Charging capacitor to power LED with a low current source

Question

I have a 10000uF 6.3V capacitor, and my source is generating 3.3V and 15uA of current. How long will it take to charge it enough so i can pulse a standard 5mm LED for lets say 1 second?.

Answer 1

I am afraid you will never reach your goal if you mean 10,000uF electrolytic capacitor. The matter is that electrolytic caps have leakage. For example, Panasonic-made caps usually quote the leakage as "0.01CV", which means that your capacitor will have 0.01 * 6V * 10,000 = 600 uA of leakage. This will make it about as 10k resistor, so 15uA into it will charge the cap to maximum 150 mV. Which won't be enough to lit a LED. Maybe you can be lucky and your cap has smaller leakage, but not by much.

AMPLIFICATION/CORRECTION: I just tried a 10,000 uF cap,

It behaves a bit better than the estimations above. When charger to 1.2V, its self-discharge goes at the rate of 1 mV/s (as measured with a >10 MOhm DMM). With formula provided here by Spehro Pefhany, I = C*dV/dT, the leakage is about 10 uA. Still it means that a 15 uA 3.3V source will charge the cap only to about 1.3V, which isn't much for a LED.

Answer 2

If you have a constant 15uA of current up to 3.3V it will charge to 3.3V in

Tc = \$C \cdot \frac{\Delta V}{I}\$ = 2200 seconds.

If you have a red LED with a Vf of 1.8V and supply it with 10mA it will discharge 1.2V (say) in 1.2 seconds.

Of course if you just buy a really good LED you'll get significant light from that 15uA with out any additional effort.