calculation · Voltage Intensifier Circuit · computed
Midpoint-diode bifilar: winding resistance damps the A–B interwinding ring but does not remove the reservoir; displacement-current path always traverses ~one full winding of R
- Prompted by Scotchn (Discord #neuralstan20-testing 2026-10-05): what if L stays ~same but each wire goes to 11.6 kΩ (stainless spec)? Extends #3477.
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- Inter-wire potential between adjacent A/B turns is set by induced EMF of one full winding (E − V_diode), shifted by I·R_winding depending on current direction; resistance does not remove it. The A–B capacitance still stores ~½·C_ab·E² at collapse. (Scotchn's point; my first answer overstated "kills it".)
- 2. Path argument — displacement current enters A at its finish, runs back to turn k, crosses to B, runs from B start to turn k: total ≈ one winding length for every k. So the fast A–B loop sees ≈ R_winding (11.6 kΩ), not 2×. With Z0 = √(0.24 mH / 1.18 nF) ≈ 450 Ω (critical ≈ 900 Ω), ζ ≈ 13: overdamped. Prediction: the six 300 kHz lobes become one smooth forward bleed after the main packet, τ up to ~R·C_ab ≈ 14 µs (lumped; distributed RC line shortens it). Resistance "cleans it up" rather than killing it.
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- Slow swing (129.3 mH vs 3.97 nF, Z0 ≈ 5.7 kΩ) with 23.2 kΩ total is also overdamped: roots ~6 µs and ~86 µs; no undershoot below the start level.
- 4. Insulation — IR drop of ~12 V per mA at 11.6 kΩ adds to inter-wire stress; cf. Scotchn's A–B breakdown (#3457).
- Proposed test (not yet run) — resistor in series with the midpoint diode (inside the A–B loop) at 470 Ω / 1 kΩ / 4.7 kΩ; check whether total step height is preserved (clean-up) or collapses (kill). Lumped estimates only; A–B C value is from the 9/29 coil.
Basis
- Method
- unversioned-legacy
- Recorded
- Published
- 5 Oct 2026
- Stanbot
- v3
- Source Ref
- Derived; Scotchn question and bench values, Discord #neuralstan20-testing 2026-10-05; extends notebook #3477, #3457, #500
- Notebook Id
- 3480
bifilar interwinding-capacitance stainless damping double-pulse scotchn