Deduce an expression for the conductivity of a p-type semiconductor.

 

Consider a block of semiconductor of length l and area of cross-section A, having electron density n_e and hole density n_h. Let, V be the the potential difference applied across the ends of the semiconductor. The magnitude of the applied electric field is given by,

                           E = $\frac{\mathrm{V}}{\mathrm{l}}$         ....... ( 1)

Due to the applied electric field, both electrons and holes move in a mutually opposite direction with drift velocities v_e and v_h and contribute current I_e and I_h. 

$\therefore$ Total current, I = I_e + I_h 
Electrons and holes in the conduction band and valence band respectivel are moving in a random fashion. Therefore, 
                      I_e = n_e A e v_e 
                      I_h = n_h A e v_h 

Thus, 
Total current, I = n_e A e v_{e +} n_h A e v_h 

                       = eA (n_ev_e + n_h v_h)
$\Rightarrow$             $\frac{\mathrm{I}}{\mathrm{A}} = e (n_e{v}_e + n_{h_{ }}{v}_h)$        ... (2) 

Let R be the resistance of the semiconductor and $\mathrm{\rho }$ is it's resistivity. 

Then, 
                        $\rho = R\frac{A}{l}$                     ... (3) 

Now, dividing 1 by 3, we get 

                   $\frac{\mathrm{E}}{\mathrm{\rho }} = \frac{V}{RA}= \frac{I}{A}$                ...(4) 

Therefore, from (2) and (4), we get

               $\frac{\mathrm{E}}{\mathrm{\rho }} = e (n_e{v}_e + n_h v_h)$

Now, mobility of electrons is defined as the drift velocity per unit electric field. Drift velocity is zero when, no electric field is applied. 

$\therefore$ mobility of electrons, ${\mathrm{\mu }}_{\mathrm{e}}$ = $\frac{{\mathrm{v}}_{\mathrm{e}}}{\mathrm{E}}$ 

$\Rightarrow$                               ${\mathrm{\nu }}_{\mathrm{e}} = {\mathrm{\mu }}_{\mathrm{e}} \mathrm{E}$ 
Similarly, 
                                    ${\mathrm{\nu }}_{\mathrm{h}} = {\mathrm{\mu }}_{\mathrm{h}} \mathrm{E}$

$\therefore$                  $\frac{\mathrm{E}}{\mathrm{\rho }} = e[n_e.{\mu }_e + n_h{\mu }_h]E$
 
$\Rightarrow$                 $\frac{1}{\mathrm{\rho }} = e[n_e.{\mu }_e + n_h{\mu }_h]$ 

We know, electrical conductivity is the reciprocal of resistivity. 

So, electrical conductivity, 

                 $\mathrm{\sigma } = \frac{1}{\mathrm{\rho }} = \mathrm{e}[{\mathrm{n}}_{\mathrm{e}}.{\mathrm{\mu }}_{\mathrm{e}} + {\mathrm{n}}_{\mathrm{h}}{\mathrm{\mu }}_{\mathrm{h}}]$ 

For a p-type semiconductor, 
n_h >> n_e  and n_h = N_A where, N_A is the number density of acceptor atoms. 

Hence, conductivity of a p-type semiconductor is given by, 
                         ${\mathrm{\sigma }}_{\mathrm{p}} = \mathrm{e} {\mathrm{N}}_{\mathrm{A}} {\mathrm{\mu }}_{\mathrm{h}}$ 
is the required result.