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Radiation Diode Detector Doping question.

  1. May 14, 2012 #1
    I'm reading up on semiconductors and there seems to be a few fundamentals I cannot find answers for if anyone can help?

    I understand that n-type substrate is doped with a donor and p-type substrate is doped with an acceptor creating excess electrons and holes. When you put the p-type and n-type together you get a diode with a depletion layer.

    My problem comes with radiation detectors. In the paper below it states that a p-type diode (not substrate) is when you dope small amounts of donor impurities into a p-type substrate. Also a n-type diode (not substrate) is when you dope small amounts of acceptor impurities into a p-type substrate. Why dope the n-type with the opposite (p-type) donor? Is this not just undoing the original doping?

    So if this is right, you dope the n-portion of a p-n junction with the opposite doping (acceptors) to form an N-type diode?

    Paper: "Modeling the instantaneous dose rate dependance of radiation diode detectors" Jie Shie and William E. Simon Med. Phys 30(9), Sept 2003

    Please clarify! Any help is appreciated!
     
  2. jcsd
  3. May 14, 2012 #2
  4. May 17, 2012 #3
    Bump! Can anyone provide any help/resources?
     
  5. May 17, 2012 #4
    The most convenient way to make a diode is taking a substrate and doping one side with different polarity.
     
  6. May 18, 2012 #5
    Thanks for the reply, to clarify:

    Do you mean taking an n-type substrate and doping acceptors on one side of the substrate with a high enough concentraton to change the majority carriers from electrons to holes?
    Thus you would have a bigger minority carrier concentration on the p-side then just joining n-type and p-type substrates together to form the junction?
     
  7. May 18, 2012 #6
    Minority carrier concentration depends on built-in potential, doping etc.
    To make a diode by joining n and p type you have to fuse the junction to make a contact. The high temperature damages the crystal.
     
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