# High Current voltage booster?

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...

## Discussions

Best Answer 6 years ago

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 totalcapacitance 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

Answer 6 years ago

Thank you, you answer is very helpful :)

I had already figured out the time the projectile would be on the rails and such (based on the desired acceleration); and i have figure out the capacitance and voltage I needed. The RC circuit page is helpful as for the time constant, I had been using a different equation (which seemed to provide the right values). The Marx generator should help a lot, so once again thank you.

But out of curiosity, if capacitance and voltage vary at the same rate (so that they will always equal the same power) won't high voltage and low capacitance be essentially the same; the initial equation I was suing (first equation: http://www.mouser.com/pdfdocs/ELNACalcuDYNACAPDisc... ) says that it wont, although if that is because i am making an error, I am not surprised... but the RC time constant says otherwise; The reason I needed the large current though is because of the equation for the force of a railgun which is:

F= (L'*I^2)/2 where L' is inductance per unit of legnth and I is current. and since L' tends to be low (something like .168 microHenries per meter) the current needs to be high

Answer 6 years ago

Energy varies linearly with capacitiance, but the SQUARE of voltage.

Minimise R

Answer 6 years ago

...or you minimise R.

6 years ago

Any semiconductor switching voltage booster will not be able to deliver the fast huge current pulse that a rail gun needs to accelerate the slug .

If small size is not a requirement, you can build that cap out of many small caps in parallel.

A

Answer 6 years ago

+1 !!!