I'm trying to determine the maximum current of a circuit while charging a capacitor but since the power supply has a maximum current of 20mA that is going to extend my charge time longer than the RC time constant suggests. I have a 235uF capacitor I'm charging to 800 VDC given 200kOhms of resistance in series to the capacitor. Does the Max current even come into play since I have so much resistance in the circuit with the capacitor?
With a 800 Volts DC supply, and a 200 kΩ resistor in series with the capacitor, the maximum current the charging will draw is 4 mA.
How to calculate this:
- Assume an ideal capacitor i.e. Zero Equivalent Series Resistance. Real capacitors and their leads can only increase this ESR.
- At start of charging the current draw is maximum, as potential difference is maximum
- At this time,
I = V / R, V = 800 Volts and R = 200 kΩ
- Thus Imax = 4 mA
- As the capacitor charges up, this current will reduce further as per the normal RC curve.
As this is lower than the 20 mA supply limit, the supply-limited maximum current will not come into play at all.
EDIT: Note that the time constant of the resistor and capacitor is 47 seconds, which means that it will take that long to reach 63% of its final voltage. It will take 3× that long (2:21) to reach 95% and 5× that long (3:55) to reach 99%. You can reduce those numbers by a factor of 5 by reducing the resistor to 40KΩ, which would just touch the source limit of 20mA at the beginning of the charge cycle. If you simply charge the capacitor with a constant current of 20mA, it will take 9.4 seconds.