Electronic – Constant power discharge of a non-ideal capacitor

In general voltage on the capacitor with respect to the current is governed by the equation:

\$v(t)= \frac{q(t)}{C} = \frac{1}{C}\int_{t_0}^t i(\tau) \mathrm{d}\tau+v(t_0)\$,

By the definition for CCS:

\$ i(\tau) = I \$,

from this we can derive that:

\$v(t)= \frac{1}{C}(I t - It_0) + v(t_0)\$

now assuming \$t_0 = 0\$ this simplifies to:

\$v(t)= \frac{1}{C}I t + v(0)\$.

What this means is simple! The voltage across capacitor will change linearly with time. The "rate" of change (or "slope") depends on the current magnitude and the capacitance:

• The bigger the capacitance the slower voltage changes.
• The bigger the current the faster voltage changes.
• The sign of the change (voltage rising or falling) depends on the sign or direction of the current. Obviously if current is flowing into capacitor voltagwe will rise if flowing out of capacitor voltage will fall.

The circuit you have included is essentially a high voltage source.

The switching frequency and capacitor impedance are not relevant for the discharge cycle. The oscillator will boost the voltage and charge the capacitor by virtue of the diode rectifier until it is charged to approximately 1.1kV. From that point on, the switching frequency of your oscillator is not relevant.

The discharge of the capacitor will take on the profile of Vcap=Vo*e^(-t/RC) The complex variable in this is the unknown discharge impedance R. You are using a gas discharge tube to discharge into the output multiplier transformer in a similar manner Tazer guns utilise spark gaps. The problem is that when the Xenon vapour/air gap ionises, its impedance is extremely dynamic and is only ionised as long as there is sufficient ionisation energy stored within the capacitor. The main limiting factor for the apparent lack of power is the ionisation characteristics of the Gas tube as it will only be conductive for a short amount of time and not deliver the full charge of the capacitor.

Also be aware, that a spark, whether in an air gap or within a ionisable vapour, is NOT a short circuit. It dissipated energy in the same manner as a resistor. All that light that comes out of the tube is energy which is NOT delivered to your prongs.

When a camera flash is fired, the storage capacitor is never fully depleted. There is always a residual hazardous voltage on the capacitor terminals because the tube vapor has a minimum ionisation energy.