I have a microcontroller that needs to be able to survive a power outage just long enough to shoot an email telling me that the power is out. I figured the best way of doing this would be a resistor capacitor circuit, like the one below that I made that works well, that keeps the LED lit for a second after pulling the plug.RC circuit for LED off delay

Ideally, I'd like to have something like this for my microcontroller! I just need to be able to keep it up a few seconds after the power goes out! (it instantly detects an outage)

I just have no clue what size resistor or capacitor I need, or if this is the right circuit type for the job.


Input Voltage: 5.15V

Minimum Voltage: 4V

Current draw: 600mA

Time needed: 5s minimum

I've run across many calculations, but I don't even know enough to say if I'm plugging in the right variables.


Let's assume for simplicity that the current is constant 600mA and the voltage is dropping from 5V to 4V during 5 seconds. The total charge over 5 sec that is drawn from the capacitor is 600mA * 5 = 3c. The initial charge of a capacitor C is 5V * C. The final charge is 4V * C. The difference is (5V-4V)C = 3. I.e. C=3F. It's a damn large capacitor.

  • \$\begingroup\$ Almost 1cm in size! ^^ \$\endgroup\$ – PlasmaHH Oct 27 '15 at 20:48
  • \$\begingroup\$ Ok, thank you so much! One more question, if you know. If i were to use three 1F capacitors in series on my circuit, would it be the same as using one 3F capacitor? \$\endgroup\$ – Skyler Oct 29 '15 at 12:01
  • \$\begingroup\$ @Skyler Capacitors in parallel are adding up the capacitance, not in series. \$\endgroup\$ – Eugene Sh. Oct 29 '15 at 12:37
  • \$\begingroup\$ @Eugene Sh. 5 Ok thank you, that's what I meant, just didn't know the lingo! \$\endgroup\$ – Skyler Oct 29 '15 at 17:49

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