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?

  • \$\begingroup\$ That makes sense. Is there a formula to plot current as it charges the capacitor? I would think it should be related to the RC time constant in some way. \$\endgroup\$ Mar 7 '13 at 20:32

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.

  • \$\begingroup\$ note that the capacitance is not relevant when calculating the maximum current! \$\endgroup\$
    – Christoph
    Mar 7 '13 at 14:39
  • \$\begingroup\$ @DaveTweed I assume that was meant to be a comment on the question not the answer? The 200k was specified in the question. \$\endgroup\$ Mar 7 '13 at 16:10
  • \$\begingroup\$ @DaveTweed Fair point... Also, the system allows you to edit an answer to incorporate inputs - answer owner can always accept or roll-back the edits. Please feel free to edit mine. :-) \$\endgroup\$ Mar 7 '13 at 16:32

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