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Old 05-14-2011, 10:47 AM - Thread Starter
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I'm trying to get a sense for how much power is stored in capacitors. I will use the old EP2500 numbers that I have for a working example.

Capacitors are rated in uF (microfarad, 1/1,000,000 of a farad) and v (voltage).

EP2500: 96,000 uF of 63v capacitors.

uF * 1,000,000 = 1 farad

1 farad * v = 1 coulomb (a unit of charge).

So, the fully charged caps in the EP2500 hold:
96,000 uF * 1,000,000 * 63 v = 6.048 coulomb

1 coulomb = 1 amp * second

So, if this is correct, the EP2500 has roughly 6 amp seconds worth of capacitance, which is enough to provide 6 amps for one second.

The question is how to translate that number to a "watts" number.

Do you just multiply the "amps" by the rail voltage to get the theoretical reserve power?

I think the Behringer has 120v rails, which would give:
[Since power = voltage * current]

reserve power = 120 v * 6 amps = 720 watts (but only available for one second).

Is this wrong, right, or incomplete?

Thanks in advance guys.

[I know that capacitors don't charge and discharge instantaneously, so let's set that piece aside.]

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Old 05-14-2011, 11:18 AM
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Not knowing the full design specs of the ep2500, I'll start the discussion by pointing out that a capacitors rated voltage (63v) in this example doesn't mean it is necessarily charged to that voltage. What is the rail voltage of this amp?

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Old 05-14-2011, 11:29 AM - Thread Starter
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iirc, +/-110v.

and there are 4 12000uF caps per channel if that helps.

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Old 05-14-2011, 03:21 PM
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A cap stores energy, rather than power. Energy is power*time.

The stored energy in a capacitor is 0.5*C*V^2, C in Farads, V in voltage across the capacitor, energy in Joules.
One Joule is 1 watt supplied for one second.

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Old 05-14-2011, 03:41 PM
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Quote:
Originally Posted by LTD02 View Post

I'm trying to get a sense for how much power is stored in capacitors. I will use the old EP2500 numbers that I have for a working example.

EP2500: 96,000 uF of 63v capacitors.

So, the fully charged caps in the EP2500 hold:
96,000 uF * 1,000,000 * 63 v = 6.048 coulomb.

So, if this is correct, the EP2500 has roughly 6 amp seconds worth of capacitance, which is enough to provide 6 amps for one second.

Correct. According to this calculator, a 96,000 uF cap charged to 63v has 6 Coulombs or 190 Joules of energy.

Quote:


The question is how to translate that number to a "watts" number. Do you just multiply the "amps" by the rail voltage to get the theoretical reserve power?

How do you intend to interpret "reserve power"? How long the amp can play after being unplugged? Or just to compare how much energy is stored on one amp's power supply vs. another? If the latter, I would use Joules. If the former, you have to know that the reserve power of the cap will not tell you that, since the amp will quit working when the DC voltage drops to maybe 75% or 50% of charge, well before the cap's energy is depleted.

The power P in watts is equal to the energy E in joules, divided by the time period t in seconds:

P = E / t
thus:
watt = joule / second

or

1 watt/volt = 1 amp = 1 coulomb/second

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Old 05-14-2011, 05:15 PM
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Maybe he's trying to figure out if this provides any insight into the burst output capability of an amp? Dunno. That is of course as limited by the output devices as it is the supply side.

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Old 05-14-2011, 07:50 PM
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Quote:
Originally Posted by bwaslo View Post

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.
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Old 05-14-2011, 08:01 PM - Thread Starter
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i was just trying to get some feel for how much "power" is stored in the amp for short bursts.

from roger's link:

watt seconds = 1/2 QV, where Q is the charge in coulomb and v is the charge voltage.

so in this case:

watt seconds = 1/2 * 6.048 * 63 = 190.5

so for short bursts of 1/10 of a second, the reserve power is about 1905 watts, which is about double the steady state power of the amp (limited of course by the output voltage of the amp).

if i have done that right, that is kind of what i was looking for. just trying to get some sense for how much capacitors assist the burstable output. i know that they are important for non smps amps, but i wanted to try to put some numbers to it.

if the numbers are roughly accurate and someone can confirm, that would be appreciated. if the numbers are completely wrong and someone can provide the correct numbers, that would be even more appreciated! :-)

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Old 05-14-2011, 08:03 PM - Thread Starter
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thanks stereo! we must have been typing at the same time, as i didn't see your note before writing mine.

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