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4hv.org :: Forums :: General Science and Electronics
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How do you calculate short circuit current of a capacitor?

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MikeT1982
Mon May 14 2007, 12:13AM Print
MikeT1982 Registered Member #621 Joined: Sun Apr 01 2007, 12:37AM
Location:
Posts: 119
I want to figure out how much current in amperes my capacitor bank would push if I dead shorted it with say a wrench or some thick wire. I can't find the formulas to do this. The cap bank is 244 uF and has a capacity of 6,400 volts. So how can I figure what it's short circuit current would be if charged to max then dead shorted? (I would NOT do this I'd have to be nuts, but want to know what would happen if I did...for conversation info when talking about my project).
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Electroholic
Mon May 14 2007, 12:21AM
Electroholic Registered Member #191 Joined: Fri Feb 17 2006, 02:01AM
Location: Esbjerg Denmark
Posts: 720
every micro ohm and micro henry counts. look into LRC circuits. find out the ESR and ESL of ur caps.
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Steve Ward
Mon May 14 2007, 01:53AM
Steve Ward Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055
As mentioned, it depends on the internal resistance and inductance of the capacitors. When you series up capacitors, these internal parasitic characteristics add up. When you parallel caps, they are reduced (usually just divide by the number of caps in parallel).

For *IDEAL* components, you can figure out the surge impedance by using Z = sqrt(L/C). L is likely on the order of a few uH (taking into account the length of your cap bank and shorting wrench). So your impedance is on the order of milliohms. 6.4kV/.01ohms puts you at 640kA, but this would assume zero resistance in your capacitors and wiring (which is never true).

Reallistically, i would take a guess and say your lytic bank could maybe do about 5-10kA. Something like a real pulse cap can easily do 200kA (thats the rating of mine, anyhow).
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IamSmooth
Mon May 14 2007, 01:53AM
IamSmooth Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
I would guess that if the capacitor was perfect without any resistance or inductance then the current would be limited by the resistance of the object shorting the terminals. The current would fall as the voltage of the capacitor fell with discharge. I worked out the second order differential once and then found someone who put together a nice computer simulation. He revised it with some suggestions that I made and now I think it is great. I think you will find this very helpful: Link2
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MikeT1982
Mon May 14 2007, 01:59AM
MikeT1982 Registered Member #621 Joined: Sun Apr 01 2007, 12:37AM
Location:
Posts: 119
Nice nice!!! Thanks guys.
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Tesladownunder
Mon May 14 2007, 06:12AM
Tesladownunder Registered Member #10 Joined: Thu Feb 02 2006, 09:45AM
Location: Bunbury, Australia
Posts: 1424
I have a fair bit of information about high current discharges into low inductance and low resistance loads here. Lots of measurements with Rogowski coils and CRO pics up to 100+kA.

(Edit. I have just read your old posts about your bank being 16 400V electrolytics. These discharge in perhaps 1000us vs 30us for a pulse cap. They won't have the inductance written on then like a proper pulse cap. Even huge pulse cap banks with electrolytics only get to say 100kA like Jason Rollette's. Cant remember what the Geek groups one gets to. So what will yours do? Hard to say. Perhaps 5 - 10kA. Note that any inductance in an capacitor discharge will induce ringing where significant energy eg 50% will end up in the reverse polarity say with can crushing. In a pulse cap this is sort of OK and the ratings will be known. In a polarised electrolytic this is bad news. Fortunately, you are only driving a flashlamp so hopefully not too much problem. )

TDU
1179123068 10 FT25185 Pulsecapssteelwool3kjhiresside Small
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