What will be the resistivity of an n-type Germanium sample at 300

SOLUTION

Concept:

In an extrinsic semiconductor, the conductivity (hence resistivity) depends on the number of carriers present and is given by:

\({σ _i} =q\left(n_0{{μ _n} + p_0{μ _p}} \right)\)

n0 = majority carrier electron concentration

p₀ = majority carrier hole concentrationμn and μp are the electron and hole mobilities respectively.

The resistivity is the inverse of conductivity, i.e.

\(ρ =\frac{1}{q\left(n_0{{μ _n} + p_0{μ _p}} \right)}\)

Calculation:

Since the intrinsic carrier concentration of Germanium is of the order of 10¹³, and the given donor density is of the order of 10²⁰, we can write:

n₀ = Nd = 1020 atoms/m3

Since there is no acceptor impurity, p₀ << n₀ and can be easily neglected and the resistivity becomes:

\(ρ =\frac{1}{qn_0{\mu _n}}\)

\(ρ =\frac{1}{(1.6\times 10^{-19})(10^{20})(0.38){}}\)

ρ = 0.164 Ωm