8XA - 4x Parallel Tubular

4 tubular cells in parallel for maximum capacitance.

From the archive's library.

The parts

The cell Tubular - 4x Parallel 3/4" x 8" — water cell tubular
Water Distilled Water — Distilled
Charging choke Positive Choke - 100 Turns AWG18 — AWG 18 Copper AWG 18
Its core Ferrite Toroid T106-2 — T106 toroid
Blocking choke Negative Choke - 100 Turns AWG18 — AWG 18 Copper AWG 18
Its core Ferrite Toroid T106-2 — T106 toroid
Transformer none

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

f Drive frequency

kHz

The pulse frequency the loop is driven at.

V Drive amplitude

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

D Pulse duty

How much of each pulse period the drive is on.

g Gate frequency

The slow gate switching the pulse train on and off.

G Gate duty

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

c Choke arrangement

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

k Coupling coefficient

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 0.2923 Ω

Series resistance, added up: Positive Choke - 100 Turns AWG18 0.0962 Ω + Negative Choke - 100 Turns AWG18 0.0962 Ω + Tubular - 4x Parallel 3/4" x 8" 0.1 Ω. Wiring and connections are not counted.
C Cell capacitance 2.189 nF

One tube of Tubular - 4x Parallel 3/4" x 8" — 0.5 in over 0.75 in, 8 in long — in distilled / deionised at 25 °C.

Coaxial cell capacitance
C Cell capacitance 8.754 nF

4 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 1.016 kΩ

The leak across one tube, from its geometry and Distilled Water at 0.781 µS/cm. 4 such cells in parallel leak 4 times as readily, so this is one 4th of that.

Leak resistance across a tubular cell

Result

f₀ Resonant frequency 121.2 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 150 Ω

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

Q Q factor 513.1

How sharp the resonance is against the series resistance.

U Resonant rise 6.158 kV

The cell voltage at resonance — Q times the drive.

ζ Damping ratio 0.000974

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

τ Decay time constant 1.347 ms

How fast the ring dies away.

I Peak current 6.646 mA

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

W Average power 3.227 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

121.2 kHz = \frac{1}{2 π \sqrt{196.9 µH \cdot 8.754 nF}}

LaTeX

121.2\,\mathrm{kHz} = \frac{1}{2\pi \sqrt{196.9\,\mathrm{µH} \cdot 8.754\,\mathrm{nF}}}

Worth knowing

Q factor and voltage rise — A Q of 513 is very high for a circuit with water in it, and implies a bandwidth of only 236.2 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 0.08× the loop's resonance (121 kHz). The net reactance of -1.81 kΩ is what limits the current here, not the resistance; on resonance the same drive would push 41.1 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 121 kHz — 0.08× resonance. The resonant rise of 6.16 kV above is what this loop would do driven on resonance; at the drive frequency the cell sees the drive through -1.81 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. 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 = \frac{1}{10} 50 µs = 100 µs \cdot \frac{50}{} 50 = ⌊ \frac{50}{} \cdot \frac{10}{} ⌋ 25 % = \frac{50 % \cdot 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}

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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

$0 mH = \pm 0 \sqrt{98.47 µH \cdot 98.47 µH} 196.9 µH = 98.47 µH + 98.47 µH + 2 \cdot 0 mH 98.47 µH = 98.47 µH + 0 mH$
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}
```
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3. Series LC resonant frequency
f 121.2 kHz

Where that inductance and the cell ring.
With these numbers

$121.2 kHz = \frac{1}{2 π \sqrt{196.9 µH \cdot 8.754 nF}}$
LaTeX
```
121.2\,\mathrm{kHz} = \frac{1}{2\pi\sqrt{196.9\,\mathrm{µH} \cdot 8.754\,\mathrm{nF}}}
```
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4. Q factor and voltage rise
Q 513.1

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

$513.1 = \frac{1}{0.2923} \sqrt{\frac{196.9}{}} 6.158 kV = 513.1 \cdot 12 V$
LaTeX
```
513.1 = \frac{1}{0.2923\,\mathrm{Ω}}\sqrt{\frac{196.9\,\mathrm{µH}}{8.754\,\mathrm{nF}}} \qquad 6.158\,\mathrm{kV} = 513.1 \cdot 12\,\mathrm{V} \qquad 236.2\,\mathrm{Hz} = \frac{121.2\,\mathrm{kHz}}{513.1}
```
- A Q of 513 is very high for a circuit with water in it, and implies a bandwidth of only 236.2 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. Damping and ringdown

ζ 0.000974

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

With these numbers

0.000974 = \frac{0.2923}{} \sqrt{\frac{8.754}{}} 1.347 ms = \frac{2 \cdot 196.9 µH}{0.2923} 121.2 kHz = \frac{1}{2 π} \sqrt{\frac{1}{196.9 µH \cdot 8.754 nF} - \frac{1}{{1.347}^{ms 2}}} 376.1 = \frac{\ln 10}{π} \cdot \frac{1}{0.2923} \sqrt{\frac{196.9}{}}

LaTeX

0.000974 = \frac{0.2923\,\mathrm{Ω}}{2} \sqrt{\frac{8.754\,\mathrm{nF}}{196.9\,\mathrm{µH}}} \qquad 1.347\,\mathrm{ms} = \frac{2 \cdot 196.9\,\mathrm{µH}}{0.2923\,\mathrm{Ω}} \qquad 121.2\,\mathrm{kHz} = \frac{1}{2\pi} \sqrt{\frac{1}{196.9\,\mathrm{µH} \cdot 8.754\,\mathrm{nF}} - \frac{1}{1.347\,\mathrm{ms}^{2}}} \qquad 376.1 = \frac{\ln 10}{\pi} \cdot \frac{1}{0.2923\,\mathrm{Ω}} \sqrt{\frac{196.9\,\mathrm{µH}}{8.754\,\mathrm{nF}}}

Open this step on its own page, with these inputs

6. Current and power under a pulsed drive
I 6.646 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

$-1.806 kΩ = 2 π 10 kHz \cdot 196.9 µH - \frac{1}{2 π 10 kHz \cdot 8.754 nF} 1.806 kΩ = \sqrt{{0.2923}^{Ω 2} + {-1.806}^{kΩ 2}} 6.646 mA = \frac{12}{} 12.91 µW = {6.646}^{mA 2} \cdot 0.2923 Ω 3.227 µW = 12.91 µW \cdot \frac{25}{} 0.6455 nJ = 12.91 µW \cdot 50 µs$
LaTeX
```
-1.806\,\mathrm{kΩ} = 2\pi 10\,\mathrm{kHz} \cdot 196.9\,\mathrm{µH} - \frac{1}{2\pi 10\,\mathrm{kHz} \cdot 8.754\,\mathrm{nF}} \qquad 1.806\,\mathrm{kΩ} = \sqrt{0.2923\,\mathrm{Ω}^{2} + -1.806\,\mathrm{kΩ}^{2}} \qquad 6.646\,\mathrm{mA} = \frac{12\,\mathrm{V}}{1.806\,\mathrm{kΩ}} \qquad 12.91\,\mathrm{µW} = 6.646\,\mathrm{mA}^{2} \cdot 0.2923\,\mathrm{Ω} \qquad 3.227\,\mathrm{µW} = 12.91\,\mathrm{µW} \cdot \frac{25\,\mathrm{%}}{100} \qquad 0.6455\,\mathrm{nJ} = 12.91\,\mathrm{µW} \cdot 50\,\mathrm{µs}
```
- The drive is at 0.08× the loop's resonance (121 kHz). The net reactance of -1.81 kΩ is what limits the current here, not the resistance; on resonance the same drive would push 41.1 A.
Open this step on its own page, with these inputs
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

$8.896 µs = 1.016 kΩ \cdot 8.754 nF 1.313e-5 = e^{- 100 µs / 8.896 µs} 12 V = 12 V \cdot \frac{1 - {1.313e-5}^{50}}{1 - 1.313e-5}$
LaTeX
```
8.896\,\mathrm{µs} = 1.016\,\mathrm{kΩ} \cdot 8.754\,\mathrm{nF} \qquad 1.313e-5 = e^{-100\,\mathrm{µs}/8.896\,\mathrm{µs}} \qquad 12\,\mathrm{V} = 12\,\mathrm{V} \cdot \frac{1 - 1.313e-5^{50}}{1 - 1.313e-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 4.38 nF to 13.1 nF with everything else held at your numbers. The dashed lines cross where you are.

The formula behind the curve

f₀ = \frac{1}{2 π \sqrt{L \cdot 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.

The formula behind the curve

f₀ = \frac{1}{2 π \sqrt{L \cdot 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.

8XA - 4x Parallel Tubular

The parts

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

From the parts

Result

Worth knowing

The working, step by step

1. Pulse train timing

2. Two chokes on one core

3. Series LC resonant frequency

4. Q factor and voltage rise

5. Damping and ringdown

6. Current and power under a pulsed drive

7. Step charging accumulation

What this looks like