I'm designing a readout circuit for an APD manifactured by Hamamatsu Photonics (datasheet). However, I've found myself lacking some information about exactly how would the device behave in my particular application. The datasheet supplies the gain M, the dark count and photon detection efficiency, but no information whatsoever about leakage and noise current, so I am unable to extrapolate what kind of current characteristics the device has. Particularly when constructing a transimpedance amplifier, I'd need to have an idea of the photodiode current at a specific light level (my goal is to work at low to no light) - how exactly could I find that out from the datasheet?

  • \$\begingroup\$ Doesn't the dark count tell you something about the noise - I'm presuming these are "false" counts? \$\endgroup\$
    – Andy aka
    Commented May 12, 2014 at 10:29
  • \$\begingroup\$ I assume the dark current is proportional to these "dark counts" - but that doesn't imply how much mA are flowing through the detector, does it? \$\endgroup\$
    – joaocandre
    Commented May 12, 2014 at 11:11

1 Answer 1


Sure it does. This photodiode works by having a detected photon kick loose an electron, which then undergoes an avalanche amplification with an amplification factor M. So each dark count produces (on average) M electrons. Multiply by the dark count rate and you have the dark current in electrons. Noise in these devices is a complex subject, and you need to do some research outside the data sheet to understand it.

  • \$\begingroup\$ Stuck at a rather trivial conversion, but how would one convert number of electrons/charge to amps (is there a thing as a tabulated electron speed?) \$\endgroup\$
    – joaocandre
    Commented May 12, 2014 at 12:53
  • \$\begingroup\$ Obviously the question should rather be: can the speed of light be used in such a conversion? I think so, but I'm not sure \$\endgroup\$
    – joaocandre
    Commented May 12, 2014 at 13:10
  • \$\begingroup\$ @joaocandre, see my recent answer on this subject. That question was about a phototransistor, but the responsivity formula for an APD is the same, even if the gain mechanism is different. \$\endgroup\$
    – The Photon
    Commented May 12, 2014 at 16:04
  • \$\begingroup\$ If you're actually working in a photon-counting regime, you need to think in terms of impulses: energy in (h x nu) results in charge out (cuolombs). There should be no speed-of-light term. The frequency response of the detector will shape the pulse and have a strong effect on the peak output current. \$\endgroup\$
    – The Photon
    Commented May 12, 2014 at 16:09
  • \$\begingroup\$ joaocandre - go find how many electrons per coulomb. One ampere current is one coulomb per second. \$\endgroup\$ Commented May 12, 2014 at 17:32

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