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Explain my capacitor discharge math discrepancy

  1. Jan 10, 2014 #1
    I'm trying to cross check mathematically the mA being discharged from a .33uF capacitor which is charged to about 1200V by a circuit (see single_circuit.jpg) 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 and get the attachment single_circuit_scope.jpg.
    X = 1 / (2 x Pi x freq x C)
    X = 1 / (6.28318 x 42 x 0.00000033)
    X = 11483 Ohms
    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:
    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. Or just see the attachment calc2.jpg. 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?

    Any help in clarifying what I am doing wrong would be appreciated.

    Attached Files:

  2. jcsd
  3. Jan 10, 2014 #2


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    I must be honest and say I haven't looked in great detail at this but it could be worth pointing out that charging a capacitor via a resistor involves energy loss in the resistor. Could this be the source of your shortfall? You will be in a better position than I, to fit that spanner into your works. Hope it helps.
  4. Jan 11, 2014 #3
    You need to keep in mind that there are losses in the transformers, the diodes, and the capacitor itself.
  5. Jan 11, 2014 #4
    Yes, but not 25W worth (35W from #1 - 10W from #3). Specially considering 2 and 3 derive their answer straight from the waveform itself, so they only care about the wave, all loss from T1 is already accounted for there yet the wattage's turn out be completely different. Someone has to know why.
  6. Jan 11, 2014 #5


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    Integrate the energy dissipated in the resistance before you wonder where it's all gone. As I wrote before, every time you charge a capacitor through a resistor from an emf of V, the energy stored is QV/2 and the energy from the supply emf is QV. You lost half. Do some actual sums for your AC circuit and your explanation should be there.
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