A silicon bar is doped with donor impurities ND = 2.25 × 1015 atoms/cm3. Given the intrinsic carrier concentration of silicon at T = 300 K is 1.5 × 1010/cm3. Assuming complete impurity ionization, the equilibrium electron and hole concentrations are respectively

Concept:

For a compensated semiconductor with donor concentration greater than the acceptor concentration, the majority carrier electron concentration is calculated as:

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

With Nd - Na ≫ ni, the above equation becomes:

n0 ≅ Nd - Na

Also, the charge carriers at thermal equilibrium follow mass-action law, i.e.

\({n_0}{p_0} = n_i^2\)

Application:

Given:

Nd = 2.25 × 1015 cm-3

ni = 1.5 × 1010 cm-3

Na = 0 cm-3

Nd - Na ≫ ni, the majority carrier electron concentration will be:

n0 ≅ Nd - Na = 2.25 × 1015 cm-3

Since the donor concentration is greater than the acceptor concentration, the material will be n-type with the majority carrier electron concentration as:

n0 ≅ Nd - Na

n0 = 2.25 × 1015 cm-3

The minority carrier hole concentration is obtained using mass action law as:

\({n_0}{p_0} = n_i^2\)

\({p_0} = \frac{{n_i^2}}{{{n_0}}}\)

\({p_0} = \frac{{{{{2.25 \times {{10}^{20}}}}}}}{{2.25 \times {{10}^{15}}}}\)

p0 = 105 cm-3