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I need a capacitor to charge using 5V in 1 seconds and discharge at 30 mA 3 Volts for 10 seconds. How do I calculate the capacitor needed.

I am sorry I don't have electrical background.

Actually in my Arduino, I am not using interrupts, it becomes unstable and I am sleeping for 8 seconds, but I want to know whether any interrupt/ high voltage occurred at analog pins.

Any help would be appreciated.

EDIT: Ok let me explain, a little bit more about my Arduino program. Using Interrupts, you can find out if the PIR motion sensor sends a pulse, when someone crosses in front of the sensor. Now consider my problem, I am not using Interrupts, as my program is already very large and interrupts make my program unstable due to less free RAM. I am anyway reading the sensors after each 8 seconds using watchdog timer. Now the problem is that the microcontroller is in sleep mode for 8 seconds, and won't poll the pins, so I was thinking of putting a correct capacitor, that can be charged by the PIR sensor and then the capacitor can keep the input pins of the microcontroller high for 8 seconds. Sorry, I don't know much about electronics, so may be this is stupid, but if this can be doable, please share your opinion.

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  • \$\begingroup\$ electronics.stackexchange.com/questions/4951/… \$\endgroup\$ – pjc50 Jun 19 '13 at 10:57
  • \$\begingroup\$ It's not clear what you are asking for. A cap can charge to 5V in very short amounts of time providing you are feeding it from a supply (via current limit) that is greater than 5V. Should the supply be exactly 5V, the cap will theoretically never reach 5V (via a resistor) and never reach 5V should you be feeding it thru a semiconductor circuit. Tell us what you want to achieve please? \$\endgroup\$ – Andy aka Jun 19 '13 at 11:07
  • \$\begingroup\$ Please clarify: Discharge down to 3 Volts in 10 seconds, or discharge by 3 Volts in 10 seconds? \$\endgroup\$ – Anindo Ghosh Jun 19 '13 at 11:18
  • \$\begingroup\$ What does the capacitor charge have to do with your Arduino interrupts? \$\endgroup\$ – baldengineer Jun 20 '13 at 19:18
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Capacitance in Farads F = Current I x time t / voltage dV

If the load current is limited by resistors and not a constant current load, the current will drop as voltage reduces, so the above calculation needs to be changed to a I x dT / dV calculus form where dT tends to zero. This becomes a bit more messy. So, let's keep things simple:

For the output case with a 3 Volt drop over 10 seconds, assuming constant current load, capacitance works out to:

F = 0.03 x 10 / 3 = 0.1 F = 100,000 microFarads.


Such "super capacitors" or "ultra capacitors" are available online fairly inexpensively these days, if the purpose is for a DIY project. For instance, on eBay, 3 pieces would cost $4.50 from this seller (plus shipping), or a single piece for $1 from a Canadian seller, if that saves you some shipping cost:

Supercap SuperCap

That is the capacitor needed.

The charging period and supply voltage have no effect on the capacitance calculation, assuming your supply can provide the required current (300mA), so ignore those.

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