# Questions regarding capacitors

Discussion in 'General Electronics Discussion' started by jackorocko, Dec 29, 2011.

1. ### jackorocko

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Apr 4, 2010
I am posting this here on request. I started a pm to ask this question since I did not want to highjack another thread at the time, but the information is to good to keep it private so we will just continue this discussion here.

http://www.engineeringtoolbox.com/capacitors-energy-power-d_1389.html

In this link it provides an example of a capacitor 10uF charged to 230V. It can dissipate in 5us 52kW of power. My question is, how much current is that? 52kW/230V=226A. But can it deliver that 226A consistent over the 5us or is it more as the capacitor discharges so does the available current? Which would make more sense. What is the peak current? Would be nice to see this graphed if anyone can help with that.

I am just trying to understand. Seems quite amazing to me that such a small capacitance value can deliver so much power, it is simply incredible and quite intriguing.

Last edited: Dec 29, 2011
2. ### Resqueline

2,848
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Jul 31, 2009
Very interesting/useful question indeed.

I'd expect the initial peak current to be higher and then fall off as it discharges.
In order to reach that current level it would need to have an ESR of 230V / 226A = 1 Ohm or less.
I'm confident that photoflash caps can easily attain this, but maybe not ordinary 80 degree filter capacitors.

Yes, capacitors can deliver amazing current pulses, 1000's or even 10000's of Amps is possible.

Small CPU MB cap's (1500uF 6.3V 0.009Ohms ESR) for example can deliver 700A peak each.

3. ### duke37

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771
Jan 9, 2011
I could not find the page quoted.

Some sums, they should be checked!

10microF 230V Energy = (C * V * V)/2 = 2.53J
52kW = 52000 J/sec
Time for constant power = 2.53/52000 = 48microsec

Therefore I cannot see how this works. In practise, there will be a resistance in the capacitor in the capacitor which will limit the current and dissipate much of the energy. If the resistance is 1 ohm, the peak current will be V/R = 230/1 = 230 A into a short circuit. The best that can be done is to match the load with the source so a total of 2 ohms giving a peak current of 115A with a peak power of 115 * 115 * 1 = 13.2 kW

The current will drop in an exponential way with a time constant of R * C = 2 * 10u = 20 microsec.

I think the claim is phantasy, what do you think?

4. ### jackorocko

1,284
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Apr 4, 2010
Fixed now!
http://www.engineeringtoolbox.com/capacitors-energy-power-d_1389.html

(C * V * V)/2 == (1/2 * C * sqr(V))
V = voltage across the capacitor

(.00001 * 230 * 230)/2 = .2645J
Time for constant power = .2645/52000 = 5uS

Seems right to me unless my calculator is broken

I have no idea how you come up with 2.53

Last edited: Dec 29, 2011
5. ### jackorocko

1,284
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Apr 4, 2010
You are talking about ESR? I was looking up HV capacitors yesterday on digikey and there was a laundry list of HV capacitors 1000uF caps that had ESR less then 1Ohm. Most where in the mOhm range. Are you saying that it would be detrimental to the capacitor if it was discharged in this manner of 5uS? If the ESR was less then 1Ohm, then the peak current would be higher and therefor the avg current would again be ~230A?

This brings me to another recent thread where the guy wanted to make a can crusher, what happens if the energy has some where to go, like an inductor. Would it still destroy the caps in that situation?

Last edited: Dec 29, 2011

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Oct 2, 2011
7. ### duke37

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Jan 9, 2011
I have no idea how I came up with 2.53 either, perhaps my brain cell is getting discharged!

1000uF capacitors will have a lower series resistance than 10uf caps. I intend to build an ESR meter this winter.

If the load resistance is equal to the internal resistance, you will get maximum power and 50% efficiency. Whether this will damage the capacitor will depend on the capacitor design. The advice which is often given is to discharge high voltage capacitors through a resistance to obviate damage.

8. ### Resqueline

2,848
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Jul 31, 2009
I gather from that site that it was a current averaged over 5µs and that the initial peak current would be higher. They didn't calculate it from the ESR but it was rather an energy / time calculation.
The pulse waveforms extracted from a capacitor can be modified to depart from the usual peaky shape and into a more rounded squarish shape with a relatively constant current over time - with the use of inductors & small cap's. These pulse shaping circuits (or Pulse Forming Networks) can be found in high power pulsed lasers etc.

A capacitor Equivalent circuit consists of 4 components; one ideal capacitor with a paralell (leakage) resistor in the MegaOhm range, one Series Resistor (electrolyte resistance) in the Ohm range, and one series inductance (leads etc.) in the µH range.
The Equivalent Series Resistance effectively limits the peak current available according to Ohms law, and it also limits the continous ripple current a capacitor can stand - due to heating.

I was not aware of a capacitor being anything but a capacitor - and that it had serious limitations until 10 years after I started dabbling with electronics. I happened to get access to an expensive HP circuit analyzer and got a "shock" when I measured electrolytics. They behaved so much worse than I had ever imagined.

The only capacitor I have ever damaged by a direct short was a relatively large oil filled mains power factor corrector capacitor (or maybe it was out of a constant voltage transformer, I don't quite remember - it was a long time ago). Opening it up I found the tabs were soldered to the foil with a thin wire. I guess it was made like that on purpose, the wire acting as a fuse - protecting the transformer against shorts in the capacitor.

Here is an interesting paper about photoflash capacitor behaviour btw..

Last edited: Dec 30, 2011
9. ### Merlin3189

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Aug 4, 2011
Well, firstly they don't say it can. They say, "IF" you assume it happens, then the value "can be calculated"

The power stored in a 10 μF capacitor charged to 230 V can be calculated ... 0.26 Joules
IF the energy is dissipated within 5 μs the potential power (generated?) can be calculated as ... 52 kW

If you assume the energy is delivered in 1microsecond, then you could calculate it to be 260kW. Or for that matter by making other assumptions, you could calculate it to be 52 MW or 52GW or 52TW etc.

So the issue is, as others are debating, how fast could we get the energy out. I'm content to accept their views on ESR, though I'm surprised that they don't think series inductance is significant. That could limit the rate of rise of current, so that some energy would have to be delivered at a lower power before you reached the current at which the power was 52kW. The Voltage (on the ideal capacitor element) would be falling during this phase, so the current needed to achieve 52kW would need to be even higher than the 400A or so needed if we could assume the capacitor simply discharged linearly.

But a second point is more general. Is it really surprising? Power is simply the RATE at which energy is delivered (or dissipated or generated or whatever.)
Energy is a conserved quantity: however you juggle it about, it stays the same and we know from experience what it is. Chuck us an apple, or pick your briefcase up and that's about a Joule or two. I've got to climb onto the table to use 1kJ and a decent mountain to use 1MJ. A capacitor storing 0.25J is hardly something to write home about: a teaspoon of water would warm up by about 0.1 degC with that.
But power is more of a mathematical abstraction, which is much less intuitive. Yes, we recognise power and can relate to it when it is measured over a reasonable timespan. If it takes me a second or two to climb on the table, that's perhaps a half kW. I doubt if I could keep that up for long, so I guess my sustained power output at less than half a kW. Climbing the mountain I'm probably putting only a few hundred W into the vertical ascent. But when we shorten the timespan, the numbers become more unreal. When I jump down off the table, it takes less than half a second to reach the ground: that's me dissipating over 2kW. If I swing a sledge hammer, say 5kg at 2m/sec = 10 Joule, and it smashes 1mm into a concrete block (2m/sec to stop over 1mm, a=2000m/sec/sec, which takes 1msec) then the 10J I put into the hammer during the swing is dissipated in 1msec at a rate of 10kW. If I could assume I was attacking a stone boulder and the hammer stopped in, say 0.1mm, then the power becomes 100kW. If the stone is hard enough to deflect by only 0.01mm, then I can dissipate 1MW during the 10 microseconds it takes the hammer to stop.
Of course, just as with the capacitor, when we start to take account of reality (stone and iron do compress a little bit, even when they don't break) our assumptions become open to question and practical limits arise.

10. ### GonzoEngineer

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Dec 2, 2011
If you want to see the latest in High Power Capacitors....google "Power Ring" capacitors.

I work in Pulsed Power applications....switching 600VDC at up to 3000A,
using IGBT's.

The inductance of capacitors is more important to me than anything else.

I have a saying about capacitors.......

"R and C are your best friends.....but L is like a damn ex-wife!: