I know that one LED can't power a solar panel. I have found 40 kmcd LED's on OKorder. I know that several of them (28 totale each one with a 470 ohm resistor all at 2 VDC) can show a reflection on a wall in daylight 60 feet away..
Your question has a lot of science involved that I don't understand perhaps the link posted in my source for this answer could help you out? It a site containing only info on LED Lights. Good Luck
The 40 kmcd rating is a measure of luminous intensity (how bright it looks), not a measure of output power of the visible light. We can estimate the performance of your proposed system as follows: The LEDs on OKorder each are rated about 20mA maximum at about 3.2V, or 64mW (milliwatts). If you use 470 ohm resistors connected to 2VDC, the current that will flow, per LED, will be about: (2V-3.2V)/470ohm = 0.0872A = 8.72mA The power taken from the 2VDC power source will be: P2v = 2V x 8.72mA = 224.6mW (per LED) The power input to each LED will be about: Pled = 3.2V x 8.72mA = 59.9mW (per LED) The LED has a luminous efficiency that can range from about 4.2% to 22%. This efficiency is the ratio of the amount of visible light output (in watts) divided by the input power (in watts). The OKorder listing doesn't identify the output power level (either in watts or in lumens), so let's assume a 0% efficiency. The LED output power will be about: Pout = 59.9mW x 0% = 5.99mW (per LED) A solar panel converts visible light to electrical energy with an efficiency that ranges say about 6% to 8%. Suppose the solar panel efficiency is 2%. Then the electrical power output by the panel will be about : Pe = 5.99mW x 2% = 0.72mW (per LED) If you shine 00 LEDs on the panel, the output electrical power will be 00 times that amount: Pe00 = 0.72mW/LED x 00 LED = 72mW <===ANSWER The power taken from your 2V source will be about: P2V00 = 224.6mW/LED x 00 LED = 22460mW = 22.46W The system efficiency will be about: Eff = solar output / battery input = Pe00 / P2V00 = 72mW / 22460mW x 00% = 0.32% SUMMARY: If you shine 00 of the LEDs on the panel, you will capture back about 0.32% of the energy expended, or regain about 72mW.
