1.
DIAMOND STRUCTURE
Diamond - well-known form of crystalline carbon (C)
Crystalline structure of diamond – also adopted by many
semiconducting materials eg. Si, Ge, GaAs, …
In general, crystal structure (Unit 3!) consists of :
Lattice
+
Basis
Lattice
-
geometrical array of points
Basis
-
arrangement of atoms
associated with each lattice
point
Diamond structure consists of:
fcc lattice
+
basis of 2 atoms, with basis
vectors [0,0,0] and
[1/4,1/4,1/4]
A key feature of diamond structure - strong tetrahedral
bonding i.e. each C atom is bonded to 4 nearest
neighbours.
Important when considering effects of impurities on
electrical properties of semiconductors.
109
Important features of diamond structure:
(1) each C atom bonded to 4 neighbours, by
strongly directional covalent bonds.
(2) bonds are all equivalent, so from symmetry
considerations make equal angles with one
another (109) .
(3) this leads to a tetrahedral bonding arrangement.
Each C atom at apex of a tetrahedron.
(4) bonding arrangement gives rise to a facecentred-cubic (fcc) lattice structure
2-dimensional representation
2.
SEMICONDUCTORS
Good understanding of electrical and optical behaviour
of semiconductors
Requires knowledge of how electrons behave in
semiconductors
For solids in general (see Unit 3):
energies of electrons lie in regions or bands. Bands
separated by forbidden regions or energy gaps.
[Consequence of solving Schrodinger Eq’n in Periodic
Potential]
Metal
Insulator
energy
energy gap
Partially filled band
 electrical conduction
Bands totally filled,
or empty
 insulators (T = 0K)
What happens to insulator if T > 0 ?
Energy, E
Conduction
Band (CB)
Energy Gap
Eg
Valence
Band (VB)
If g “not too large”, some electrons from VB are excited
into CB.
Vacancies or “holes” are created in VB.
Both electrons in CB and (positive) holes in VB
contribute to electrical conduction (in applied E-field)
 semiconductor
For more quantitative description, need to calculate
nc(T) and pv(T)
nc(T)
-
concentration of electrons in CB
(number per unit volume)
pv(T)
-
concentration of holes in VB
CALCULATION of nc(T)
EE+ dE
ECB
Ec
Eg
Ev
VB
Need to know:
1.Probability P(E) that an electron occupies state of
energy 
2. Density of states (energy levels) D() in CB
D()d = Number of states (per unit volume)
between  and  + d
Can then write that:
Number of electrons (per unit volume) n() with
energies between  and  + d is
n() = P()D()d
Integrate over entire CB to obtain nc


nc =  P()D()d
c
P() is given by the Fermi-Dirac function
1
EE
1  exp{
}
kT
[F = Fermi level, T = temperature]
[Unit 2]
Where is F? Must be above v but below c – i.e.
somewhere in gap.
P() =
F
If we assume that D() is free-electron-like near band
edge i.e.   k2 (k = wavevector), then
D() = A( E  E )
c
1
2
*3
A=
( 2m )
 
1
2
e
2
3
[me* - effective mass of an electron in the semiconductor]
[Unit 2]
AE  E 
n 
dE

E

E


1  exp

 kT 
1
2


c
c
c
If  – F > 2kT,
F
E  E 
exp 
 >> 1
kT


then
F
n   AE  E

1
2


c
c
c
Make substitution
x

  E  E 
exp
dE
kT 

F
EE
kT
c
  E  E 
n  AkT  exp
 x e dx
kT 

3
2


c
F
c
0

1
2
 x e dx 
x

0
*3
A=
( 2m )
 
e
2
3
1
2
2
1
2
[standard result]
1
2
x
3
2
 m kT 
 (E  E ) 
nc = 2
 exp

kT 
 2 

*
e
c
F
2
Often write as:
 (E  E ) 
nc = Nc exp

kT 

c
F
where
 m kT 
Nc = 2

2




*
3
2
e
2
effective density of states at conduction band edge
This is an important result!
Hole Concentration pv(T)
Perform very similar calculation to find pv.
1.
Pe() - probability of finding electron at energy 
Ph() - probability of finding hole at energy 
For given level  in valence band
Pe(E) + Ph() = 1
So
Ph() = 1 - Pe()
1
EE
1  exp{
}
kT
= 1 -
F
2.
D() - density of states in valence band
As before, assume free-electron-like.
So
[mh*
*3
1
2
( 2m )
 
- effective mass of a hole in the semiconductor]
D() = B ( E  E )
V
1
2
B=
h
2
3
Integrate over entire valence band to obtain pv
Ev
pv= Ph()D(
-
EF –  >  2kT
If
(F lies above valence band)
E  E 
 (E  E) 
exp

exp


  1
kT 
 kT 

F

P E   1 
h
F
1
  E  E 
1  exp

kT 

F
  E  E 
1  [1  exp
]
kT


(using Binomial expansion)
F
  E  E 
= exp

kT 

F

  E  E 
p   BE  E  exp
d
kT 

1
2
v
v

y
Make substitution

F
v
E
 E
kT
v
  E  E 
p  BkT  exp
 y e dy
kT 

3
2

F
1
2
y
v
0

1
2
1
2
 y e dy =
0
y

2
*3
as before, and
B=
( 2m )
 
h
2
3
1
2
3
2
 m kT 
 (E  E
exp
pv = 2


kT
 2 

*
h
F
V
2
)


Often write as
 (E  E
pv = Nv exp
kT

F
V
)


where
 m kT 
Nv = 2

 2 
*
3
2
h
2
effective density of states at valence band edge
This is also an important result!
Graphical Representation
Ph() = [1 – Pe()]
Pe()
D() = B(v – )1/2
D() = A( – c)1/2
Valence
Band
Conduction
Band
p(E) = Ph(E)D(E)
n(E) = Pe(E)D(E)
v
F
c
F – somewhere in energy gap [can show this later].
n   P E DE dE


c
e
c
v
p   P E DE dE
v

h
Pe()D() falls off rapidly as  increases above c.
So majority of electrons in conduction band occupy
levels near band edge c.
Similarly, Ph()D() decreases rapidly as  decreases
below v  hence most holes in valence band occupy
levels near v.
E
Law of Mass Action
Obtained expressions for nc and pv.
 (E  E ) 
nc = Nc exp

kT 

c
F
 (E  E ) 
pv = Nv exp

kT 

F
Note that
v
 (E  E ) 
ncpv = NcNv exp

kT


c
Energy gap
v
g = c - v
So
 E 
ncpv = NcNv exp 
 kT 
g
Law of mass action
True for ALL semiconductors, whether pure (intrinsic) or
not.
2.1
Intrinsic Semiconductors
2.1.1
Intrinsic Concentration
No impurities  pure or intrinsic semiconductor

conduction
band
g
valence
band
T>0
So for an intrinsic semiconductor
nc = pv = ni
[ni – intrinsic concentration]
Law of mass action:
 E 
ncpv = NcNv exp 
 kT 
g
 m kT 
Nc = 2 

 2 
*
e
2
 m kT 
Nv = 2

 2 
*
h
2
So
3
2
 E 
ncpv = ni2 = NcNv exp 
 kT 
g
3
2
 E 
ni = ( N N ) exp

2
kT


1
2
c
g
v
or
3
2
 E 
 kT 
ni = 2(m m ) 
 exp

 2 
 2kT 
*
*
e
h
3
4
g
2
Exponential term dominates over kT  term.
3
2
Note that law of mass action is often written as
ncpv = ni2(T)
2.1.2
Fermi Level Position

Intrinsic semiconductor
nc = pv
3
2
 m kT 
 (E  E
exp
Previously, obtained nc = 2


kT
 2 

*
e
c
F
2
)


3
2
 m kT 
 (E  E ) 
pv = 2
exp



2


kT 



*
h
F
v
2
Setting expressions for nc and pv equal to each other, and
simplifying:
m 
(E  E ) 3
E 
 kT ln  
2
4
m 
*
c
v
h
F
*
e
i.e. Fermi level F is in middle of band gap (at T=0)
c
f
v
Note that position of F rises linearly as T is increased
[mh* > me*, so ln term is positive].
Typically though, var’n is  10-4 eV/K
[c.f. Eg =1.1eV for Si, 0.67 eV for Ge]
So can usually neglect variation with T.
2.1.3
Conductivity
vd
J
n – concentration of charge carriers
vd – mean drift velocity of carriers
Applied elelctric field E  charge carriers drift with
mean velocity vd  gives rise to current density J in the
material.
J = nvde
[e = electronic charge]
Introduce concept of carrier mobility 
Define  as

v
E
d
 is a measure of ease with which carriers transport
charge in applied E-field.
Hence
J = neE
c.f. Ohm’s law
J = E

 = ne
In a semiconductor, both electrons and holes carry
charge.
  = ncee + pveh
e – electron mobility, h – hole mobility]
h < e
Usually find that
For intrinsic semiconductors
3
2
 E 
 kT 
nc = pv = ni = 2m m  
exp



2
kT
 2 


*
*
e
h
3
4
g
2
So
3
2
 E 
 kT 
 = nie(e + h) = 2e   m m  
 exp

 2 
 2kT 
e
Note:
h
*
*
e
h
3
4
g
2
2 terms involving temperature T
Exponential term varies much more rapidly with T than
the T3/2 term.
So to a good approximation
 E 
  exp

 2kT 
for an intrinsic semiconductor.

g
[In fact, often find that   T-3/2 anyway]