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Dual switchover

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?

Loading a circuit fills the slots; change any of them afterwards and the result is your variation on it. All saved circuits.

The parts

The cell the legs fire into. About this part.

What is in it — sets the capacitance and the leak.

The drive

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.

How many pulses to follow the accumulation for.

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 100 nF

    Film Capacitor 100nF, as recorded on the part.

  • 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.

Result

U Cell voltage after one pulse 99.26 V

How far one leg's pulse charges the cell from empty.

e Effective on-time 490 µs

The on-time less the dead time.

I Peak current 100 mA

The instant a leg closes.

J Average supply current 9.926 mA

Charge per pulse times the rate.

W Average power 992.6 mW

What the supply delivers.

S After the burst 908.6 V

Where the cell voltage stands after n pulses against the leak.

r Retention per pulse 0.9802

Fraction of the cell voltage surviving to the next pulse.

With your numbers
99.26 V=100 V(1−e−490 µs/(1000 Ω·100 nF))
LaTeX 
99.26\,\mathrm{V} = 100\,\mathrm{V} \left( 1 - e^{-490\,\mathrm{µs} / (1000\,\mathrm{Ω} \cdot 100\,\mathrm{nF})} \right)

This result is a link — the address bar holds your numbers, so it can be pasted into a post and opened to the same answer.

The working, step by step

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

    U 99.26 V

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

    With these numbers
    100 µs=1000 Ω·100 nF99.26 V=100 V(1−e−490 µs/100 µs)9.926 µC=100 nF·99.26 V9.926 mA=9.926 µC·1 kHz992.6 mW=100 V·9.926 mA
    LaTeX 
    100\,\mathrm{µs} = 1000\,\mathrm{Ω} \cdot 100\,\mathrm{nF} \qquad 99.26\,\mathrm{V} = 100\,\mathrm{V} \left( 1 - e^{-490\,\mathrm{µs} / 100\,\mathrm{µs}} \right) \qquad 9.926\,\mathrm{µC} = 100\,\mathrm{nF} \cdot 99.26\,\mathrm{V} \qquad 9.926\,\mathrm{mA} = 9.926\,\mathrm{µC} \cdot 1\,\mathrm{kHz} \qquad 992.6\,\mathrm{mW} = 100\,\mathrm{V} \cdot 9.926\,\mathrm{mA}

    Open this step on its own page, with these inputs

  2. 2. Step charging accumulation

    U 908.6 V

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

    With these numbers
    50 ms=500 kΩ·100 nF0.9802=e−1000 µs/50 ms908.6 V=99.26 V·1−0.9802101−0.9802
    LaTeX 
    50\,\mathrm{ms} = 500\,\mathrm{kΩ} \cdot 100\,\mathrm{nF} \qquad 0.9802 = e^{-1000\,\mathrm{µs}/50\,\mathrm{ms}} \qquad 908.6\,\mathrm{V} = 99.26\,\mathrm{V} \cdot \frac{1 - 0.9802^{10}}{1 - 0.9802}

    Open this step on its own page, with these inputs

What this looks like

Cell voltage after one pulse against supply voltage Supply voltage swept from 50 V to 150 V with everything else held at your numbers. The dashed lines cross where you are.
Cell voltage after one pulse against supply voltageCell voltage after one pulse rises from 49.6 V to 149 V as supply voltage rises from 50 V to 150 V. At your supply voltage of 100 V it is 99.3 V.50751001251505075100125150100 V99.3 VSupply voltage (V)Cell voltage after one pulse (V)
The formula behind the curve
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)
What moves the answer Each input moved 10% either way, with the others held still, and the effect on cell voltage after one pulse.
What moves the answerCell voltage after one pulse is most sensitive to Supply voltage, which moves it by about 10% for a 10% change. It is least sensitive to Pulses in a burst, at about 0%.Change in the answer when each input moves by 10%-10%-5%0%5%10%Supply voltage±10Duty±0.487Pulse frequency±0.432Load resistor±0.421Cell capacitance±0.421Dead time±0.00754Cell leak resistance±0Pulses in a burst±0
The formula behind the curve
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)

Open the bare numbers — the same simulation on its own page, every derived value editable.