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this time I'd like to ask a few questions about capacitors, all noob questions again;

  • What's a better idea, having one capacitor at 5x farads or five capacitors at x farads each? I want to get a certain amount of voltage and want to discharge the direct current as fast as possible.
  • I'm trying to build an amateur railgun, and intend to shoot projectiles that weigh about 10 grams. Using the formula that someone helped me with in my other question, in order to give that a speed of 100m/s at a total of 20.000 uF assuming 2% efficiency I'd need a little less than 112V's. The question is, is this amount of voltage I need my capacitor or series of capacitors to provide for the duration of the shooting?
  • What determines how much voltage a capacitor can hold? This may come off stupid as I am aware that there is a voltage rating when you buy them, but better safe than sorry.
  • I was only assuming 2% efficiency, and depending on my rail length it could be higher. The question is, what determines how fast the capacitor will discharge and how it can be prevented from going above the rated voltage?
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    \$\begingroup\$ If you are asking these questions, I would question the safety of doing this project. \$\endgroup\$ – Brian Carlton May 15 '13 at 19:23
  • \$\begingroup\$ I'm trying to get started, why is that a bad thing? Thank you for your concern though. \$\endgroup\$ – lychnus May 15 '13 at 19:25
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    \$\begingroup\$ a good way to learn is to blow things up, just don't do it stupidly. \$\endgroup\$ – placeholder May 15 '13 at 19:43
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    \$\begingroup\$ @pjc50 suggested your 16,000uF bank of caps would need to be charged to 176V and be capable of discharging a current of 280A in order to get the energy. Then he quite rightly mentioned that perhaps only 20% of the stored charge would be useful and I suspect this is because of laws of diminishing returns - as you discharge the cap the current rapidly and progressively gets smaller due to the terminal voltage dropping. In short you may be looking at purchasing 250V rated electrolytic caps with a total capacity of over 50,000uF. I think this should be your starting point. \$\endgroup\$ – Andy aka May 15 '13 at 19:59
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    \$\begingroup\$ I've seen this kmestore.com/470uF-250V-Electrolytic-Capacitor - it's a 470uF 250V cap and it costs $10 - you'll need over one-hundred in parallel. Then you'll need to find something that can handle a 1000A switching that will connect this large bank of caps to the rails. Just one ohm impedance at say 200V will limit the current to 200A so you'll need something that is significantly less than 0.5ohm and preferably 0.1ohm - I don't know where you'd get one from. \$\endgroup\$ – Andy aka May 15 '13 at 20:05
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  1. Having a bunch of caps in parallel will add their capacitance but also puts all their internal resistances in parallel, giving you less overall internal resistance and more current dumping capability.
  2. Capacitors don't hold voltage, they hold charge. You can take a set amount of charge and discharge it in a certain time period. You need to rethink of your problem in terms of energy you need and discharge rate of that energy.
  3. Caps have a maximum rated voltage above which they'll be damaged. However caps store charge, not just a certain "voltage". They hold energy. 20 uF capacitor can be charged with a 3V rail, or a 6V rail or any rail up to it's maximum rated value. You put a certain amount of charge into it and get a certain amount of charge out of it. The useful thing here is that caps don't like the voltage across them to change quickly. They exhibit inertia when it comes to voltage whereas inductors exhibit inertia when it comes to current. When you discharge a cap you are draining power out of it very quickly. You are dumping out a lot of "energy"/"power". Since P = IV will always be true and caps don't like sudden change of voltage across them they will instead dump a lot of current -- i.e. show sudden change in current to keep the V part of that equation dropping smoothly (or following a discharge curve).
  4. A cap's current capability is determined by it's internal resistance + resistance of the load.
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  • \$\begingroup\$ That was helpful, but what exactly is the 112V I've calculated? Will that amount be "spent" when I let the capacitor discharge? Or am I supposed to calculate the charge that will be applied to each shot? I am asking this in order to figure out how much electrical power I can channel per shot, and knowing what is what will allow me to design a gun with semi automatic action as well. Basically I want to shoot x amount of times using up all the charge stored in the capacitor before leaving it to charge again, which should prevent it being a hazard too. I just want to know how to calculate it. \$\endgroup\$ – lychnus May 15 '13 at 19:56
  • \$\begingroup\$ I'll need to see that formula to be able to help you more. From what you have Q=CV therefore the charge you are calculating is 20uF * 112V. Current coming out of your capacitor is proportional to rate of discharge and C. i = C * dV/dt. Higher the initial voltage and higher the capacitance the higher your current. \$\endgroup\$ – EEToronto May 15 '13 at 20:07
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    \$\begingroup\$ Regarding point 2, capacitors don't "hold" charge, they store energy. The total number of electrons in a capacitor is the same before and after "charging." The charges are just arranged differently (positive charge on one plate and negative on the other). And separated charge is really just another name for voltage. \$\endgroup\$ – Eric May 15 '13 at 20:11
  • \$\begingroup\$ Hold and store really mean the same thing here. I don't think anyone is going to read that and assume that caps somehow hold the charge in some kind of stasis field. Even the Wikipedia article on caps uses the term "holds charge". Voltage is potential difference between two charges. \$\endgroup\$ – EEToronto May 15 '13 at 20:17
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    \$\begingroup\$ I just think that is sloppy language. If an equal amount of electrons enters and leaves a cap, in what sense does it store charge? It doesn't. It stores energy. We would say something stores charge if it collected way more electrons or protons than it started out with. But that's not what a capacitor does. I think the confusion is that each individual plate in a typical capacitor does store charge but the capacitor as a whole doesn't. \$\endgroup\$ – Eric May 15 '13 at 20:58

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