The trim that takes the drift with it

What matching does about temperature solved a bipolar pair whose two transistors differ in saturation current by a ratio $m$ and found that the input offset is exactly $V_T \ln m$ — proportional to the thermal voltage and so to absolute temperature. It drifts in proportion to itself, 3,333 parts per million per kelvin at 300 K, whatever the mismatch, and that essay drew from the proportionality a consequence it put in bold: trimming the offset to zero at one temperature trims the drift to zero with it. A quantity proportional to $T$ with its constant of proportionality set to zero is zero at every temperature.

The sentence is true of the offset that essay modelled. It is not true of every trim, and it is not true of every offset, and the two failures are the same failure seen from two sides. Whether a trim takes the drift with it depends on where the trim is applied — inside the junction law or outside it — and on what the offset is made of. A pair whose offset has two causes with two different temperature laws cannot be trimmed by one knob at all, only traded between them, and the knob that is right depends on which cause is larger. This essay solves the pair again with both causes in it and trims it three ways.

A trim inside the logarithm and a trim after it #

The pair here is a bipolar pair at 10 µA a side from a 20 µA tail, with collector loads of 10 kΩ. The second transistor has $m$ times the first’s saturation current. At each of thirty-one temperatures from 250 to 400 K the pair is solved from the junction laws — each collector current by Newton’s method on its own base-emitter equation, the shared emitter node by bisection on the tail current — and then the differential input that brings the two collector voltages level is found by bisection on the output. Nothing in that solve is the closed form; the closed form is what it is compared with.

There are two places to put a trim. One is after the pair: subtract a constant equal to the offset at 300 K, as an offset adjustment summed into a later stage or a nulling voltage applied at an input does. The other is in the pair: make the collector loads unequal so that the pair itself balances at 300 K. A load ratio $r = R_{c1}/R_{c2}$ makes the nulled output demand collector currents in the ratio $1/r$, and the input that achieves it is

$$V_{os} = V_T \ln\frac{m}{r}.$$

The load ratio sits inside the same logarithm as the mismatch. Set $r = m$ and the offset is $V_T \ln 1$, which is zero at every temperature, because no temperature appears anywhere except in the factor in front.

At 2% mismatch the untrimmed pair needs 511.9 µV at 300 K and 682.6 µV at 400 K. The constant trim makes it zero at 300 K, and from there it drifts exactly as it did before trimming, 1.706 µV/K, reaching 170.6 µV at 400 K and −85.3 µV at 250. The figure’s solved residual follows $V_{os}(300)\,(T/300 - 1)$ to a nanovolt at every one of the thirty-one temperatures. The load trim lands on a ratio of 1.02000 — the bisection finds the mismatch itself, to five figures — and leaves the pair inside a nanovolt of zero across the whole 150 K.

The slider runs the mismatch from half a per cent to ten. At ten per cent the untrimmed offset is 2.464 mV and the constant trim leaves 8.213 µV/K; the load trim is still exact, because the argument never depended on the mismatch being small. So the bold sentence in the earlier essay needs a qualifier it did not carry. Trimming the offset trims the drift when the trim is a ratio applied inside the junction law. Trimming with a voltage — anywhere outside the pair, or at its input as a fixed source — removes the offset at one temperature and leaves every microvolt per kelvin of its drift.

That is why precision amplifiers whose offset can be adjusted route the adjustment to the input pair’s collector or drain loads rather than to a summing node, and why their data sheets warn that nulling the offset by any other route degrades the drift by about 3.3 µV/K for every millivolt nulled. The 3.3 is the 1/T of the earlier essay, and the warning is this figure.

When the offset is not all one thing #

The earlier essay’s list of what it left out named the first complication. A bipolar input draws base current, and when each input is driven through a source resistance the base currents drop a voltage across it. If the two transistors’ current gains differ, the two drops differ, and that difference is an offset with a temperature law of its own.

Current gain rises with temperature, roughly as $T^{1.5}$ in an ordinary small-signal transistor. At a fixed collector current the base current falls as $T^{-1.5}$, and so does the offset it causes: it is largest cold and smallest hot, the opposite sense to the saturation-current share, and nothing about it is proportional to absolute temperature.

Give the pair current gains of 200 and 220 at 300 K — a ten per cent mismatch, ordinary between two devices that were not selected for it — and drive both inputs through 20 kΩ. The untrimmed offset is now 602.4 µV at 300 K, of which 90.5 µV is the base currents’. The constant trim leaves 1.256 µV/K: the saturation-current share’s 1.706 less the base-current share’s own fall of about 0.45. The load trim lands on a ratio of 1.02273 instead of 1.02000 — it is now nulling both shares — and leaves −0.574 µV/K, reaching −47.2 µV at 400 K.

The load trim’s residue is the interesting one, because it is not small for lack of trying. The trim set a term proportional to absolute temperature equal and opposite to the base-current share at 300 K. That term rises as $T$; the share it opposes falls as $T^{-1.5}$. Away from 300 K they part company in opposite directions, so the trim has not merely failed to remove the base-current share’s drift; it has added a second drift of the same sign on top of it.

What the load trim converts #

That reasoning gives the residual drift in one line. After a load trim at 300 K, the pair’s offset is a PTAT term pinned at $-V_b$ plus the base-current share $V_b$, and its slope is

$$\frac{dV_{os}}{dT} = -\frac{V_b}{T} + \frac{dV_b}{dT} = -(1 + 1.5)\,\frac{V_b}{T}.$$

The solved pair agrees with that law to within two per cent at each of four source resistances. At 5 kΩ the base currents’ share of the trimmed pair’s offset is 17.8 µV and the residual drift −0.148 µV/K; at 10 kΩ, 35.2 µV and −0.292; at 20 kΩ, 69.1 µV and −0.574; at 50 kΩ, 164.1 µV and −1.364. At 50 kΩ the pair that was trimmed to zero at 300 K is 112 µV out at 400 K. The 2.5 is the sum of two exponents: the 1 belongs to the trim, which is PTAT, and the 1.5 to the current gain.

A subtlety in the numbers is worth stating, because it catches the eye: the base-current share in the trimmed pair is smaller than in the untrimmed one, 69.1 against 90.5 µV at 20 kΩ. The trim moves the collector currents apart by two per cent to balance the loads, and the base currents move with them, so a little of the β mismatch’s effect is offset by the collector-current mismatch the trim introduced. The law is right in terms of the share that is actually there after trimming, which is the share the figure reports.

Which knob, then #

If the load trim converts the base-current share and the constant trim leaves the saturation-current share, the right single knob is the one that matches the larger share.

Sweep the source resistance from zero to 150 kΩ. With no source resistance the load trim leaves nothing and the constant trim 1.706 µV/K. As the source resistance rises the load trim’s residue grows steadily — 0.292 µV/K at 10 kΩ, 1.110 at 40, 3.502 at 150 — and the constant trim’s falls, because the base-current share’s negative drift is eating into the saturation-current share’s positive one: 1.481 at 10 kΩ, 0.806 at 40, and nearly zero near 80 kΩ, where the two shares’ drifts happen to cancel, before it rises again to 1.671 at 150 kΩ. The two cross at 33.8 kΩ, where the base currents contribute 152.7 µV of the untrimmed offset against the saturation-current mismatch’s 511.9.

The crossing is not at equal shares, and the reason is the 2.5. Each microvolt of base-current share costs the load trim 2.5 units of drift but costs the constant trim only 1.5, while each microvolt of saturation-current share costs the constant trim 1 and the load trim nothing. So the load trim stays ahead until the base-current share is about three-tenths of the saturation-current share, which is what the crossing shows.

The near-zero at 80 kΩ is a trap rather than a design point. It is a cancellation between a positive drift and a negative one, and the negative one depends on the sign of the current-gain mismatch. Swap which transistor has the higher gain and the two drifts add at every source resistance. A population of pairs has both signs, so across a batch the constant trim at 80 kΩ has no special virtue: the pairs with one sign of mismatch land near zero and the pairs with the other land at the sum of the two drifts.

Two knobs, set for value and slope #

The general fix is not to choose a knob but to use two. Two trims with two different temperature laws — one inside the logarithm, proportional to absolute temperature, and one a constant — have two degrees of freedom, and two conditions can be imposed at 300 K: zero offset and zero drift. Set the load ratio so that the slope of the offset at 300 K is zero, then subtract a constant equal to the offset that remains.

At 20 kΩ the load ratio for zero drift is 1.01554 rather than the 1.02273 that zeroes the offset, and the constant that then nulls the offset is 189.0 µV. The pair is left 4.8 µV out at 250 K and 11.2 µV at 400 K, against −47.2 µV for the load trim alone and 139.0 µV for the constant alone. Both ends have the same sign, which identifies what is left: it is curvature. The base currents’ $T^{-1.5}$ is not a straight line, and two knobs set at one temperature can remove a value and a slope, not a bend.

That is the shape of every two-temperature trim procedure. A part is measured at two temperatures, one knob absorbs the offset and the other the drift, and the specification that follows is a box that the residual curvature has to fit inside rather than a line. The knobs need different temperature laws for the procedure to be possible at all: two constants, or two ratios inside the logarithm, give one degree of freedom twice and cannot null a slope and a value together.

A trim is a term with a temperature law #

The general statement is short. An offset is a sum of shares, each with its own temperature law. A trim is one more term, and it has a temperature law too — PTAT when it acts inside the junction law, constant when it acts outside it, and whatever the trimming element’s own coefficient is when that element is a resistor or a reference. Trimming at one temperature sets the trim’s term equal and opposite to the sum of the shares there. Everywhere else, the residue is the shares’ drift minus the trim’s, and it is zero only when every share and the trim follow one law.

What matching does about temperature had one share and chose, without needing to say so, the trim that followed its law. The pair here has two shares, and no single trim follows both; the best one can do is match the larger share, or use two trims and pay only the curvature. The same reasoning reaches the other shares that essay listed. Load resistors that do not track each other add a share with their own coefficient, which is a ratio drift of the kind The small leg of a large ratio followed through a divider. Package stress adds one that depends on assembly. Each needs a trim with its law, or it is converted.

How the numbers were obtained #

Every offset is the input that nulls the solved pair’s output: Newton’s method on $v_s - V_e = V_T \ln(I_c/I_s) + R_s I_c/\beta$ in the logarithm of the collector current, which is monotonic and converges from any start; bisection on the emitter node until the emitter currents sum to the tail; and bisection on the differential input until the collector drops are equal. Saturation current scales with temperature as $T^3 e^{-E_g/kT}$ with a 1.12 V gap, identically for both transistors, so it changes the junction voltages but not their difference. Current gain scales as $(T/300)^{1.5}$. The trims are found the same way — the load ratio by bisection on the offset at 300 K, or on its slope for the two-knob trim, the slope by a central difference two kelvin either side — and then the trimmed pair is re-solved at every temperature. The closed forms in the text are compared with those solves: the load trim’s ratio against $m$ to five figures, the constant trim’s residue against $V_{os}(300)(T/300 - 1)$ to a nanovolt, and the converted drift against $-2.5\,V_b/T$ to two per cent.

What it leaves out #

The tail current’s own temperature law. The tail here is a constant 20 µA. Many pairs are biased with a current proportional to absolute temperature so that their transconductance holds still, and that changes the base-current share’s law — its collector current then rises as $T$, so its base current falls as $T^{-0.5}$, and the 2.5 becomes 1.5. The method is unchanged; the numbers are not.

Bias-current cancellation. Precision bipolar inputs often supply each base’s current from an internal source, leaving only the mismatch between the source and the base to flow in the external resistance. That shrinks $V_b$ and changes its law to whatever the cancellation’s error follows.

Resistor coefficients. The collector loads here are ideal, so the load trim’s ratio does not move with temperature. A real ratio moves by the two loads’ coefficient mismatch, and the load trim then carries that drift with it — a share entering through the trim itself.

A population. One pair is solved, with one sign of current-gain mismatch. The conclusion about which trim to choose depends on the ratio of the two shares, which in a population is itself spread; the crossing at 34 kΩ is for this pair.

Still open: curvature, a population, and the trim’s own drift #

The curvature budget. Two knobs leave 11.2 µV of curvature at 400 K for a 20 kΩ source. Whether a third term — a trim proportional to $T^2$, or a second load ratio set at a second temperature — can absorb it, and what the specification box looks like after a three-point trim, is the next measurement on this pair, and it is the same question a band-gap reference asks of its own curvature.

A population of pairs. Every pair in a production run has its own saturation-current mismatch and its own current-gain mismatch, of either sign. The distribution of drift after each trim strategy — and the yield against a drift specification — would turn the crossing here into a decision rule over a batch rather than one device, and it is a correlated-tolerance problem of the kind the divider essays have been measuring.

The trimming element. A load trim made with a resistor of a different film, or a fuse-linked resistor network, has its own coefficient, and so the trim adds a share with that coefficient’s law. How closely the trim network has to track the loads for the load trim’s advantage to survive is a number this model can give by adding one more coefficient, and it is the number a layout is built against.

The thread runs back through the pair. What a pair cancels, and what it only halves found that the pair’s symmetry removes even harmonics exactly; Two millivolts a kelvin, and the wrong sign measured the drift a single junction has and the pair removes; The offset that brings the even orders back showed what any residual offset does to that harmonic cancellation; and The base current the mirror counts twice met the base current as an error in a copy rather than an offset. The rejection the parts have is where the next stage’s matching, rather than the pair’s, sets a floor, and The mismatch that cancels itself separated a mirror’s three mismatches the way this essay separates the pair’s two.