VIC5 - 8x Extreme HV Plate

From the parts

Every number handed to the simulation that you did not type, and where it came from. Each derivation that ran a calculation links to that calculation's page with these numbers in, so it can be checked alone.

• σ On the way 390.6 µS/cm

  Tap Water (Hard)'s conductivity, from its dissolved solids reading.

  Dissolved solids to conductivity

• L₁ Charging choke 38.54 mH

  The charging choke's inductance: 100 turns on PC40 Ferrite Pot Core (μᵣ 2300, 0.8 cm², 6 cm path).

  Inductance of a winding on a core — 1 note on that page

• L₂ Blocking choke 38.54 mH

  The blocking choke's inductance: 100 turns on PC40 Ferrite Pot Core (μᵣ 2300, 0.8 cm², 6 cm path).

  Inductance of a winding on a core — 1 note on that page

• R Series resistance 0.4123 Ω

  Series resistance, added up: Bifilar Choke 2 - 100T AWG18 (Finish-Finish) 0.0962 Ω + Bifilar Choke 2 - 100T AWG18 (Finish-Finish) 0.0962 Ω + Plate - 8x Series (Extreme HV) 0.22 Ω. Wiring and connections are not counted.

• C Cell capacitance 3.39 nF

  One pair of plates of Plate - 8x Series (Extreme HV) — 309.6768 cm² at a 6.35 mm gap — in tap water (typical).

  Flat-plate cell capacitance and field

• C Cell capacitance 423.7 pF

  8 such cells in series. The inductance handed to the array calculation is the two chokes added plainly, for its resonance figure only; the capacitance does not depend on it.

  Multi-cell array — 2 notes on that page

• ℓ Cell leak resistance 41.99 Ω

  The leak across one pair of plates, from its geometry and Tap Water (Hard) at 391 µS/cm. 8 such cells in series: 8 times that.

  Leak resistance across a plate cell — 1 note on that page

• V Drive amplitude 60 V

  The drive amplitude of 12 V stepped up through Forward Mode 1:5 Step-Up (100:500). The 1 A primary current given to the ratio calculation is for its current figures only; the voltage ratio does not depend on it.

  Transformer step-up to the cell

• Both chokes are recorded as bifilar windings, but the arrangement is set to independent. If they share a core, choose an aiding or opposing arrangement and a coupling coefficient.

The working, step by step

1. 1. Pulse train timing

   n 50

   The drive as timing: how long each pulse is, how many fit inside one gate, and what fraction of the time the drive is on.

   $$100 \mathrm{\mu s}=\frac{1}{10}50 \mathrm{\mu s}=100 \mathrm{\mu s}·\frac{50}{ }50=\lfloor \frac{50}{ }·\frac{10}{ }\rfloor 25 \mathrm{\%}=\frac{50 \mathrm{\%}·50 \mathrm{\%}}{100}$$

   LaTeX

   100\,\mathrm{µs} = \frac{1}{10\,\mathrm{kHz}} \qquad 50\,\mathrm{µs} = 100\,\mathrm{µs} \cdot \frac{50\,\mathrm{%}}{100} \qquad 50 = \left\lfloor \frac{50\,\mathrm{%}}{100} \cdot \frac{10\,\mathrm{kHz}}{100\,\mathrm{Hz}} \right\rfloor \qquad 25\,\mathrm{%} = \frac{50\,\mathrm{%} \cdot 50\,\mathrm{%}}{100}

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2. 2. Two chokes on one core

   L 77.07 mH

   The two chokes as the loop sees them — their sum, plus or minus the mutual term if they share a core.

   $$0 \mathrm{mH}=\pm 0\sqrt{38.54 \mathrm{mH}·38.54 \mathrm{mH}}77.07 \mathrm{mH}=38.54 \mathrm{mH}+38.54 \mathrm{mH}+2·0 \mathrm{mH}38.54 \mathrm{mH}=38.54 \mathrm{mH}+0 \mathrm{mH}$$

   LaTeX

   0\,\mathrm{mH} = \pm 0 \sqrt{38.54\,\mathrm{mH} \cdot 38.54\,\mathrm{mH}} \qquad 77.07\,\mathrm{mH} = 38.54\,\mathrm{mH} + 38.54\,\mathrm{mH} + 2 \cdot 0\,\mathrm{mH} \qquad 38.54\,\mathrm{mH} = 38.54\,\mathrm{mH} + 0\,\mathrm{mH}

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3. 3. Series LC resonant frequency

   f 27.85 kHz

   Where that inductance and the cell ring.

   $$27.85 \mathrm{kHz}=\frac{1}{2\pi \sqrt{77.07 \mathrm{mH}·423.7 \mathrm{pF}}}$$

   LaTeX

   27.85\,\mathrm{kHz} = \frac{1}{2\pi\sqrt{77.07\,\mathrm{mH} \cdot 423.7\,\mathrm{pF}}}

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4. 4. Q factor and voltage rise

   Q 32712

   How sharply, against everything resistive in the loop, and how far the cell voltage rises above the drive on resonance.

   $$32712=\frac{1}{0.4123}\sqrt{\frac{77.07}{ }}1963 \mathrm{kV}=32712·60 \mathrm{V}$$

   LaTeX

   32712 = \frac{1}{0.4123\,\mathrm{Ω}}\sqrt{\frac{77.07\,\mathrm{mH}}{423.7\,\mathrm{pF}}} \qquad 1963\,\mathrm{kV} = 32712 \cdot 60\,\mathrm{V} \qquad 0.8514\,\mathrm{Hz} = \frac{27.85\,\mathrm{kHz}}{32712}

   • A Q of 32712 is very high for a circuit with water in it, and implies a bandwidth of only 0.8514 Hz. Confirm the series resistance includes the cell’s own losses — using just the choke’s winding resistance is the usual way to arrive at a Q like this on paper and not on the bench.
   • A predicted 1963 kV across the cell is past where insulation, standoffs and the water itself hold up. Check the field strength across the gap before trusting this.

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5. 5. Damping and ringdown

   ζ 1.528e-5

   What happens after each pulse: whether the loop rings or sags, and for how long.

   $$1.528e-5=\frac{0.4123}{ }\sqrt{\frac{423.7}{ }}373.9 \mathrm{ms}=\frac{2·77.07 \mathrm{mH}}{0.4123}27.85 \mathrm{kHz}=\frac{1}{2\pi }\sqrt{\frac{1}{77.07 \mathrm{mH}·423.7 \mathrm{pF}}-\frac{1}{{373.9}^{ }}}23976=\frac{\ln 10}{\pi }·\frac{1}{0.4123}\sqrt{\frac{77.07}{ }}$$

   LaTeX

   1.528e-5 = \frac{0.4123\,\mathrm{Ω}}{2} \sqrt{\frac{423.7\,\mathrm{pF}}{77.07\,\mathrm{mH}}} \qquad 373.9\,\mathrm{ms} = \frac{2 \cdot 77.07\,\mathrm{mH}}{0.4123\,\mathrm{Ω}} \qquad 27.85\,\mathrm{kHz} = \frac{1}{2\pi} \sqrt{\frac{1}{77.07\,\mathrm{mH} \cdot 423.7\,\mathrm{pF}} - \frac{1}{373.9\,\mathrm{ms}^{2}}} \qquad 23976 = \frac{\ln 10}{\pi} \cdot \frac{1}{0.4123\,\mathrm{Ω}} \sqrt{\frac{77.07\,\mathrm{mH}}{423.7\,\mathrm{pF}}}

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6. 6. Current and power under a pulsed drive

   I 1.834 mA

   What the drive is asked for at the frequency it actually runs at — which is the resonance only if the two agree.

   $$-32.72 \mathrm{k\Omega }=2\pi 10 \mathrm{kHz}·77.07 \mathrm{mH}-\frac{1}{2\pi 10 \mathrm{kHz}·423.7 \mathrm{pF}}32.72 \mathrm{k\Omega }=\sqrt{{0.4123}^{ }+{-32.72}^{ }}1.834 \mathrm{mA}=\frac{60}{ }1.386 \mathrm{\mu W}={1.834}^{ }·0.4123 \mathrm{\Omega }0.3466 \mathrm{\mu W}=1.386 \mathrm{\mu W}·\frac{25}{ }0.06932 \mathrm{nJ}=1.386 \mathrm{\mu W}·50 \mathrm{\mu s}$$

   LaTeX

   -32.72\,\mathrm{kΩ} = 2\pi 10\,\mathrm{kHz} \cdot 77.07\,\mathrm{mH} - \frac{1}{2\pi 10\,\mathrm{kHz} \cdot 423.7\,\mathrm{pF}} \qquad 32.72\,\mathrm{kΩ} = \sqrt{0.4123\,\mathrm{Ω}^{2} + -32.72\,\mathrm{kΩ}^{2}} \qquad 1.834\,\mathrm{mA} = \frac{60\,\mathrm{V}}{32.72\,\mathrm{kΩ}} \qquad 1.386\,\mathrm{µW} = 1.834\,\mathrm{mA}^{2} \cdot 0.4123\,\mathrm{Ω} \qquad 0.3466\,\mathrm{µW} = 1.386\,\mathrm{µW} \cdot \frac{25\,\mathrm{%}}{100} \qquad 0.06932\,\mathrm{nJ} = 1.386\,\mathrm{µW} \cdot 50\,\mathrm{µs}

   • The drive is at 0.36× the loop's resonance (27.9 kHz). The net reactance of -32.7 kΩ is what limits the current here, not the resistance; on resonance the same drive would push 146 A.

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7. 7. Step charging accumulation

   U 60 V

   The pulse train against the cell and its leak: does the charge accumulate into a staircase, or drain away between pulses?

   $$0.01779 \mathrm{\mu s}=41.99 \mathrm{\Omega }·423.7 \mathrm{pF}0=e^{-100 \mathrm{\mu s}/0.01779 \mathrm{\mu s}}60 \mathrm{V}=60 \mathrm{V}·\frac{1-0^{50}}{1-0}$$

   LaTeX

   0.01779\,\mathrm{µs} = 41.99\,\mathrm{Ω} \cdot 423.7\,\mathrm{pF} \qquad 0 = e^{-100\,\mathrm{µs}/0.01779\,\mathrm{µs}} \qquad 60\,\mathrm{V} = 60\,\mathrm{V} \cdot \frac{1 - 0^{50}}{1 - 0}

   • Only 0% of the voltage survives between pulses, so there is no staircase — each pulse is very nearly starting from nothing. The cell is discharging through the water faster than it is being charged.
   • The burst reaches 2% of what perfect retention would give. Raising the drive frequency shortens the gap between pulses and helps; raising the water resistance helps far more.

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