# Lighting and LED with a capacitor- the ins and outs.?

I'm looking to light an LED with a charged capacitor for a decent amount of time, but we haven't quite gotten to that point in my circuits class.

What all do I need to know? Just how does the capacitance relate to the amount of time the LED will stay lit? What about voltage? How do I avoid killing my LEDs?

I'm fairly proficient with the book knowledge part of capacitors, the RC time constant, ect. so feel free to use these when answering if need be,

I've seen this instructable here, and saw that he had some pretty big capacitors. Two 220 F in series, to be exact. That's 40 USD worth of parts. I know I can combine capacitor in parallel to increase capacitance, but how will this affect other things, such as voltage output and disharcharge time?

What all do I need to know? Just how does the capacitance relate to the amount of time the LED will stay lit? What about voltage? How do I avoid killing my LEDs?

I'm fairly proficient with the book knowledge part of capacitors, the RC time constant, ect. so feel free to use these when answering if need be,

I've seen this instructable here, and saw that he had some pretty big capacitors. Two 220 F in series, to be exact. That's 40 USD worth of parts. I know I can combine capacitor in parallel to increase capacitance, but how will this affect other things, such as voltage output and disharcharge time?

yourcat8 years agoReply

one farad is one amp-second (think amp-hours like for a battery) per volt. example: if you hook up a 3V 20ma LED to a 1F capacitor charged to 3V, then in 50 seconds the capacitor will be at 2V. (actually a little over 2V, because at that point not as much current will be flowing.)

kjordaan5 years agoReply

actually no (yourcat) it is the rc time constant is you don't know as you may know the rc time constant is (resistance) x (capacitance), and the answer of that is the amount of time for the voltage, ampere to fall by 2/3 so say you have a capacitor of 4700 uf and a resistance of 20k you will have a rc constant of 94 this means that every 94 seconds the voltage output of the capacitor will fall by 33.3%. so using that knowledge if you charge the capacitor by 9v till it is full then and have a led that requires 3v min to operate it would help to make a graph this way you could just say 9*(2/3) then the answer of that 6*(2/3) and so on then finally when it reaches a value bellow 3v which would almost be 4 so say 94 * 4 and you get 376 seconds. this will work but also cross reference current. if you want i have built a device that will fit your problem perfectly the device has a led that glows during charging as well as 2 led output and allows you to control discharge time as well as brightness of led . if you want to check it out go over to http://www.ljlabs.co.za/capacitors