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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 470 nF

    Film Capacitor 470nF, 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 64.74 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 30.43 mA

Charge per pulse times the rate.

W Average power 3.043 W

What the supply delivers.

S After the burst 635.2 V

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

r Retention per pulse 0.9958

Fraction of the cell voltage surviving to the next pulse.

With your numbers
64.74 V=100 V(1−e−490 µs/(1000 Ω·470 nF))
LaTeX 
64.74\,\mathrm{V} = 100\,\mathrm{V} \left( 1 - e^{-490\,\mathrm{µs} / (1000\,\mathrm{Ω} \cdot 470\,\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 64.74 V

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

    With these numbers
    470 µs=1000 Ω·470 nF64.74 V=100 V(1−e−490 µs/470 µs)30.43 µC=470 nF·64.74 V30.43 mA=30.43 µC·1 kHz3.043 W=100 V·30.43 mA
    LaTeX 
    470\,\mathrm{µs} = 1000\,\mathrm{Ω} \cdot 470\,\mathrm{nF} \qquad 64.74\,\mathrm{V} = 100\,\mathrm{V} \left( 1 - e^{-490\,\mathrm{µs} / 470\,\mathrm{µs}} \right) \qquad 30.43\,\mathrm{µC} = 470\,\mathrm{nF} \cdot 64.74\,\mathrm{V} \qquad 30.43\,\mathrm{mA} = 30.43\,\mathrm{µC} \cdot 1\,\mathrm{kHz} \qquad 3.043\,\mathrm{W} = 100\,\mathrm{V} \cdot 30.43\,\mathrm{mA}

    Open this step on its own page, with these inputs

  2. 2. Step charging accumulation

    U 635.2 V

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

    With these numbers
    235 ms=500 kΩ·470 nF0.9958=e−1000 µs/235 ms635.2 V=64.74 V·1−0.9958101−0.9958
    LaTeX 
    235\,\mathrm{ms} = 500\,\mathrm{kΩ} \cdot 470\,\mathrm{nF} \qquad 0.9958 = e^{-1000\,\mathrm{µs}/235\,\mathrm{ms}} \qquad 635.2\,\mathrm{V} = 64.74\,\mathrm{V} \cdot \frac{1 - 0.9958^{10}}{1 - 0.9958}

    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 32.4 V to 97.1 V as supply voltage rises from 50 V to 150 V. At your supply voltage of 100 V it is 64.7 V.204060801005075100125150100 V64.7 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±6.11Pulse frequency±6.07Load resistor±5.96Cell capacitance±5.96Dead time±0.116Cell 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.