Symmetry and a resistor together

Two boundaries removed, and one moved put a differential pair beside a single transistor and asked where each stops being small-signal. The pair’s transfer is odd, so two of the single transistor’s three one-per-cent boundaries, which three amplitudes, all of them one per cent separated — the shift in its operating point and its second harmonic — do not exist for a pair at all. The third, the loss of gain, moves outward by $\sqrt2$ in the limit of a tight criterion and by 1.426 at one per cent: 10.42 mV against 7.30.

A pair is rarely used bare. Its two emitters usually carry resistors, which degenerate each device by the ratio of the resistor to its own $1/g_m$, and a designer reaches for both at once because both are said to linearise. That essay’s question was whether they compound. The $\sqrt2$ from symmetry and the factors that what a resistor in the emitter buys measured for feedback might multiply, or one might swamp the other. The degenerated junction has no closed form forwards, so every waveform below is solved sample by sample and every harmonic read from its transform.

The two transfers, with the resistors in #

Write the drive in thermal voltages and the degeneration as $D = g_m R_e$, each device’s emitter resistor over its own $1/g_m$. A single transistor’s fractional change in current, $y$, then satisfies $\ln(1 + y) + D\,y = w$ for a drive of $w$ thermal voltages. A pair’s normalised differential output $x$, the fraction of its tail steered to one side, satisfies $\operatorname{atanh} x + D\,x = u$, with $u$ the drive over two thermal voltages because the pair divides it between two junctions. Both are solved by Newton’s method at every sample of a sinusoid, and the fundamental of each solved waveform is compared with its small-signal value.

With a degeneration of one — each emitter resistor equal to its device’s $1/g_m$ — the pair’s gain is one per cent out at 29.2 mV and the single transistor’s at 29.9 mV. The pair’s $\sqrt2$ advantage has not been multiplied. It has gone. The slider on the figure at the head of the page steps the degeneration through 0, ½, 1, 3 and 10, and the ratio moves through all of 1.43, 0.27, 0.97, 1.56 and 1.84: no single factor describes what degeneration does to the pair’s advantage, because it does quite different things to the two circuits.

The pair’s own law #

The pair is the simpler of the two. Its transfer has only odd terms, and expanding $\operatorname{atanh} x = x + x^3/3 + \dots$ against the resistor’s $D\,x$ gives the fundamental’s compression at a peak drive $U$ as $U^2/(4(1 + D)^3)$. One per cent is then

$$v_{1\%} = 0.4\,V_T\,(1 + g_m R_e)^{3/2},$$

which at no degeneration is the 0.4 thermal voltages the earlier essay found.

The pair follows it from no degeneration up to about three — 0.403 thermal voltages bare, 3.13 at $g_m R_e = 3$, each within three per cent of the law — and then falls below it: 13.36 at ten, where the law says 14.6. The departure is not a failure of the algebra but of its assumption. At a degeneration of ten, one per cent of compression needs a drive that steers more than half of the tail to one side, and a cubic expansion of the tanh’s inverse is no description of a pair that is most of the way to switching. The three-halves power is the law of a pair that is still a small signal on its own tail, and heavy degeneration makes the one-per-cent point a large signal on it.

The single transistor’s cancellation, and why the pair has none #

The single transistor’s boundary is the curve with the spike in it. Bare, its gain expands with drive and reaches one per cent at 0.282 thermal voltages. At $g_m R_e = 1/2$ its boundary leaps to 2.78, ten times further. Past that its gain compresses and its boundary settles on half the pair’s law. Where the two exponents come from found the same singularity by reverting the degenerated junction’s series, as a gain edge that is infinite at a degeneration factor of three halves — which is $g_m R_e = 1/2$ in the units used here — and the solved waveforms put it where the series did.

The cubic coefficient says why, and says why the pair cannot share it. A single transistor’s third-order term is the sum of two contributions of opposite sign: the junction’s own cubic, which compresses, and its square returned through the emitter resistor, which expands — the second term is what makes a bare transistor’s gain rise with drive. The two are $-1/3$ and $1/(2(1 + D))$ of the same factor, and they cancel at $D = 1/2$. The difference quotients on the solved junction follow that closed form to one per cent at every degeneration, and at a half they sit at $2 \times 10^{-7}$.

The pair’s coefficient is $-1/(12(1 + D)^3)$, one term, negative everywhere. A pair has no square to return: its transfer is odd, so the term that competes with the cubic in the single transistor does not exist. That is the same symmetry that removes its second harmonic, and here it has a cost. The single transistor can be degenerated to a point where its gain error is fifth-order, and the pair never can. Near $g_m R_e = 1/2$ the single transistor holds its gain to one per cent up to 71.8 mV where the pair manages 19.0, and at $g_m R_e = 1$ the single transistor still edges it.

From root two to two, and why #

The ratio of the two boundaries collects all of this into one curve.

Bare it is the 1.426 of the earlier essay. Near a degeneration of a half it falls to 0.265, the single transistor’s cancellation. Past about one it rises again, and at heavy degeneration it approaches two: 1.950 at $g_m R_e = 50$.

The two at the far end is not $\sqrt2$ grown, and it does not come from the junctions at all. With a large emitter resistor each circuit is linear until it runs out of current. A single transistor biased at $I_c$ cuts off when the drive pulls its current to zero, at a drive of $I_c R_e$, which is $(1 + D)$ thermal voltages in these units; the solved boundary at $g_m R_e = 50$ is 50.3, within a per cent of that limit’s 51. A pair with the same current in each device has steered its whole tail to one side when the differential current reaches $I_c$, and since the drive falls across both emitter resistors in series that takes $2 I_c R_e$ — twice as much. The pair’s boundary at fifty is 98.0 thermal voltages, and the ratio of the two is a ratio of resistances.

So the three regimes of the ratio have three different causes. Bare, the $\sqrt2$ is the pair dividing its drive between two exponentials and losing its square. At light degeneration the ratio is dominated by a cancellation that only the single transistor can have. At heavy degeneration it is the current each circuit has available against the resistance its drive meets. None of them is “the pair’s linearity” in the sense the phrase is usually meant.

What symmetry does buy #

Headroom in gain is not what a pair is for, and the last figure is the reminder.

At half a thermal voltage of drive, 12.9 mV, a bare single transistor puts out 12 per cent of second harmonic. Degeneration takes it down as the square of the factor — 3.1 per cent at $g_m R_e = 1$, 0.10 at ten, 0.013 at thirty — and it remains the single transistor’s largest distortion product throughout. The pair’s second harmonic is zero to $10^{-12}$ at every degeneration, and its third starts at 0.51 per cent and falls as the cube of the factor. What a pair cancels measured the even-order cancellation for a bare pair; with the resistors in, it holds unchanged, because a resistor in each emitter does not break the symmetry.

That is the clean division of labour. Feedback buys headroom and buys it for both circuits by nearly the same law. Symmetry buys the even orders and buys them outright. A designer choosing between a degenerated single stage and a degenerated pair for its gain accuracy is choosing between two things that are within a few per cent of each other over the useful range; choosing for its second harmonic, between a small number and zero.

A stage, worked #

Take a pair with a milliampere in each device. Its $1/g_m$ is 26 Ω at 27 °C, so a 26 Ω emitter resistor is a degeneration of one and a 78 Ω resistor a degeneration of three. Bare, the pair holds its gain to one per cent up to 10.4 mV peak. With 26 Ω in each emitter it holds to 29.2 mV and its gain is halved; with 78 Ω, to 80.9 mV with a quarter of the gain. A stage asked to take fifty millivolts with one per cent of gain accuracy needs $(1 + D)^{3/2} = 50/10.35$, a degeneration of about 1.86, or 48 Ω.

The single transistor built from the same device and the same resistor does as well or better up to a degeneration of about one, and the reason to use the pair anyway is the second harmonic in the last figure: 3.1 per cent from the single transistor at 12.9 mV with $g_m R_e = 1$, and nothing from the pair. The operating point is the same story. A single transistor’s mean current shifts with the drive, which is the boundary the point the device is never at measures, and a degenerated one shifts less; a pair’s mean does not shift at all.

The boundary that stops moving with the room #

The two regimes of the pair’s boundary have one more difference, and it matters to anything that must work over temperature. At light degeneration the boundary is $0.4\,V_T(1 + g_m R_e)^{3/2}$, and both $V_T$ and $g_m$ depend on temperature: the thermal voltage rises in proportion to the absolute temperature, and at a fixed current $g_m = I_c/V_T$ falls in the same proportion, so the degeneration factor falls as the device warms. The boundary moves with the room, as the edges that move with the room found for every junction-limited edge.

At heavy degeneration the boundary is set by $2 I_c R_e$, a current and a resistance, and the thermal voltage has left it. A pair degenerated deeply enough has a one-per-cent boundary that is as stable as its tail current and its resistors — which is to say it has exchanged a junction’s temperature coefficient for a resistor’s. That is one more sense in which heavy degeneration turns a stage limited by its device into one limited by its components, the change what a resistor in the emitter buys described for the gain itself, arriving here at the edge of the gain’s range.

What a designer should take #

For a degenerated pair, the one-per-cent gain boundary is $0.4\,V_T(1 + g_m R_e)^{3/2}$ while $g_m R_e$ is below about three, and beyond that it is set by the tail current: the drive can approach $2 I_c R_e$ before the gain fails, and no further. Do not expect the pair to add a factor of $\sqrt2$ to whatever the degeneration buys; at moderate degeneration it adds nothing, and a single transistor near $g_m R_e = 1/2$ holds its gain to larger drives than the pair. Choose the pair for its even orders, which it removes at every degeneration, and for its common-mode behaviour, not for its gain range.

A rule of thumb that takes a pair’s input range as $\sqrt2$ times a transistor’s, times whatever the degeneration buys, multiplies two numbers measured under different conditions. The first belongs to a bare pair and a tight criterion; the second to a single degenerated transistor. The combined circuit has neither: its boundary is its own curve, and the curve is cheap to solve and expensive to guess.

The general point is that two linearising mechanisms do not multiply unless they act on the same term. Symmetry removes the even terms; degeneration divides all of them and, in the single transistor, sets two odd contributions against each other. Applied together they act on different things, and the only honest way to know the combined boundary is to solve the combined circuit.

How the numbers were obtained #

Each circuit’s instantaneous transfer is solved at 1,024 points of a sinusoid by Newton’s method on its implicit equation, $\ln(1 + y) + D y = w$ for the single transistor, with its iteration held above cut-off, and $\operatorname{atanh} x + D x = u$ for the pair, held inside its tail. The fundamental and harmonics are read from a discrete transform of the solved waveform. The small-signal gains are $1/(1 + D)$ in each circuit’s own units. Every boundary is bisected in log drive, seventy iterations, on the magnitude of the fundamental’s departure. The cubic coefficients are fourth-order central differences at a drive of a hundredth of a thermal voltage. The temperature is 27 °C throughout.

What it leaves out #

The base resistance and the transistor’s finite β. Both add to the effective emitter degeneration, a base resistance divided by β being indistinguishable from part of $R_e$ at low frequency. They move $D$ rather than the argument.

Mismatched emitter resistors. Two emitter resistors that differ break the pair’s symmetry exactly as a mismatch between the transistors does, and bring back a second harmonic in proportion. The offset that brings the even orders back measures that for the transistors’ own mismatch; a resistor mismatch of $\epsilon$ is an offset of about $\epsilon$ times the signal’s share of the drop across the resistors, which grows with the drive rather than being fixed.

A field-effect pair. A MOS pair’s transfer is not a hyperbolic tangent but a square law that ends where one device turns off, and it is odd for the same reason, so the even orders vanish and the argument about symmetry carries over whole. The light-degeneration law does not: the cubic term of a square-law pair has a different coefficient and a different dependence on the overdrive, which the exponent that is a square is the place to start from. The heavy-degeneration limit, $2 I R_e$, is a statement about current and resistance and should hold for either device.

The tail’s own impedance. Every pair here has an ideal tail. A finite tail impedance turns a common-mode input into a differential error and, under large drive, lets the tail current change with the signal, which moves the current limit the heavy-degeneration regime is set by.

Still open: the mismatched resistors, the degenerated tail, and the fifth order at ½ #

Emitter resistors that do not match. The even orders a mismatch in the emitter resistors restores should grow with the drive rather than stay in a fixed ratio to the fundamental, since the offset they create is proportional to the current through them. Measuring whether that makes a resistor mismatch worse or better than a transistor mismatch of the same size, at a stated drive, would complete the pair’s matching budget.

A tail that is a resistor. Many pairs are fed from a resistor rather than a current source. The tail current then rises with the common-mode level and, under large differential drive, with the signal itself; where the current limit of the heavily degenerated pair goes, and whether the ratio of two survives, is a solve on the same transfer with one more equation.

What is left at ½. A single transistor degenerated to $g_m R_e = 1/2$ has no cubic, so its gain error is fifth order and its boundary is set by a coefficient nobody has drawn. That coefficient, and how sharply the boundary falls away either side of the cancellation — how precisely the resistor must be set to keep most of the benefit — would say whether the cancellation is a design point or a curiosity.