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 452.7 pF

  One sphere-in-sphere of Spherical - 2x Series 2" & 4" — 2 in inside 4 in — in distilled / deionised at 20 °C.

  Concentric-sphere cell capacitance

• C Cell capacitance 226.4 pF

  2 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.

   $$0.2264 \mathrm{\mu s}=1000 \mathrm{\Omega }·226.4 \mathrm{pF}100 \mathrm{V}=100 \mathrm{V}(1-e^{-490 \mathrm{\mu s}/0.2264 \mathrm{\mu s}})22.64 \mathrm{nC}=226.4 \mathrm{pF}·100 \mathrm{V}22.64 \mathrm{\mu A}=22.64 \mathrm{nC}·1 \mathrm{kHz}2.264 \mathrm{mW}=100 \mathrm{V}·22.64 \mathrm{\mu A}$$

   LaTeX

   0.2264\,\mathrm{µs} = 1000\,\mathrm{Ω} \cdot 226.4\,\mathrm{pF} \qquad 100\,\mathrm{V} = 100\,\mathrm{V} \left( 1 - e^{-490\,\mathrm{µs} / 0.2264\,\mathrm{µs}} \right) \qquad 22.64\,\mathrm{nC} = 226.4\,\mathrm{pF} \cdot 100\,\mathrm{V} \qquad 22.64\,\mathrm{µA} = 22.64\,\mathrm{nC} \cdot 1\,\mathrm{kHz} \qquad 2.264\,\mathrm{mW} = 100\,\mathrm{V} \cdot 22.64\,\mathrm{µA}

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2. 2. Step charging accumulation

   U 100 V

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

   $$113.2 \mathrm{\mu s}=500 \mathrm{k\Omega }·226.4 \mathrm{pF}0.000146=e^{-1000 \mathrm{\mu s}/113.2 \mathrm{\mu s}}100 \mathrm{V}=100 \mathrm{V}·\frac{1-{0.000146}^{10}}{1-0.000146}$$

   LaTeX

   113.2\,\mathrm{µs} = 500\,\mathrm{kΩ} \cdot 226.4\,\mathrm{pF} \qquad 0.000146 = e^{-1000\,\mathrm{µs}/113.2\,\mathrm{µs}} \qquad 100\,\mathrm{V} = 100\,\mathrm{V} \cdot \frac{1 - 0.000146^{10}}{1 - 0.000146}

   • 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 10% 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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