Three amplitudes, all of them one per cent

How small is small signal put a number on a phrase. Linearising an exponential replaces a curve by its tangent, which is exact at a point and progressively wrong away from it, and the amplitude at which it is one per cent wrong is 7.3 millivolts at room temperature — 28 per cent of the thermal voltage, not a small fraction of it, and a good deal smaller than “small signal” suggests.

That number is correct and it answers one of three questions that the phrase could have meant.

A sinusoid on an exponential produces a whole spectrum. Its fundamental is not the tangent’s prediction, which is the departure that essay measured. Its harmonics are not zero, which is a different statement. And its mean is not the quiescent value, which is a third one — the operating point the model was linearised about has moved, and the linearisation is therefore about a point the device is no longer at.

Three quantities, three “one per cent wrong” amplitudes, and they are seven times apart.

One function, three orders #

Driving $\exp(v/V_T)$ with $v = a V_T\sin\theta$ gives $\exp(a\sin\theta)$, which has a classical expansion:

$$e^{a\sin\theta} = I_0(a) + 2\sum_{n\ge1} I_n(a)\,\{\dots\}$$

where $I_n$ is the modified Bessel function of the first kind. The mean is $I_0(a)$, the fundamental is $2I_1(a)$, and the $n$-th harmonic is $2I_n(a)$. Three questions about one waveform, three orders of one function, and every one of them equals one at $a = 0$ and departs from it as the drive grows.

Every value drawn here is computed twice. Once from the series, which converges in a dozen terms for the arguments used; and once by transforming $\exp(a\sin\theta)$ numerically over four thousand points and reading the lines out. The two routes share nothing but the arithmetic of the sine, and they agree to nine parts in $10^{11}$ over six orders.

That is more care than a textbook identity needs and it is the right amount, because the essay’s claim is that three numbers derived from this one function are different, and a reader is entitled to suspect a slip in the derivation rather than a fact about transistors.

Where each of them reaches one per cent #

The gain. The fundamental’s amplitude is $2I_1(a)$ against a tangent’s prediction of $a$, so the ratio is $2I_1(a)/a$, which expands to $1 + a^2/8$. One per cent of that is at $a = 0.28237$, which is 7.30 mV at 27 °C. That is that essay’s number and it is the largest of the three.

The operating point. The mean is $I_0(a)$ against a quiescent value of one, expanding to $1 + a^2/4$. One per cent is at $a = 0.2$, which is 5.17 mV. The collector current is one per cent higher than it was with no signal applied, from a signal that the gain measurement still calls small.

The distortion. The second harmonic is $2I_2(a)$ against a fundamental of $2I_1(a)$, and the ratio expands to $a/4$. One per cent is at $a = 0.04$, which is 1.03 mV — a fifth of the bias boundary and a seventh of the gain one.

Every one of those three is honestly described as “the amplitude at which the small-signal model is one per cent wrong”. The one usually quoted is the largest.

The root two between two of them, and it is exact #

The ratio between the gain boundary and the bias one is $\sqrt2$, and the reason is one line of series arithmetic.

Both quantities are second order in the amplitude: $I_0 = 1 + a^2/4 + \dots$ and $2I_1/a = 1 + a^2/8 + \dots$. A criterion $\varepsilon$ is therefore reached at $a = 2\sqrt\varepsilon$ for the first and $a = 2\sqrt{2\varepsilon}$ for the second, and the ratio of those is $\sqrt2$ identically, at every criterion.

The figure measures it rather than substituting. At a criterion of one part in ten thousand the ratio of the two bisected amplitudes is 1.414192 against $\sqrt2 = 1.414214$, four parts in a hundred thousand apart. At one per cent it is 1.41363, which is four parts in ten thousand below $\sqrt2$ — measurably not the limit, and the figure states that departure as a fact rather than rounding it away. The leading term is not the whole answer at one per cent, and a collection that quotes one-per-cent boundaries should know that its own tidy ratios are asymptotic.

The spacing is not a property of the device #

The three boundaries stand in the ratio $1 : 5 : 5\sqrt2$ at a criterion of one per cent, and a reader who took that away would have a rule that is wrong at every other criterion.

The second harmonic is first order in the amplitude and the other two are second. So the distortion boundary moves as $\varepsilon$ and the other two as $\sqrt\varepsilon$, and the gap between them goes as $1/2\sqrt\varepsilon$:

Tighten the requirement and the three boundaries do not move together — they spread, and the distortion one runs away from the other two. A stage that has to be one per cent linear in gain and one per cent in distortion faces a factor of seven between its two limits; one that has to be a part in ten thousand in both faces a factor of seventy.

That is the design consequence and it is the reason the three deserve separate names. For a gain stage, the gain boundary binds and the distortion is comfortably inside it. For a mixer or a low-distortion oscillator, the distortion boundary binds by two orders, and a “small-signal” design rule taken from the gain boundary is wrong by that much.

Which is the general form: a first-order departure and a second-order departure cannot be summarised by one amplitude, and the ratio between their thresholds depends on the threshold.

What each boundary is actually a requirement on #

Three numbers are useful only if it is clear which one a given design should be reading, and the answer follows from what the stage is for rather than from what it contains.

A gain stage whose output is fed back. The gain boundary binds, because a gain error inside a loop is divided by the loop gain and a distortion is too — both are reduced by the same factor, so their ratio is unchanged and the one that was larger stays larger. At the drive amplitudes an ordinary amplifier’s input stage sees, a few millivolts, the gain error is a per cent and the loop makes it a part in ten thousand. The gain that is exactly one is the limit of that argument.

A mixer, a modulator or a logarithmic converter. The distortion boundary binds by two orders, because the whole function of the stage is the shape of the curve and an error in the shape is the output. A logarithmic amplifier driven at a millivolt is at its distortion boundary and at a seventh of its gain one, and the logarithm is in the collector is where the curve is used deliberately.

An oscillator or anything with a settled amplitude. The bias boundary binds, and it binds in a way the other two do not: the shifted operating point changes the transconductance, which changes the loop gain, which changes the amplitude, which changes the shift. That is a fixed point rather than an error, and it is the mechanism by which an oscillator’s amplitude settles at all — what the limiter charges for is the same loop seen from the amplitude’s side.

So the three boundaries are not three grades of one requirement. They belong to three different kinds of circuit, and a designer reading the wrong one is not being conservative or optimistic — they are answering a question about a stage they are not building.

The boundary nobody counts #

Of the three, the operating point is the one that does not appear in any treatment of small-signal validity, and it is the one with the strangest consequence.

At a drive of one thermal voltage the mean collector current is $I_0(1) = 1.266$ times the quiescent value — 26.6 per cent more current than the bias network was designed for. Everything downstream of that follows: the transconductance is 26.6 per cent higher, the collector resistor drops 26.6 per cent more, the operating point has moved down the load line, and the linearisation was performed about a point the device is no longer at.

The distinction is worth being precise about. A gain error is the model predicting the wrong output for a given input. A distortion is the model predicting the wrong shape. A bias shift is the model having been built about the wrong point — the model is not making a wrong prediction so much as being the wrong model, and refitting it about the new operating point makes both other errors different.

It is also the one that has a memory. Gain error and distortion are instantaneous properties of a waveform; a bias shift is an average and appears over whatever time constant the bias network has. A stage driven by a burst of signal shifts its operating point during the burst and returns afterwards, with a settling time that belongs to the coupling capacitors rather than to the signal — which is a thump, and is why an audio stage that measures well on a steady tone misbehaves on transients.

This puts it in the same family as the flux that walks and the node that is at ground for a while: a quantity that is supposed to be a constant of the circuit, moved by the signal, with a recovery governed by something unrelated to the signal.

The one measurement that finds all three #

A bench check that distinguishes the three takes one instrument and one sweep, and it is worth writing down because the three departures are separable in the data rather than only in the algebra.

Drive the stage with a sinusoid of slowly increasing amplitude and record, at each amplitude, three numbers from the output: the direct component, the fundamental, and the second harmonic. That is one spectrum analyser and one sweep.

The direct component against amplitude is $I_0(a)$ and is a parabola through the quiescent value; its curvature gives the bias boundary. The fundamental divided by the amplitude is $2I_1(a)/a$ and is a parabola too, with half the curvature; the ratio of the two curvatures is that $\sqrt2$ squared, and measuring it is a check on the device being exponential that needs no absolute calibration at all. The second harmonic over the fundamental is a straight line through the origin of slope $1/4$ in units of $a$, and its slope gives the thermal voltage directly.

Three fits, three coefficients, and between them they give the thermal voltage, the ideality and a test of the model — from an ordinary two-tone bench setup with no thermometry and no curve tracer. The exponential’s signature is the pair of curvatures being in the ratio two to one, and a device whose two parabolas are not in that ratio is not obeying the law the small-signal model was derived from.

That is the practical use of having three boundaries rather than one. A single departure measured against a single criterion tests nothing — any smooth curve produces one — while three departures in fixed ratios are a fingerprint, and the ratios are the part that does not depend on temperature, on bias current, or on the amplitude the measurement happened to stop at.

What one criterion cannot settle #

That the bench procedure above has been performed. It is a consequence of the three expansions and it is stated because it follows, not because it has been run here — everything measured above is computed from the model rather than from a device, which is what the distortion a linear model cannot have is careful about too.

That the second harmonic is the right measure of distortion. It is the largest one and it is what $a/4$ describes. Total harmonic distortion sums all of them and is larger by a factor that depends on the amplitude — a few per cent at these levels, because the third harmonic is $a^2/24$ of the fundamental and is negligible where the second is one per cent. Using the second alone understates the total slightly and gets the exponent right, which is what the argument needs.

That a real transistor is an exponential. It is, to within an ideality factor and a series resistance and the Early effect, and the constant that is a window measures how far. What is being priced here is the consequence of the linearisation, which is the same shape for any smooth curve: the gain and the mean depart at second order and the second harmonic at first.

That the bias shift always matters. In a stage with strong direct-current feedback — an emitter resistor, a servo loop — the shift is divided by the loop gain and may be negligible. What survives is that the device has moved even if the stage has not, so the transconductance used in the small-signal model is wrong by the shift’s fraction whatever the bias network does.

That the criterion should be one per cent. It is a convention. Everything above is stated as a function of the criterion precisely because the convention is arbitrary and the three boundaries do not scale with it in the same way.

Bessel’s series against a transform of the waveform, at six orders #

Every order is computed twice — Bessel series and numerical transform of the waveform — over six orders, and required to agree to a part in a billion. They come out at nine parts in $10^{11}$.

The $\sqrt2$ is checked as a limit and as a departure. The ratio at a criterion of $10^{-4}$ is checked against $\sqrt2$ to a part in a hundred thousand, and the ratio at one per cent is checked to be measurably not $\sqrt2$ — between one part in ten thousand and one in a thousand away from it.

The spacing law is checked at three criteria. Five at one per cent, fifty at a part in ten thousand, and $1/2\sqrt\varepsilon$ in between to half a per cent, so that “the three do not keep their spacing” is a measured law rather than two points.

And the refusal is the contradiction between two of them: at the amplitude where the gain is one per cent low the second harmonic is already seven per cent, so no single amplitude can serve as the boundary for both.

A boundary is a question before it is a number #

A boundary is a model and a tolerance established the general point this essay is a worked case of: an edge is not a property of a model, it is a property of a model and a stated tolerance, and changing the tolerance moves the edge.

This adds the part that is easy to miss. It is not only the tolerance that has to be stated — it is the quantity the tolerance is on. One model, one device, one temperature, one criterion of one per cent, and three different amplitudes, because there are three different things that could be one per cent wrong. Pick the wrong one and the answer is out by seven, which is larger than most of the uncertainties anybody worries about.

The three are not independent — they come from three orders of one function, so knowing one determines the others — and that is what makes the situation tractable rather than hopeless. The useful form of the result is not the three numbers but the two exponents: the gain and the operating point depart at second order in the drive, and the harmonics at first, so a requirement on distortion always binds harder than a requirement on gain, and the tighter the requirement the more it binds.

Which is a better sentence to carry than 7.3 mV, and it is the sentence the essay before it could not produce because it was measuring one of the three.

Still open: the third harmonic, the model refitted about the shifted point, and the same three for a square law #

The third harmonic, which is second order. $2I_3/2I_1$ goes as $a^2/24$, so its boundary moves as $\sqrt\varepsilon$ like the gain’s and unlike the second harmonic’s. That gives a fourth amplitude on the same axis, and — more interestingly — a criterion at which the second and third harmonic boundaries cross, above which the third is the binding one. Solving for it would say whether that crossing is anywhere near a real requirement.

A model refitted about the point the device has actually reached. The bias shift moves the operating point; a linearisation about the shifted point has a different transconductance and therefore different gain and distortion errors. Whether refitting improves the gain error or merely moves it is a computation this figure’s apparatus could do directly, and it would say whether “large signal” is a different model or the same model about a different point.

And the same three boundaries for a square law. A field-effect device’s transfer is a square rather than an exponential, so its second harmonic is exactly proportional to the amplitude with no higher terms, its third harmonic is exactly zero, and its bias shift is exactly $a^2/4$ of the quiescent current with no series at all. Every boundary is then a closed form rather than a Bessel ratio, and the three would stand in ratios that are exact rather than asymptotic — which would make the square law the cleaner worked example and is a reason to do it second rather than first.