# High Current voltage booster? Answered

So I have been doing a lot of math for a rail gun, and I finally found out what statistics I need for the capacitor to get the energy needed.
But I may have ran into a problem. The voltage for the system is rather low, but the current is high, so I need a capacitor with something like 60-ish Farads, and 20 V, and of course the ones I find are outrageously expensive. (\$2,400!!!)
But I was able to find ones with 70 F but only 2.1 V, but since I need 20 volts I was thinking that I could use a voltage booster, but I am not sure if that would work, or if I can find/make a Voltage booster that can withstand that kind of current (would be something like 67.5 KA)
the other option is find ones with higher Capacitance, which I did, but they are still a bit on the expensive side (\$70, and since i need to it comes up to be \$140 or so (not as much but still))

So what I wanted to know is whether it would be less expensive to try to find/make a voltage booster, or just buy the \$70 capacitors?
I suppose I should also be asking is it even possible to get a voltage booster that can that that many amps, and how easy would it be to make?

Oh almost forgot, another question (which would skip having to solve the others) is: Is there a way I can increase the voltage that a capacitor can take, or increase the voltage in a capacitor bank, or some variation of those? Because if there is, that makes everything easier, but I couldn't find anything on doing that so yeah...

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## Discussions

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I think for a rail gun, or any other pulsed power application, you are going to want a capacitor bank that discharges quickly, and to do that you will want the total capacitance to be as small as possible.

Of course you also want to store a large amount of energy. The energy stored in a capacitor is (ideally) U = 0.5*C*V^2,

http://en.wikipedia.org/wiki/Capacitor#Energy_of_e...

where C is capacitance, in farads. V is the voltage across the capacitor, in volts. U is the total stored energy, in joules.

Just from that equation, there are two ways to increase U, namely increase C or increase V. Increasing C, by using a larger capacitor, will make the discharge happen more slowly, so the only choice left for storing more energy is to use large V; i.e. use higher voltage.

Another trick, is to use a Marx generator type circuit.

http://en.wikipedia.org/wiki/Marx_generator

Roughly speaking, the way that works is you have several capacitors which get charged in parallel. Then the capacitors are rearranged (somehow) into a series circuit, and discharged in series. The neat thing about that is while charging the capacitor bank looks like a very large capacitance, low voltage, capacitor, (e.g. 10*C at 1*V) but when it discharges it looks like a small capacitance, high voltage, capacitor (C/10 at 10*V).

Also there is a very clever trick for doing the parallel to series rearrangement, by using spark gaps that fire almost simultaneously, and I think the Wiki article linked above explains how that is done. As a alternative to spark gaps, it is also possible to build some kind of mechanical switching, for to rearrange the capacitors from parallel to series, prior to discharging.

Anyway, to summarize, I kind of think you're going in the wrong direction by choosing large capacitance and low voltage, and the usual way this is done is with small capacitance and high voltage.

By the way, if you are wondering how much time, how fast or how slow, it takes for the discharge to happen, I think you can sort of guestimate that by modeling (pretending) your circuit as an ideal RC circuit:

http://en.wikipedia.org/wiki/RC_circuit

or RLC circuit:

http://en.wikipedia.org/wiki/RLC_circuit