Photodiode : ask help too find 10x calculation error

In summary, Roger's calculations for light falling on a photodiode are correct, but the area of the photodiode (7.34mm2) affects the output voltage. A data sheet for the BPW21 states that when exposed to standard light A, 10nA/lux is produced, while a data sheet for the S7686 states that 3,74nA/lux is produced. When a 3.74nA/lux photodiode is exposed to 100 lux, 37mV is produced instead of 0.38V as stated in an article on the Net.
  • #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
 
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  • #2
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?
 
  • #3
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.
 

FAQ: Photodiode : ask help too find 10x calculation error

1. What is a photodiode?

A photodiode is a type of semiconductor device that converts light into an electrical current. It is commonly used in various electronic devices, such as cameras, solar cells, and optical communication systems.

2. How does a photodiode work?

A photodiode works by absorbing photons of light and converting them into electron-hole pairs. These pairs then create an electrical current that can be measured and used in various applications.

3. What is the principle behind a photodiode?

The principle behind a photodiode is the photoelectric effect, where photons of light cause the emission of electrons from a material. This creates a flow of current in the photodiode, which is proportional to the intensity of light.

4. What is the difference between a photodiode and a regular diode?

A photodiode is designed specifically to convert light into an electrical current, while a regular diode is used for controlling the flow of electricity in a circuit. Additionally, a photodiode is sensitive to light, while a regular diode is not.

5. How can I calculate the efficiency of a photodiode?

The efficiency of a photodiode can be calculated by dividing the output current by the incident light power. This value is typically expressed as a percentage and can be used to determine the performance of the photodiode. If there is a 10x calculation error, double check your values and calculations to ensure accuracy.

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