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Carrier concentration in a semiconductor

  1. Oct 9, 2015 #1
    1. The problem statement, all variables and given/known data
    The question is in the picture attached. When Na=0, that means the acceptor concentration(i.e. open holes?) is zero?



    2. Relevant equations
    (Nd-Na)/2 + sqrt((Nd-Na/2)^2+ni^2)
    (Na-Nd)/2 + sqrt((Na-Nd/2)^2+ni^2)

    3. The attempt at a solution
    Solving for Nd=10^8:

    majority carrier =
    (Nd-Na)/2 + sqrt((Nd-Na/2)^2+ni^2)
    (10^8)/2 + sqrt((10^8/2)^2+1.5*10^10^2)= 1.505*10^10... almost same as ni?
    minority carrier=
    (-10^8)/2 + sqrt((-10^8/2)^2+1.5*10^10^2)=1.495*10^10... close to ni


    Log10(10^8) =
    8... is this all they want?

    log10(1.495*10^10) =
    10.17

    Fermi energy difference
    (Efi-Ev)
    8.612*10^-5*300ln(1.505*10^10/1.5*10^10)=8.598*10-5


    Do these answers look right for the first one? Is it the same the whole way down?
     

    Attached Files:

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  2. jcsd
  3. Oct 12, 2015 #2
    Ok I calculated the rest the same way, here's how I did the Nd=10^14


    majority carrier =
    (Nd-Na)/2 + sqrt((Nd-Na/2)^2+ni^2)
    (10^14)/2 + sqrt((10^14/2)^2+1.5*10^10^2)= 1.00*10^14
    minority carrier=
    (-10^14)/2 + sqrt((-10^14/2)^2+1.5*10^10^2)
    2.24*10^6

    log(10^14) = 32.32 donor concentration
    log(1.00*10^14)=32.23 majority carrier

    So does this make sense? My donor conentration is now equation to my majority carrier concentration
    Where as my minority carrier is a lot lower

    log(2.24*10^6)=14.62
     
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