Determining the maximum number of minority electrons in semiconductor

Discussion in 'Homework Help' started by tahayassen, Sep 16, 2013.

  1. tahayassen

    Thread Starter New Member

    Apr 1, 2013
    2
    0
    1. The problem statement, all variables and given/known data

    P-type silicon can be doped in the range from 5x10[SUP]14[/SUP] cm[SUP]-3[/SUP] to 10[SUP]20[/SUP] cm[SUP]-3[/SUP]. Determine the maximum possible number of minority electrons in a neutral P-type region if the device area is limited to A[SUB]D[/SUB] = 1 cm x 1 cm and the thickness of the P-type region is limited to t[SUB]P[/SUB]=100 μm. Assume room temperature and full acceptor ionization. n[SUB]i[/SUB] = 1.02 x 10[SUP]10[/SUP] cm[SUP]-3[/SUP]

    2. Relevant equations



    3. The attempt at a solution

    Solution:

    Maximum\quad concentration\quad of\quad minority\quad carriers\quad is\quad obtained\quad for\quad the\quad minimum\quad doping\quad level:\\ n=\frac { { { n }_{ i } }^{ 2 } }{ { N }_{ A } } =\frac { { (1.02*{ 10 }^{ 10 }) }^{ 2 } }{ 5*{ 10 }^{ 14 } } =2.1*{ 10 }^{ 11 }\quad { m }^{ -3 }\\ The\quad maximum\quad volume\quad is:\\ V={ A }_{ D }{ t }_{ p }={ (0.01) }^{ 2 }(100)({ 1 }0^{ -6 })={ 10 }^{ -8 }\quad { m }^{ 3 }\\ N=nV\\ =2.1*{ 10 }^{ 3 }

    My confusion is how they got this equation:

    n=\frac { { { n }_{ i } }^{ 2 } }{ { N }_{ A } }
     
  2. anhnha

    Active Member

    Apr 19, 2012
    773
    47
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