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Diode and doping on N and P regions

  1. Mar 28, 2013 #1
    Dear forum,

    in a diode semiconductor, the N region is rich in electrons. The P region is rich in holes.

    Is the number of extra electrons in the N region equal, larger or smaller than the number of holes in the P region? If so, why?


    Also, in the N region, it is said that the majority carriers are the electrons. I get that. The minority carriers are the holes....but there are no holes in the N region so how can there be any minority carriers at all?


    Thanks,
    fisico30
     
  2. jcsd
  3. Mar 28, 2013 #2

    marcusl

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    The number of electrons/holes is a function of doping and can be controlled by the device designer. The doping saturates at around 10^19 cc-1, which is called "extrinsic" doping. It gives the highest conductivity possible which is useful near the electrode connections.

    You can inject holes into the n region, where they live for a time (the mean recombination time) before meeting electrons and annihilating. Transistor action is caused by minority carriers injected across the base region into the collector.
     
  4. Mar 28, 2013 #3

    DrDu

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    No, at non-zero temperature, even in equilibrium there are both electrons and holes present in both the n and p region. They fulfill a equilibrium constant equation ##n_en_h=K(T)##. On the n side ##n_e=n_{0e}-n_h\approx n_{0e}## and thus ##n_h\approx K/n_{0e}## where ##n_e## is the number density of electrons ##n_h## the number density of holes and ##n_{0e}## the number density of donor atoms.
    A similar equation holds for the p side.
     
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