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Dual switchover into a cell: charge, leak, accumulate?

With the legs firing in turn through their load resistors, how far does each pulse charge this cell, what does the supply deliver, and does anything build up against the water's leak?

The formula
U=V(1−e−e/(R·C))
U
Cell voltage after one pulse, V
V
Supply voltage, V
e
Effective on-time, µs
R
Load resistor, Ω
C
Cell capacitance, nF
LaTeX 
U = V \left( 1 - e^{-e / (R \cdot C)} \right)

Work it out

Fill from a part the library

V

The B+ the legs switch onto the cell.

kHz

How often a leg fires.

%

How much of the period the firing leg is on.

µs

The gap between one leg turning off and the other turning on.

Ω

In series with each leg.

nF

The cell as a capacitor.

kΩ

The water's resistance across the gap.

How many pulses to follow the accumulation for.

or see a worked example with typical numbers

Method

  1. Run one leg's pulse: RC charging of the cell through the load resistor for the on-time less the dead time.
  2. Feed the voltage that pulse reaches, and the period, into the step-charging calculation with the cell's leak resistance, and follow n pulses.

Assumptions

  • The two steps disagree on purpose about one thing and the page says so: the first assumes the cell is empty when each leg fires, the second lets charge survive between pulses. The truth for a given rig is between them, and depends on whether the second leg drains the cell or adds to it — which is a question about the wiring, not the arithmetic.
  • Every assumption of each step applies.