How can I calculate capacitor for LED powered by solar panel

LED with Solar Panel

I'm planning to power a Led with Solar Panel.
The circuit should be : solar panel - capacitor - sensor - led.
The led will be turn on when dark.
How can I calculate the value of the capacitor? :idea:

Re: LED with Solar Panel

For the specific circuit ..."The circuit should be : solar panel - capacitor - sensor - led. "...

-> The solar panel should be greater than 2 V ; preferred at 5,5V, its power capability multiplied by the light exposed hours should be greater than the power consumed by the LED multiplied by the hours you want it to be on when dark.

-> The capacitor voltage rating should be greater than the solar panel and its capacity should be calculated depending on the current you want the LED to conduct and the time you want it to last; Probably above 2 Farads.

-> The sensor should have a low dark resistance value equal to the requiered resistor value for limiting the current of the charged capacitor as per the LED consumption and of the highest light resistance available.
I do not know if a optoJFET is available.

-> The LED should be high efficiency type, capable of shining at a wide range of current while the capacitor discharges.

Miguel

Re: LED with Solar Panel

You will have a much cheaper circuit for long light times if you use a rechargeable battery (NiCD or NiMH). If you only want the light to run for a few minutes then a capacitor is fine. If you want it to run all night

The battery will hold a lot more charge and run your light a lot longer than an expensive capacitor, although the life of the capacitor circuit will be many, many years whilst the battery may only last a couple before needing replacement.

Another thing to remember is that a capacitor's voltage will decrease as it is discharged, needing an active regulator to the LED. A battery is much less troublesome untill it suddenly runs out.

As to the calculation...

a) Charge = Capacitance x Voltage
b) Current = Charge Flowing per Second

So, if you have a 1 farad capacitor and charge it to 5 volts, (a) says it will have a charge of 1x5 = 5 coulombs.

Now, if we draw a current of 30mA from it for the LED, (b) says that's a charge flowing of 0.03 coulombs every second. So, the capacitor will be emptied in 5/0.03 = 166 seconds.

See what I mean about it not lasting long? Remember too that not all that current will be useful as the voltage sags linearly over the 33 seconds from 5V to 0V.

Go with batteries

Check here for some circuits for solar chargers:

http://www.discovercircuits.com/S/solar-cell.htm

Cheers,
FoxyRick.