VIC4 - Meyer 9x Tubular

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 390.6 µS/cm

  Tap Water (Hard)'s conductivity, from its dissolved solids reading.

  Dissolved solids to conductivity

• L₁ Charging choke 195.4 µH

  The charging choke's inductance: 100 turns on Ferrite Toroid T200-2 (μᵣ 10, 0.79 cm², 5.08 cm path).

  Inductance of a winding on a core

• L₂ Blocking choke 195.4 µH

  The blocking choke's inductance: 100 turns on Ferrite Toroid T200-2 (μᵣ 10, 0.79 cm², 5.08 cm path).

  Inductance of a winding on a core

• R Series resistance 0.3923 Ω

  Series resistance, added up: Bifilar Choke 1 - 100T AWG18 (Start-Start) 0.0962 Ω + Bifilar Choke 1 - 100T AWG18 (Start-Start) 0.0962 Ω + Tubular - 9x Series (Meyer Style) 0.2 Ω. Wiring and connections are not counted.

• C Cell capacitance 4.627 nF

  One tube of Tubular - 9x Series (Meyer Style) — 0.75 in over 1 in, 12 in long — in tap water (typical) at 20 °C.

  Coaxial cell capacitance — 1 note on that page

• C Cell capacitance 514.1 pF

  9 such cells in series. 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 34.61 Ω

  The leak across one tube, from its geometry and Tap Water (Hard) at 391 µS/cm. 9 such cells in series: 9 times that.

  Leak resistance across a tubular cell — 1 note on that page

• V Drive amplitude 120 V

  The drive amplitude of 12 V stepped up through Forward Mode 1:10 Step-Up (100:1000). The 1 A primary current given to the ratio calculation is for its current figures only; the voltage ratio does not depend on it.

  Transformer step-up to the cell

• Both chokes are recorded as bifilar windings, but the arrangement is set to independent. If they share a core, choose an aiding or opposing arrangement and a coupling coefficient.

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.

   $$100 \mathrm{\mu s}=\frac{1}{10}50 \mathrm{\mu s}=100 \mathrm{\mu s}·\frac{50}{ }50=\lfloor \frac{50}{ }·\frac{10}{ }\rfloor 25 \mathrm{\%}=\frac{50 \mathrm{\%}·50 \mathrm{\%}}{100}$$

   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 390.8 µH

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

   $$0 \mathrm{mH}=\pm 0\sqrt{195.4 \mathrm{\mu H}·195.4 \mathrm{\mu H}}390.8 \mathrm{\mu H}=195.4 \mathrm{\mu H}+195.4 \mathrm{\mu H}+2·0 \mathrm{mH}195.4 \mathrm{\mu H}=195.4 \mathrm{\mu H}+0 \mathrm{mH}$$

   LaTeX

   0\,\mathrm{mH} = \pm 0 \sqrt{195.4\,\mathrm{µH} \cdot 195.4\,\mathrm{µH}} \qquad 390.8\,\mathrm{µH} = 195.4\,\mathrm{µH} + 195.4\,\mathrm{µH} + 2 \cdot 0\,\mathrm{mH} \qquad 195.4\,\mathrm{µH} = 195.4\,\mathrm{µH} + 0\,\mathrm{mH}

   Open this step on its own page, with these inputs

3. 3. Series LC resonant frequency

   f 355.1 kHz

   Where that inductance and the cell ring.

   $$355.1 \mathrm{kHz}=\frac{1}{2\pi \sqrt{390.8 \mathrm{\mu H}·514.1 \mathrm{pF}}}$$

   LaTeX

   355.1\,\mathrm{kHz} = \frac{1}{2\pi\sqrt{390.8\,\mathrm{µH} \cdot 514.1\,\mathrm{pF}}}

   Open this step on its own page, with these inputs

4. 4. Q factor and voltage rise

   Q 2223

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

   $$2223=\frac{1}{0.3923}\sqrt{\frac{390.8}{ }}266.7 \mathrm{kV}=2223·120 \mathrm{V}$$

   LaTeX

   2223 = \frac{1}{0.3923\,\mathrm{Ω}}\sqrt{\frac{390.8\,\mathrm{µH}}{514.1\,\mathrm{pF}}} \qquad 266.7\,\mathrm{kV} = 2223 \cdot 120\,\mathrm{V} \qquad 159.7\,\mathrm{Hz} = \frac{355.1\,\mathrm{kHz}}{2223}

   • A Q of 2223 is very high for a circuit with water in it, and implies a bandwidth of only 159.7 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.
   • A predicted 266.7 kV across the cell is past where insulation, standoffs and the water itself hold up. Check the field strength across the gap before trusting this.

   Open this step on its own page, with these inputs

5. 5. Damping and ringdown

   ζ 0.000225

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

   $$0.000225=\frac{0.3923}{ }\sqrt{\frac{514.1}{ }}1.993 \mathrm{ms}=\frac{2·390.8 \mathrm{\mu H}}{0.3923}355.1 \mathrm{kHz}=\frac{1}{2\pi }\sqrt{\frac{1}{390.8 \mathrm{\mu H}·514.1 \mathrm{pF}}-\frac{1}{{1.993}^{ }}}1629=\frac{\ln 10}{\pi }·\frac{1}{0.3923}\sqrt{\frac{390.8}{ }}$$

   LaTeX

   0.000225 = \frac{0.3923\,\mathrm{Ω}}{2} \sqrt{\frac{514.1\,\mathrm{pF}}{390.8\,\mathrm{µH}}} \qquad 1.993\,\mathrm{ms} = \frac{2 \cdot 390.8\,\mathrm{µH}}{0.3923\,\mathrm{Ω}} \qquad 355.1\,\mathrm{kHz} = \frac{1}{2\pi} \sqrt{\frac{1}{390.8\,\mathrm{µH} \cdot 514.1\,\mathrm{pF}} - \frac{1}{1.993\,\mathrm{ms}^{2}}} \qquad 1629 = \frac{\ln 10}{\pi} \cdot \frac{1}{0.3923\,\mathrm{Ω}} \sqrt{\frac{390.8\,\mathrm{µH}}{514.1\,\mathrm{pF}}}

   Open this step on its own page, with these inputs

6. 6. Current and power under a pulsed drive

   I 3.879 mA

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

   $$-30.93 \mathrm{k\Omega }=2\pi 10 \mathrm{kHz}·390.8 \mathrm{\mu H}-\frac{1}{2\pi 10 \mathrm{kHz}·514.1 \mathrm{pF}}30.93 \mathrm{k\Omega }=\sqrt{{0.3923}^{ }+{-30.93}^{ }}3.879 \mathrm{mA}=\frac{120}{ }5.904 \mathrm{\mu W}={3.879}^{ }·0.3923 \mathrm{\Omega }1.476 \mathrm{\mu W}=5.904 \mathrm{\mu W}·\frac{25}{ }0.2952 \mathrm{nJ}=5.904 \mathrm{\mu W}·50 \mathrm{\mu s}$$

   LaTeX

   -30.93\,\mathrm{kΩ} = 2\pi 10\,\mathrm{kHz} \cdot 390.8\,\mathrm{µH} - \frac{1}{2\pi 10\,\mathrm{kHz} \cdot 514.1\,\mathrm{pF}} \qquad 30.93\,\mathrm{kΩ} = \sqrt{0.3923\,\mathrm{Ω}^{2} + -30.93\,\mathrm{kΩ}^{2}} \qquad 3.879\,\mathrm{mA} = \frac{120\,\mathrm{V}}{30.93\,\mathrm{kΩ}} \qquad 5.904\,\mathrm{µW} = 3.879\,\mathrm{mA}^{2} \cdot 0.3923\,\mathrm{Ω} \qquad 1.476\,\mathrm{µW} = 5.904\,\mathrm{µW} \cdot \frac{25\,\mathrm{%}}{100} \qquad 0.2952\,\mathrm{nJ} = 5.904\,\mathrm{µW} \cdot 50\,\mathrm{µs}

   • The drive is at 0.03× the loop's resonance (355 kHz). The net reactance of -30.9 kΩ is what limits the current here, not the resistance; on resonance the same drive would push 306 A.

   Open this step on its own page, with these inputs

7. 7. Step charging accumulation

   U 120 V

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

   $$0.01779 \mathrm{\mu s}=34.61 \mathrm{\Omega }·514.1 \mathrm{pF}0=e^{-100 \mathrm{\mu s}/0.01779 \mathrm{\mu s}}120 \mathrm{V}=120 \mathrm{V}·\frac{1-0^{50}}{1-0}$$

   LaTeX

   0.01779\,\mathrm{µs} = 34.61\,\mathrm{Ω} \cdot 514.1\,\mathrm{pF} \qquad 0 = e^{-100\,\mathrm{µs}/0.01779\,\mathrm{µs}} \qquad 120\,\mathrm{V} = 120\,\mathrm{V} \cdot \frac{1 - 0^{50}}{1 - 0}

   • 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