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The easiest way to measure total power output is to position the LED very near to a large photodetector.
You should be able to get large area photodiodes from: Vactec, EG&G, Hammamatsu, IPI Centronic, Optodiode, Silonix, among others. The last one is a Canadian co. Solar cells OK, but you should probably do a rough calibration. Here's one way to do it: Noonday sun at your latitude irradiates the surface of the earth with about 800 Watts/sq-m. This is equivalent to 800/10,000 = 80mW/sq-cm. Since Si efficacy for sunlight is about 0.5Amps/Watt, you should get around 40milliAmps for each square cm of solar cell. Supposing that you get a smaller value, then you can assume that your particular solar cell is less sensitive than 0.5A/W and calculate a calibration value appropriately.
Example: you obtain a 1"x1" solar cell, i.e., about 6.5sq-cm. At midday on a clear day, you measure about 200mA short circuit current max as you orient it toward the sun. You calculate 80 * 6.5 = 520mW incident on the cell, so you were expecting more like 260mA. This means that your cell has more like 0.38A/W efficacy.
I've glossed over some of the fine points, but all the above info is reasonably accurate.
The easiest way to measure total power output is to position the LED very near to a large photodetector.
You should be able to get large area photodiodes from: Vactec, EG&G, Hammamatsu, IPI Centronic, Optodiode, Silonix, among others. The last one is a Canadian co. Solar cells OK, but you should probably do a rough calibration. Here's one way to do it: Noonday sun at your latitude irradiates the surface of the earth with about 800 Watts/sq-m. This is equivalent to 800/10,000 = 80mW/sq-cm. Since Si efficacy for sunlight is about 0.5Amps/Watt, you should get around 40milliAmps for each square cm of solar cell. Supposing that you get a smaller value, then you can assume that your particular solar cell is less sensitive than 0.5A/W and calculate a calibration value appropriately.
Example: you obtain a 1"x1" solar cell, i.e., about 6.5sq-cm. At midday on a clear day, you measure about 200mA short circuit current max as you orient it toward the sun. You calculate 80 * 6.5 = 520mW incident on the cell, so you were expecting more like 260mA. This means that your cell has more like 0.38A/W efficacy.
I've glossed over some of the fine points, but all the above info is reasonably accurate.
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