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Question on Reducing Dark Noise in Photodiodes

  1. Jul 4, 2003 #1
    [SOLVED] Question on Reducing Dark Noise in Photodiodes

    I realize that this is more of a technology question than a physics question, but our technology forum seems to have been overrun by computer nerds. Anyway, I am wondering if it is possible to reduce the dark current of a diode by placing a lens in front of it. Let's say that in theory, you could take the illumination normally impinged on a diode of a certain surface area and shrink that same intensity down to a smaller area. Does intensity per unit area affect the dark noise in a photodiode?

    Please show me some equations.


    Last edited by a moderator: Jul 4, 2003
  2. jcsd
  3. Jul 4, 2003 #2
    As I remember the dark current is one of the parameters caracterizing a photodiode. So if you wanna reduce the dark current you'd have to change the technology in which the photodiode is produced. But I'll take a look at my electronic device course tonight and get back to you.
    http://www.ee.bgu.ac.il/~Orly_lab/publictions/OE_Empirical_CMOS_APS_DC_Modeling_Final.pdf [Broken]
    Last edited by a moderator: May 1, 2017
  4. Jul 4, 2003 #3
    The dark noise

    is largely caused by thermally created hole-electron pairs, to reduce them cool the diode.
  5. Jul 5, 2003 #4
    Re: The dark noise

    Great... except your avoiding my question. Can you increase the signal to noise ratio by making the area smaller?

  6. Jul 5, 2003 #5
    Re: Re: The dark noise

    Sorry, didn't mean to do that! Sounds probable, maybe the mfr. will supply noise data on the diodes, I'm sure a lot of that stuff is on line now. Also, reverse biasing the diode slightly might reduce the thermal effect. At least it's simple to try it. The result may depend somewhat on whether you using a curent or voltage amplifier for the signal.
  7. Jul 5, 2003 #6


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    I know this might sound simple but ever think about adding a 1k resistor in series to reduce its current. better yet, install a variable resistor to limit the current (like an extra intensity button). Just an idea, also don't forget to take into consideration the wattage suppiled dont want you using a 1/4W when you should be using a 1W resistor.
    Dx :wink:
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