# Capacitor discharge math discrepancy

I'm trying to cross check mathematically the mA being discharged from a .33uF capacitor which is charged to about 1200V by a circuit

which is provided 37W of power to it. I do it three ways.

1) The circuit draws 35W total. 13V @ 5A = 65W. But the power supply drops the 13V to 7V to maintain the 5A so the effective wattage is 7V x 5A = 35W. I just assumed this 35W carries through the entire circuit and out T2 because energy cannot be created and destroyed and there is little to no resistance from what I can see. So the wattage coming out of the capacitor is still roughly < 35W. We should use Vavg I think, not Vpeak, so 35W / 600Vavg = 58mA.

2) Now (1) makes sense, but I was hoping there was a cross check. I noticed on a page like this: http://cnx.org/content/m42427/latest/?collection=col11406/latest (see Example 2) you can calculate the mA from a capacitor via it's reactance. X = 1 / (2 x Pi x freq x C). As shown there I do the same. I put the gas discharge tube on the scope

$X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{6.28318 × 42 × 330\cdot 10^{-9}} = 11483 \Omega$

720Vrms / 11483 = 0.06270A or 62mA Great, this works. Essentially the same as (1). So (1) and (2) validate each other. This also means the wattage then is 720Vrms x .0627 = 45W. 45W is a little too high and not valid I think. After all Vrms is for AC, I have changing DC. So if we used Vavg its 600/11483 = 52mA, 52mA x 600Vavg = 31W. This makes more sense and probably also accounts for the wattage loss from stepping up the voltage @ T1. (35W - 31W = 4W loss).

3) But then I found E = (V² x C) / 2 which is used to calculate the energy per pulse from a capacitor discharge. Also conveniently calculated at: http://www.vishay.com/resistors/pulse-energy-calculator/ E = (1200Vpeak² x 0.00000033) / 2 E = .2376 Joules per pulse Joules per second = .2376 x frequency Joules per second = .2376 x 42.4Hz Joules per second = 10.07. Joules per second is the same as watts. Watts = 10W. This is shown easily and proven with the Vishay pulse calculator at that URL above. 10W? Why the sudden massive drop in wattage?

So this is where I am right now, I want to know why (3) is only giving me 10W. How can 1 and 2 work out, but 3 be so far off. Is (3) not an applicable formula to use in this case?

• Please check mathjax to polish up your question, I inserted an example. The text is very hard to read. Also insert the images in the text where you refer to them, not at the end. Jan 12, 2014 at 11:41
• Also, please try to narrow this down to a specific question. You wander all over the place and you seem to have many implied questions here. If your question is about charging a capacitor, then don't throw in every other part of your system. Jan 12, 2014 at 13:03
• You are responding to a thread that died out almost 10 years ago. Nov 9, 2023 at 1:41

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.

In addition to what Martin said, that circuit is inefficient unless T1 is a high leakage type that acts like it has an inductor in series with the o/p. Until the capacitor is charged, the FETs are driving into a partial short circuit, dropping part of the i/p voltage across themselves, like when you charge a capacitor through a resistor (but via the transformer). With the low impedance of those FETs, some of the loss would also be in the transformer winding resistance and some in the internal resistance (ESR (effective series res.)) of the power supply o/p capacitor. A flyback type inverter where energy is stored in the core of the transformer then fed out when the FET is turned off would be more efficient.

Incidentally, the residual voltage left on a camera flash cacacitor is only about 20V, from about 300. That is about 0.444% of the energy, but what Martin said about the light out the tube being wasted energy is certainly correct.