If we have a large area, we capture more optical power which increases the short circuit current and consequently the open circuit voltage. This is the main reason of making photodetectors large. Why else would we be making large solar panels? :)  Now coming to the next part of the question..about figures or merits..well it depends on which FoM you are interested in. Everything comes at a price. For example, the direct price you pay by increasing area is that you have a big capacitance, which means it can affect bandwidth. Plus a large area also means greater chances of trap-induced effects (higher dark current, flicker noise etc.). In essence, what is important is how large is the effective capture area of the detector (which can vary with how the illumination is provided) by keeping the physical area as low as possible.

Lots of reasons for large photodiodes. Better signal to noise when imaging a dim diffuse source.  Maybe want to do photometry on a large beam.  Maybe have an imaging problem with a jittering point source that you can't stabilize.

Dear Ali,

I agree with Satadal increasing the area of the photo detector will increase the harvested current for the incident photon flux which is the number of incident photons/cm^2/s such that Iph= Jph x A , where Jph is the photocurrent density and A is the area of the diode. As the current increases it can be easily handled and measures.

However, the open circuit voltage will not increase as by definition the open circuit voltage Voc= n Vt Ln (Jph/ J0) where J0 is the reverse saturation current density.

Yes there are two bad effects of increasing the area:

-Increasing the dark current= J0 A,

- slowing the switching of the diode because of the larger diode capacitance.But for static applications this effect is irrelevant .

Best wishes

Thank you all for your replies, I appreciate it. Dear @Abdelhalim Zekry, I think these large area diodes operate at high voltages and so effect of capacitance is not

problematic for switching applications.

best regards,

Yes Ali,

If you operate the diode at higher voltage its junction capacitance will be smaller and therefore it switching time will be smaller. However the small area diodes can also be operated at higher reverse bias voltage and therefore their junction capacitance will be smaller also compared to that at lower operating voltage. This means you can speed up both diodes by operating them at higher voltage.

Best wishes

Thanks Abdelhalim for pointing out the msitake in my answer. Voc indeed does not increase.