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Why some photodetectors (specifically Si pin PDs) have large areas? and how this large area affects on the figure of merits of a photodetector? Excuse me if the question is and general.

I could not find good (answer and) references for my question. Thank you.

Mr. Nobody
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  • Andy has given you some reasons. But from experience, I can mention another. (1) Realistic optical systems have variations; and, (2) even if they were perfectly the same, one to another, the image on the focal plane depends upon the source and acceptance angles too. So you need a large enough detector to accept the image and to accept realistic variations in the optical system. If you want reproducable results, anyway. In short, you want one that is as large as necessary and no larger. And that's application dependent. – jonk Jul 09 '17 at 16:25

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A larger area means it can collect more light power and hence it becomes more sensitive. The down side is that capacitance is increased and the response time falls off making a large area photodiode quite unsuitable for high speed comms applications.

It's a the same reasoning behind a satellite dish - a bigger dish collects more power from a remote transmission and produces a bigger signal. Hence why voyager 1 and 2 had quite large dishes for comms with earth.

Andy aka
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  • Thank you @Andy aka for your answer. I saw a formula that states sensitivity is proportional to (Area^0.5). But it is not clear and intuitive for me, why collecting more light power means more sensitivity? Again many thanks. – Mr. Nobody Jul 09 '17 at 14:58
  • @ObjectOrientedMan, never mind sensitivity. If your light beam diameter is 2 cm and your photodiode diameter is 0.1 cm, what fraction of the available light are you going to be able to collect? – The Photon Jul 09 '17 at 15:04
  • @The Photon, thank you for the answer . In this case, you are right. But is there any case that light beam diameter is smaller than two photodiodes with different diameters? Now both collects same number of photons. – Mr. Nobody Jul 09 '17 at 15:15
  • Even if the beam is smaller, you still have to put the the detector in the right place to capture the beam (or steer the beam to find the detector). That can be easier with a large detector than a small one. – The Photon Jul 09 '17 at 15:17
  • @ObjectOrientedMan If the spot is larger than the diode, the spot may have differential intensity over its spot area and therefore different alignments will get different currents, making it very sensitive to alignment -- and reproducability goes to hell. Regardless of that, it's still dependent on the exact diode area which also can vary slightly part to part. If the spot is smaller than the diode, misalignments can be tolerated assuming the diode has a uniform spatial response. – jonk Jul 09 '17 at 17:38
  • @jonk I had to run one system where the spot was bigger than the photodiode area because of vibration misaligning the two and the need to run at over 600 Mbps. We couldn't get closer because of the temperature variations in the machine would have collided static sensor and rotating laser. The spot had about ten times the area of the photodiode and although we were a little close to losing the signal in noise we had zero misalignment problems. Bottom line: it worked! – Andy aka Jul 09 '17 at 19:46
  • @Andyaka Probably because you were ten times the area! There's not a lot of variation that much in the center of the field (the "center of confusion" I sometimes imagine it) -- most rays are parallel and the distribution is mostly uniform. It may also be the precision you needed to hit, too. Did you find that all diodes in the batches performed exactly the same without any calibration? Or did you have to do a one-point calibration to deal with variations of size and sensitivity from one diode to another? (By the way, try a largish spot size with a flexible optical fiber system.) – jonk Jul 09 '17 at 21:03