Devices, and the amplitude they stop being linear at

The distortion a linear model cannot have

A small-signal model's output is a scaled copy of its input by construction, so it has no second harmonic and asking it for one is not a hard question but a meaningless one. Measured on the curve itself, an exponential produces one per cent of harmonic distortion at 1.03 mV of drive — seven times sooner than the 7.30 mV at which its gain is one per cent wrong.

There is a question a small-signal model cannot be asked. Not one it answers badly — one that does not exist for it.

A linear model’s output is, by its definition, a scaled copy of its input. Feed it a sinusoid and a sinusoid comes out, at the same frequency, always. It has no second harmonic, not because the second harmonic is small but because the model has no term that could produce one. Every distortion figure in every datasheet describes a quantity that lives entirely outside the model most of the analysis is done with.

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. 1 An exponential driven ten millivolts either side of its bias. A sinusoid goes in and something with taller peaks than troughs comes out; the bars are what it is made of. Each one is measured by transforming five hundred and twelve samples of one period and is predicted independently by a Bessel series that the measurement knows nothing about. The slider is the drive amplitude.

Where the harmonics come from

The mechanism is easier than it looks. Expand a device’s transfer characteristic about its bias point as a power series:

i = a₀ + av + av² + av³ + …

The small-signal model is the first two terms and nothing else; a₁ is the transconductance. Drive that series with v = cos ωt and each power produces its own harmonics: the square of a cosine is a constant plus a cosine at twice the frequency, the cube is a cosine at the fundamental plus one at three times it, and so on.

Two consequences fall straight out, and they explain everything measured below.

The second harmonic is first order in the drive. It comes from a²/2, so its amplitude is proportional to ², and the ratio of second harmonic to fundamental — which is what “distortion” means — is proportional to . Double the drive, double the distortion.

The gain error is second order. The fundamental picks up a correction from the cubic term, 3a³/4, so the gain’s fractional departure from a₁ goes as ². Double the drive, and the gain error quadruples from a much smaller starting point.

That difference in order is why the two boundaries in the title are seven times apart rather than close together, and it is not a property of this particular device.

The check that shares no arithmetic

An exponential driven by a sinusoid has harmonic amplitudes that are known exactly, in closed form, and they are the modified Bessel functions of the first kind: the nth harmonic of exp( cos θ / V_T) is proportional to I_n(/V_T). So the ratio of the second harmonic to the fundamental is I₂(x)/I₁(x) with x = /V_T, and nothing else.

Nothing in the measurement knows that. The figure samples the curve five hundred and twelve times, transforms the samples, and reads off the ratios. The prediction sums a power series. They share the drive amplitude and the thermal voltage and no arithmetic at all.

At ten millivolts of drive the measured second harmonic is 9.6107% of the fundamental and I₂/I₁ at x = 0.3868 is 9.6107%. The worst relative disagreement over the harmonics the arithmetic can resolve is at the level of 10⁻¹⁴.

That last phrase is doing real work. At a millivolt of drive the sixth harmonic sits at 3.8×10⁻¹² of the fundamental, and a transform of samples of order one carries about fifteen digits, so comparing it is comparing rounding error. The check therefore runs only on harmonics above 10⁻⁸ of the fundamental — a floor set by measurement rather than taste: the relative error of a harmonic of size m runs at about 1.2×10⁻¹⁵/m across four decades of drive, so 10⁻⁸ buys agreement to better than 1.2×10⁻⁷, two decades inside the tolerance asserted. Harmonics below the floor are still drawn, with their measured values; they are simply not claimed about.

An exponential driven 26.0 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 24.15% of the fundamental, measured by transforming 512 samples and predicted independently as I₂(1.006)/I₁(1.006) = 24.15%. The two routes agree to 5e-12 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 224.15%harmonic 33.96%harmonic 40.49%harmonic 50.05%harmonic 64.1e-5agreement: 5e-12 relativesolved, then checked — a transform against a seriessecond harmonic 24.1% at 26.0 mV
Fig. 2 The same device at twenty-six millivolts — one thermal voltage of drive, which is where most readers’ intuition puts the boundary of “small”. The second harmonic is 24.1% of the fundamental and the third is 4.0%. The output waveform’s peaks are now about three times the depth of its troughs, and the ghost line showing a symmetric waveform of the same fundamental is nowhere near it.

Why a transform rather than a series

The measurement could have been done by differentiating the transfer characteristic three times and reading a₂ and a₃ off, and it deliberately is not. The reason is generality, and it is the same reason the rest of the site solves networks rather than manipulating expressions.

A transform takes samples. It does not care where the samples came from, so the identical routine measures the distortion of an exponential written in closed form, of a hyperbolic tangent, and of a network solved point by point with the Newton loop from a bias point is a solution. A series expansion cares a great deal: it needs symbolic derivatives, it needs them about the right point, and it stops being available the moment the characteristic is something a solver produced rather than something an author wrote down.

The cost of the transform is that it has a floor and a truncation, which the previous section had to spend two paragraphs on. That is a fair trade — a stated floor on a general method is worth more than exactness on a method that only works for the cases where the answer was already known.

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 Where the series above is expanded. Every coefficient in a₀ + av + av² + … is a derivative of the characteristic at the operating point, so all of the distortion numbers on this page are properties of a point as much as of a device. Move the bias and every coefficient changes; for the exponential the ratios happen not to, which is a special property of that curve and not a general one.

For the exponential there is a striking consequence of that last point. Every derivative of exp(v/ V_T) is the function itself divided by a power of V_T, so the ratios a₂/a₁ and a₃/a₁ do not depend on the bias current at all. The distortion of an exponential transconductor at a given drive voltage is the same at ten microamperes and at ten milliamperes, which is why the boundary in this essay is quoted as a voltage with no current beside it, and why the site’s existing measurement found the same 28.2% of V_T at every bias current tested.

That is unusual. For almost any other characteristic the bias point matters enormously, and a stage biased near a point of inflection has a third-order coefficient that passes through zero — which is the basis of every “sweet spot” in a device datasheet and of a family of biasing tricks that trade one harmonic for another.

Two boundaries, seven times apart

The site already carries one number for the edge of the small-signal model. How small is small signal measures the drive at which the linearised gain is one per cent optimistic, and gets 7.30 mV — 28.2% of the thermal voltage, a fraction that turns out to be the same at every temperature and every bias current.

That number is about the gain. Measuring the same device’s distortion gives a completely different answer.

An exponential: distortion arrives seven times sooner than gain errorcomputed by solving, not by drawing at 61 amplitudes. Total harmonic distortion reaches one per cent at 1.03 mV and the gain falls one per cent short of its small-signal value at 7.30 mV. They are 7.1 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 1.03 mV1% gain error at 7.30 mVsolved, then checked — both edges bisected7.1× apart
Fig. 4 Harmonic distortion and gain error against drive amplitude, on the same axes, with both one-per-cent crossings bisected on the measured curves. Distortion reaches one per cent at 1.03 mV; the gain reaches one per cent at 7.30 mV, which is the number the limits field already carries. The two curves have visibly different slopes, and the difference in slope is the difference between first and second order.

One per cent of total harmonic distortion arrives at 1.0341 mV. One per cent of gain error arrives at 7.2999 mV. The ratio is 7.06.

Both numbers describe the same device and the same word — small — and they differ by a factor of seven, so any statement of the form “the signal is small enough” has to say small enough for what. A stage biased to run comfortably inside its gain-error budget at five millivolts of drive is producing 4.8% of second harmonic, which for an audio amplifier is a catastrophe and for a limiter is irrelevant.

The general shape holds beyond this device, because it follows from the orders. Wherever a nonlinearity is smooth, distortion grows as the first power of the drive and gain error as the second, so the distortion boundary always arrives first, and the ratio between the two boundaries grows as the tolerance tightens.

That last sentence is a prediction rather than a summary, and it can be checked. If distortion is first order and gain error second, then tightening the tolerance by a decade should move the distortion boundary down by ten and the gain boundary down by only √10 — so the ratio between them should grow by √10 = 3.162 per decade of tolerance.

Bisecting both boundaries at four tolerances gives 2.190, 7.059, 22.357 and 70.710 for ten per cent, one per cent, a tenth and a hundredth. The successive ratios are 3.223, 3.167 and 3.163. The last two are √10 to three figures, and the first is a little high because at ten per cent the drive is large enough that the higher terms of the series are no longer negligible — which is exactly the condition under which the argument was derived to fail.

A prediction that comes out right, and comes out wrong in the place its own derivation says it should, is worth more than either half alone.

Linearising an exponential at 27 °C, and what it costsThe linear model understates the gain by 1% at 7.30 mV and by 10% at 22.8 mV. The thermal voltage at this temperature is 25.9 mV, so "small compared with V_T" is not the criterion — 28% of V_T is already 1% wrong.1.0m10m1.0e+2m110100100m110100drive amplitude (millivolts)how much the linear model understates the gain (per cent)1% understated10% understated1% at 7.3 mVV_T = 25.9 mVsolved, then checked — the Bessel ratio from its seriesthe tangent is 1% wrong above 7.3 mV
Fig. 5 The gain-error boundary as the limits field draws it: the true gain of an exponential falling away from its small-signal value as the drive grows, with the one-per-cent crossing marked. This figure and the distortion curve above are measurements of the same curve at the same amplitudes, and they disagree about where “small” ends by a factor of seven.

What total harmonic distortion actually totals

The figures quote a single number for distortion, and it is worth saying exactly what goes into it, because the definition contains two choices.

Total harmonic distortion here is the root-sum-square of every harmonic above the fundamental, divided by the fundamental. At ten millivolts that is 9.631%, against a second harmonic alone of 9.611% — so the second harmonic is very nearly the whole of it, and it stays that way until the drive is comparable with the thermal voltage. At fifty millivolts the second harmonic is 42.2% and the total is 44.2%, with the third contributing 12.7%.

The first choice is that the direct-current term is excluded. That is the usual convention and it is a choice: a bias shift is not distortion of the signal in the sense meant here, and including it would make a stage’s distortion figure depend on where its output happened to sit. It is worth noticing that the exponential does shift its average as the drive grows — that is what I₀(x) > 1 means — and that shift is a real effect which this measure deliberately does not report.

The second choice is the harmonic count. Truncating at twelve is fine here because the series falls away quickly, and the check that it is fine comes free: Parseval’s identity requires the mean square of the samples to equal the sum of the squares of the harmonic amplitudes, and any energy in harmonics beyond the count shows up as a discrepancy. On a square wave truncated at twenty harmonics that discrepancy is 1.01% and correctly so; on every waveform in this field it is at the level of the arithmetic.

The topology that removes half of them

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. 6 One exponential and one differential pair, driven identically. The pair’s characteristic is odd, so its even harmonics vanish — the measured second harmonic is at 10⁻¹⁶ of the fundamental, which is the arithmetic’s floor rather than a small physical residue. What it costs is on the same picture: the third harmonic is exactly half the single stage’s, which is a reduction and not a cancellation.

That figure is the argument of what a pair cancels and is placed here for one reason: it is the answer to the problem this essay poses, and it is a structural answer rather than a numerical one. The even harmonics do not become small; they become zero, because an odd function driven symmetrically cannot produce them. No amount of biasing, trimming or matching is involved, and no tolerance is attached.

It is the strongest kind of result available in this area, and it is worth contrasting with the alternative approach of driving the device less hard. Halving the drive halves the second harmonic. Making the characteristic odd removes it.

What the numbers mean for a stage that has to work

It is worth turning the boundaries into the constraint they actually impose, because the arithmetic is not encouraging and the way out of it is the whole of amplifier design.

A stage whose distortion must stay below a tenth of a per cent may be driven to 103 µV. A single bipolar transconductor with a load resistance chosen to give a gain of a hundred therefore has an output swing of about ten millivolts, which is not a useful amplifier. Ask for a hundredth of a per cent and the drive falls to 10 µV and the swing to a millivolt.

Three things are done about that, and each is a different part of this collection.

Make the characteristic odd, which removes the second harmonic entirely and buys a factor of about eighteen in allowable drive at one per cent — the differential pair reaches one per cent of distortion at 18.2 mV against the single stage’s 1.03 mV. That is the largest single improvement available and it costs a second device.

Degenerate the stage, which is to say put a resistance in series with the emitter so that most of the input voltage falls across something linear. The device sees a fraction of the drive, its distortion falls with that fraction, and the gain falls by the same factor. This is the trade nobody escapes: linearity bought at exactly the price of gain.

Apply feedback, which is the feedback field and reduces distortion by the loop gain — the same loop gain that sets the bandwidth, the margin and the ringing. A stage with forty decibels of loop gain has a hundredth of the distortion, and has spent its forty decibels.

All three are the same transaction seen from different sides: gain is the currency and linearity is what it buys. The reason the boundaries in this essay are worth measuring rather than assuming is that they set the price.

What the field’s rule looks like here

Every figure on this site carries the frequency, amplitude or size at which the model in it stops being true, and this page is one of the few where the answer is squarely an amplitude and where there are two of them.

The honest statement of the small-signal model’s domain therefore has two numbers in it: excursions below about a millivolt if harmonic content matters, and below about seven millivolts if only the gain does. Quoting one without saying which is being measured is how a specification comes to mean two different things to the person who wrote it and the person who reads it.

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 boundaries the collection has measured, on one frequency axis — with the amplitude boundaries stated beside it rather than drawn on it, because they are not frequencies. That separation is the point of putting this essay in a field of its own: the semiconductor limits are not “the circuit is too fast” but “the signal is too big”, and no amount of slowing anything down moves them.