Semiconductor Equilibrium
Equilibrium
No external forces (voltages, electric fields, temp.gradients)
First
Consider pure crystal
Then
Consider addition of dopants
Semiconductor Equilibrium
Charge carriers
Electrons in conductance
n(E) = gc(E)fF(E)
n(E) - prob. dens. of electrons
gc(E) - conductance density
fF(E) - Fermi-Dirac prob.
function
Holes in valence
p(E) = gV(E)(1 - fF(E))
p(E) - prob. dens. of holes
gv(E) - valence density
fF(E) - Fermi-Dirac prob.
function
Semiconductor Equilibrium
Charge carriers(cont.)
n0 
 g (E) f
c
F
(E)dE
(E C  E F )
n 0  N c exp



kT
2m*n kT 3 / 2
NC  2

2
 h



p0 
 g (E)(1 f
(E))dE

(E F  E v )
p0  N v exp



kT
v
F



2m *p kT 3 / 2
N v  2

2
 h

Semiconductor Equilibrium
Charge carriers(cont.)
Example
Find the probability that a state in the conduction band is occupied
and calculate the electron concentration in silicon at T = 300K.
Assume Fermi energy is .25 eV below the conductance band
(E c  E F )
5
f F (E)  exp

exp(0.25/.0259
)

6.4310



kT


(E  E F ) 
19
5
15
3
n 0  N c (E)exp c

(2.8
10
)(6.4310
)

1.8
10
cm



kT
Note low probability per state but large number of states implies
reasonable concentration of electrons
Semiconductor Equilibrium
Charge carriers(cont.)
For intrinsic semiconductor, concentration of electrons in
conductance band is equal to holes in the valence band. Thus,
E g 
(E  E v )
ni2  ni pi  N c N v exp c

N
N
exp



 kT
 c v
 kT 

Semiconductor Equilibrium
Dopant Atoms (n-type semiconductor)
Phosphorous has 5 valence electrons
Energy-band diagram
Semiconductor Equilibrium
Dopant Atoms (p-type semiconductor)
Boron has 3 valence electrons
Energy-band diagram
Semiconductor Equilibrium
The Extrinsic Semiconductor
n-type
p-type
Semiconductor Equilibrium
The Extrinsic Semiconductor
Example
Consider doped silicon at 300K. Assume that the
Fermi enery is .25 eV below the conduction band
and .87 eV above the valence band. Calculate the
thermal equilibrium concentration of e’s and holes
(E c  E F ) 
19
15
3
n 0  N c (E)exp

(2.8
10
)exp(0.025/0.0259
)

1.8
10
cm



kT


(E F  E v ) 
p0  N v (E)exp
 (1.041019 )exp(0.087/0.0259)  2.7 104 cm3



kT
Semiconductor Equilibrium
The Extrinsic Semiconductor
The n0p0 product
E g  2
(E c  E F ) (E F  EV ) 
n0 p0  N c N v exp
exp
 N c N v exp
 ni



 

kT
kT
 kT 

That is, the product of n0 and p0 is a constant for a given
semiconductor at a given temperature.
Semiconductor Equilibrium
Statistics of donors and acceptors
Ratio of electrons in donor state total electrons
nd

no  nd
1
(E c  E d ) 
Nc
1
exp 



2N d
kT
Example

Consider phosporous doped silicon at T = 300K and
at a concentration of Nd = 1016 cm-3. Find the
fraction of electrons in the donor state.
nd

no  nd

1
0.045 
2.8 10
exp

0.0259 

2 1016
19
1
 .41%
Semiconductor Equilibrium
Compensated semiconductors
Formed by adding both donor and acceptor impurities
in the same region
Energy-band diagram
Semiconductor Equilibrium
Compensated semiconductors (cont.)
With the assumption of charge neutrality, we can derive
N d  N a 2
(N d  N a )
2
n0 
 
  n i
 2 
2

N a  N d 2
(N a  N d )
2
p0 
 
  ni
 2 
2
Example

Consider a silicon semiconductor at T = 300K in
which Na = 1016 cm-3 and Nd = 3 1015 cm-3.
Assume ni = 1.5 1010 cm-3 and find p0 and n0.
1016  31015 2
(1016  31015 )
10 2
15
3
p0 
 
  (1.5 10 )  7 10 cm
2
2



n i2
n0 
 3.21104
p0
Semiconductor Equilibrium
Position of Fermi energy level
As a function of doping levels
As a function of temperature
for a given doping level