Q factor and voltage rise

How sharp is the resonance, and how much voltage appears across the cell?

Q = \frac{1}{R}\sqrt{\frac{L}{C}} \qquad U = Q \cdot V \qquad B = \frac{f}{Q}

Method

1. Convert to SI: henries, farads, ohms.
2. Find the characteristic impedance, √(L ÷ C). At resonance the inductor and the capacitor each present this reactance, equal and opposite, so between them they cancel and only the resistance is left.
3. Divide that impedance by the total series resistance. The result is Q — literally how many times larger the circulating reactive current is than the resistive loss implies.
4. Multiply the drive voltage by Q for the voltage that appears across the cell. This is the voltage rise the VIC is built to produce, and it is the same Q that produced it.
5. For the bandwidth, divide the resonant frequency by Q. A high Q is a narrow window: at Q = 100 a ten kilohertz circuit holds its rise over only a hundred hertz, so the drive has to track it.

Assumptions

• All the loss is in one series resistance. Real loss is distributed and frequency-dependent: winding resistance rises with skin effect, core loss rises with frequency, and the water’s conduction loss falls with it. A single R is a snapshot at one frequency.
• The resistance is constant while the circuit rings. The cell’s effective resistance depends on the field across it, so a cell at 3 kV is not the same resistor it was at 300 V.
• The circuit is driven at exactly resonance. Off it, the rise falls away over the bandwidth calculated here.
• Nothing saturates. A choke pushed into saturation loses inductance, which moves the resonance and collapses the Q at the same moment.
• The voltage rise is real but it is not free energy. The power delivered is still set by the drive; Q trades current for voltage, and a high-Q circuit takes many cycles to build up to the rise this predicts.