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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 979.7 pF

    One tube of Tubular - Single 1/2" x 6" — 0.25 in over 0.5 in, 6 in long — in distilled / deionised at 20 °C.

    Coaxial cell capacitance

  • 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 100 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 97.97 µA

Charge per pulse times the rate.

W Average power 9.797 mW

What the supply delivers.

S After the burst 114.9 V

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

r Retention per pulse 0.1299

Fraction of the cell voltage surviving to the next pulse.

With your numbers
100 V=100 V(1−e−490 µs/(1000 Ω·979.7 pF))
LaTeX 
100\,\mathrm{V} = 100\,\mathrm{V} \left( 1 - e^{-490\,\mathrm{µs} / (1000\,\mathrm{Ω} \cdot 979.7\,\mathrm{pF})} \right)

Worth knowing

  • Step charging accumulation — Only 13% 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.
  • Step charging accumulation — 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.

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 100 V

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

    With these numbers
    0.9797 µs=1000 Ω·979.7 pF100 V=100 V(1−e−490 µs/0.9797 µs)97.97 nC=979.7 pF·100 V97.97 µA=97.97 nC·1 kHz9.797 mW=100 V·97.97 µA
    LaTeX 
    0.9797\,\mathrm{µs} = 1000\,\mathrm{Ω} \cdot 979.7\,\mathrm{pF} \qquad 100\,\mathrm{V} = 100\,\mathrm{V} \left( 1 - e^{-490\,\mathrm{µs} / 0.9797\,\mathrm{µs}} \right) \qquad 97.97\,\mathrm{nC} = 979.7\,\mathrm{pF} \cdot 100\,\mathrm{V} \qquad 97.97\,\mathrm{µA} = 97.97\,\mathrm{nC} \cdot 1\,\mathrm{kHz} \qquad 9.797\,\mathrm{mW} = 100\,\mathrm{V} \cdot 97.97\,\mathrm{µA}

    Open this step on its own page, with these inputs

  2. 2. Step charging accumulation

    U 114.9 V

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

    With these numbers
    489.9 µs=500 kΩ·979.7 pF0.1299=e−1000 µs/489.9 µs114.9 V=100 V·1−0.1299101−0.1299
    LaTeX 
    489.9\,\mathrm{µs} = 500\,\mathrm{kΩ} \cdot 979.7\,\mathrm{pF} \qquad 0.1299 = e^{-1000\,\mathrm{µs}/489.9\,\mathrm{µs}} \qquad 114.9\,\mathrm{V} = 100\,\mathrm{V} \cdot \frac{1 - 0.1299^{10}}{1 - 0.1299}
    • Only 13% 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.

    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 50 V to 150 V as supply voltage rises from 50 V to 150 V. At your supply voltage of 100 V it is 100 V.50751001251505075100125150100 V100 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±10Pulse frequency±0Duty±0Dead time±0Load resistor±0Cell capacitance±0Cell 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.