Question

If resistivity of pure silicon is 3000$\Omega$m and the electron and hole mobilities are $0.12m^2{v}^{-1}{s}^{-1}$ and $0.045m^2{v}^{-1}{s}^{-1}$ respectively, determine. The resistivity of a specimen of the material when ${10}^{19}$ atoms of phosphorous are added per $m^3$
Given e=$1.6\times {10}^{-19}$c, p=3000$\Omega$m, ${\mu }_e=0.12m^2{v}^{-1}{s}^{-1}$, ${\mu }_h=0.045m^2{v}^{-1}{s}^{-1}$

Answer 1

The resistivity of pure Si given by
$\rho =\frac{1}{\sigma }=\frac{1}{e(n_e{\mu }_e+n_h{\mu }_h)}=\frac{1}{{en}_i({\mu }_e+{\mu }_h)}$
or $n_i=\frac{1}{e\rho ({\mu }_e+{\mu }_h)}=\frac{1}{1.6\times {10}^{-19}\times 3000(0.12+0.045)}=1.26\times {10}^{16}{m}^{-3}$
When ${10}^{19}$ atoms of phosphorous (donor atoms of valence five) are added per $m^3$, the semiconductor becomes n-type semiconductor.
$\therefore n_e-n_h\approx n_e=N_d={10}^{19}\because n_h=1.26\times {10}^{16}$
Resistivity $\rho =\frac{1}{n_e{\mu }_{e}e}=\frac{1}{1.6\times {10}^{-19}\times {10}^{19}\times 0.12}=5.21\omega m$