Dual switchover

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.

• C Cell capacitance 2.306 nF

  One pair of plates of Plate - 3x Series 4" x 4" — 103.2256 cm² at a 3.175 mm gap — in distilled / deionised.

  Flat-plate cell capacitance and field

• C Cell capacitance 768.6 pF

  3 such cells in series. The array calculation's inductance is a placeholder for its resonance figure; the capacitance does not depend on it.

  Multi-cell array — 2 notes on that page

• ℓ Cell leak resistance 500 kΩ

  Leak resistance assumed at 500 kΩ: the water's conductivity is not known. Open the bare-numbers page to set it yourself.

• No water chosen, so the cell is taken as full of distilled water at 20 °C. Pick a water to change that.

The working, step by step

1. 1. Dual switchover: one leg's pulse into the cell

   U 100 V

   One leg's pulse: how far the cell charges through the load resistor, and what that costs the supply.

   $$0.7686 \mathrm{\mu s}=1000 \mathrm{\Omega }·768.6 \mathrm{pF}100 \mathrm{V}=100 \mathrm{V}(1-e^{-490 \mathrm{\mu s}/0.7686 \mathrm{\mu s}})76.86 \mathrm{nC}=768.6 \mathrm{pF}·100 \mathrm{V}76.86 \mathrm{\mu A}=76.86 \mathrm{nC}·1 \mathrm{kHz}7.686 \mathrm{mW}=100 \mathrm{V}·76.86 \mathrm{\mu A}$$

   LaTeX

   0.7686\,\mathrm{µs} = 1000\,\mathrm{Ω} \cdot 768.6\,\mathrm{pF} \qquad 100\,\mathrm{V} = 100\,\mathrm{V} \left( 1 - e^{-490\,\mathrm{µs} / 0.7686\,\mathrm{µs}} \right) \qquad 76.86\,\mathrm{nC} = 768.6\,\mathrm{pF} \cdot 100\,\mathrm{V} \qquad 76.86\,\mathrm{µA} = 76.86\,\mathrm{nC} \cdot 1\,\mathrm{kHz} \qquad 7.686\,\mathrm{mW} = 100\,\mathrm{V} \cdot 76.86\,\mathrm{µA}

   Open this step on its own page, with these inputs

2. 2. Step charging accumulation

   U 108 V

   Whether those pulses accumulate against the water's leak between them, or each starts from nothing.

   $$384.3 \mathrm{\mu s}=500 \mathrm{k\Omega }·768.6 \mathrm{pF}0.07411=e^{-1000 \mathrm{\mu s}/384.3 \mathrm{\mu s}}108 \mathrm{V}=100 \mathrm{V}·\frac{1-{0.07411}^{10}}{1-0.07411}$$

   LaTeX

   384.3\,\mathrm{µs} = 500\,\mathrm{kΩ} \cdot 768.6\,\mathrm{pF} \qquad 0.07411 = e^{-1000\,\mathrm{µs}/384.3\,\mathrm{µs}} \qquad 108\,\mathrm{V} = 100\,\mathrm{V} \cdot \frac{1 - 0.07411^{10}}{1 - 0.07411}

   • Only 7% 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 11% 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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