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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 µF

    Electrolytic 100µF, 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 488.8 mV

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 48.88 mA

Charge per pulse times the rate.

W Average power 4.888 W

What the supply delivers.

S After the burst 4.888 V

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

r Retention per pulse 1

Fraction of the cell voltage surviving to the next pulse.

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

Worth knowing

  • Dual switchover: one leg's pulse into the cell — The on-time of 490 µs is shorter than the RC time constant of 100 ms, so the cell reaches only 0% of the supply before the leg opens. A smaller load resistor charges it faster, and draws more current doing so.

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 488.8 mV

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

    With these numbers
    100 ms=1000 Ω·100 µF488.8 mV=100 V(1−e−490 µs/100 ms)48.88 µC=100 µF·488.8 mV48.88 mA=48.88 µC·1 kHz4.888 W=100 V·48.88 mA
    LaTeX 
    100\,\mathrm{ms} = 1000\,\mathrm{Ω} \cdot 100\,\mathrm{µF} \qquad 488.8\,\mathrm{mV} = 100\,\mathrm{V} \left( 1 - e^{-490\,\mathrm{µs} / 100\,\mathrm{ms}} \right) \qquad 48.88\,\mathrm{µC} = 100\,\mathrm{µF} \cdot 488.8\,\mathrm{mV} \qquad 48.88\,\mathrm{mA} = 48.88\,\mathrm{µC} \cdot 1\,\mathrm{kHz} \qquad 4.888\,\mathrm{W} = 100\,\mathrm{V} \cdot 48.88\,\mathrm{mA}
    • The on-time of 490 µs is shorter than the RC time constant of 100 ms, so the cell reaches only 0% of the supply before the leg opens. A smaller load resistor charges it faster, and draws more current doing so.

    Open this step on its own page, with these inputs

  2. 2. Step charging accumulation

    U 4.888 V

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

    With these numbers
    50 s=500 kΩ·100 µF1=e−1000 µs/50 s4.888 V=488.8 mV·1−1101−1
    LaTeX 
    50\,\mathrm{s} = 500\,\mathrm{kΩ} \cdot 100\,\mathrm{µF} \qquad 1 = e^{-1000\,\mathrm{µs}/50\,\mathrm{s}} \qquad 4.888\,\mathrm{V} = 488.8\,\mathrm{mV} \cdot \frac{1 - 1^{10}}{1 - 1}

    Open this step on its own page, with these inputs

What this looks like

Cell voltage after one pulse against duty Duty swept from 25 % to 75 % with everything else held at your numbers. The dashed lines cross where you are.
Cell voltage after one pulse against dutyCell voltage after one pulse rises from 240 mV to 737 mV as duty rises from 25 % to 75 %. At your duty of 50 % it is 489 mV.2004006008002040608050 %489 mVDuty (%)Cell voltage after one pulse (mV)
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 Pulse frequency, which moves it by about 11.3% 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%-20%-10%10%20%Pulse frequency±11.3Load resistor±11.1Cell capacitance±11.1Duty±10.2Supply voltage±10Dead time±0.204Cell 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.