Determining the maximum number of minority electrons in semiconductor

[Turion]

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 A_D = 1 cm x 1 cm and the thickness of the P-type region is limited to t_P=100 μm. Assume room temperature and full acceptor ionization. n_i = 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:

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

 

[ehild]

My confusion is how they got this equation:

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

It follows from the Mass Action Law n*p=n_i². As the acceptors are fully ionized, the concentration of the holes can be taken equal to N_A. See:

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

ehild

 

[Turion]

It follows from the Mass Action Law n*p=n_i². As the acceptors are fully ionized, the concentration of the holes can be taken equal to N_A. See:

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

ehild

What about holes created because of thermal energy when temperature is larger than 0K?

 

[ehild]

Their number is usually negligible with respect to N_A.

ehild