VIC5 - 6x Parallel 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 125 µS/cm

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

  Dissolved solids to conductivity

• L₁ Charging choke 31.42 mH

  The charging choke's inductance: 100 turns on MnZn Ferrite E-Core (μᵣ 2000, 1 cm², 8 cm path).

  Inductance of a winding on a core — 1 note on that page

• L₂ Blocking choke 31.42 mH

  The blocking choke's inductance: 100 turns on MnZn Ferrite E-Core (μᵣ 2000, 1 cm², 8 cm path).

  Inductance of a winding on a core — 1 note on that page

• R Series resistance 0.2423 Ω

  Series resistance, added up: Bifilar Choke 2 - 100T AWG18 (Finish-Finish) 0.0962 Ω + Bifilar Choke 2 - 100T AWG18 (Finish-Finish) 0.0962 Ω + Tubular - 6x Parallel (High Capacity) 0.05 Ω. Wiring and connections are not counted.

• C Cell capacitance 4.627 nF

  One tube of Tubular - 6x Parallel (High Capacity) — 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 27.76 nF

  6 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 2.003 Ω

  The leak across one tube, from its geometry and Tap Water (Soft) at 125 µS/cm. 6 such cells in parallel leak 6 times as readily, so this is one 6th of that.

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

• V Drive amplitude 60 V

  The drive amplitude of 12 V stepped up through Forward Mode 1:5 Step-Up (100:500). 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 62.83 mH

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

   $$0 \mathrm{mH}=\pm 0\sqrt{31.42 \mathrm{mH}·31.42 \mathrm{mH}}62.83 \mathrm{mH}=31.42 \mathrm{mH}+31.42 \mathrm{mH}+2·0 \mathrm{mH}31.42 \mathrm{mH}=31.42 \mathrm{mH}+0 \mathrm{mH}$$

   LaTeX

   0\,\mathrm{mH} = \pm 0 \sqrt{31.42\,\mathrm{mH} \cdot 31.42\,\mathrm{mH}} \qquad 62.83\,\mathrm{mH} = 31.42\,\mathrm{mH} + 31.42\,\mathrm{mH} + 2 \cdot 0\,\mathrm{mH} \qquad 31.42\,\mathrm{mH} = 31.42\,\mathrm{mH} + 0\,\mathrm{mH}

   Open this step on its own page, with these inputs

3. 3. Series LC resonant frequency

   f 3.811 kHz

   Where that inductance and the cell ring.

   $$3.811 \mathrm{kHz}=\frac{1}{2\pi \sqrt{62.83 \mathrm{mH}·27.76 \mathrm{nF}}}$$

   LaTeX

   3.811\,\mathrm{kHz} = \frac{1}{2\pi\sqrt{62.83\,\mathrm{mH} \cdot 27.76\,\mathrm{nF}}}

   Open this step on its own page, with these inputs

4. 4. Q factor and voltage rise

   Q 6209

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

   $$6209=\frac{1}{0.2423}\sqrt{\frac{62.83}{ }}372.5 \mathrm{kV}=6209·60 \mathrm{V}$$

   LaTeX

   6209 = \frac{1}{0.2423\,\mathrm{Ω}}\sqrt{\frac{62.83\,\mathrm{mH}}{27.76\,\mathrm{nF}}} \qquad 372.5\,\mathrm{kV} = 6209 \cdot 60\,\mathrm{V} \qquad 0.6138\,\mathrm{Hz} = \frac{3.811\,\mathrm{kHz}}{6209}

   • A Q of 6209 is very high for a circuit with water in it, and implies a bandwidth of only 0.6138 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 372.5 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 8.053e-5

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

   $$8.053e-5=\frac{0.2423}{ }\sqrt{\frac{27.76}{ }}518.6 \mathrm{ms}=\frac{2·62.83 \mathrm{mH}}{0.2423}3.811 \mathrm{kHz}=\frac{1}{2\pi }\sqrt{\frac{1}{62.83 \mathrm{mH}·27.76 \mathrm{nF}}-\frac{1}{{518.6}^{ }}}4551=\frac{\ln 10}{\pi }·\frac{1}{0.2423}\sqrt{\frac{62.83}{ }}$$

   LaTeX

   8.053e-5 = \frac{0.2423\,\mathrm{Ω}}{2} \sqrt{\frac{27.76\,\mathrm{nF}}{62.83\,\mathrm{mH}}} \qquad 518.6\,\mathrm{ms} = \frac{2 \cdot 62.83\,\mathrm{mH}}{0.2423\,\mathrm{Ω}} \qquad 3.811\,\mathrm{kHz} = \frac{1}{2\pi} \sqrt{\frac{1}{62.83\,\mathrm{mH} \cdot 27.76\,\mathrm{nF}} - \frac{1}{518.6\,\mathrm{ms}^{2}}} \qquad 4551 = \frac{\ln 10}{\pi} \cdot \frac{1}{0.2423\,\mathrm{Ω}} \sqrt{\frac{62.83\,\mathrm{mH}}{27.76\,\mathrm{nF}}}

   Open this step on its own page, with these inputs

6. 6. Current and power under a pulsed drive

   I 17.78 mA

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

   $$3.375 \mathrm{k\Omega }=2\pi 10 \mathrm{kHz}·62.83 \mathrm{mH}-\frac{1}{2\pi 10 \mathrm{kHz}·27.76 \mathrm{nF}}3.375 \mathrm{k\Omega }=\sqrt{{0.2423}^{ }+{3.375}^{ }}17.78 \mathrm{mA}=\frac{60}{ }76.6 \mathrm{\mu W}={17.78}^{ }·0.2423 \mathrm{\Omega }19.15 \mathrm{\mu W}=76.6 \mathrm{\mu W}·\frac{25}{ }3.83 \mathrm{nJ}=76.6 \mathrm{\mu W}·50 \mathrm{\mu s}$$

   LaTeX

   3.375\,\mathrm{kΩ} = 2\pi 10\,\mathrm{kHz} \cdot 62.83\,\mathrm{mH} - \frac{1}{2\pi 10\,\mathrm{kHz} \cdot 27.76\,\mathrm{nF}} \qquad 3.375\,\mathrm{kΩ} = \sqrt{0.2423\,\mathrm{Ω}^{2} + 3.375\,\mathrm{kΩ}^{2}} \qquad 17.78\,\mathrm{mA} = \frac{60\,\mathrm{V}}{3.375\,\mathrm{kΩ}} \qquad 76.6\,\mathrm{µW} = 17.78\,\mathrm{mA}^{2} \cdot 0.2423\,\mathrm{Ω} \qquad 19.15\,\mathrm{µW} = 76.6\,\mathrm{µW} \cdot \frac{25\,\mathrm{%}}{100} \qquad 3.83\,\mathrm{nJ} = 76.6\,\mathrm{µW} \cdot 50\,\mathrm{µs}

   • The drive is at 2.62× the loop's resonance (3.81 kHz). The net reactance of 3.37 kΩ is what limits the current here, not the resistance; on resonance the same drive would push 248 A.

   Open this step on its own page, with these inputs

7. 7. Step charging accumulation

   U 60 V

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

   $$0.0556 \mathrm{\mu s}=2.003 \mathrm{\Omega }·27.76 \mathrm{nF}0=e^{-100 \mathrm{\mu s}/0.0556 \mathrm{\mu s}}60 \mathrm{V}=60 \mathrm{V}·\frac{1-0^{50}}{1-0}$$

   LaTeX

   0.0556\,\mathrm{µs} = 2.003\,\mathrm{Ω} \cdot 27.76\,\mathrm{nF} \qquad 0 = e^{-100\,\mathrm{µs}/0.0556\,\mathrm{µs}} \qquad 60\,\mathrm{V} = 60\,\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