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Re: calculations of capacitance



Original poster: Mike <megavolts61@xxxxxxxxx>

Hi James,
Your total capitance would indeed be 0.164uF (1/6.1). I'm not sure if you have any real advantage adding the 10nF cap to the string, unless you want slightly LESS capacitance and a slightly higher voltage rating. Without the 10 nF cap (assuming 'perfect' caps lol) your six 1nF caps in series would be 0.167nF and with the 10nF, as stated above, you reduce it to 0.164nF. Also, due to the fact that each cap in series aquires the same charge, the voltage across the 10nF cap will only be 1/10th of the voltage across the other in the string. This is based on the equation C = Q/V or V=Q/C . It might be useful to use the 10nF cap for just a little higher voltage rating, but it almost seems like it would be better used elsewhere.
Hope this helps,
Mike
PS. also, because the voltage is only 1/10th of the others, the energy (= 1/2 * C *V^2) will actually only be 1/10 of the energy of the others. for example a one Farad cap charged to 100V would have .5*1*10000 or 5000J. a 10F cap charged to 10V would have 0.5*10.100 J or 500J. I used huge values for caps just to avoid dividing both by 10^9 for nanofarads, but it still works the same.



Original poster: "James Howells" <james@xxxxxxxxxxxxxxxxxxxxxxx>

I don't know why but I have got myself completely flummoxed trying to
work out a primary cap from the components I have at hand

I don't have enough capacitors of the same value to do a normal MMC
and want to mix and match what I have got But I have got myself in a
real error loop


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