at 29.2 millivolts the answer is 3.1336, 10.8 per cent high, and r² is 0.99967.

This matters because of where a real measurement has to be taken. An analyser cannot see a product at $10^{-6}$ of the fundamental, so a two-tone measurement is made where the product is visible — tens of decibels below the tones rather than a hundred and twenty. That is exactly the drive range in which the answer is going wrong while the fit still looks perfect, and it is why intercept figures on two data sheets for the same part can differ by several decibels with neither of them wrong about anything they measured.

The same shape appears whenever this collection fits a line to extrapolate it. The straight lines, and where they are not the curve measures an asymptotic sketch against the response it stands in for, and the number that was wrong is about a shipped measurement whose fault was in where it was taken rather than in what it computed. The general rule is that a fit’s quoted goodness reports the noise the fit had to fight and not the model error it was asked to ignore.

Which leaves the fix somewhere else

If degeneration is the wrong tool for this product, the question is what the right one is, and the three curves here answer it between them.

The second-order product is a symmetry problem rather than a feedback problem. Local feedback divides every coefficient by a power of the loop factor and cannot distinguish an even term from an odd one, which is why it takes $D^{-2}$ off one and $D^{-3}$ off the other and can never remove either. An odd characteristic removes the even terms outright — not by a factor, at the arithmetic’s floor — and no amount of feedback reproduces that. The measured contrast is 6.159 × 10⁻¹⁶ against 5.373 × 10⁻⁴ at the same drive, which is twelve orders of magnitude and is not a difference of degree.

So the two techniques are not alternatives with different exchange rates. They act on disjoint halves of the series, and a stage that needs both needs both: a differential pair for the even orders, and degeneration inside it for the odd ones. That is what the arrangement in every low-distortion front end actually is, and this is the measurement that says why it is not one or the other.

What is not in this

Every curve here is memoryless. The products are the same at a tone spacing of one bin and at fifty-nine, to nine figures — which the figure asserts, because that identity is the assumption the whole two-tone method rests on. A real device has reactances, so its products depend on the spacing, and the second-order difference product is where that shows first: it sits near direct current, where a coupling capacitor, an emitter bypass and a bias loop all have poles. Nothing here can say what happens to it there.

The tone placement is a condition on the measurement rather than on the device, and it is checked rather than assumed. The two tones sit at bins 60 and 65 of one record, and the four product bins are 5, 55, 70 and 125 — all inside the lines the transform keeps. Placed carelessly, one half of a product pair falls outside and the measurement silently halves; that is not a small risk, since a spacing of sixty-five with the same lower tone does exactly it.

And a differential pair is only as odd as it is matched. The floor here is the arithmetic’s, because the two halves are the same function; a real pair has an offset, and the error that is a distribution is what a mismatch does to results of this kind. The second-order product of a real pair is set by that offset and is not $10^{-16}$ of anything.

The number worth carrying

A hundred and three to one at a millivolt, on the device this field is built out of — and $4V_T/a$ for the ratio, with nothing in it but the drive and the temperature. A differential pair takes the larger of the two products to the arithmetic’s floor and halves the smaller. Degeneration takes the larger as $D^{-2}$ and the smaller as $D^4/|3 - 2D|$ , so it makes the balance between them worse by a factor of $D$ .

Two habits go with it. The first is that a linearisation has to be measured against every order it is claimed to improve, because a technique that attacks one coefficient is under no obligation to touch another, and the two here share no term at all — which the null at three halves makes visible by moving one of them and not the other.

The second is about extrapolated specifications. A number obtained by fitting two lines and crossing them is a number about the window the lines were fitted over, and the goodness-of-fit statistic that comes with it is not evidence: it stayed above 0.99 while the answer went 89 per cent wrong. The evidence a fit like that needs is a second window, and the difference between the two answers — which is what how small is small signal does for an amplitude boundary and what every model has an edge asks of every model here. One window and an $r^2$ is a confident number about nothing.