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Exactly what the title asks. "Diode" comes with it the ideas of depletion layers and forward/reverse biasing and electron-hole recombination, but SPAD physics doesn't seem to be dependent on any of that. You just put a giant electric field on, and as soon as a photon comes and promotes a charge to the conduction band, it gets so much energy that it can knock other electrons out of their atoms, creating the avalanche.

In other words the fact that it's "reverse-biased" relative to the doping of the semiconductors seems irrelevant, as does the fact that the reverse-bias is above the breakdown voltage.

Could I take a piece of undoped silicon and use it as a SPAD if I put a sufficiently large voltage on it? Or else why is it important that it's a diode?

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  • $\begingroup$ Well, they are reverse biased p-n junctions - Wikipedia even states "SPADs are semiconductor devices based on a p–n junction reverse-biased at an operating voltage that exceeds the junctions breakdown voltage". $\endgroup$
    – Jon Custer
    Commented Jun 30, 2023 at 15:32
  • $\begingroup$ I understand. My question is, why is it important that it be a junction with high reverse-bias, as opposed to any old piece of silicon or something that you put a big electric field on? $\endgroup$
    – flevinBombastus
    Commented Jun 30, 2023 at 15:35
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    $\begingroup$ There are other devices (wire chambers, Geiger counters, etc.) that act more along the lines you are thinking, so while a detector does not have to be a diode, a SPAD is a diode implementation of one. $\endgroup$
    – Jon Custer
    Commented Jun 30, 2023 at 15:38

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The essential problem is that you can't really have perfectly pure silicon. A reverse-biased junction is the trick used to clear the carriers out of a layer of silicon so that you can have a high field across it with minimal "dark current".

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  • $\begingroup$ This sounds promising... can you explain more about why using a reverse-biased junction clears out the carriers? Are you just referring to the formation of the depletion region? $\endgroup$
    – flevinBombastus
    Commented Jun 30, 2023 at 15:35
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    $\begingroup$ @flevinBombastus Yes, a depletion region is a region depleted of carriers. $\endgroup$
    – John Doty
    Commented Jun 30, 2023 at 15:36
  • $\begingroup$ I see. So you're saying a regular old piece of silicon has impurities that lead to excess charges floating around (in the conduction band), so if you tried to make a SPAD from it, you would get messed up by those charges. But if you form a p-n junction, you form a region of no excess charge, so that you can actually be sensitive to a single photon producing a single electron. I think I understand. Thanks!! $\endgroup$
    – flevinBombastus
    Commented Jun 30, 2023 at 15:39
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    $\begingroup$ @flevinBombastus Electrons in the conduction band or holes in the valence band. $\endgroup$
    – John Doty
    Commented Jun 30, 2023 at 15:41
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    $\begingroup$ @flevinBombastus It doesnt even need to be silicon with impurities. Perfect intrinsic silicon still has 10^10/cm^3 electrons and holes at room temperature. $\endgroup$
    – Matt
    Commented Jun 30, 2023 at 18:41

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