Conductivity of Extrinsic Semiconductor

In an extrinsic semiconductors, there are two types of materials n type and p type. Let us obtain the expressions for the conductivity of n type and p type materials.
1.1 Conductivity of n Type Material
It is known that in n type material, the free electrons are majority carriers while the holes are minority carriers.
Let              nn = Concentration of free electrons in n type
pn = Concentration of holes in n type
ND = Concentration of donor atoms
Note : In the symbol, main letter n or p indicates concentration of type of charge carrier electron or whole while the suffix indicates the type of material i.e. n type or p type. Thus nn indicates electron concentration in n type material while np indicates electron concentration in p type material and so on.
From the basic equation of conductivity, the conductivity of n type material can be expressed as,
σn = (nn μn + pn μp) e                        ............(1)
But pn << nn   as holes are in minority hence,
σn ≈ nn μn e                                 ................. (2)
The number of free electrons is dominantly controlled by donor atoms added than the thermal generation at room temperature. Hence concentration of donor atoms ND added can be approximately assumed to be equal to the concentration of free electrons nn in n type materials.
Thus as ND >> ni we can write,
nn  ≈ ND                       ................ (3)
...                             σn ≈ ND μn e                  ..................(4)
1.2 Conductivity of p Type Material
For a type material, holes are in majority and electrons are in minority.
Let      np = Concentration of free electrons in p type
pp = Concentration of holes in p type
NA = Concentration of acceptor atoms
Thus the conductivity of p type material can be expressed as,
σp = (np μn + pp μp) e                   ................ (5)
But np << pp as free electrons are in minority hence,
σp  ≈ pp μp e                                     ........... (6)
The number of holes is dominantly controlled by added acceptor impurity than the thermal generation. Each added impurity atom creats a hole hence NA >> ni. Thus all the holes generated pp can be approximately assumed to be equal to the concentration of acceptor NA. Thus,
pp  ≈ NA                                             ............... (7)
...                    σp ≈ NA μp e                                        ............. (8)