Electronics > Electronics

capacitor discharge

**vipulan12**:

how can i determine the current flowing out of a capacitor?

i have a 330 v 80 uf cap

if you multiply them you get 26400 columns

**jkerns**:

I = C* dV/dt

You would also want to account for any resistance in the circuit and, potentially, the internal resistance of the capacitor.

**jwatte**:

With "ideal" components, "dt" will go towards zero, so "current" will go towards infinity.

Note that the 330V rating has little to do with the actual current -- what matters is the actual voltage you put across the capacitor, and the delta in that voltage.

Your calculation is off by a factor of 1,000,000, btw, because you multiply micro-Farads with Volts. This yields micro-Columbs, not Columbs.

If your circuit is fully resistive, then add your resistive load to the ESR of the capacitor for an estimate of R, and calculate the time constant RC. The time constant RC tells you how long it takes to drop voltage by about 63% across that resistive load. From that you can calculate the current.

Also, the ESR of the capacitor becomes very important here -- multiply current squared times ESR to get the power that the capacitor will see as heat. If this is too high, then the capacitor will overheat and blow up. This is why electrolytic capacitors often have a "ripple current" rating, which is roughly speaking the max changing current they can see during normal operation without overheating.

**vipulan12**:

then formula can i use exactly?

**jkerns**:

If your circuit is resistive, and if my memory is correct,

I = (V0 -e^(-t/(RC)))/R

Where R is the total resistance including the equivalent resistance of the capacitor, V0 is the initial voltage, and t is time, and e is ... well .... e.

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