
In all practical calculations, the
effective density of state
N_{eff } is used instead of
D(E). N_{eff } is just a
number, lets see how we can this from the free electron gas model. 


Lets just look at electrons in the conduction
band; for holes everything is symmetrical as usual. We want to get an idea
about the distribution of the electrons in the conduction band on the available
energy states (given by D(E)). 

We have in
fulll generality for
n(E) = density of electrons in the conduction band 


n(E') 
= 
E*
ó
õ
E_{c} 
D(E') · f(E',T) ·
dE' 




With f(E', E_{F},
T) = FermiDirac distribution, and the integration running from the
bottom of the conduction band to the top of the band at E*, (or
to infinity in practice). The dash at the symbol for
the energy, E' just clarifies that the zero point of the energy scale is
not yet the bottom of the conduction band. 


Of course we use the Boltzmann approximation for
the tail end of theFermi distribution and obtain 


n(E') 
= 
¥
ó
õ
E_{c} 
D(E') · exp – 
(E' – E_{F})
kT) 
· dE 



If we now take the bottom of the
conduction band as the zero point of the energy scale for
D(E), we have E = E' 
E_{C} with E_{C} = energy of the
conduction band. Insertion in the formula from above gives 


n(E') 
= 
exp – 
– E_{C} – E_{F}
kT 
· 
¥
ó
õ
0 
D(E) · exp – 
E(k)
kT) 
· dE 



Inserting the density of states from
above with the abbreviation 


N_{0} 
= 
1
2p^{2} 
æ
ç
è 
2m
^{2} 
ö
÷
ø 
3/2 




gives a final formula for computing 


n(E) 
= 
exp – 
– E_{C} – E_{F}
kT 
· N_{0} · 
¥
ó
õ
0 
E^{1/2} · exp– 
E
kT 
· dE 




The definite integral (E^{1/2} ·
exp–[E/kT)dE can be found in integral tables; its
value is (1/2) · (p^{1/2}) ·
(kT)^{3/2} 




Insertion, switiching from
to
h, and some juggling of the terms gives the final result defining the
effective density of states N_{eff } 


n(E) 
= 2· 
æ
ç
è 
2p · m
· kT

ö
÷
ø 
3/2 
· exp – 
E_{C} – E_{F}

:= 
N_{eff
} · exp – 
E_{C} – E_{F}

h^{2} 
kT 
kT 



We now have the final result 


N_{eff } 
= 2 · 
æ
ç
è 
2p · m
· kT

ö
÷
ø 
3/2 
h^{2} 




And this is the formula we
used in the
backbone. 

What about numbers
and the dimension? We have 


[N_{eff }] 
= 
kg^{3/2} · eV^{3/2} · eV^{–3}
· s^{–3} = kg^{3/2} ·
eV^{–3/2} · s^{–3} 




From before we
have 1 kg^{3/2} = 1,559 · 10^{28} eV^{3/2}
· m^{–3} · s^{3}. Inserting this finally
gives (for masses given in kg, length in m and energies in
eV): 


N_{eff } 
= 
4.59 · 10^{15}
· T^{3/2} cm^{–3} 



= 
2.384 · 10^{19}
cm^{–3} 

T = 300 K 

= 
2.384 · 10^{25}
m^{–3} 





And those are very useful numbers  in
particular, becasue they are quite close to the "real" (i.e.
measured) values for Si. 


