Capacitor bank for Coilgun help please!

I havetwo 680 uF capacitors, rated for 200 volts.. And I want to use a disposable camera charger to charge them both. The camera charger stops at 300 something volts, and when I wire them up in series, I get 400 volts, right?.. I want to charge them and discharge them into a coil of wire to launch a projectile at a fair velocity. , I can work with up to 10k wraps of wire, and i have only two capacitors and a major one..

First 2:

680 uF 200v
680 uF 200v

Major one:
180 uF 400v

Now I imagine I'd get more joules with the two smaller ones than the big one because of the microfarads, right?
These specific type of capacitors are made from Elite, and are electrolytic I think, and I got the major one from a computer CRT monitor, and the two smaller ones from a computer power supply.... I also have a 3.3uF 50v capacitor.

Now the question is: Can someone help me add these together to increase maximum joule output? I want to hook up the two 680 uF 200v capacitors in series to get a maximum voltage of 400. Then I want to add the major capacitor, which is 180 uF in parallel with the two other capacitors to increase microfarads, in this case, it'd be:
680
680   +
-------
= 680 (because they are in series)
------- +
180
 = 860

I want two capacitors in series, and the major in parallel, increasing voltage to 400 volts, and the major capacitor doing nothing but adding capacitance (IF possible)

Sorry, this is alot to ask for you guys to answer, If none of the above can work, What would work better, the two 680 uF's in series, or one big 180 uF 400 volt capacitor?

I apologize if I cannnot explain my question properly.... it's very hard to understand capitance and series/parallel blah blah mumbo jumbo.

Thank you.

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When you place caps in series, the total is determined by the same math as when resistors are placed in parallel, or
C_equivalent= C1*C2/ (C1 + C2)

In your case, the two 680 uFs in series will result in 340uF.

This assembly should withstand being charged to 400V (the ratings are added, as one half the voltage will appear across the two caps, as they are the same value).

If you then put the 180uF in parallel with the series connected caps, the result will be 340uF + 180 uF =520uF. The capacitance values add when capacitors are placed in parallel.

The energy stored in a capactor is E = 0.5 * C * V^2.
In this equation, capacitance is in Farads and voltage is in volts.

So, if your three caps are connected as I describe, and the resulting assembly of 520 uF is charged to 400V, it will store about 41.6 Joules.

It is best not to charge caps to their rating, use a value somewhat lower to give yourself some safety margin.

All this having been said, it would be well worth your while to learn a bit more, and be carefull.
Justdoofus (author)  LargeMouthBass6 years ago
I knew it, However, the way I calculated it, I was impressed at the joules I supposedly would get, (21.6 joules), or somewhere between there, But dang, I would've never calculated 41.6 joules, that's ALOT for what I'm asking, Thank you, sir. Just the answer I needed...

So if I put the two 680's, I will get 340 uF, but increased voltage, right?, And for the 180 uF, the voltage is NOT added to the total of the two 680's, for example

two 680 uf caps, in series, come out to be 340, but the voltage is 400, right? And I add the other 200 volt, 180 uf capacitor, and it does nothing but increase the capacity of the bank, right?

But thank you, I really needed this answer, as this is my first time with HV electronics, and I am doing very extensive noting and research before I attempt any kind of experiment, I will do it properly, The caps are rated 200, two of those in series will give me a 400 volt, The camera charger outputs somewhere between 340+ volts, but not over the "danger" rating when the two are added in series.
By placing lower rated capacitors in series, you can create an equivalent capacitor with a higher voltage ratings, but you do have to be carefull. The actual voltage across each capacitor must not exceed its rating. In your example, the math is easy because you have two caps of the same capacitance and votlage rating, so the voltage rating of the equivalent capacitor is the sum of the two. However, if you connect capacitors of different sizes in series, the ratings do not simply add.

If you are placing two different capacitance values in series, the voltage dropped across each will not be equal. There will be a greater voltage dropped across the lower value capacitor.

In general, for two capacitors in series, the voltage across each cap will be:
Voltage across C1 = Vsource*( C2/(C1+C2 ))
Voltage across C2 = Vsource*( C1/(C1+C2 ))

Note that if your capacitance values are equal, the voltage across each cap will be equal, but only in that special case.
Justdoofus (author) 6 years ago
@LargeMouthBass

Hello, Thank you for your reply, I understand fully now. I am glad to have found this community. Great people. Great answers..


Heh, For some reason, I can not reply to your last response.. So I made a new comment. Thanks again.
caitlinsdad6 years ago
That's Justdoofus trying to build a coilgun without a full understanding of electronics. You'll put an eye out.
Justdoofus (author)  caitlinsdad6 years ago
Haha good point, But I am very experienced in safety, and plus, I would do more than take an eye out. .. :)