Hello, I already worked with supercapacitor to use them as small battery. I am now trying to use a smaller capacitor to work on a very low power consumption project. When charging the supercapacitor of 4F the V keeps rising up normally and when using it as a battery everything works fine. But when trying to charge up that smaller capacitor the voltage don't rise up and it doesn't power up any sort of led. Do you have any clue why this capacitor doesn't seem to work like the superone.
Yes, in terms of energy a 100uF capacity has very little charge in it. You would only be talking of milliseconds worth of charge.
The energy in a capacitor Q (charge in coulombs) is equal to Voltage * Capacitance.
Correction :- And as a 100uF is E1-4 of a Farad that is very small.
Why when I posted this it was reply 2 and now I find it at reply 4?
Just look at the post times, it was the newest reply.
No. A 100uF is like a soda you drink and you're not thirsty for a couple of hours. But try use it as the supply for the combined water supply of 1000 people.
How much energy are you trying to get out of the capacitor? How many milliamps or microamps for how many seconds?
A 100uF capacitor might have enough energy to flash an LED, or the discharge may be so fast that you can't see it.
Are you checking the voltage while charging? It may be charging and then discharging so quickly that you can't tell.
It's easy to charge the 100uF capacitor by directly connecting to a battery or through a resistor. If you connect a multimeter, and nothing else, you should be able to watch it discharge.
Capacitors make terrible batteries. An "ideal battery" would hold it's voltage until it's discharged. Real batteries aren't perfect that way, but the voltage holds-up pretty-well at through most of the charge-life and then falls-off near the end.
Capacitor voltage drops rapidly at the beginning of the discharge, and then levels-off more the more the voltage drops. Capacitor charge/discharge curves
