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Determining the maximum number of minority electrons in semiconductor

  1. Sep 16, 2013 #1
    1. The problem statement, all variables and given/known data

    P-type silicon can be doped in the range from 5x1014 cm-3 to 1020 cm-3. Determine the maximum possible number of minority electrons in a neutral P-type region if the device area is limited to AD = 1 cm x 1 cm and the thickness of the P-type region is limited to tP=100 μm. Assume room temperature and full acceptor ionization. ni = 1.02 x 1010 cm-3

    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. jcsd
  3. Sep 17, 2013 #2

    ehild

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    It follows from the Mass Action Law n*p=ni2. As the acceptors are fully ionized, the concentration of the holes can be taken equal to NA. See:

    http://en.wikipedia.org/wiki/Mass_action_law_(electronics)

    ehild
     
  4. Sep 17, 2013 #3
    What about holes created because of thermal energy when temperature is larger than 0K?
     
  5. Sep 17, 2013 #4

    ehild

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    Their number is usually negligible with respect to NA.

    ehild
     
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