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

 

[Nickie]

This equation is derived from the law of mass action, which states that the product of the concentration of majority carriers (n or p) and the concentration of minority carriers (p or n) is equal to the square of the intrinsic carrier concentration (ni). In a neutral region, the concentration of majority carriers is equal to the concentration of ionized acceptors (NA), and the concentration of minority carriers is equal to the concentration of ionized donors (ND). Therefore, the equation can be rewritten as:

$$np={n}_{i}^{2}$$

Since we are looking for the maximum concentration of minority carriers, we can assume that the concentration of majority carriers is negligible compared to the concentration of ionized acceptors. Thus, we can approximate the equation as:

$$n\approx\frac{{n}_{i}^{2}}{N_A}$$

This gives us the equation used in the solution:

$$n=\frac{{n}_{i}^{2}}{N_A}$$