Series

Differential pair — the series

3 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. One exponential and one pair, both driven 20.0 mV. computed by solving, not by drawing. The pair's characteristic is odd, so its even harmonics vanish: the second comes out at 1.5e-16 of the fundamental against 18.88% for the single stage. It is not a small residue but the floor of the arithmetic. The price is the third harmonic, 1.202% against 2.404%, and total distortion of 1.202% against 19.03%.

    What a pair cancels, and what it only halves

    A differential pair's transfer characteristic is an odd function, and an odd function driven symmetrically produces no even harmonics at all. Measured, the second harmonic comes out at 10⁻¹⁶ of the fundamental — the arithmetic's own floor, not a small physical residue. The third harmonic is a different story, and it comes out at exactly half the single stage's, which is a reduction and not a cancellation.

    part 2 · semiconductors
  2. What a 5% mismatched pair leaves behind, across 150 K. computed by solving, not by drawing. A saturation-current mismatch of 5.0% appears as an input offset of Vₜ·ln(m) — 1.2613 mV at 300 K — which is proportional to absolute temperature and therefore drifts at 4.2044 µV/K, exactly the offset divided by the temperature. That is 3333 ppm per kelvin at every mismatch on the slider, because the ratio is 1/T and contains nothing about the device. One junction on its own drifts 1.828 mV/K, 435 times harder.

    What matching does about temperature

    A pair cancels the 1.8 mV/K that broke the previous essay, and what it leaves behind is exact: a saturation-current mismatch of m shows up as an offset of the thermal voltage times ln(m), which is proportional to absolute temperature and therefore drifts in proportion to itself. 3 333 parts per million per kelvin, at every mismatch on the slider, because the ratio is 1/T and contains nothing about the device at all.

    part 3 · semiconductors
  3. A pair with 2.0% saturation-current mismatch, nulled at 300 K: trimmed by a constant it drifts 1.706 µV/K, trimmed by its load ratio 0.000 µV/K. computed by solving, not by drawing, at 31 temperatures — each a Newton solve of both junctions, a bisection for the tail current and a bisection for the input that nulls the output — a bipolar pair at 10 µA a side whose second transistor has 1.02 times the first's saturation current. Untrimmed (grey) it needs 511.9 µV at 300 K and 682.6 µV at 400 K. Trimmed at 300 K by subtracting a constant (dashed) it is zero there and 170.6 µV at 400 K, drifting 1.706 µV/K — the untrimmed offset's drift, untouched. Trimmed at 300 K by making the collector loads 1.02000 to one (solid) it is 0.0 µV at 400 K, drifting 0.000 µV/K. The load ratio acts inside the same logarithm the mismatch does, so it cancels the mismatch at every temperature, not at one.

    The trim that takes the drift with it

    A bipolar pair's offset from a saturation-current mismatch is proportional to absolute temperature, so trimming it to zero at one temperature can trim its drift to zero too — but only if the trim acts inside the same logarithm the mismatch does. Trimmed by its collector-load ratio, a 2% mismatched pair is zero from 250 to 400 K; trimmed by subtracting a constant after it, it drifts the same 1.706 µV/K it drifted untrimmed. Add 20 kΩ of source resistance and a 10% current-gain mismatch, and a fifth of the offset follows the current gain instead: the load trim now leaves −0.574 µV/K, the constant trim 1.256, and above 34 kΩ the constant trim is the better one. A trim cancels the drift of the share whose temperature law it shares and converts the rest; two knobs, set for value and slope, leave 11 µV of curvature.

    part 4 · semiconductors

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