Field

Devices, and the amplitude they stop being linear at

An operating point is where a transcendental equation and a linear network agree, and finding it is Newton's method on the whole netlist. Past that, the thing a linear model cannot express at all: distortion. It arrives seven times sooner than gain error does, its harmonics are Bessel functions of the drive, and a differential pair removes every even one of them exactly.
02.5057.5000.2000.4000.6000.800voltage across the diode (volts)current (milliamperes)load line: (5 V − v)/1.0 kΩ123450.6925 V, 4.307 mAthe circuit5 V1.0ksolved, then checked — two Newtons, no shared arithmeticthe drop moves 60 mV per decade

A bias point is a solution, not a choice

The phrase "the diode drops 0.7 volts" is a constant standing in for the root of a transcendental equation. Solved properly, from a five-volt supply through a kilohm, it drops 0.692544 V — and from forty-eight volts through the same kilohm it drops 0.754459 V, because the drop moves about sixty millivolts for every decade of current through it.

the drive: a sinusoidthe current out, and a symmetric one for comparisonone cyclemeasured, against the Bessel ratioharmonic 29.61%harmonic 30.62%harmonic 40.03%harmonic 51.2e-5harmonic 63.7e-7agreement: 5e-10 relativesolved, then checked — a transform against a seriessecond harmonic 9.6% at 10.0 mV

The distortion a linear model cannot have

A small-signal model's output is a scaled copy of its input by construction, so it has no second harmonic and asking it for one is not a hard question but a meaningless one. Measured on the curve itself, an exponential produces one per cent of harmonic distortion at 1.03 mV of drive — seven times sooner than the 7.30 mV at which its gain is one per cent wrong.

-101-0.100-0.05000.0500.100differential drive (volts)output, normalisedthe pair: odd, and it saturatesone exponentialharmonics, as a fraction of the fundamentalsingle, h218.876%pair, h21e-16single, h32.404%pair, h31.202%single, h40.231%pair, h46e-17single, h50.018%pair, h50.017%solved, then checked — evenness measured, not assumedthe pair's second harmonic is 1e-16

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.

-20020401001k10k100k1M10M100M1G10Gfrequency (hertz)gain (decibels)midband 43.2 dBsolved: 503 kHzMiller says 643 kHzsecond pole 336 MHzzero at g_m/C_μsolved, then checked — two networks, one measurementMiller is 22% optimistic

The frequency a device sets for itself

A common-emitter stage's bandwidth is decided by two picofarads between its collector and its base. The Miller approximation says how — lump it at the input, multiplied by one plus the gain — and predicts 643 kHz where the solved network gives 503 kHz. Twenty-two per cent optimistic, and it has no room at all for the second pole or for the zero in the right half-plane that the network also has.

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