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8XA - Film Cap Test

Test profile using standard film capacitor instead of water cell.

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.

  • 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 0.2423 Ω

    Series resistance, added up: Positive Choke - 100 Turns AWG18 0.0962 Ω + Negative Choke - 100 Turns AWG18 0.0962 Ω + Film Capacitor 100nF 0.05 Ω. Wiring and connections are not counted.

  • 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

f₀ Resonant frequency 35.86 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 44.38 Ω

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

Q Q factor 183.2

How sharp the resonance is against the series resistance.

U Resonant rise 2.198 kV

The cell voltage at resonance — Q times the drive.

ζ Damping ratio 0.00273

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

τ Decay time constant 1.626 ms

How fast the ring dies away.

I Peak current 81.75 mA

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

W Average power 404.9 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 571.5 V

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

With your numbers
35.86 kHz=12π196.9 µH·100 nF
LaTeX 
35.86\,\mathrm{kHz} = \frac{1}{2\pi \sqrt{196.9\,\mathrm{µH} \cdot 100\,\mathrm{nF}}}

Worth knowing

  • Current and power under a pulsed drive — The drive is at 0.28× the loop's resonance (35.9 kHz). The net reactance of -147 Ω is what limits the current here, not the resistance; on resonance the same drive would push 49.5 A.
  • The drive is at 10 kHz and the loop rings at 35.9 kHz — 0.28× resonance. The resonant rise of 2.2 kV above is what this loop would do driven on resonance; at the drive frequency the cell sees the drive through -147 Ω 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 35.86 kHz

    Where that inductance and the cell ring.

    With these numbers
    35.86 kHz=12⁢π⁢196.9 µH·100 nF
    LaTeX 
    35.86\,\mathrm{kHz} = \frac{1}{2\pi\sqrt{196.9\,\mathrm{µH} \cdot 100\,\mathrm{nF}}}

    Open this step on its own page, with these inputs

  4. 4. Q factor and voltage rise

    Q 183.2

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

    With these numbers
    183.2=10.2423 Ω⁢196.9 µH100 nF2.198 kV=183.2·12 V
    LaTeX 
    183.2 = \frac{1}{0.2423\,\mathrm{Ω}}\sqrt{\frac{196.9\,\mathrm{µH}}{100\,\mathrm{nF}}} \qquad 2.198\,\mathrm{kV} = 183.2 \cdot 12\,\mathrm{V} \qquad 195.8\,\mathrm{Hz} = \frac{35.86\,\mathrm{kHz}}{183.2}

    Open this step on its own page, with these inputs

  5. 5. Damping and ringdown

    ζ 0.00273

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

    With these numbers
    0.00273=0.2423 Ω2100 nF196.9 µH1.626 ms=2·196.9 µH0.2423 Ω35.86 kHz=12π1196.9 µH·100 nF−11.626 ms2134.2=ln10π·10.2423 Ω196.9 µH100 nF
    LaTeX 
    0.00273 = \frac{0.2423\,\mathrm{Ω}}{2} \sqrt{\frac{100\,\mathrm{nF}}{196.9\,\mathrm{µH}}} \qquad 1.626\,\mathrm{ms} = \frac{2 \cdot 196.9\,\mathrm{µH}}{0.2423\,\mathrm{Ω}} \qquad 35.86\,\mathrm{kHz} = \frac{1}{2\pi} \sqrt{\frac{1}{196.9\,\mathrm{µH} \cdot 100\,\mathrm{nF}} - \frac{1}{1.626\,\mathrm{ms}^{2}}} \qquad 134.2 = \frac{\ln 10}{\pi} \cdot \frac{1}{0.2423\,\mathrm{Ω}} \sqrt{\frac{196.9\,\mathrm{µH}}{100\,\mathrm{nF}}}

    Open this step on its own page, with these inputs

  6. 6. Current and power under a pulsed drive

    I 81.75 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
    -146.8 Ω=2π⁢10 kHz·196.9 µH−12π⁢10 kHz·100 nF146.8 Ω=0.2423 Ω2+-146.8 Ω281.75 mA=12 V146.8 Ω1.619 mW=81.75 mA2·0.2423 Ω404.9 µW=1.619 mW·25 %10080.97 nJ=1.619 mW·50 µs
    LaTeX 
    -146.8\,\mathrm{Ω} = 2\pi 10\,\mathrm{kHz} \cdot 196.9\,\mathrm{µH} - \frac{1}{2\pi 10\,\mathrm{kHz} \cdot 100\,\mathrm{nF}} \qquad 146.8\,\mathrm{Ω} = \sqrt{0.2423\,\mathrm{Ω}^{2} + -146.8\,\mathrm{Ω}^{2}} \qquad 81.75\,\mathrm{mA} = \frac{12\,\mathrm{V}}{146.8\,\mathrm{Ω}} \qquad 1.619\,\mathrm{mW} = 81.75\,\mathrm{mA}^{2} \cdot 0.2423\,\mathrm{Ω} \qquad 404.9\,\mathrm{µW} = 1.619\,\mathrm{mW} \cdot \frac{25\,\mathrm{%}}{100} \qquad 80.97\,\mathrm{nJ} = 1.619\,\mathrm{mW} \cdot 50\,\mathrm{µs}
    • The drive is at 0.28× the loop's resonance (35.9 kHz). The net reactance of -147 Ω is what limits the current here, not the resistance; on resonance the same drive would push 49.5 A.

    Open this step on its own page, with these inputs

  7. 7. Step charging accumulation

    U 571.5 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
    50 ms=500 kΩ·100 nF0.998=e−100 µs/50 ms571.5 V=12 V·1−0.998501−0.998
    LaTeX 
    50\,\mathrm{ms} = 500\,\mathrm{kΩ} \cdot 100\,\mathrm{nF} \qquad 0.998 = e^{-100\,\mathrm{µs}/50\,\mathrm{ms}} \qquad 571.5\,\mathrm{V} = 12\,\mathrm{V} \cdot \frac{1 - 0.998^{50}}{1 - 0.998}

    Open this step on its own page, with these inputs

What this looks like

Resonant frequency against cell capacitance Cell capacitance swept from 50 nF to 150 nF with everything else held at your numbers. The dashed lines cross where you are.
Resonant frequency against cell capacitanceResonant frequency falls from 50.7 kHz to 29.3 kHz as cell capacitance rises from 50 nF to 150 nF. At your cell capacitance of 100 nF it is 35.9 kHz.304050605075100125150100 nF35.9 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.