# Explain my capacitor discharge math discrepancy

• kenw232
In summary, the conversation discusses the process of cross-checking the mA discharged from a .33uF capacitor charged to 1200V by a circuit with 37W of power. Three methods are used to calculate the mA, with the first two methods validating each other, but the third method resulting in a drastic decrease in wattage. It is suggested that the energy loss in the resistors and other components may account for the discrepancy in the results. Further calculations are recommended to fully understand the energy dissipation in the circuit.
kenw232
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:
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. 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.

#### Attachments

• single_circuit.jpg
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• single_circuit_scope.jpg
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• calc2.jpg
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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.

You need to keep in mind that there are losses in the transformers, the diodes, and the capacitor itself.

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.

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.

## 1. What is a capacitor discharge math discrepancy?

A capacitor discharge math discrepancy refers to a difference between the calculated and measured values of the discharge time of a capacitor. This can occur due to various factors such as circuit resistance, stray capacitance, and non-ideal components.

## 2. How do I calculate the discharge time of a capacitor?

The discharge time of a capacitor can be calculated using the formula t = RC, where t is the time in seconds, R is the resistance in ohms, and C is the capacitance in Farads. However, this calculation may not always match the measured value due to external factors.

## 3. What causes a capacitor discharge math discrepancy?

Some of the common reasons for a capacitor discharge math discrepancy include variations in resistance due to temperature changes, parasitic capacitance in the circuit, and non-linearity of the components used in the circuit.

## 4. How can I minimize capacitor discharge math discrepancy?

To minimize the discrepancy between calculated and measured values, it is important to use high-quality components, minimize the length of wires and traces in the circuit, and account for any parasitic capacitance. Additionally, considering the effects of temperature on resistance and using a more accurate measurement device can also help reduce the discrepancy.

## 5. Can a capacitor discharge math discrepancy be completely eliminated?

While it is not possible to completely eliminate a capacitor discharge math discrepancy, it can be minimized by following good design practices and using high-quality components. It is also important to keep in mind that some level of discrepancy is expected due to external factors that cannot be controlled.

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