A block of Silicon is doped with a donor atom density N D = 3 × 1

SOLUTION

Concept:

For an extrinsic semiconductor with the given donor and acceptor concentration, the majority carrier electron concentration is given by:

\({n_0} = \frac{N_d-N_a}{2}+ \sqrt {{{\left( {\frac{{{N_d-N_a}}}{2}} \right)}^2} + n_i^2} \)

N_d = Donor concentration

N_a = Acceptor concentration

With Nd - N_a ≫ ni, the above equation can be simplified as:

n0 ≅ Nd - N_a

Calculation:

For a silicon semiconductor, the intrinsic carrier concentration is 1.5 × 10¹⁰ cm^-3 at room temperature.

Given ND = 3 × 1014 atoms/cm3 and NA = 0.5 × 1014 atoms/cm3.

Since Nd - Na ≫ ni, the majority carrier electrons will be:

n0 ≅ Nd - Na

n0 ≅ 3 × 1014 - 0.5 × 1014 electrons/cm³

n₀ = 2.5 × 10¹⁴ electrons/cm³