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SPAD (Single Photon Avalanche Detector) works beyond Breakdown Voltage which is mentioned as Geiger Mode. But how this actually working. When bias voltage aceede breakdown voltage, generally high current flow started.But case of SPAD , this is not the case. Why ? Should not we found a high current without any photon impact as bias voltage crossed breakdown voltage?

I'm bit of confused here. I searched to understand how this is operating over breakdown voltage but found nothing helpful . They just mention that it operate beyond breakdown voltage and when any photon impact , it gives high current . But Though it's beyond the breakdown voltage, when no photon impact occur, why don't we found anything? Why no significant amount of current flow without photon impact ? My knowledge in this topic is not that much , so detail explanation will be appreciated. Thanks in advance .

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So, lets step back a bit. First lets look at avalanche breakdown. Electrons are constantly scattering, off atoms and other electrons, with some average scattering rate under given conditions. In a semiconductor, avalanche breakdown occurs when the field is strong enough that a free conduction electron gains (through accelerating in the field) a threshold level of energy. Then, in the next scattering event there is a probability of transferring enough energy to promote another electron from the valence band to the conduction band. Now there are two free electrons, in a field strong enough such that at the next scattering event both of them can promote two more electrons. You know what happens next...

So, at a high enough reverse bias, you can get avalanche breakdown, and this is commonly referred to as the breakdown voltage. OK, you take a nice diode, and reverse bias it to that point, or beyond. Nothing happens. Why not? Well, you need that first free electron. If the material in the depletion layer of the diode is really good, you won't have a high intrinsic carrier generation rate, so there are no free electrons. But, wait for that photon to come along and create an electron-hole pair, and the avalanche can start (note that it does not have to start).

Or, you just wait long enough, and an electron-hole pair will be generated spontaneously in the depletion layer and you get a false positive. This is why one usually doesn't leave the device biased all the time, but would pulse it into the right bias condition when you expect a photon, and then reduce the bias when experimental conditions indicate you won't get a photon.

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  • $\begingroup$ May I ask why I need that first free electron ? beyond breakdown voltage , there is no depletion region or I can say I can say depletion region become bigger then the diode so that you can't identify the junction area from p type or n type doped area. So why the source couldn't supply the free electron ? $\endgroup$ – Anklon Jun 13 '16 at 20:03
  • $\begingroup$ Of course there is a depletion region in an avalanche device. If there were lots of free carriers you could not build up a large potential in the first place. You need that first free electron to accelerate and be able to transfer energy (scatter) and make more free electrons, which make more... $\endgroup$ – Jon Custer Jun 13 '16 at 20:20
  • $\begingroup$ Don't we create potential difference using external voltage source ? correct me if I'm being silly or wrong . But if we create voltage difference using external source , how free carriers stop building up a large potential. they should just gives high current spontaneously. My knowledge in this section is not that deep I guess. $\endgroup$ – Anklon Jun 13 '16 at 20:25
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    $\begingroup$ In simple theory, a reverse-biased diode has no reverse current. In reality, the reverse current is caused by carrier generation in the depletion region. If you have a low carrier generation rate, you will be able to trigger an avalanche from a photon-induced electron-hole pair before an intrinsic avalanche. $\endgroup$ – Jon Custer Jun 13 '16 at 22:59

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