Two decibels of matching for one of gain

The offset that brings the even orders back found the closed form for what an input offset does to a differential pair: the second harmonic over the fundamental is $(u/2)\tanh(V_{os}/2V_T)$, with $u$ the drive over $2V_T$ — a single transistor’s second harmonic times the fraction of the tail the offset steers aside. It read the result backwards as a matching specification: −60 dB of second harmonic at 10 mV of drive allows 0.535 mV of offset. And it named the next measurement: with degeneration the second harmonic an offset restores should fall faster than the fundamental, by some power of $1 + g_mR_e$. Measuring it would say whether degenerating a pair relaxes its matching specification, and by how much for each decibel of gain given up.

Symmetry and a resistor together had already degenerated a balanced pair and found its third harmonic falling as the cube of $1/(1 + g_mR_e)$ and its second at zero at every setting. The present question is what the resistor does to the second harmonic an imbalance puts back.

The pair with both #

The pair is the one the earlier essays solved: two matched transistors at 27 °C sharing a tail, with an emitter resistor in each arm whose value is stated as the degeneration $D = g_mR_e$. Its output, as a fraction of the tail, solves $\operatorname{atanh}(x) + Dx = u + d$ exactly, where $d = V_{os}/2V_T$ is the offset in the same units as the drive — Newton on that one line gives the waveform, and the harmonics are read from it by a discrete Fourier transform, with nothing assumed about their size.

With a millivolt of offset and $g_mR_e = 1$, both harmonics are lower than the bare pair’s at every drive, and both curves have moved to the right by about the factor the gain has fallen — twice, here. The second harmonic still rises with drive, still peaks as the pair saturates and still falls again beyond; its peak is now 0.406 per cent at 177 mV of drive, against the bare pair’s 0.736 per cent at 76 mV.

The law, at small drive #

Where the drive is small the second harmonic has a closed form that includes the degeneration exactly. The transfer is the inverse of $g(x) = \operatorname{atanh}(x) + Dx$, and a small cosine of amplitude $u$ about the offset’s own operating point $x_0$ has a second harmonic, over its fundamental, of

$$\frac{h_2}{h_1} = \frac{u}{4}\,\frac{g''(x_0)}{g'(x_0)^2}, \qquad g' = \frac{1}{1-x_0^2} + D, \quad g'' = \frac{2x_0}{(1-x_0^2)^2}$$

At $D = 0$ this is $(u/2)\tanh d$, the earlier essay’s result. For a small offset $x_0 \approx d/(1 + D)$, and the expression becomes

$$\frac{h_2}{h_1} \approx \frac{u}{2}\cdot\frac{d}{(1+D)^3}$$

The figure checks the exact form against the solved waveform at the smallest drive drawn, to two per cent, at each degeneration.

Two factors of $1 + D$ are easy to see. The offset’s operating point is $d/(1 + D)$ rather than $d$, because the resistors reduce the imbalance the offset produces in the tail. And the curvature at that point is divided by the square of the slope, which the resistors have raised by $1 + D$. The third factor is the one that changes the specification.

Fixed input, and fixed output #

The cube is the fall at a fixed input. But degeneration also lowers the gain by $1 + D$, and a designer who degenerates a pair usually wants the same output as before and drives it harder to get it. At a fixed output the drive $u$ rises by $1 + D$ and one of the three factors is given back.

The measured slopes are −3.00 at fixed input and −2.00 at fixed output, between $g_mR_e = 1$ and 10, and the figure requires each to within 0.15. At a fixed output the second harmonic falls as $(1 + D)^{-2}$, and since the gain given up is $20\log_{10}(1 + D)$ decibels, every decibel of gain given up lowers the offset’s second harmonic by two decibels.

Read the other way, it is the matching specification relaxing. For a stated second harmonic at a stated output, the allowed offset rises as $(1 + D)^2$: the earlier essay’s 0.535 mV for −60 dB at 10 mV becomes 2.14 mV at $g_mR_e = 1$ and 8.6 mV at 3, twelve decibels of gain spent for a sixteenfold relaxation in matching. The two decibels per decibel is the exchange rate, and it holds wherever the drive is small against the degenerated pair’s own linear range — which, since that range has itself grown by $1 + D$, is a wider set of drives the more the pair is degenerated. The law is most useful exactly where it is most accurate.

Why the feedback gives back one factor and keeps two #

The exchange rate is better than the usual account of feedback predicts, and the difference is instructive. The standard statement — how much of the amplifier gets through makes it precisely — is that feedback divides a forward element’s variation by one plus the loop gain, and that at a fixed output a distortion generated inside the loop is divided by the same factor once. On that account the offset’s second harmonic should fall as $(1 + D)^{-1}$ at a fixed output, one decibel per decibel of gain.

It falls as $(1 + D)^{-2}$ because the offset is not a distortion generated inside the loop. It is an input to it. The resistors feed back the imbalance itself: the offset’s operating point is $d/(1 + D)$ rather than $d$, so the asymmetry that produces the second harmonic is reduced by the loop before it produces anything. The distortion that asymmetry then generates is reduced by the loop a second time, as any distortion is. One factor for the cause and one for the effect.

That is also why the single degenerated transistor does worse. Where the two exponents come from found its distortion edge moving as the square of the degeneration factor, which is its second harmonic falling as $(1 + D)^{-2}$ at a fixed input — one power fewer than the pair’s with an offset — and so as $(1 + D)^{-1}$ at a fixed output, the textbook rate. A single transistor’s second harmonic is intrinsic to its exponential; there is no imbalance for the resistor to reduce first. A pair with an offset has borrowed its second harmonic from an asymmetry, and asymmetries are what negative feedback on the tail’s division reduces.

Why the third harmonic does not change the answer #

The balanced pair’s own third harmonic falls under degeneration too, and it is worth checking that it does not overtake the second in the regime where the exchange rate applies — if it did, relaxing the matching would buy nothing, because the distortion would be the third harmonic’s.

At a fixed input the pair’s third harmonic falls as $(1 + D)^{-3}$, like the second; at a fixed output, where $u$ has risen by $1 + D$, it falls as $(1 + D)^{-1}$, since it grows with the square of the drive. So under degeneration at fixed output the second harmonic falls faster than the third, by one power of $1 + D$. A pair whose distortion was set by its offset, with the second above the third, moves towards being set by the third as it is degenerated, and the offset at which the two are equal rises with it. That is the same statement from the other side: degeneration relaxes the matching by more than it relaxes the pair’s intrinsic linearity.

The worst case over every drive #

The exchange rate is a small-drive law. The earlier essay’s other number was a large-signal one: the offset below which the second harmonic never reaches one per cent at any drive, because the harmonic rises, peaks as the pair saturates, and falls.

That bound relaxes too, and much more slowly: from 1.36 mV bare to 2.46 at $g_mR_e = 1$ and 4.39 at 3, roughly as the 0.85 power of $1 + D$ rather than the square. The reason is where the peak sits. The worst case over drive is at the edge of saturation, and at the edge of saturation the emitter resistors are carrying the whole tail in one arm and none in the other; the feedback that made the small-signal law work is not operating on a signal any more, it is operating on a switch. Degeneration moves the saturation edge outwards — from 76 mV to 328 mV of drive — and lowers the peak, but not by the small-signal factor.

So the two numbers that describe a pair’s matching tolerance diverge under degeneration. The small-signal specification — what offset for what distortion at a working amplitude — relaxes as $(1 + D)^2$. The large-signal one — what offset keeps the second harmonic under one per cent whatever happens — relaxes as roughly $(1 + D)^{0.85}$. A design that is judged at its working level gains a great deal; one that must survive overload gains much less.

A correction to the earlier boundary #

Measuring the large-signal bound at every degeneration meant searching drives out to where each pair saturates, and that search found an error in the undegenerated number the earlier essay printed.

The earlier figure located the peak of the second harmonic on a grid of drives that stopped at 55 mV, and its own description said the peak sits near 76 mV. The peak was outside the grid; the search read the highest point on its rising side instead, and the offset at which that point touched one per cent came out as 1.44 mV. With the grid extended past the peak — the same grid, continued to 318 mV — the bound is 1.36 mV. The difference is small and in the flattering direction: a pair with 1.40 mV of offset does reach one per cent, at 76 mV of drive, which the earlier figure said it never would.

The crossover at 3.14 mV, where the second harmonic reaches one per cent before the third, is unchanged: it lies at drives of about 18 mV, inside the grid either way. The earlier essay’s text and figure now carry the corrected number. The shape of the mistake is worth carrying: a search that is required to find a maximum must be shown to have bracketed it, and the new figure requires, at every degeneration, that the peak it reports lies inside the drives it searched.

What the exchange rate is worth #

For a pair designed to a distortion at a working level, degeneration is the cheapest matching there is. Two decibels of matching tolerance for each decibel of gain, where the alternative is a trimmed or a larger, better-matched pair, is a good exchange whenever the gain can be spent — and what a resistor in the emitter buys is the linearity that comes with it.

For a pair that has to behave in overload, the exchange rate is poor, and the matching has to be bought directly. The large-signal bound relaxes as a weak power of the degeneration because the resistors cannot linearise a pair that has switched.

For a pair inside a larger loop, the exchange compounds. An operational amplifier’s input pair sits inside the overall feedback loop, which divides whatever second harmonic the pair produces by its own loop gain at the signal frequency; the local degeneration has already divided it by $(1 + D)^2$ before the global loop sees it. The two are independent, so they multiply, and the global loop’s contribution falls with frequency as its gain does while the local one does not — which is why a degenerated input pair keeps an amplifier’s even-order distortion low at high frequencies where the overall loop has little left to give.

And for a pair whose offset is trimmed, the relevant offset is the residual after trimming, and the exchange rate multiplies it: a trim to 0.3 mV on a pair degenerated to $g_mR_e = 2$ behaves, for second harmonic at a fixed output, like 33 µV on a bare pair — a factor of nine.

A population of pairs, not one #

A matching specification is written for a population: pairs come off a process with offsets spread about zero, and the specification says what fraction of them meets the distortion target. The exchange rate acts on that population directly.

With the second harmonic proportional to the offset, a population whose offsets have a standard deviation $\sigma$ has second harmonics with a proportional spread, and the fraction meeting a target is the fraction with $|V_{os}|$ below the allowed offset. Degenerating to $g_mR_e = 1$ multiplies the allowed offset by four at a fixed output, which for a normal spread moves the target from, say, one standard deviation — 68 per cent of pairs passing — to four, where one pair in sixteen thousand fails. The error that is a distribution is the arithmetic of that tail for a current mirror’s mismatch, and it applies unchanged, since the second harmonic is linear in the offset at a working level.

The other thing a population brings is the choice between buying the tail and trimming it. The mismatch that cancels itself finds arrangements in which a mismatch’s effect is removed by construction; the exchange rate here is the cheaper route when gain is available, and the only route when trimming is not.

How the harmonics were measured #

Every waveform is solved: the pair’s transfer $\operatorname{atanh}(x) + Dx = u\cos\theta + d$ by Newton at 1,024 points of a cycle, and the harmonics taken from it by a discrete Fourier transform. The small-drive law is checked against it at the smallest drive in each figure, to two per cent. The slopes are measured between $g_mR_e = 1$ and 10 on solved harmonics, the fixed-output drive found by bisection on the solved fundamental.

The large-signal bound is the offset, found by bisection, at which the largest second harmonic over a grid of 170 drives, scaled with the degeneration so that every grid reaches past its pair’s saturation, is one per cent. The figure requires that each maximum lies inside its grid, and that the bound rises with every added degeneration.

An offset that is only a voltage at the input #

The offset is a voltage at the input. A real pair’s offset comes from its transistors’ saturation currents and its load resistors, and the two sources place the imbalance differently; an offset from the loads is not reduced by degeneration at all, since it is after the transistors.

The emitter resistors match. A mismatch between the two emitter resistors is itself an offset — a current offset rather than a voltage one — and it grows with the degeneration it was added to provide.

And the transistors are ideal exponentials. Their base resistance and the Early effect add second-order terms of their own, and where the mechanisms are one mechanism is how the site treats several of those at once.

Still open: the single-ended drive, the mismatched gains, and the resistor mismatch #

A single-ended drive with a real tail. Driving one input and holding the other moves the tail node, and through the tail’s finite impedance the tail current follows the square of the drive. That puts a second harmonic into a perfectly matched pair; its size against the offset’s contribution, as a function of the tail impedance and now of the degeneration, decides whether matching or the tail limits a single-ended stage.

Current gains that differ. A difference in β between the two devices loads a source unequally and becomes an offset that depends on the signal. Whether that restores even orders faster or slower than a fixed offset, and whether degeneration buys the same two decibels against it, is a solve with the source resistance in it.

The emitter resistors’ own mismatch. At $g_mR_e = 3$ each emitter resistor carries three quarters of the arm’s voltage-to-current conversion, so a one-per-cent mismatch between them is an imbalance comparable to a small $V_{BE}$ mismatch. The degeneration that relaxes the transistors’ matching tightens the resistors’, and the crossover between the two is a number.