- #1
Roger44
- 80
- 1
Hello
Can anybody help me find my mistake in the following reasoning for light falling on a photodiode :
"When 0.1464 watts/m2 fall on a 7.34 mm2 window, 1.07E-06 watts will penetrate and for a 0.38 A/W photodiode will produce 4.08E-07 amps which, across a 100k resistance will produce 0.0408 volts.
If the light is monochromatic at 550, then we can say that 100 lux (683x0.1464 ) produce 0.0484 volts.
If the photodiode has a near perfect eye response, 100 lumens, whatever their wavelength in the visible range, will produce the same voltage".
Something's wrong, 100 lux on visible light photodiodes produces far more volts across 100k. A document I found on the Net for a S7686 says 100 lux produces 0.38V, and that seems to more correspond to my experimental results.
Thanks if you can help me fathom this out.
Roger
Can anybody help me find my mistake in the following reasoning for light falling on a photodiode :
"When 0.1464 watts/m2 fall on a 7.34 mm2 window, 1.07E-06 watts will penetrate and for a 0.38 A/W photodiode will produce 4.08E-07 amps which, across a 100k resistance will produce 0.0408 volts.
If the light is monochromatic at 550, then we can say that 100 lux (683x0.1464 ) produce 0.0484 volts.
If the photodiode has a near perfect eye response, 100 lumens, whatever their wavelength in the visible range, will produce the same voltage".
Something's wrong, 100 lux on visible light photodiodes produces far more volts across 100k. A document I found on the Net for a S7686 says 100 lux produces 0.38V, and that seems to more correspond to my experimental results.
Thanks if you can help me fathom this out.
Roger