Photodiode : ask help too find 10x calculation error

AI Thread Summary
The discussion revolves around a calculation error related to the output voltage of a photodiode when exposed to light. The original poster calculated that 100 lux would produce 0.0484 volts, but noted discrepancies with other sources indicating higher voltages, specifically 0.38V for the S7686 photodiode. A contributor clarified that their calculations, which considered various factors including spectral response and sensitivity, indicated a much lower output of 37mV for the S7686 at 100 lux. They also provided experimental results showing a consistent ratio between the BPW21 and S7686 outputs under standard lighting conditions. The conversation highlights the importance of accurate photodiode specifications and calculations in determining expected performance.
Roger44
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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
 
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Roger44 said:
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

"7.34 mm2" would be a huge photodiode. Where did you get this area from?
 
7.34mm2 for the BPW21, 6.72mm2 for the almost perfect 100$ S7686.

My calculations are correct finally. Taking into account :
- a table of spectral distribution of a standard light A;
- a table of CIE photopic eye sensitivity;
- a table of spectral response of the industry classic BPW21(had to numerize a curve manually, rather tedious)
- the sensitivity of this photodiode at 550nm

Excel calculated that the BPW21 would give 12.2 nA/lux bathed in standard light A. The data sheets quote a typical value of 10nA/lux in this light. I felt the relative closeness of thes two figure validated my approach.

Doing the same thing for the S7686 gives 3,74nA/lux. No figure of nA/lux appears on the data sheet.

12.2/3.74 = 3.27
This value is simple to confirm. I stuck the BPW21 under a 116W halogene, got 329mV across a 100k, then replaced it by the S7686 which gave 116mV.
329/116 = 2.84
Again sufficiently close for me to say that for a 3,74nA/lux S7686, 100 lux produces 37mV and not 0,38V stated in an article on the Net.

This value of 3.74nA/lux will hold what ever the source of light because the S7686 has a spectral response very close to the CIE curve.
 
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