Electrical – Solar powered LED via capacitor

capacitorcapacitor chargingledphotoresistorsolar cell

I am building a large garden peice in wood, which I hope to illumine from within, powered by a small, (perhaps 90×65, 6v, 0.6w 100mA as space is limited) solar panel.
After testing, I noticed that this small power supply is insuficient to even enable a charge controller, let alone charge the USB (1×18650 battery) power stick, I suspect that the current is too low to "wake up" the charge controller?

Whilst considering smaller batteries, or replacing charge controller with a diode, I had a "capacitor" moment, and wondered whether this could be an alternative, if even even better solution?

The ultimate goal is to be able to maintain enough charge during the day to illuminate an LED of any colour (pure white requires a lot more power and pale yellow is more than adequate tho I can use any colours which are more efficient, but I need two of high contrast). I have a photoresistor switch which will also take a little power from the solar panel.
Furthermore… I wondered if a combo cap/battery might work well?

My guess is that this is essentially the same circuit used by a solar garden path/accent light… space within the item is not an issue, only the location for the solar collector mounted on the smallest, topmost panel of the assembly, approx 100mm dia.

My reseach so far indicates that a 6v 1F cap can light 12ohm LED for approx 8 hours, but I need far more understanding to know if this calculation is correct, and am very unsure about the resitance of the LED

Any help would be great?

Best Answer

My reseach so far indicates that a 6v 1F cap can light 12ohm LED for approx 8 hours, but I need far more understanding to know if this calculation is correct

6 volts and 1 farad is an energy storage of 18 joules (\$C\cdot V^2/2\$). Spread over 8 hours that's a power of 18/(3600 x 8) watts. So, if you consumed a power of 625 micro watts from your capacitor you would take an energy of 625 micro joules every second and over an hour that would be 2.25 joules and over 8 hours that would be 18 joules.

You also need to consider that only about 75% of that energy is usable because as the capacitor voltage droops below 3 volts your LED driver circuit may falter and switch off the LED.

It's nowhere near enough for even a standard red LED at 1 mA and 2.0 volts.

Any help would be great?

Glad to oblige but I would recommend dismantling one of those garden lights to see what they use then scale up your PV panel and battery/supercap accordingly to match your needs.