3.4. TRANSPORT UNDER AN ELECTRIC FIELD 109
You may assume the following values:
μn(Si) = 1000 cm^2 /V·s
μp(Si) = 350 cm^2 /V·s
μn(GaAs) = 8000 cm^2 /V·s
μp(GaAs) = 400 cm^2 /V·s
In the doped semiconductors, the electron density is (50 % of 1017 cm−^3 )
nn 0 =5× 1016 cm−^3
and hole density can be found from
pn 0 =
n^2 i
nn 0
For silicon we have
pn 0 =
2. 25 × 1020
5 × 1016
=4. 5 × 103 cm−^3
which is negligible for the conductivity calculation.
The conductivity is
σn=nn 0 eμn+pn 0 eμp=8 (Ωcm)−^1
In the case of undoped silicon we get(n=ni=p=1. 5 × 1010 cm−^3 )
σundoped=nieμn+pieμp=3. 24 × 10 −^6 (Ω cm)−^1
For GaAs we get
σn=5× 1016 × 1. 6 × 10 −^19 ×8000 = 64 (Ω cm)−^1
For undoped GaAs we get (ni=1. 84 × 106 cm−^3 )
σundoped=nieμn+pieμp=2. 47 × 10 −^9 (Ω cm)−^1
You can see the very large difference in the conductivities of the doped and undoped
samples. Also there is a large difference between GaAs and Si.
Example 3.6Consider a semiconductor in equilibrium in which the position of the Fermi
level can be placed anywherewithinthebandgap.
What is the maximum and minimum conductivity for Si and GaAs at 300 K? You can use
the data given in the problem above.
The maximum carrier density occurs when the Fermi level coincides with the conduction
bandedge ifNc>Nvor with the valence bandedge ifNv>Nc.IfNc>Nv;the
Boltzmann approximation gives
nmax=Nc
