3. Intrinsic Semiconductor
A perfect semiconductor crystal with no
impurities or lattice defects is called an Intrinsic
semiconductor.
In such material there are no charge carriers at
0ºK, since the valence band is filled with
electrons and the conduction band is empty.
5. Intrinsic Semiconductor
If we denote the generation rate of EHPs
as and the recombination rate
as equilibrium requires that:
)
(T
gi
)
( 3
s
cm
EHP
ri
i
i g
r =
Each of these rates is temperature dependent.
For example, increases when the temperature
is raised.
)
( 3
s
cm
EHP
gi
i
i
r
r
i g
n
p
n
r =
=
= 2
0
0 α
α
6. Donors and Acceptors
The conductivity of a pure
(intrinsic) s/c is low due to
the low number of free
carriers.
For an intrinsic semiconductor
n = p = ni
n = concentration of electrons per unit volume
p = concentration of holes per unit volume
ni = the intrinsic carrier concentration of the semiconductor under consideration.
The number of carriers are
generated by thermally or
electromagnetic radiation
for a pure s/c.
7. n.p = ni
2
n = p
number of e-’s in CB = number of holes in VB
This is due to the fact that when an e- makes
a transition to the CB, it leaves a hole behind
in VB. We have a bipolar (two carrier)
conduction and the number of holes and e- ‘s
are equal.
n.p = ni
2
This equation is called as mass-action law.
8. The intrinsic carrier concentration ni depends on;
the semiconductor material, and
the temperature.
n.p = ni
2
For silicon at 300 K, ni has a value of 1.4 x 1010 cm-3.
Clearly , equation (n = p = ni) can be written as
n.p = ni
2
This equation is valid for extrinsic as well as
intrinsic material.
9. To increase the conductivity, one can
dope pure s/c with atoms from column lll
or V of periodic table. This process is
called as doping and the added atoms
are called as dopants impurities.
What is doping and dopants impurities ?
There are two types of doped or extrinsic s/c’s;
n-type
p-type
Addition of different atoms modify the conductivity of
the intrinsic semiconductor.
10. p-type doped semiconductor
Si + Column lll impurity atoms
Boron (B)
has three valance
e-’ s
Have
four
valance
e-’s
Si
Si Si
Si
B
Electron
Hole
Bond
with
missing
electron
Normal
bond with
two
electrons
Boron bonding in Silicon
Boron sits on a lattice side
p >> n
11. Boron(column III) atoms have three valance
electrons, there is a deficiency of electron or
missing electron to complete the outer shell.
This means that each added or doped boron
atom introduces a single hole in the crystal.
There are two ways of producing hole
1) Promote e-’s from VB to CB,
2) Add column lll impurities to the s/c.
12. Energy Diagram for a p-type s/c
Ec = CB edge energy level
Ev = VB edge energy level
EA= Acceptor energ level
Eg
CB
VB
acceptor
(Column lll) atoms
The energy gap is forbidden only for pure material, i.e. Intrinsic material.
Electron
Hole
13. The impurity atoms from column lll occupy at an energy level
within Eg . These levels can be
1. Shallow levels which is close to the band edge,
2. Deep levels which lies almost at the mid of the band gap.
If the EA level is shallow i.e. close to the VB edge, each added
boron atom accepts an e- from VB and have a full
configuration of e-’s at the outer shell.
These atoms are called as acceptor atoms since they accept
an e- from VB to complete its bonding. So each acceptor
atom gives rise a hole in VB.
The current is mostly due to holes since the number of holes are
made greater than e-’s.
p-type semiconductor
14. Holes = p = majority carriers
Electrons = n = minority carriers
Majority and minority carriers in a p-type semiconductor
t2
t1
t3
Electric field direction
Holes movement as a
function of applied
electric field
Hole movement direction
Electron movement
direction
15. Ec
Ev
Ea
Eg
Electron
Hole
Shallow acceptor in silicon
Si
Si Si
Si
P
Electron
Weakly bound
electron
Normal
bond with
two
electrons
Phosporus bonding in silicon
n-type semiconductor
16. Extra e- of column V atom is weakly attached to its host atom
n-type semiconductor
Si
Si
Si
As
Si
Si + column V (with five valance e- )
ionized (+ve)
donor centre
Ec
Ev
ED = Donor energy level (shallow)
Band gap is 1.1 eV for silicon
Hole
Electron
Eg
n - type semiconductor
17. Ec
Ev
Ed
Eg
Electron
Valance band
Conduction band
Band gap is 1.1 eV for silicon
Shallow donor in silicon
Donor and acceptor charge states
Electron
Hole
Neutral donor
centre
İonized (+ve)
donor centre
Neutral
acceptor
centre
İonized (-ve)
acceptor
centre
Ec
Ev
Ea
Ec
Ev
Ea
18. np , pn
n-type , n >> p ; n is the majority carrier
concentration nn
p is the minority carrier
concentration pn
p-type , p >> n ; p is the majority carrier
concentration pp
n is the minority carrier
concentration np
np pn
Type of semiconductor
22. Extrinsic Semiconductor
We can calculate the binding energy by
using the Bohr model results, considering
the loosely bound electron as ranging about
the tightly bound “core” electrons in a
hydrogen-like orbit.
r
K
n
h
K
mq
E ε
ε
π 0
2
2
4
4
,
1
;
2
=
=
=
Example 3-3:
Calculate the approximate donor binding
energy for Ge(εr=16, mn
*=0.12m0).
24. calculation
Calculate the hole and electron densities in a piece of silicon that has
been doped with 5 x 1016 acceptor atoms per cm3 .
ni = 1.4 x 1010 cm-3 ( at room temperature)
Undoped
n = p = ni
p-type ; p >> n
n.p = ni
2 NA = 5 x 1016 p = NA = 5 x 1016 cm-3
3
3
16
2
3
10
2
10
9
.
3
10
5
)
10
4
.
1
(
x
cm
x
cm
x
p
n
n i
=
=
= −
−
electrons per cm3
p >> ni and n << ni in a p-type material. The more holes you put in
the less e-’s you have and vice versa.
25. Fermi level , EF
This is a reference energy level at which the probability of occupation
by an electron is ½.
Since Ef is a reference level therefore it can appear anywhere in the
energy level diagram of a S/C .
Fermi energy level is not fixed.
Occupation probability of an electron and hole can be determined
by Fermi-Dirac distribution function, FFD ;
EF = Fermi energy level
kB = Boltzman constant
T = Temperature
)
exp(
1
1
T
k
E
E
F
B
F
FD
−
+
=
26. E is the energy level under investigation.
FFD determines the probability of the energy level E being
occupied by electron.
determines the probability of not finding an
electron at an energy level E; the probability of finding a
hole .
)
exp(
1
1
T
k
E
E
F
B
F
FD
−
+
=
FD
FD
F
f
f
E
E
if
−
=
+
=
→
=
1
2
1
0
exp
1
1
Fermi level , EF
27. Fermi level , EF
kT
f
E
E
e
E
f )
(
1
1
)
( −
+
=
k : Boltzmann’s constant
f(E) : Fermi-Dirac distribution function
Ef : Fermi level
28. Fermi level , EF
28
2
1
1
1
1
1
1
)
( )
( =
+
=
+
= −
kT
f
E
f
E
e
E
f f
Ef
f(E)
1
1/2
E
T=0ºK
T1>0ºK
T2>T1
29. Fermi level , EF
Ev
Ec
Ef
E
f(E)
0
1/2
1
≈
≈
f(Ec) f(Ec)
[1-f(Ec)]
Intrinsic
n-type
p-type
30. Electron and Hole Concentrations at Equilibrium
30
∫
∞
=
C
E
dE
E
N
E
f
n )
(
)
(
0
The concentration of electrons in the conduction band is
N(E)dE : is the density of states (cm-3) in the energy range dE.
The subscript 0 used with the electron and hole concentration
symbols (n0, p0) indicates equilibrium conditions.
The result of the integration is the same as that obtained if we
represent all of the distributed electron states in the conduction
band edge EC.
)
(
0 C
C E
f
N
n =
31. Electron and Hole Concentrations at Equilibrium
31
EC
EV
Ef
E
Holes
Electrons
Intrinsic
n-type
p-type
N(E)[1-f(E)]
N(E)f(E)
32. Electron and Hole Concentrations at Equilibrium
32
kT
F
E
C
E
kT
F
E
C
E e
e
E
f C
)
(
)
(
1
1
)
(
−
−
− ≅
+
=
kT
F
E
C
E
e
N
n C
)
(
0
−
−
=
2
3
)
2
(
2 2
*
h
kT
m
N n
C
π
=
33. Electron and Hole Concentrations at Equilibrium
33
)]
(
1
[
0 V
V E
f
N
p −
=
kT
V
E
F
E
kT
F
E
V
E e
e
E
f V
)
(
)
(
1
1
1
)
(
1
−
−
− ≅
+
−
=
−
kT
V
E
F
E
e
N
p V
)
(
0
−
−
=
2
3
)
2
(
2 2
*
h
kT
m
N
p
V
π
=
34. Electron and Hole Concentrations at Equilibrium
34
kT
v
E
i
E
e
N
p V
i
)
( −
−
=
kT
i
E
c
E
e
N
n C
i
)
( −
−
=
kT
g
E
kT
v
E
c
E
e
N
N
e
N
N
p
n v
c
v
c
−
−
−
=
=
)
(
0
0
kT
g
E
e
N
N
p
n v
c
i
i
−
= kT
g
E
e
N
N
n v
c
i
2
−
=
2
0
0 i
n
p
n =
kT
F
E
i
E
e
n
p i
)
(
0
−
=
kT
i
E
F
E
e
n
n i
)
(
0
−
=
35. Electron and Hole Concentrations at Equilibrium
35
Example 3-4:
A Si sample is doped with 1017 As Atom/cm3. What is
the equilibrium hole concentration p0 at 300°K? Where
is EF relative to Ei?
With: kT=0.0259eV; ni (for Si)=1.5×1010 cm-3;
36. Electron and Hole Concentrations at Equilibrium
36
3
3
17
20
0
2
0 10
25
.
2
10
10
25
.
2 −
×
=
×
=
= cm
n
n
p i
Answer:
Since Nd»ni, we can approximate n0=Nd and
kT
i
E
F
E
e
n
n i
)
(
0
−
=
eV
n
n
kT
E
E
i
i
F 407
.
0
10
5
.
1
10
ln
0259
.
0
ln 10
17
0
=
×
=
=
−
37. Electron and Hole Concentrations at Equilibrium
37
Answer (Continue) :
Ev
Ec
EF
Ei
1.1eV
0.407eV
38. Carrier concentration equations
The number density, i.e., the number of electrons
available for conduction in CB is
3/ 2
*
2
2
2 exp ( )
exp ( ) exp( )
p F V
F V i F
V i
m kT E E
p
h kT
E E E E
p N p n
kT kT
π
−
= −
− −
=
− =
3/ 2
*
2
2
2 exp ( )
exp ( ) exp( )
n C F
C F F i
C i
m kT E E
n
h kT
E E E E
n N n n
kT kT
π
−
= −
− −
=
− =
The number density, i.e., the number of holes available
for conduction in VB is
39. How does the position of the Fermi Level change with
(a) increasing donor concentration, and
(b) increasing acceptor concentration ?
Worked example
(a) We shall use equation
İf n is increasing then the quantity EC-EF must be decreasing
i.e. as the donor concentration goes up the Fermi level moves
towards the conduction band edge Ec.
exp ( )
C F
C
E E
n N
kT
−
= −
40. Worked example
But the carrier density equations such as;
aren’t valid for all doping concentrations! As the fermi-level comes
to within about 3kT of either band edge the equations are no longer
valid, because they were derived by assuming the simpler Maxwell
Boltzmann statics rather than the proper Fermi-Dirac statistic.
−
=
−
−
=
kT
E
E
n
p
and
kT
E
E
h
kT
m
n
F
i
i
F
c
n
exp
exp
2
2
2
3
2
*
π
42. Worked example
(b) Considering the density of holes in valence band;
It is seen that as the acceptor concentration increases, Fermi-level
moves towards the valance band edge. These results will be used in
the construction of device (energy) band diagrams.
−
−
=
kT
E
E
N
p V
F
v exp
43. Donors and acceptors both present
Both donors and acceptors present in a s/c in
general. However one will outnumber the other
one.
In an n-type material the number of donor
concentration is significantly greater than that
of the acceptor concentration.
Similarly, in a p-type material the number of
acceptor concentration is significantly greater
than that of the donor concentration.
A p-type material can be converted to an n-
type material or vice versa by means of adding
proper type of dopant atoms. This is in fact how
p-n junction diodes are actually fabricated.
44. Donors and acceptor both present
n D A
n N N
= −
2) Similarly, when the number of shallow acceptor concentration is signicantly
greater than the shallow donor concentration in a piece of a s/c, it can be
considered as a p-type s/c and
• In general, both donors and acceptors are present in a piece of a
semiconductor although one will outnumber the other one.
• The impurities are incorporated unintentionally during the growth of the
semiconductor crystal causing both types of impurities being present in a piece
of a semiconductor.
• How do we handle such a piece of s/c?
1) Assume that the shallow donor concentration is significantly greater than
that of the shallow acceptor concentration. In this case the material behaves as
an n-type material and
45. For the case NA>ND , i.e. for p-type material
Donors and acceptor both present
2
2
2 2
.
0
0 ( ) 0
p p i
p A D P p D p A
i
p p D A p D A p i
p
n p n
n N N p p N n N
n
p p N N p N N p n
p
− +
=
+ = + ⇒ + − − =
⋅ + − − = ⇒ + − − =
46. Donors and acceptor both present
( )
2
2 2
1,2
1
2 2
2
2
4
( ) 0 , solving for ;
2
1
4
2
p D A p i p
p A D A D i
i
p
p
b b ac
p N N p n p x
a
p N N N N n
n
n
p
− −
+ − −
= =
= − + − +
=
majority
minority
47. Donors and acceptor both present
For the case ND>NA , i.e. n-type material
( )
2
2
2
2 2
2
1,2
1
2 2
2
2
.
0
0 ( ) 0
4
solving for n ;
2
1
4
2
i
n n i n
n
n A D n n A n D
i
n n A D n A D n i
n
n
n D A D A i
i
n
n
n
n p n p
n
n N N P n N p N
n
n n N N n N N n n
n
b b ac
x
a
n N N N N n
n
p
n
= ⇒ =
+ = + ⇒ + − − =
⋅ + − − = ⇒ + − − =
− −
=
= − + − +
=
