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Homework Help: How to calculate the carrier concentration

  1. Sep 26, 2007 #1
    Can any please help me in solving the following two questions

    A Si sample is doped with 10^16 per cm cube boron atoms and a certain
    number of shallow donors. The fermi level (Ef) is 0.36 eV above Ei
    (intrinsic energy level) at 300K. What is the donor concentration Nd?

    For Si at 300K ni(intrinsic carrier concentration) = 1.5 x 10^10 per
    cm cube

    A Si sample contains 10^16 per cm cube In(indium) acceptor atoms and
    a certain number of shallow donors. The In (indium) acceptor level is
    0.16 eV above Ev(Valence band edge), and Ef is 0.26eV above Ev at
    300K. How many in atoms in cm per cube are unionized (i.e. neutral)?

    For Si at 300K ni(intrinsic carrier concentration) = 1.5 x 10^10 per
    cm cube

    There is no additional information available. Please state the
    question number when answering and indicate any formulas used.

    The following equations may prove useful

    n(o) x p(o) = ni^2

    n(o) = ni x e((Ef-Ei))/KT)

    p(o) = ni x e((Ei-Ef)/KT)
    Last edited: Sep 26, 2007
  2. jcsd
  3. Sep 27, 2007 #2


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    Homework Helper

    What have you done so far? If you want help, than you should say what you think and what you have tried.

    Is there any more relations that you know of? How about "carge neutralisty condition of doped semi conductor" ? And Law of mass action?
  4. Sep 27, 2007 #3
    This is what I have came up with so far for question number 1

    using the relation n(o) = ni x e((Ef-Ei))/KT)

    with Ef-Ei=0.36 x 1.6 x 10^-19 , ni=1.5 x 10^10, T=300k , K=1.38 x 10^-23

    we get n(o) = 1.654 x 10^16 per cm cube

    However there are 10^16 B atoms to neutralize these charges

    Hence Nd = 1.654 x 10^16 - 10^16 = 6.5 x 10^15 per cm cube
  5. Sep 27, 2007 #4
    For question number 2

    unionized atoms are left at the acceptor level



    Ea-Ef=0.16-0.26= -0.10eV

    Using fermi-driac statistics f(E)=1/(1+e((E-Ef)/KT)))

    for E=Ea,T=300 and substituting all the constants

    f(E)=1/(1+e((Ea-Ef)/KT))), gives = 0.9794

    However fermi driac statistics give the probablity of occupance of an electron in an Energy state E. hence 0.9794 is the probablity of occupance of an electron.

    hole probablity of occupance=1-electron probablity of occupancy

    hole probablity of occupance of the energy state Ea = 1-0.9794 = 0.02053

    unionized atoms are left at the acceptor level(Ea)

    Hence, number of IN(indium) atoms left unionized = hole probabilty of occupancy of Energy state Ea x number of In acceptor atoms

    =0.02053 x 10^16

    =2.05 x 10^14
  6. Sep 27, 2007 #5
    Anyway there are still questions that remain unanswered

    For instance, what role does shallow donor impurities have to play in question number 2.

    Any suggestions to the proposed solution above will be highly appreciated
    Last edited: Sep 27, 2007
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