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Determining the maximum number of minority electrons in semiconductor

  • Thread starter Turion
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  • #1
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Homework Statement



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

Homework Equations





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:

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

Answers and Replies

  • #2
ehild
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  • #4
ehild
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Their number is usually negligible with respect to NA.

ehild
 

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