Photodiode current vs. irradiance and finding upper limit current

1.) What equation could be used to calculate, or make a decent estimate of the current produced by a photo-diode from incident irradiance and its operating parameters. For example for this PIN photo-diode BPW 34 FS from Osram

BPW 34 FS

here is the plot of the photo-current vs. irradiance 2.) How to calculate (or estimate) an upper limit current produced by the photo-diode when the incident irradiance rises.

• For part 2, clearly, with only the graph to go on, your answer is 70 uA. For part 1 it is a log-log relationship but this is hardly an EE question. – Andy aka May 1 at 11:52
• Maybe 70 uA is the end of the linear region but I do not understand why an upper limit. I did some practical test, the photodiode directed towards the sun and measured the current directly with a multimeter. I think I got around 500 uA (but I am not shure how the multimeter influence in measurments) – miquo May 1 at 12:54
• I will think about moving the part 1 to physics. – miquo May 1 at 12:56

part 1) - A straight line on a log-log plot means a simple power law, with the slope giving the power. As the slope is very close to unity, the power is very close to unity. That is, the relationship is approximately linear, or $$\I_p \approx kE_e\$$. You can try and read the slope very carefully, and write down the power law if you like.