If the circuit is something overly simple, like just a resistor and a LED in series, then it will only be able to use a fraction of the capacitor's stored energy; i.e that available between the initial voltage, and the minimum forward voltage drop for the LED. Also for a simple circuit, the light output will not be constant. It will be bright at first, then dim as the capacitor voltage drops to the LEDs minimum forward voltage.

For a circuit containing some kind of power converter, you should be able to get more run time. To get an estimate for that, just divide the capacitors stored energy (U = 0.5*C*V^2) by the rate at which your circuit is using energy, i.e. dissipating power, in watts = joules/second. For a truthful estimate, include the power being wasted by the power converter circuit. For example, if your load draws 1 watt, 60% efficient converter would imply load and converter together are drawing 1/0.6 = 1.67 watt.

How much power (or how many mA's and at what voltage) do the LEDs draw? There are a lot of useful formulas that you can use to figure it out. To get accurate numbers though, we need to know what LEDs, and what driving circuitry.

I can tell you that it would probably be a VERY long time, if you are talking about a single super bright LED driven and like 0.5mA @ 2.5V with a high efficiency buck regulator

So for example if your LED needs 20mA at 2V your capacitors should be charged at a higher voltage (4V for example) since the voltage drops when discharging and they will be able to run your LED for about 13 hours. But thats only an estimation and I don't even have much knowledge in this domain so I can be wrong. The best would be to try it and you'll see. Check out the other thread for further details.

## Comments

6 years ago

1.) Depends on the charged voltage.

2.) Depends on the discharge current to the LED.

3.) Depends on the LED driver circuit

6 years ago

It depends on what the circuit looks like.

If the circuit is something overly simple, like just a resistor and a LED in series, then it will only be able to use a fraction of the capacitor's stored energy; i.e that available between the initial voltage, and the minimum forward voltage drop for the LED. Also for a simple circuit, the light output will not be constant. It will be bright at first, then dim as the capacitor voltage drops to the LEDs minimum forward voltage.

For a circuit containing some kind of power converter, you should be able to get more run time. To get an estimate for that, just divide the capacitors stored energy (U = 0.5*C*V^2) by the rate at which your circuit is using energy, i.e. dissipating power, in watts = joules/second. For a truthful estimate, include the power being wasted by the power converter circuit. For example, if your load draws 1 watt, 60% efficient converter would imply load and converter together are drawing 1/0.6 = 1.67 watt.

6 years ago

How much power (or how many mA's and at what voltage) do the LEDs draw? There are a lot of useful formulas that you can use to figure it out. To get accurate numbers though, we need to know what LEDs, and what driving circuitry.

I can tell you that it would probably be a VERY long time, if you are talking about a single super bright LED driven and like 0.5mA @ 2.5V with a high efficiency buck regulator

6 years ago

According to answer in this thread https://www.instructables.com/answers/How-many-mAh... one Farad = 0,2777mAh/V

So for example if your LED needs 20mA at 2V your capacitors should be charged at a higher voltage

(4V for example)

since the voltage drops when discharging and they will be able to run your LED for about 13 hours. But thats only an estimation and I don't even have much knowledge in this domain so I can be wrong. The best would be to try it and you'll see. Check out the other thread for further details.