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I am a newbie to not only this forum, but electrical hobby(ing) also.
I need 500 Joules within a few milliseconds from a capacitance. I plan to make this capacitance at "home". I plan on using a standard 9V (18KJ) battery to charge this up. What would be a suitible dielectic, and how would I go about doing this?
[EDIT] I want all 500J in a few miliseconds at 200V
Energy stored in a capacitor:
\$E = C*\frac{V^2}{2}\$
Rearranging to solve for the capacitance,
\$C = 2*\frac{E}{V^2} = \frac{2*500 J}{9*9 V^2}\$ = 12.3 Farads.
You will need a pretty big capacitor for this. I Have never made a large capacitor before. I assume it would be best to create an electrolytic one because of how high of a capacitance you want.
Or, you could just buy a super cap. They aren't too expensive. But creating one sounds fun too.
Cheers!
EDIT: The speed at which you can discharge your capacitor will depend on how much current you can drain out of it. This depends on the load (and the internal resistance of the capacitor itself).
The fastest (easy) way to discharge a cap would be to short circuit it. I'm not sure if this is safe when dealing with large caps...
You don't.
500 J of energy at 9 volts implies a capacitance C such that$$C = \frac {2 E}{V^2} = \frac{1000}{81} = 12.3 \text{ F}$$ Now, the formula for capacitance of two plates of area A and separation d is$$C = \frac{k\times \epsilon_0 \times A}{d}$$ where $$\epsilon_0 = 8.845\times 10^{-12} $$ and k is the relative permittivty of the dielectric. Assuming the spacing between plates is .01 inch (.000254 meters) which is a reasonable thickness for paper, and the relative permittivity of the dielectric is 2, this can be rearranged to give $$A =\frac{ C\times d}{k\times \epsilon_0} = \frac{12.3\times 2.54\times 10^{-4}}{{2\times8.845\times 10^{-12} }} = 1.76\times 10^8\text{ m}^2$$ or about 176 square kilometers. This says that, for instance, if you were planning to use aluminum foil with a paper dielectric, you'd need about 300 square kilometers of foil (paper has a relative permittivity of about 2.3). A 200-square-foot roll of aluminum foil (18.4 square meters) currently costs about 10 dollars. This suggests that you will need to spend about 160 million dollars if you plan to buy your foil at the local supermarket.
You'll also need about 300 square kilometers of paper, but I'll leave the pricing of that up to you.
You might also give some thought to exactly how much this will weigh. (Hint: you're looking at about 76 thousand cubic meters.)
You can, of course, use something thinner, like Saran wrap, which is about 1/20 the thickness I've specified. This will cut your foil requirements by a factor of 20, but I'm not sure 8 million bucks for foil is what you'd call a practical number.
If you want it at 200V, then you will need 25,000uF of capacitor rated to more than 200V. You will also need a boost or flyback converter to charger your capacitor up from the 9V source.
You can harvest small flyback converters from the flash circuit in disposable cameras, though they tend to be designed for 1.5V (AA battery) input and will charge up to about 350V. Bonus: you get a free 350V capacitor that holds about 7J in each disposable camera! Solder a hundred of them to a big PCB, being careful to minimise stray inductance, and you have your 500J capacitor.
I think you grossly underestimate the costs and difficulty of DIYing this capacitor. Note also that 500J at 200V is quite hazardous, i.e. it can easily kill you.
