Diode and doping on N and P regions

  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. marcusl

    marcusl 2,109
    Science Advisor
    Gold Member

    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. DrDu

    DrDu 4,157
    Science Advisor

    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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