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Depleted region in Semiconductor and leakage current

  1. Sep 26, 2013 #1

    Iam reading the book Glen F. Knoll Radiation Detection and Measurements, and Iam having a trouble understanding a certain part of the chapter about semiconductor, he writes:

    Electron-hole pairs that are created within the depletion region by the passage of radiation
    will be swept out of the depletion region by the electric field, and their motion constitutes
    a basic electrical signal.

    The thermal generation of charge carriers will continue to take place in the depletion
    region, contributing a component sometimes called the generation current to the observed
    leakage current. These charges are swept away typically within a few nanoseconds, however,
    a time that is many orders of magnitude shorter than the time required to establish thermal
    Thus, the steady-state concentration of carriers is strongly reduced in the
    depletion region because the removal of charges is a much faster process than their

    I dont understand the sentences in bold, why does it matter how fast they are swept away? As long as they are created, and move under an electric field, they constitute an electric signal. Sure, the depletion region is depleted of mobile charge carriers and that greatly reduces the leakage current, but this second benefit of the short time they are swept away?

    Iam certain that Iam missing something here. Can someone who understand this clarify this for me?

    Thank you very much!
  2. jcsd
  3. Sep 26, 2013 #2


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    It appears that you are talking about photodiode detectors? The quote agrees with your statement that thermally generated electron-hole pairs constitute a current--specifically, a leakage current that will look like a noisy background to the photocurrent. I think the author is simply pointing out that the charges do not remain and build up in the depletion zone because they are rapidly swept away,
  4. Sep 27, 2013 #3
    Thank you for your reply.
    Yes, photodiode detectors.

    I still dont get it. Thermally generation is a random process, it happens all the time and you could say that an equilibrium would be created, where these thermally generated carriers contribute to a current of e.g 1mA. Why is time important?
    Its like saying you have a material where a radionuclide with short half life is constantly created and because the half life is so short, you do not detect this with e.g. GM tube. Of course you would. However, if the radionuclide is in a limited amount, then sure, time is important, if the radionuclide decays away then there would be no additional counts.

    Please help me understand this!
  5. Sep 27, 2013 #4


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    It's hard for me to know what the point is from a short snippet of text. Here is my guess: You are focusing on the leakage current, but there are other effects in a diode that can be important. Charges building up in the depletion zone, or in the I region of a PIN photodiode, can change the device's effective capacitance and conductance, and could potentially alter its response to photo-generated carriers. (If the depletion zone disappears, for instance, then photo-carriers would have to diffuse to the terminals instead of being swept at the drift velocity. Many photo-carriers would recombine, hurting sensitivity. The diode response, furthermore, would be orders of magnitude slower.) The back bias that is implied in your short quote keeps the depletion zone free of thermally generated carriers, however. This is because they are generated at a slow rate but are swept away nearly instantaneously.
  6. Sep 27, 2013 #5
    This description is what I remember from yo those many years ago of study....


    first sentence:
    due to N and P doped regions adjacent which induce an electric field. All I read into this is that thermally generated carriers move along 'promptly' and constitute part of leakage current.

    Maybe your particular description is a 'warm up' for a later discussion of thermal/shot noise??

    If you don't get an answer here, try 'electrical engineering'..... they study this for sure.
  7. Sep 27, 2013 #6
    If the thermally generated charge carriers in the depletion region persisted for a time on the order of the time required for the E field to establish thermal equilibrium(in the Gibbs sense), then the thermal noise signal would swamp out the radiation signal. The average E field amplitude would be a random function of time and the response of the system would be highly non linear.
  8. Sep 28, 2013 #7
    If the thermally generated electron-hole pairs were created much faster than they are swept out, then you would not get a depletion effect. You'd have a normal thermal equilibrium electron and hole concentration and hence basically a low DC conductivity. The thermal background current would be quite large, and you would be very limited in the bias voltage you can apply.

    If on the other hand the sweeping out is fast and the thermal generation slow, then you have a depleted region without any free carriers and occasionally a electron-hole pair produces a very small and short current spike. Now since even in a small and thin photodiode you still have a lot of atoms, this still happens very often and thus produces a noisy, fluctuating background current.
  9. Sep 28, 2013 #8
    I don't understand the implication of the prior two posts. It would seem the thermal noise current is what it is.

    It seems counter-intuitive that a slower removal of thermally generated charge carriers, that is, more mobile charge carriers in the depletion region, would increase such a current. Is that what is being implied?
  10. Oct 8, 2013 #9
    Thank you for your answer. Iam beginning to understand what the author mean.
    The author just made it confusing I think. He propably meant this simple statement. Like you said:
    If they were to remain there would be a rising leakage current (and the response of the diode would be non-linear) rather than a constant leakage current (enabling a linear response).

    Thank you, this helped me understand it a bit more.

    Thank you all for the replies!
    Last edited: Oct 8, 2013
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