Re: capacitor charge rate/power

[Terry Fritz <twftesla-at-uswest-dot-net>]

Hi Pholp,

	You should check out:

http://www.sweethaven-dot-com/acee/forms/frm1002.htm
http://othello.mech.nwu.edu/ea3/book00/elec2/RC.html
Web searching on "capacitor time constant" will find many pages...

Basically, V = 1 / C x Intergral I  Dt

For a resistor and capacitor being charged from a fixed Vo source:

Vc = Vo x (1 - e^(-1/RC))

The energy in joules stored in a capacitor is E = 1/2 x C x V^2

The time it takes for a resistor to discharge a cap to a safe value (less
than 20 volts starting with 21kV)is about:

	T = 7 x R x C   or R = T / ( 7 x C )

So if you want the cap to discharge in say 10 seconds:

	R = 10 / ( 7 x 0.06 x 10^-6) = 23.8M Ohm.

The power dissipated is:

	P = V^2 / R   or   15000 x 15000 / 23.8E6 = 9.45 watts.

10 watts of loss (0.27%) will not hurt your coil's performance but will
discharge the primary cap fast enough to save you.  So you need to string
resistors in series to add to around 24Meg at 10 watts.


An NST charging a 0.1uF cap is messy because the input AC will go through
the cycle and back to negative before the cap is charged.  However, if you
could maintian 60mA, the voltage would be Vc = T/C x 0.060  
In 1/120sec, that gives 5000 volts.

I hope I hit on the answer you wanted.

Cheers,

	Terry


At 11:12 PM 7/29/00 +0000, you wrote:
>
>
>
>can anyone give me a formula as to how long it takes to charge a capacitor 
>depending on size and power? i.e.: using a 12kv-at-60mA transformer, how long 
>would it take to charge a .1mF capacitor? Can someone tell me and show me 
>the math behind it? And how much power is actually stored and let out in a 
>discharge. Also i'd like to know how to determing the size of a bleeder 
>resistor for capacitors.
>
>thanks
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