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:

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

 

Hi,
It is from Law of Mass Action for semiconductor.
Check this slide, especially page 14.
http://web.mit.edu/6.012/www/SP07-L2.pdf