Circuit Response
every model has an edge, and the figure carries it
Where the models stop

How small is small signal

Linearising an exponential replaces a curve by its tangent, which is exact at a point and progressively wrong away from it. 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.
14 min read 4 figures Every model has an edgeRefused, not extrapolatedThe constants decide

A transistor’s collector current is an exponential function of its base-emitter voltage. A diode’s current is an exponential function of the voltage across it. Neither is linear anywhere, and both are analysed everywhere as though they were — by replacing the curve with its tangent at the operating point and calling the slope a transconductance or a dynamic resistance.

That is a perfectly good model with a stated range. The range is an amplitude, and it is smaller than the name suggests.

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. 1 How much a linearised exponential understates the gain, against how hard it is driven. One per cent at 7.3 mV, ten per cent at 22.8 mV, against a thermal voltage of 25.9 mV. The slider is the temperature, because the thermal voltage is kT/q and the boundary moves with it.

The measurement

Drive an exponential with a sinusoid of amplitude v̂ and ask what comes out at the fundamental frequency. The exponential of a sinusoid has a standard expansion, and the amplitude of the fundamental component works out as

2 · I₁(v̂/V_T) / (v̂/V_T)

times the small-signal answer, where I₁ is the modified Bessel function of the first kind of order one. That ratio is 1 in the limit of no drive — a tangent is exact at a point — and grows as the drive increases, because the exponential curves upward more than it curves downward.

So the linear model understates the gain, always, and by an amount that depends only on the drive amplitude relative to the thermal voltage. The Bessel function is computed here from its series rather than looked up, and the boundaries are found by bisection on the ratio.

The numbers, and why they are surprising

At room temperature the thermal voltage kT/q is 25.9 millivolts, and:

  • the linear model is 1% wrong at 7.30 mV
  • and 10% wrong at 22.8 mV

The first of those is 28% of the thermal voltage, not a tenth or a hundredth of it. The commonly taught guidance — “keep the signal small compared with kT/q” — points in the right direction and gives no scale, and a reader who takes “small compared with” to mean a factor of ten will be at 2.6 mV, which is four times more conservative than necessary. A reader who takes it to mean “well under 25 mV” and settles on 15 mV is already 4% wrong.

Neither of those is a disaster and both are worth knowing, which is the point of computing the number rather than repeating the guidance.

What the boundary does not depend on

The figure’s slider is a temperature, and the choice is deliberate: it makes visible which parameters move the boundary and which do not.

The bias current does not. The transconductance is I_C/V_T, so a transistor at 100 mA has a transconductance ten thousand times that of one at 10 µA. The boundary is identical: 7.30 mV at both, because the ratio being measured depends on the drive relative to the thermal voltage and the current cancels out. That is worth knowing because it contradicts a plausible intuition — that a device running at more current is somehow more robust to a large signal. It is not.

The temperature does. The thermal voltage is proportional to absolute temperature, so at −40 °C it is 20.1 mV and the boundary is 5.67 mV; at 125 °C it is 34.3 mV and the boundary is 9.69 mV. The ratio between them is constant — the boundary is always 28.2% of the thermal voltage — which is asserted in the figure at every position of the slider.

That constancy is itself the useful fact. The small-signal boundary is a fixed fraction of kT/q, so it can be carried as one number and converted whenever the temperature is known.

What crossing it produces

Understating the gain by a per cent or two is not, by itself, why anyone cares. What matters is what accompanies it.

An exponential driven beyond its linear range does not merely change gain; it generates harmonics — frequencies that were not present at the input. A pure sinusoid in produces a fundamental plus a second harmonic plus a third and so on, with amplitudes that grow rapidly with drive. The second harmonic of an exponential driven at 7.3 mV is around 3.5% of the fundamental, which is audible distortion by any standard, and it is present at precisely the amplitude the gain error is only one per cent.

So the gain error is the mildest symptom of crossing the boundary, and a model checked only for gain accuracy will pass a circuit that is producing several per cent of distortion.

There is a second consequence that is easier to miss. Two signals through the same nonlinearity produce sum and difference frequencies as well as harmonics — intermodulation — and the difference frequencies can land inside the band of interest where harmonics would not. That is why distortion in a receiver’s front end is specified with two tones rather than one, and it is a failure mode with no equivalent in the linear world at all.

Why the model is used anyway

None of this is an argument against small-signal analysis, and it is worth saying why.

Below the boundary the linear model is not merely convenient but complete. Superposition holds, so a frequency response is a full description; poles exist; a transfer function can be written; and every technique in the first five fields of this collection applies. Above it, none of that is true, and the only general method is to integrate the equations forward in time and take what comes out.

That is an enormous difference in what can be known, and it is why circuits are designed to stay inside the boundary wherever possible — by using enough feedback that the device’s own nonlinearity is divided down, by biasing so the signal swing at the nonlinear element is small, or by arranging two devices so their curvatures cancel.

The last of those is worth naming since it is how most of the analogue world is actually built. A differential pair driven symmetrically has its even-order distortion cancel between the two halves, leaving only odd orders, which are much smaller at a given drive. That is why nearly every amplifier input stage is a differential pair, and it is a design decision made entirely in response to the boundary this essay measures.

Why it is an amplitude and not a frequency

Every other boundary in this collection is a frequency, and the reason this one is not is worth setting out, because it changes how it has to be looked for.

A frequency boundary arises when a model neglects something whose effect grows with frequency: a parasitic inductance, a finite gain–bandwidth product, a propagation time. In every such case the model is linear on both sides of the boundary — it is simply the wrong linear model above it — and the fix is a better linear model with a wider range.

An amplitude boundary arises when a model neglects curvature. A linearisation is exact at a point and its error grows with the square of the excursion, and the excursion is an amplitude, not a frequency. Nothing about repeating the signal faster makes the tangent a better fit, and nothing about repeating it slower makes it a worse one; the boundary is at the same seven millivolts at direct current and at a gigahertz.

That has a practical consequence for how a circuit gets checked. A frequency boundary can be respected by looking at a plot and staying to the left of a mark. An amplitude boundary is invisible on every plot a linear analysis produces, because such a plot has no amplitude axis at all. The only way to find it is to know the number and compare it against the signal — which is why the number is worth carrying, and why it appears in this collection’s summary figure with an explicit note that it cannot go on the axis with the others.

The boundary that moves the other way

There is a second amplitude boundary in every real circuit, and it sits at the opposite end.

Below some amplitude a signal disappears into noise: thermal noise in every resistance, shot noise in every junction, and the amplifier’s own input noise. That is not a model boundary in the sense this collection uses — the linear model remains exactly correct — but it is a limit on what a measurement can say, and it bounds the useful range from below in the same way the linearisation bounds it from above.

Between them the two define a window, and the ratio of its ends is the dynamic range. What is worth noticing is that both ends are set by the same physical constant: the thermal energy kT. The noise floor is kT per unit bandwidth into a matched load; the small-signal boundary is 28% of kT/q. A circuit’s usable range in amplitude is therefore bracketed at both ends by temperature, which is a tidier statement than either end makes alone, and it is the reason cooling an instrument’s front end buys range at both ends at once.

The other exponential in the same circuit

The transistor is not the only place this appears, and the second instance is worth noting because it is the one that decides a circuit’s bias rather than its signal.

A diode’s dynamic resistance — the slope of its curve at a current — is kT/(qI), which is 25.9 ohms at a milliampere and 25.9 kilohms at a microampere. That is the same linearisation, of the same exponential, with the same boundary in millivolts, and it is used constantly: in the small-signal model of a diode detector, in the emitter resistance of a transistor stage, and in the temperature compensation networks that exist precisely because the quantity moves with temperature.

The last of those is the interesting one. Because the boundary is proportional to absolute temperature, so is the dynamic resistance and so is the gain of any stage that depends on it — a circuit whose transconductance is I_C/V_T loses about a third of its gain between −40 °C and 125 °C with nothing else changing. The standard remedy is to make the stage’s gain depend on a ratio of resistances instead, by putting a resistor in series with the emitter that is large compared with the dynamic resistance, which is the same “make the interface impedance large” argument that runs through the whole of the first field of this collection.

What “1%” is measuring, exactly

The threshold used throughout this essay deserves one paragraph of precision, since a per cent of a gain is not the same claim as a per cent of a waveform.

The quantity plotted is the ratio of the true fundamental component to the one the tangent predicts, so “1% wrong” means the fundamental is 1% larger than the linear model says. It says nothing directly about the harmonics, which is why the essay quotes them separately. And it is a steady-state statement about a sinusoid: the same device driven by a step or a pulse departs at a different point, because the departure depends on the whole waveform rather than on its amplitude alone.

That is not a caveat so much as an instruction about how to read every number of this shape on the site. A boundary is defined by a quantity, a threshold and a test signal, and changing any of the three moves it. What makes a boundary useful is not that it is the only one but that the three are stated — so a reader who cares about a different quantity, threshold or signal knows exactly which of them to change.

The refusal

The small-signal model here declines a large-signal question. Asked for its answer at fifty millivolts, it raises an error naming the amplitude and the boundary and saying that a linearisation is not a large-signal answer with a wider error bar but a different claim.

That last phrase is the important one. The temptation with any model outside its range is to report its answer with a caveat, and for most models that is reasonable — an ideal amplifier at twice its bandwidth is wrong by a factor that could in principle be quoted. A linearisation outside its range is different in kind: it does not predict the harmonics, the intermodulation products or the shift in operating point, and those are not errors in its answer but phenomena it has no vocabulary for.

The diode, where the same curve is visible directly

The exponential in question is easier to see in a diode than in a transistor, because a diode’s whole behaviour is the curve rather than a curve controlling something else.

The 0.7 volt constant, solved over eight decades of currentThe forward voltage moves 59.5 mV for every factor of ten in current, so over the range drawn here it runs from 0.298 V to 0.774 V. The three marked points are solutions for 1 V, 5 V and 12 V through a kilohm, found by Newton's method; they span 88 mV.0.400.600.8011.21e-61e-51e-41m10m100m110100current through the diode (milliamperes)forward voltage across it (volts)the constant everybody is taught1 V in: 0.629 V5 V in: 0.693 V12 V in: 0.717 Vsolved, then checked — Newton's method on the exponentialthe constant moves 59.5 mV per decade
Fig. 2 The same exponential, plotted as the voltage needed for each current. The “0.7 volt drop” moves 59.5 millivolts for every factor of ten in current, so over eight decades it runs from 0.298 V to 0.774 V. The three marked points are operating points solved by Newton’s method for 1 V, 5 V and 12 V through a kilohm — they span 88 mV.

The two essays are about the same equation read in two directions. Here it is the slope of the curve and how far the tangent can be trusted; there it is the curve itself and how far the “constant” moves. Both are consequences of current being exponential in voltage with a scale of kT/q, and both are numbers rather than cautions.

A 9 V source with 500 mΩ inside itThe ideal source is the flat line. The solved terminal voltage leaves it at a rate set entirely by the internal resistance: 1% low at 180 mA, half gone at 9.0 A.024681010m100m110current drawn from the source (amperes)terminal voltage, solvedthe ideal source: 9 V at any current1% low at 180 mAthe model9 Vrloadsolved, then checked — the load swept over four decadesthe ideal source is 1% wrong above 180 mA
Fig. 3 A linear model’s boundary for comparison. An ideal source fails above a current, computably, and the failure is a smooth departure that a better linear model repairs. A linearised exponential fails above an amplitude and produces frequencies that were not in the input, which no linear model repairs.

Where this sits among the boundaries

The small-signal boundary is the one most likely to be crossed without noticing, and the reason is structural: it is in the wrong units.

Every other boundary in this collection is a frequency, and a frequency can be avoided by looking at a plot. This one applies at every frequency there is, does not appear on any frequency response, and is not visible in a schematic. A circuit operating at a hundred hertz with fifty millivolts on a base is outside its model, and every figure anybody draws of it will look entirely correct.

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. 4 The four boundaries that are frequencies, on one axis. This essay’s is the fifth and cannot go on it — which is exactly why it is worth its own essay, and why the caption of that figure ends by naming it anyway.

Where this goes next

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 Every model has an edge Four assumptions this collection runs on, with the frequency at which each stops being true, on one axis. The ordering is not the one most readers would guess — an ordinary amplifier circuit runs out of model at 1.42 kHz, three thousand times sooner than a ten-centimetre circuit board does. 1101001k10k100k1M10M100M1Gfrequency (hertz)impedance looking into 10.0 cm of track (ohms)the lumped model: one L, one C1° long at 3.97 MHza tenth of a wavelength at 143 MHzthe 200 Ω at the far endsolved, then checked — the line against a two-element modelKirchhoff's laws run out at 143 MHz Kirchhoff's own frequency The current law says the current entering a node equals the current leaving it at the same instant, which assumes the signal crosses the circuit in no time. It crosses at about two-thirds the speed of light, so the law has a frequency of its own — set by nothing but the physical size of the board. 00.2500.5000.750105101520time (microseconds)output, divided by the size of its own step20 mV step1.0e+2 mV step5.0e+2 mV step2 V step8 V steplinear below 79.6 mVsolved, then checked — integrated with the rate limitscaling fails above a 79.6 mV step The step that is too big A linear circuit scales — double the input and the output doubles, exactly. A real amplifier does not, because its output can only move at a fixed rate, and the amplitude at which the two stop agreeing is about eighty millivolts for an ordinary part. No transfer function contains that number, because no transfer function can.