Originally Posted by bwaslo
A cap stores energy, rather than power. Energy is power*time.
bwaslo is right, caps store energy, in Joules, but you can convert it to Watts if you can calculate how much the voltage has dropped down to within a certain period of time.
I found a schematic of the EP2500 at http://diybanter.com/electronic-...tic-p-l-s.html
and it shows it is a "class H" amp, which has two supply rails. (±55V with another ±55V stacked on top to provide ±110V) Calculating power draw from caps in a class H amp complicates things considerably, but if we make a few assumptions we can come up with a few numbers.
- According to the online-shashki "Measuring Amplifiers" link /avs-vb/showt...3#post10753603
the EP2500 can provide 450W/channel into 8Ω at bass frequencies.
- 450W into 8Ω requires 60V RMS, or 85V peak for sine waves.
- So let's assume that the positive rail can drop from 110V down to 85V before the output will clip.
- Let's assume the higher voltage rail caps will provide all the required power. (they won't, but ignore that for now)
- The higher voltage rail has a total supply capacitance of two x 12,000µF caps in series, which means the total value is only half that, which = 6,000µF. (or 6.0mF)
- The energy stored in a cap is 0.5*C*V^2, as bwaslo stated.
- So, the energy removed from the high voltage rail caps is (0.5*6.0mF*110V^2) - (0.5*6.0mF*85V^2) = 14.625 Joules.
- The caps get recharged 120 times a second by the positive AND negative peaks of the 60Hz AC power source, so the caps will be discharged for 1/120 of a second, or 8.33mS.
- Power taken from the caps is P = J/S, so 14.625Joules/8.33mSec = 1755W.
Now 1755W is a lot more than the 900W going to the speakers, but even Class H amps aren't 100% efficient so some of that stored power will have to go towards heating up the room. All the transistors will have voltage drops, so the 85V minimum will need to be raised. The AC transformer will have higher losses at full load, so the 110V rail will probably sag. All these deficiencies combine to rob the theoretical power available from the caps to practically half the original value.
So, to answer your question, to calculate the power stored in a cap you first have to calculate the energy stored in it at its highest voltage and then subtract the energy remaining in it at the lower voltage to determine how much energy was removed. You then take that removed energy and divide it by the amount of time it took to discharge it to the lower level in order to determine the power the cap provided.