Figure: A broken covalent bond resulting to a hole and a free electron in a semiconductor crystal.

The hole may be filled by an electron from another valence bond. The hole thus moves in a direction opposite to that of the electron and we have a mechanism for the conduction of electricity without involving free electrons. The hole behaves like a charge carrier which carries a positive charge equal in magnitude to the electronic charge. Quantum mechanical studies support the behavior of the hole as a free charge carrier.

For an intrinsic semiconductor, the number of holes (p) is equal to the number of free electrons (n).

i.e.,         $p=n=n_i$

where, ni is the intrinsic concentration.

Extrinsic Semiconductor

If a small percentage of trivalent or pentavalent atoms is added to an intrinsic semiconductor, an extrinsic semiconductor is formed.

If the impurity added is pentavalent, n-type semiconductor is formed. The impurity of atoms replaces some of the germanium or silicon atoms. Four of the five valence electrons will form covalent bonds and the fifth will be unbound and available as a carrier of electric current. The energy required to detach the fifth electron from the pentavalent impurity is 0.01 eV for germanium and 0.05 eV for silicon. As these impurities donate excess negative charge carrier (i.e, electron) they are known as n-type impurities or donors. Phosphorus, antimony, and arsenic are important donors. The donor energy level is 0.01 eV below the conduction band for germanium and for silicon the donor energy level is 0.05 eV below the conduction band.

Addition of a trivalent impurity, e.g., boron, gallium or indium to an intrinsic semiconductor result to a p-type semiconductor. As these impurities have 3 valence electrons only three of the covalent bonds will be filled and a hole is created which can accept electrons. So, these impurities are known as acceptors or p-type impurities. The acceptor energy level is 0.01 eV above the P-valence bond for germanium and 0.05eV  above the valence bond for silicon.

 Electrical conductivity and current d Current density,

Current density, $J=(n\mu_n+p\mu_p)q\varepsilon=\sigma\varepsilon——-(E1-1)$

where,

n = magnitude of free electron

p = magnitude of hole concentration

$\mu_n$ = mobility of electrons

$\mu_p$ = mobility of holes

q = electronic charge

e = electric field applied

 $\sigma$= electrical conductivity.

Law of mass action

$$np=n_i^2—–(E1-2)$$

Charge density

If $N_D$ is the concentration of the donor atoms $N_A$ that of acceptor atoms, then.

$$N_D+p=N_A+n——–(E1-3)$$

For N-type semiconductor, $N_A=0$

And $n_n>>p_n$( the subscript n representing n-type)

From eq(E1-3)

$$n_n\approx N_D——-(E1-4)$$

From eq(E1-2)

$$n_np_n=n_i^2$$

$$p_n=\frac{n_i^2}{N_D}$$

Similarly, for a p-type semiconductor,

$$n_p=\frac{n_i^2}{N_A}$$