Pure Si at 500 K has equal number of electron $(n_e)$ and hole $(n_h)$ concentrations of $1.5×10^{16}m^{-3}$. Doping by indium increases $n_h$ to $4.5×10^{22}m^{-3}$. The doped semiconductor is of: |
n – type with electron concentration $n_e=5×10^{22}m^{-3}$ p – type with electron concentration $n_e=2.5×10^{10}m^{-3}$ n – type with electron concentration $n_e=2.5×10^{23}m^{-3}$ p – type having electron concentration $n_e=5×10^{9}m^{-3}$ |
p – type having electron concentration $n_e=5×10^{9}m^{-3}$ |
As $n_i^2=n_e×n_h$ $(1.5×10^{16}m^{-3})^2=n_e×(4.5×10^{22}m^{-3})$ $∴n_e=\frac{(1.5×10^{16}m^{-3})^2}{(4.5×10^{22}m^{-3})}=5×10^{9}m^{-3}$ As $n_h >> n_e$, so semiconductor is p – type and $n_e=5×10^{9}m^{-3}$ |