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8XA - 2x Parallel Plate

Simple 2-cell parallel plate for doubled capacitance.

From the archive's library.

The parts

The drive no drive was saved with it; these are the engine's defaults

kHz

The pulse frequency the loop is driven at.

V

The peak voltage applied across the loop — after the transformer, if there is one.

%

How much of each pulse period the drive is on.

Hz

The slow gate switching the pulse train on and off.

%

How much of each gate period the pulses are let through.

Separate, or on one core with fields aiding or opposing.

How much of one choke's flux threads the other. Zero for separate cores.

or change the parts in the simulator

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.

  • σ On the way 0.7813 µS/cm

    Distilled Water's conductivity, from its dissolved solids reading.

    Dissolved solids to conductivity

  • L₁ Charging choke 98.47 µH

    The charging choke's inductance: 100 turns on Ferrite Toroid T106-2 (μᵣ 10, 0.21 cm², 2.68 cm path).

    Inductance of a winding on a core

  • L₂ Blocking choke 98.47 µH

    The blocking choke's inductance: 100 turns on Ferrite Toroid T106-2 (μᵣ 10, 0.21 cm², 2.68 cm path).

    Inductance of a winding on a core

  • R Series resistance 0.3923 Ω

    Series resistance, added up: Positive Choke - 100 Turns AWG18 0.0962 Ω + Negative Choke - 100 Turns AWG18 0.0962 Ω + Plate - 2x Parallel 8" x 10" 0.2 Ω. Wiring and connections are not counted.

  • C Cell capacitance 5.764 nF

    One pair of plates of Plate - 2x Parallel 8" x 10" — 516.128 cm² at a 6.35 mm gap — in distilled / deionised.

    Flat-plate cell capacitance and field

  • C Cell capacitance 11.53 nF

    2 such cells in parallel. The inductance handed to the array calculation is the two chokes added plainly, for its resonance figure only; the capacitance does not depend on it.

    Multi-cell array — 2 notes on that page

  • ℓ Cell leak resistance 787.4 Ω

    The leak across one pair of plates, from its geometry and Distilled Water at 0.781 µS/cm. 2 such cells in parallel leak 2 times as readily, so this is one half of that.

    Leak resistance across a plate cell

Result

f₀ Resonant frequency 105.6 kHz

Where the loop — both chokes and the cell — actually rings. Compare with the drive frequency.

L Loop inductance 196.9 µH

Both chokes together, with the mutual term if they are coupled.

M Mutual inductance 0 mH

Signed: positive aiding, negative opposing, zero if independent.

Z Characteristic impedance 130.7 Ω

√(L/C) — what the loop looks like at resonance before resistance.

Q Q factor 333.2

How sharp the resonance is against the series resistance.

U Resonant rise 3.998 kV

The cell voltage at resonance — Q times the drive.

ζ Damping ratio 0.001501

Below 1 the loop rings after each pulse; above 1 it sags.

τ Decay time constant 1.004 ms

How fast the ring dies away.

I Peak current 8.771 mA

What the drive pushes at the drive frequency, not at resonance.

W Average power 7.545 µW

Dissipated in the resistance, averaged over pulses and gate.

n Pulses per gate 50

How many pulses arrive before the gate closes.

S Staircase peak 12 V

Where the step-charge staircase tops out against the leak, after n pulses.

With your numbers
105.6 kHz=12π196.9 µH·11.53 nF
LaTeX 
105.6\,\mathrm{kHz} = \frac{1}{2\pi \sqrt{196.9\,\mathrm{µH} \cdot 11.53\,\mathrm{nF}}}

Worth knowing

  • Q factor and voltage rise — A Q of 333 is very high for a circuit with water in it, and implies a bandwidth of only 317 Hz. Confirm the series resistance includes the cell’s own losses — using just the choke’s winding resistance is the usual way to arrive at a Q like this on paper and not on the bench.
  • Current and power under a pulsed drive — The drive is at 0.09× the loop's resonance (106 kHz). The net reactance of -1.37 kΩ is what limits the current here, not the resistance; on resonance the same drive would push 30.6 A.
  • Step charging accumulation — Only 0% 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 2% of what perfect retention would give. Raising the drive frequency shortens the gap between pulses and helps; raising the water resistance helps far more.
  • The drive is at 10 kHz and the loop rings at 106 kHz — 0.09× resonance. The resonant rise of 4 kV above is what this loop would do driven on resonance; at the drive frequency the cell sees the drive through -1.37 kΩ of reactance instead. Either move the drive or change a choke so the two agree.

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. Pulse train timing

    n 50

    The drive as timing: how long each pulse is, how many fit inside one gate, and what fraction of the time the drive is on.

    With these numbers
    100 µs=110 kHz50 µs=100 µs·50 %10050=⌊50 %100·10 kHz100 Hz⌋25 %=50 %·50 %100
    LaTeX 
    100\,\mathrm{µs} = \frac{1}{10\,\mathrm{kHz}} \qquad 50\,\mathrm{µs} = 100\,\mathrm{µs} \cdot \frac{50\,\mathrm{%}}{100} \qquad 50 = \left\lfloor \frac{50\,\mathrm{%}}{100} \cdot \frac{10\,\mathrm{kHz}}{100\,\mathrm{Hz}} \right\rfloor \qquad 25\,\mathrm{%} = \frac{50\,\mathrm{%} \cdot 50\,\mathrm{%}}{100}

    Open this step on its own page, with these inputs

  2. 2. Two chokes on one core

    L 196.9 µH

    The two chokes as the loop sees them — their sum, plus or minus the mutual term if they share a core.

    With these numbers
    0 mH=±0⁢98.47 µH·98.47 µH196.9 µH=98.47 µH+98.47 µH+2·0 mH98.47 µH=98.47 µH+0 mH
    LaTeX 
    0\,\mathrm{mH} = \pm 0 \sqrt{98.47\,\mathrm{µH} \cdot 98.47\,\mathrm{µH}} \qquad 196.9\,\mathrm{µH} = 98.47\,\mathrm{µH} + 98.47\,\mathrm{µH} + 2 \cdot 0\,\mathrm{mH} \qquad 98.47\,\mathrm{µH} = 98.47\,\mathrm{µH} + 0\,\mathrm{mH}

    Open this step on its own page, with these inputs

  3. 3. Series LC resonant frequency

    f 105.6 kHz

    Where that inductance and the cell ring.

    With these numbers
    105.6 kHz=12⁢π⁢196.9 µH·11.53 nF
    LaTeX 
    105.6\,\mathrm{kHz} = \frac{1}{2\pi\sqrt{196.9\,\mathrm{µH} \cdot 11.53\,\mathrm{nF}}}

    Open this step on its own page, with these inputs

  4. 4. Q factor and voltage rise

    Q 333.2

    How sharply, against everything resistive in the loop, and how far the cell voltage rises above the drive on resonance.

    With these numbers
    333.2=10.3923 Ω⁢196.9 µH11.53 nF3.998 kV=333.2·12 V
    LaTeX 
    333.2 = \frac{1}{0.3923\,\mathrm{Ω}}\sqrt{\frac{196.9\,\mathrm{µH}}{11.53\,\mathrm{nF}}} \qquad 3.998\,\mathrm{kV} = 333.2 \cdot 12\,\mathrm{V} \qquad 317\,\mathrm{Hz} = \frac{105.6\,\mathrm{kHz}}{333.2}
    • A Q of 333 is very high for a circuit with water in it, and implies a bandwidth of only 317 Hz. Confirm the series resistance includes the cell’s own losses — using just the choke’s winding resistance is the usual way to arrive at a Q like this on paper and not on the bench.

    Open this step on its own page, with these inputs

  5. 5. Damping and ringdown

    ζ 0.001501

    What happens after each pulse: whether the loop rings or sags, and for how long.

    With these numbers
    0.001501=0.3923 Ω211.53 nF196.9 µH1.004 ms=2·196.9 µH0.3923 Ω105.6 kHz=12π1196.9 µH·11.53 nF−11.004 ms2244.2=ln10π·10.3923 Ω196.9 µH11.53 nF
    LaTeX 
    0.001501 = \frac{0.3923\,\mathrm{Ω}}{2} \sqrt{\frac{11.53\,\mathrm{nF}}{196.9\,\mathrm{µH}}} \qquad 1.004\,\mathrm{ms} = \frac{2 \cdot 196.9\,\mathrm{µH}}{0.3923\,\mathrm{Ω}} \qquad 105.6\,\mathrm{kHz} = \frac{1}{2\pi} \sqrt{\frac{1}{196.9\,\mathrm{µH} \cdot 11.53\,\mathrm{nF}} - \frac{1}{1.004\,\mathrm{ms}^{2}}} \qquad 244.2 = \frac{\ln 10}{\pi} \cdot \frac{1}{0.3923\,\mathrm{Ω}} \sqrt{\frac{196.9\,\mathrm{µH}}{11.53\,\mathrm{nF}}}

    Open this step on its own page, with these inputs

  6. 6. Current and power under a pulsed drive

    I 8.771 mA

    What the drive is asked for at the frequency it actually runs at — which is the resonance only if the two agree.

    With these numbers
    -1.368 kΩ=2π⁢10 kHz·196.9 µH−12π⁢10 kHz·11.53 nF1.368 kΩ=0.3923 Ω2+-1.368 kΩ28.771 mA=12 V1.368 kΩ30.18 µW=8.771 mA2·0.3923 Ω7.545 µW=30.18 µW·25 %1001.509 nJ=30.18 µW·50 µs
    LaTeX 
    -1.368\,\mathrm{kΩ} = 2\pi 10\,\mathrm{kHz} \cdot 196.9\,\mathrm{µH} - \frac{1}{2\pi 10\,\mathrm{kHz} \cdot 11.53\,\mathrm{nF}} \qquad 1.368\,\mathrm{kΩ} = \sqrt{0.3923\,\mathrm{Ω}^{2} + -1.368\,\mathrm{kΩ}^{2}} \qquad 8.771\,\mathrm{mA} = \frac{12\,\mathrm{V}}{1.368\,\mathrm{kΩ}} \qquad 30.18\,\mathrm{µW} = 8.771\,\mathrm{mA}^{2} \cdot 0.3923\,\mathrm{Ω} \qquad 7.545\,\mathrm{µW} = 30.18\,\mathrm{µW} \cdot \frac{25\,\mathrm{%}}{100} \qquad 1.509\,\mathrm{nJ} = 30.18\,\mathrm{µW} \cdot 50\,\mathrm{µs}
    • The drive is at 0.09× the loop's resonance (106 kHz). The net reactance of -1.37 kΩ is what limits the current here, not the resistance; on resonance the same drive would push 30.6 A.

    Open this step on its own page, with these inputs

  7. 7. Step charging accumulation

    U 12 V

    The pulse train against the cell and its leak: does the charge accumulate into a staircase, or drain away between pulses?

    With these numbers
    9.078 µs=787.4 Ω·11.53 nF1.644e-5=e−100 µs/9.078 µs12 V=12 V·1−1.644e-5501−1.644e-5
    LaTeX 
    9.078\,\mathrm{µs} = 787.4\,\mathrm{Ω} \cdot 11.53\,\mathrm{nF} \qquad 1.644e-5 = e^{-100\,\mathrm{µs}/9.078\,\mathrm{µs}} \qquad 12\,\mathrm{V} = 12\,\mathrm{V} \cdot \frac{1 - 1.644e-5^{50}}{1 - 1.644e-5}
    • Only 0% 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 2% 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

Resonant frequency against cell capacitance Cell capacitance swept from 5.76 nF to 17.3 nF with everything else held at your numbers. The dashed lines cross where you are.
Resonant frequency against cell capacitanceResonant frequency falls from 149 kHz to 86.2 kHz as cell capacitance rises from 5.76 nF to 17.3 nF. At your cell capacitance of 11.5 nF it is 106 kHz.8010012014016057.51012.51517.511.5 nF106 kHzCell capacitance (nF)Resonant frequency (kHz)
The formula behind the curve
f₀=12πL·C
f₀
Resonant frequency, Hz
L
Loop inductance, mH
C
Cell capacitance, nF
LaTeX 
f₀ = \frac{1}{2\pi \sqrt{L \cdot C}}
What moves the answer Each input moved 10% either way, with the others held still, and the effect on resonant frequency.
What moves the answerResonant frequency is most sensitive to Cell capacitance, which moves it by about 5.41% for a 10% change. It is least sensitive to Cell leak resistance, at about 0%.Change in the answer when each input moves by 10%-10%-5%5%10%Cell capacitance±5.41Charging choke±2.6Blocking choke±2.6Coupling coefficient±0Series resistance±0Drive frequency±0Drive amplitude±0Pulse duty±0Gate frequency±0Gate duty±0Cell leak resistance±0
The formula behind the curve
f₀=12πL·C
f₀
Resonant frequency, Hz
L
Loop inductance, mH
C
Cell capacitance, nF
LaTeX 
f₀ = \frac{1}{2\pi \sqrt{L \cdot C}}

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