Capacitor calibration question

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knujknuj

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15000uf Cap
0.1% 1.2K ohm 20W resistor
charged to 300V

Can someone help me with this? I'm trying to figure out how to make software to calibrate my charge circuit to compensate for the +- 15% cap that I'm using. So formulas would help. My plan is to charge the cap to say 300V, discharge it using the above resistor and measure the time. The time it takes to discharge should give me the information I need to see how much uf the cap actually is and using the formula, figure out how much above or below the 300V I need to charge the cap to to always get the same output between each batch. Now, can someone help me with the formula? Thank you.
 

hi,
Are you asking for the Cap discharge formula, look at this link
Capacitor Discharging
 
if you want to know the time it takes to dischagre the cap completely it will be infinity, but as a practice you can say for 80% voltage drop you can measure the time it takes, then use the following formula to evaluate the capacitance value.

V(t) = V(0) e^(-RCt)
in your case V(t) = 0.2x300 (for 80% drop)
V(0) = 300, R & t are known
so you can calculate the value of C

you will get ln(0.2)= - RCt
C= -ln(0.2) / Rt ( t is the time you measure for the voltage to drop to 60V in your case)
 
Last edited:
mbarazeen said:
if you want to know the time it takes to dischagre the cap completely it will be infinity,...
Click to expand...
In theory it takes infinity, but in practice it's the time it takes to discharge to the noise level of the capacitor, at which point the electrons left are passing in and out of the cap due to the noise. For a 15,000µF cap the noise level is about 0.5nV at room temp. To get to this level from 300V takes about 27 time constants. That would be 8.1 minutes for the 15,000µF cap and 1.2kΩ discharge resistor.
 
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