Question

A semiconductor is known to have an electron concentration of $5\times {10}^{12}{cm}^{-3}$ and a hole concentration $8\times {10}^{13}{cm}^{-3}$.
(a) Is the semi-conductor n-type or p-type?
(b) What is the resistivety of the sample. If the electron, mobility is $23000{cm}^{-2}{v}^{-1}{s}^{-1}$ and hole mobility is$100{cm}^{-2}{v}^{-1}{s}^{-1}$? Take charge on electron, e=$1.6\times {10}^{-19}$c.

Answer 1

$n_e=5\times {10}^{12}{cm}^{-3}=5\times {10}^{18}{m}^{-3}$
$n_h=5\times {10}^{13}{cm}^{-3}=8\times {10}^{19}{m}^{-3}$
${\mu }_e=23000{cm}^{-2}{v}^{-1}{s}^{-1}={2.3m}^{-2}{v}^{-1}{s}^{-1}$
${\mu }_e=100{cm}^{-2}{v}^{-1}{s}^{-1}=0.01m^{-2}{v}^{-1}{s}^{-1}$
(a) Since the semiconductor has greater hole density hence it is p-type.
(b) Now$\left(1/p\right)=e(n_e{\mu }_e+n_h{\mu }_h)=1.6\times {10}^{-19}[5\times {10}^{18}\times 2.3+8\times {10}^{19}\times 0.01]$
$=1.6\times {10}^{-19}[1.15\times {10}^{19}+0.08\times {10}^{19}]$
or p=(1/1.968)=0.508$\Omega$-m