The resistor mismatch that is only an offset

Two decibels of matching for one of gain found the exchange rate a designer most wants for a differential pair. An input offset brings back the second harmonic the pair’s symmetry removed; emitter resistors divide that second harmonic by the square of one plus $g_mR_e$ at a fixed output; so every decibel of gain given up buys two decibels of distortion, or allows two decibels more offset at the same distortion. A matching specification that allowed 0.535 mV at −60 dB undegenerated allows 2.14 mV at $g_mR_e = 1$.

The offset in that calculation was a property of the transistors — a mismatch in their saturation currents, which is a fixed number of millivolts at the input whatever surrounds them. The emitter resistors were taken as equal. They are parts with a tolerance like any other, and at $g_mR_e = 3$ each carries three quarters of its arm’s input voltage, so a one per cent difference between them is not obviously small beside a transistor mismatch. The question the earlier essay left was how a resistor mismatch enters, and whether degeneration relaxes it the way it relaxes the transistors’.

It enters in the simplest way there is, and that simplicity is what makes it behave differently. A mismatch between the emitter resistors is an input offset — exactly, with no slope term, no drive dependence and nothing a trim leaves behind — but an offset whose size is proportional to the degeneration that is meant to suppress it. The two decibels per decibel survive for the transistors and become one for the resistors.

The arithmetic of one arm #

Write the pair with a tail current $I$, arm currents $i_1$ and $i_2$ that sum to it, and emitter resistors $R_{1,2} = R(1 \pm \delta/2)$. Around the input loop the differential input is the difference of the two base–emitter voltages plus the difference of the two resistor drops:

$$v = V_T\ln\frac{i_1}{i_2} + i_1R_1 - i_2R_2.$$

The resistor drops expand into two parts:

$$i_1R_1 - i_2R_2 = R(i_1 - i_2) + \frac{R\delta}{2}(i_1 + i_2).$$

The first is the balanced pair’s degeneration term. The second contains the sum of the arm currents, and an ideal tail fixes that sum at $I$ at every instant, whatever the input. So the second part is a constant, $IR\delta/2$, and the mismatched pair’s transfer is the balanced pair’s with its input moved by that much. In the units the earlier essays used, where $g_m = I/2V_T$ for each device at balance, the offset is

$$V_{os} = \frac{I}{2}R\,\delta = V_T\,g_mR_e\,\delta.$$

There is no slope term. It is natural to expect one: a resistor mismatch makes one arm’s degeneration stronger than the other’s, and that sounds like a difference of gain between the halves, which would distort asymmetrically. It does not, because the two arms are not separate amplifiers. Whatever current leaves one arm arrives in the other, and the total voltage the resistors drop on a difference of current is set by their sum, $R_1 + R_2 = 2R$, which the mismatch leaves alone. What the mismatch changes is the drop on the common current, and the tail holds that still.

Two routes to one waveform #

That derivation forms the offset algebraically, so the check has to avoid doing the same. Every waveform in this essay is solved arm by arm: for each instant of the drive, one arm’s current is bisected in amperes until the input loop closes with the two resistors at their own values, and the other arm takes the rest of the tail. Nothing in that solve knows that the mismatch is an offset.

At $g_mR_e = 3$ on a one-milliamp tail, the resistors are 155 Ω each, and a one per cent difference between them is 1.55 Ω. The prediction is an offset of $0.5\,\text{mA} \times 1.55\,\Omega = 0.776$ mV. The arm-by-arm solve and the balanced pair moved by 0.776 mV give transfers that differ by $4.4\times10^{-16}$ of the tail anywhere between ±0.4 V of input — the last bit of a double — and second harmonics that agree to a part in ten thousand at every drive from one millivolt to four hundred, where the pair is saturated.

The decisive test is the trim. If the mismatch had any slope component, subtracting the offset at the input would remove the DC shift and leave an even-order residue that grew with the drive. Subtract exactly $V_T g_mR_e\delta$ and the second harmonic drops below $10^{-9}$ at every drive, which is the arithmetic’s floor for a harmonic read off 512 samples. Nothing survives the trim.

That gives a matching budget its simplest possible form. The transistors contribute an offset $\Delta V_{BE}$, which what matching does about temperature wrote as $V_T\ln$ of the saturation-current ratio; the resistors contribute $V_T g_mR_e\delta$; and the pair behaves exactly as the balanced pair with the sum at its input. Everything the offset that brings the even orders back and its successor measured about an offset applies to that sum unchanged.

The offset that grows with what divides it #

The difference from a junction’s offset is in the factor $g_mR_e$. A transistor mismatch is so many millivolts at any degeneration. A resistor mismatch is zero with no degeneration — there are no resistors to mismatch — and proportional to the degeneration after that. Meanwhile the degeneration divides an offset’s effect.

At a fixed input the earlier essay found the second harmonic of a fixed offset falling as $(1 + g_mR_e)^{-3}$. For the resistors’ offset the same law gives a second harmonic proportional to

$$\frac{g_mR_e}{(1 + g_mR_e)^3},$$

which starts at zero, rises, and falls again.

The peak is where the derivative of $D/(1+D)^3$ vanishes, $(1 + D) = 3D$, so at $g_mR_e = \tfrac12$ exactly, and the solved harmonics put it there. At 5 mV of drive a one per cent mismatch peaks at 0.358 parts in ten thousand of second harmonic there, far below the junction’s contribution at that degeneration — a half-millivolt $V_{BE}$ mismatch is a much larger offset than the 0.13 mV a one per cent resistor mismatch makes at $g_mR_e = \tfrac12$. The point of the figure is the shape, not the size: degenerating a pair first creates a resistor-mismatch distortion that was not there, and only later begins to suppress it, and it suppresses it one power of $(1 + g_mR_e)$ more slowly than it suppresses the transistors’.

One decibel for one, at a fixed output #

The comparison that matters for a design is at a fixed output, since a degenerated stage is driven harder to make up its gain. There the earlier essay’s law for a fixed offset is $(1 + g_mR_e)^{-2}$, and the resistors’ offset turns it into $g_mR_e/(1 + g_mR_e)^2$ — a curve whose local slope against $1 + g_mR_e$ on logarithmic axes is $1/g_mR_e - 1$. That tends to −1: each decibel of gain given up buys one decibel of distortion, not two.

The hero figure measures both. Between $g_mR_e = 5$ and 20 the junction’s second harmonic falls with slope −2.00 and the resistors’ with −0.89, the latter still on its way to −1, as the local slope says it should be at $g_mR_e$ near ten. The total, for a pair carrying both, follows the junction’s line at light degeneration and the resistors’ at heavy, and bends between them at the degeneration where the two offsets are equal:

$$g_mR_e^{*} = \frac{\Delta V_{BE}}{V_T\,\delta}.$$

For half a millivolt of $V_{BE}$ mismatch and one per cent resistors that is 1.93. Below it, the transistors set the even orders and the two-for-one exchange of the earlier essay holds. Above it, the resistors set them and the exchange is one for one.

What a resistor in the emitter buys priced degeneration as gain traded for linearity; symmetry and a resistor together found what it does to the pair’s odd-order boundary. Neither saw this, because both took the resistors as equal, and with equal resistors there is nothing for the resistors to contribute to the even orders at all.

Why the arms cannot disagree about their gain #

The expectation of a slope term is worth taking apart, because it is the expectation most people bring to the circuit and it fails for a reason that generalises.

Treat each arm as a degenerated single transistor and the mismatch looks like a gain error: the arm with the larger resistor has the smaller transconductance, $g_m/(1 + g_mR_1)$ against $g_m/(1 + g_mR_2)$, and a pair whose halves have different gains is a pair whose output is not an odd function of its input. That reasoning would be right for two transistors driven from two separate sources and summed afterwards. It is wrong for a pair, because a pair has one input voltage and one current to divide. Any change in $i_1$ is an equal and opposite change in $i_2$, so the voltage the resistors develop against a change of current is always $R_1 + R_2$ times the change — the series resistance of the two, which does not care how it is split. The split matters only to the voltage each resistor develops against the current both carry in common, and in a pair with an ideal tail that current never changes.

The same reasoning predicts the case where the slope term does appear. Drive the two transistors from one input but give them separate current sources, so that each arm’s quiescent current is fixed independently: the arms then no longer trade current, each is a single degenerated transistor, and a resistor mismatch is a mismatch of two gains. The pair removes the slope term by forcing its arms to share one current, which is the same constraint that removed its even orders in what a pair cancels — and it removes it for the same reason, so long as the constraint holds exactly.

Where the resistors take over #

The crossing is a ratio of two offsets and contains nothing about the drive, which makes it a design number rather than an operating condition.

The crossings were found without the closed form: at each resistor tolerance and junction mismatch, the degeneration was bisected on the condition that the two solved second harmonics are equal, at a drive raised in proportion to $1 + g_mR_e$. They sit on $\Delta V_{BE}/(V_T\delta)$ to two per cent across two decades of resistor tolerance and a decade of junction mismatch. A pair of well-matched transistors, 0.3 mV, with one per cent resistors hands control of its even orders to the resistors at $g_mR_e = 1.16$ — a modest degeneration, the kind used in any stage that wants its gain set by resistors. Tightening the resistors to 0.1 per cent moves that to 11.6, and with a millivolt of transistor mismatch to 38.7, at which point the resistors never matter in practice.

The pattern is the one a ladder is not a cascade and the tolerance essays found in filters: the component that sets a specification depends on how the network distributes its sensitivity, and a design change that desensitises one part can expose another. Here degeneration desensitises the pair to its transistors by moving the voltage from the junctions into the resistors, and the resistors inherit the matching problem in proportion.

What a distortion specification asks of the resistors #

The earlier essay turned the exchange rate into a matching specification for the transistors. The same can be done for the resistors, with one change: their allowance is the total offset budget divided by $V_T g_mR_e$, so it carries the degeneration in its denominator.

For −80 dB of second harmonic at the output a bare pair gives with 10 mV of drive, the total input offset allowed is 0.054 mV undegenerated, 0.213 at $g_mR_e = 1$, 0.842 at 3 and 6.32 at 10 — growing as $(1 + g_mR_e)^2$, the earlier essay’s law at a tenth of its specification. Divided by $V_T g_mR_e$, the resistor mismatch that would by itself use that budget is not defined undegenerated, since there is no resistor; it is 0.823 per cent at $g_mR_e = 1$; and 1.08 per cent at 3 and 2.44 at 10. The tightest point is where $(1 + D)^2/D$ is least, at $D = 1$ exactly, and there the allowance is four times the bare pair’s offset allowance over the thermal voltage.

That has a practical edge. A designer who adds degeneration to relax a pair’s matching, and specifies one per cent resistors because resistors are cheap to match, has a stage at $g_mR_e = 1$ whose resistors alone take more than the whole −80 dB budget even with perfect transistors. Share the budget — give 0.04 mV of it to the junctions — and the resistors get nothing at all below $g_mR_e = 0.25$ and 0.623 per cent at their tightest. The loosening beyond $g_mR_e = 1$ is real but only linear: a factor of ten more degeneration buys a factor of three in resistor tolerance, where it bought a factor of thirty in transistor tolerance.

A trim that works, and the conditions it assumes #

Because the mismatch is exactly an offset, an offset trim removes it exactly — not only its DC shift but every even harmonic it causes, at every drive, as the second figure measured. That is better news than the expectation of a slope term would have allowed, and it is the reason a pair with poorly matched emitter resistors can still be trimmed to low distortion.

Three assumptions carry that conclusion, and each is a place where a real stage can depart from it.

The tail is ideal. The whole argument rests on $i_1 + i_2$ being constant. A tail made from a resistor, or a current source with finite output impedance, lets the sum move — with the common-mode level, and under a large differential drive with the signal’s own even harmonics, since the tail node rides up and down at twice the signal frequency. The mismatch term then becomes $(R\delta/2)(i_1 + i_2)$ with a varying sum: a drive-dependent offset, which a fixed trim cannot remove, and whose size is the resistor mismatch times the tail’s own second-harmonic current.

The base currents are zero. With finite β the emitter currents are $(1 + 1/\beta)$ times the collector currents, and the emitter currents are what flow in the resistors. A β mismatch between the two devices then turns the equal collector currents of balance into unequal emitter currents, which the resistors convert into one more offset. It is still a constant for an ideal tail, and still trimmed with the rest.

The resistors do not drift apart. A trim made at one temperature removes the offset at that temperature. The resistors’ offset is $(I/2)R\delta$, which drifts with the tail current and with any difference between the two resistors’ temperature coefficients; the junctions’ is $V_T$ times a logarithm, which drifts as absolute temperature. The trim that takes the drift with it found that a trim cancels a junction’s drift only when it acts inside the same logarithm; a trim that cancels a resistor mismatch’s offset by moving the junctions’ does not share its temperature law at all.

The pair’s matching, in one line #

With an ideal tail, a pair whose emitter resistors differ by δ is exactly the balanced pair with an input offset of $V_T g_mR_e\delta$: 0.776 mV for one per cent at $g_mR_e = 3$, confirmed to $4\times10^{-16}$ of the tail by a solve that never formed it, and removed completely by a trim of that size. Its second harmonic is largest at $g_mR_e = \tfrac12$ for a fixed input and falls only as $(1 + g_mR_e)^{-1}$ at a fixed output, against a junction offset’s $(1 + g_mR_e)^{-2}$, so the resistors take over above $g_mR_e = \Delta V_{BE}/(V_T\delta)$ — 1.93 for half a millivolt and one per cent. At −80 dB and a fixed output, perfect transistors still need resistors matched to 0.823 per cent at $g_mR_e = 1$.

Still open: a tail that moves, a single-ended drive, and the resistors’ own heating #

A tail that moves. With a resistor in the tail the sum of the arm currents follows the common mode and the drive’s own second harmonic, and the resistor mismatch becomes an offset that moves with the signal. The size of the second harmonic that leaves after a perfect DC trim, as a function of the tail resistance and the degeneration, would say how good a tail has to be before the trim result above is worth having.

A single-ended drive. Two boundaries removed, and one moved measured the pair driven differentially. Driving one input and holding the other moves the tail node at the signal frequency, so the tail’s finite impedance converts the drive into common-mode current directly; whether the resistors’ mismatch then contributes a second harmonic proportional to the drive rather than a fixed offset is the next solve on the same arm-by-arm model.

The resistors heating unequally. The arm currents swing in antiphase, so under drive each emitter resistor dissipates a power that follows the signal. If the two have different thermal resistances to their surroundings, their values separate at the signal’s own rate and the mismatch δ becomes a function of time. Whether that thermal distortion, which how small is small signal would count as another departure from the small-signal model, ever exceeds the static mismatch’s for a practical layout is a measurement with a thermal time constant in it.