Scotchn diode test: doubling C_t (2 → 3.97 nF) left V_C ≈ V_D ≈ 87 V unchanged; simple capacitive-divider model FALSIFIED (amends #3446)

Bench: Scotchn, Discord #neuralstan20-testing 2026-09-30. Same rig as #3445/#3446 (bifilar A/B, diode between A end and B start, CH1 = 100x differential with + on the anode, CH2 = 100x differential across C_t). Supply 4.00 V, 20 kHz, 50 %.

What changed: added two Y2 discs (1.864 nF measured), giving C_t = 3.97 nF (LCR, 100 kHz).

Diode zero-bias capacitance on the LCR (100 kHz, small signal): C_jd = 1.336 nF. That is a zero-bias maximum; the meter's test signal may partly forward-bias the diode. Junction capacitance falls steeply under reverse bias (roughly 1/sqrt(1+V/Vbi)), so at 87 V reverse it is likely to be of order 0.1 nF (my estimate; datasheet curve not yet checked).

Result: CH1 and CH2 are still both about 86-89 V deep, the same double-dip shape as the 2 nF run. The photo was too blurry to read the exact measurement values.

My prediction was V_C ≈ 59 V and V_D ≈ 115 V, from C_b = C_t·V_C/V_D with C_b ≈ 2 nF. It FAILED: the split does not follow the capacitance ratio. Charge check: Q_t per hump ≈ 3.97 nF × 87 V ≈ 345 nC. Cj (at most 1.336 nF × 87 V ≈ 116 nC) cannot carry that. The interwire C (1.18 nF) would need a swing of about 200 V between the A and B wires to pass the rest.

Working hypothesis (untested): the equal split reflects the symmetry of the two identical windings. Each wire's EMF is about 11.6 × 4 V ≈ 46 V, doubled by the LC step overshoot to about 90 V, so each element carries about one winding's EMF, largely independent of C_t.

Next tests:
(a) Put a series shunt in the C_t leg and integrate the current to get Q per hump.
(b) Differential probe from A start to B start to measure the interwire voltage directly.
(c) Supply 4 → 5 V: both V_C and V_D should rise about 25 %.
(d) C_t = 1 nF alone: V_C should stay near 87 V if C_t truly does not matter.