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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.

specs:

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.

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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.

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  • \$\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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