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Solid state question

  1. Mar 30, 2010 #1
    a silicon crystal is doped with 10^15 cm^-3 atoms of antimony
    and its also doped with 10^16 cm^-3 atoms of boron

    will the crystal be type p or type n

    what will be the concentration of minory charge carriers and majority charge carriers
    ?

    Boron has 3 electrons in outer circle
    and Sb has 5 on the outer circle
    so its type "p" because there there is more +3 then +5
    thats as far as could go
    how to find the concentration of minory charge carriers and majority charge carriers
     
  2. jcsd
  3. Mar 30, 2010 #2

    berkeman

    User Avatar

    Staff: Mentor

    But the two doping densities are not the same...
     
  4. Apr 4, 2010 #3
    boron is an acceptor donates 1 hole per atom.
    Sb is a donor, donates 1 electon per atom

    10^16 boron atoms vs 10^15 Sb means that this is a P type.

    n0p0=ni^2

    minority negative charge carriers : Npo=ni^2/p0 = 10^20/10^16 = 10^4 cm^-3

    major charge carriers = 10^16
     
  5. Apr 4, 2010 #4
    what are n0 p0 and ni
    ?

    what N represents in Npo=ni^2/p0 ?
     
  6. Apr 4, 2010 #5
    n0 is thermal equibrium density of electrons
    p0 is thermal equilibruim density of holes

    ni = electron density in intrinsic semiconductor

    Npo is the thermal equilibrium minority carrier density.
     
  7. Apr 4, 2010 #6
    how you get that the "major charge carriers = 10^16 "
    ?
     
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