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Chapter 3
Introduction to Quantum
Theory of Solids
W.K. Chen
Electrophysics, NCTU
1
Outline
„
„
„
„
„
Allowed and forbidden energy bands
Electrical conduction in solid
Extension to three dimension
Density of state
Statistical mechanics
W.K. Chen
Electrophysics, NCTU
2
3.1 Allowed and forbidden energy bands
„
In one-electron atom
‰ Only discrete values of energy
are allowed
‰ The probability function shows that
the electron is not localized at a
given radius
‰ We can extrapolate these singleatoms results to a crystal and
quantitatively derive the concepts of
allowed and forbidden energy
bands
− mo e 4
En =
(4πε o ) 2 2h 2 n 2
W.K. Chen
Electrophysics, NCTU
3
3.1.1 Formation of energy bands
„
„
Fig. (a) shows the radial probability density function for the lowest
electron energy state of the single, non-interacting hydrogen atom Fig. (b)
shows the radial probability curves for two atoms that are in close proximity
to each other
Due to Pauli exclusion principle, no two electrons may occupy the same
quantum state, the interaction of two electrons give each discrete quantized
energy level splitting into two discrete energy levels
W.K. Chen
Electrophysics, NCTU
4
„
„
„
„
By pushing the numbers of hydrogen type atoms together with a regular
periodic arrangement, the initial quantized energy level will split into a band
of discrete energy levels
The Pauli exclusion principle states that the joining of atoms to form a
system (crystal) does not alter the total number of quantum states
The discrete energy must split into a band of energies in order that each
electron can occupy a distinct quantum state
For a system with 1019 one-electron atoms with a width of allowed energy
band of 1 eV, if the discrete energy states are equidistant, then the energy
levels are separated by 10-19 eV, which is extremely small
W.K. Chen
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5
Example 3.1 Small change of kinetic energy
Consider an electron traveling at a velocity of 107 cm/s. Assume the velocity
increases by a value of 1 cm/s. Find the increase in kinetic energy
1
1
4.56 ×10 −21
2
−31
5 2
− 21
E = mυ = (9.11× 10 )(10 ) = 4.56 ×10 J =
= 2.85 ×10 − 2 eV
−19
2
2
1.6 ×10
1
1
ΔE = mυ 22 − mυ12
2
2
Let υ 2 = υ1 + Δυ
⇒ υ 22 = (υ1 + Δυ ) 2 = υ12 + 2υ1Δυ + (Δυ ) 2
1
m(2υ1Δυ ) = mυ1Δυ
2
ΔE ≈ mυ1Δυ = (9.11× 10 −31 )(105 )(0.01) = 9.11×10 −28 J
⇒ ΔE ≈
9.11×10 − 28
ΔE =
= 5.7 ×10 −9 eV
−19
1.6 ×10
W.K. Chen
Electrophysics, NCTU
6
Discrete energy levels for a hydrogen atom
− mo e 4
En =
(4πε o ) 2 2h 2 n 2
E4
E3
E2
E1 = −13.6eV
E1
= −3.4eV
22
E
E3 = 21 = −1.51eV
3
E
E4 = 21 = −0.85eV
4
ΔE = 0.66 eV
ΔE = 1.89 eV
E2 =
W.K. Chen
„
E1
7
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For a atom contains electrons up through the n=3 energy level, when we
bought these atoms close together from initially far apart
‰
„
ΔE = 10.2 eV
The n=3 energy shell will begin to interact initially, followed by n=2 energy shell
interaction and finally the innermost n=1 shell interaction
If ro is the equilibrium interatomic distance, then at this distance we have
‰
‰
Bands of allowed energies (electrons may occupy)
Bands of forbidden energies
allowed band
forbidden band
W.K. Chen
Electrophysics, NCTU
8
„
For silicon atoms, having 14
electrons,
‰
‰
‰
Isolated silicon atom
„
For silicon crystal
‰
‰
3s states corresponds to n=3 and
l=0 ( containing 2 quantum states)
(n=3, l=0, s=-1/2, +1/2)
3p states corresponds to n=3 and
l=1 ( containing 6 quantum states)
n=3, l=+1, m=1, s=-1/2, +1/2
m=0, s=-1/2, +1/2
m=-1, s=-1/2, +1/2
W.K. Chen
„
„
„
„
2 electrons occupy n=1 energy
level
8 electrons occupy n=2 energy
level
4 remaining valence electrons
occupying n=3 energy level are
involved in chemical reactions
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silicon crystal
9
As the interatomic distance decreases, the 3s and 3p states and
gradually merge together to form broad band
At the equilibrium interatomic distance, the broad band has again
split with 4 quantum states per atom in the lower band and 4
quantum states per atom in the upper band
The upper band is assigned as
conduction band
the lower band is call valence band
Since each Si atom has only 4
valence electrons
‰
At 0 K, electrons are in the lowest
wnergy states
all states in the valence band will be
full, and all states in the conduction
band will be empty
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Electrophysics, NCTU
10
3.1.2 The Kronig-Penney Model
„
„
„
„
Fig. (a) shows potential function of a
single, noninteracting, one-electron atom
Fig. (b) shows potential function for the
case when several atoms are in close
proximity arranged, one-dimensional
array
Fig. (c) is the net potential function in Fig.
(b)
One-dimensional square periodic
potential function is used in K.P model
W.K. Chen
11
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Kronig-Penny Model
⎧0
V ( x) = ⎨
⎩Vo
„
„
0< x<a
−b < x < 0
Block Theorem
The theorem states that all one-electron
wave functions involving periodically
varying potential energy functions must be
of the form
ψ ( x ) = u ( x )e
u ( x) : periodic function with period(a + b)
jkx
e jkx :Q traveling wave
The general time-independent wave function
Ψ ( x, t ) = ψ ( x)φ (t )
= u ( x)e jkx ⋅ e − jωt
W.K. Chen
ω=
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E
h
( E = hω )
12
Time-independent Schrodinger wave equation
∂ 2ψ ( x) 2m
+ 2 ( E − V ( x))ψ ( x) = 0
h
∂x 2
ψ ( x) = u ( x)e jkx
∂ 2ψ ( x) 2mE
Region I :
+ 2 ψ ( x) = 0
h
∂x 2
(0 ≤ x ≤ a )
∂ 2ψ ( x) 2m( E − Vo )
Region II :
+
ψ ( x) = 0
h2
∂x 2
(−b ≤ x ≤ 0)
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13
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Region I:
du ( x)
d 2u1 ( x)
+ 2 jk 1 − (k 2 − α 2 )u1 ( x) = 0
2
dx
dx
α2 =
2mE
h2
u1(x): the amplitude of the wave function in region I
if Vo > E
Region II :
d 2u 2 ( x )
du ( x)
+ 2 jk 2
− (k 2 − β 2 )u2 ( x) = 0
2
dx
dx
β=
u2(x): the amplitude of the wave function in region II
u1 ( x) = Ae j (α − k ) x + Be − j (α + k ) x
for 0 < x < a
u2 ( x) = Ce j ( β − k ) x + Be − j ( β + k ) x
for − b < x < 0
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Electrophysics, NCTU
2m( E − Vo )
= jγ
h2
if E >Vo
β2 =α2 −
2mVo
h2
14
u1 ( x) = Ae j (α − k ) x + Be − j (α + k ) x
for 0 < x < a
u2 ( x) = Ce j ( β − k ) x + Be − j ( β + k ) x
for − b < x < 0
4 unknowns, 4 BCs
ψ
( x) = u ( x)e jkx
The wave function ψ(x) and its derivative ∂ψ/∂x
∂ψ ( x) ∂u ( x)
must be continuous
⇒u(x) and its derivative ∂u/∂x must be continuous ∂x
−
=
∂x
e jkx + jku ( x)e jkx
+
ψ ( x − ) = ψ ( x + ) ⇒ u ( x − )e jkx = u ( x + )e jkx ⇒ u ( x − ) = u ( x − )
∂ψ ( x)
∂ψ ( x)
=
∂x
⇒
x
∂x
−
x+
−
−
+
+
∂u ( x)
∂u ( x)
e jkx + jku ( x − )e jkx =
e jkx + jku ( x + )e jkx
∂x x −
∂x x +
⇒
∂u ( x)
∂u ( x)
=
∂x x −
∂x x +
W.K. Chen
Electrophysics, NCTU
u1 ( x) = Ae j (α − k ) x + Be − j (α + k ) x
for 0 < x < a
u2 ( x) = Ce j ( β − k ) x + Be − j ( β + k ) x
for − b < x < 0
B.C.1 :
u2 (0 − ) = u1 (0 + ) ⇒
B.C.2 :
∂u2
∂x
B.C.3 :
=
0
−
∂u2
∂x
W.K. Chen
∂u1
∂x
=
x=a
∂u2
∂x
A+ B = C + D
⇒ (α − k ) A − (α + k ) B − ( β − k )C + ( β + k ) D = 0
0
+
u1 (a) = u2 (−b)
B.C.4 :
15
Ae j (α − k ) a + Be − j (α + k ) a = Ce − j ( β − k )b + Be + j ( β + k )b
(α − k )e j (α − k ) a + (α + k )e − j (α + k ) a
x =−b
= ( β − k )e − j ( β − k ) b − ( β + k )e + j ( β + k ) b
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From the above 4 eq, derived from 4 BCs, we obtain the parameter k in
relation of the total energy E (through the parameter α)and the potential
function Vo (through the parameter β)
Kronig-Penny Model
− (α 2 + β 2 )
(sin α a )(sin β b) + (cos α a )(cos β b) = cos k (a + b)
2αβ
⎧0
V ( x) = ⎨
⎩Vo
0< x<a
−b < x < 0
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17
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3.1.3 The k-space diagram
− (α 2 + β 2 )
(sin α a)(sin β b) + (cos α a )(cos β b) = cos k (a + b)
2αβ
α=
2mE
h2
β = jγ = j
2m(Vo − E )
, γ >0
h2
Case 1: Free particle:
bVo is finite value, and
b=0
Vo=0
V(x)
2mE
β =α =
h2
W.K. Chen
E
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18
0
1
− (α + α )
(sin α a)(sin α b) + (cos α a )(cos α b) = cos k (a + b)
2α 2
2
2
cos αa = cos ka (free particle)
⇒
α =k
2mE
2m( 12 mυ 2 ) mυ p
2mE
⇒
=k
=
=
=
=
k
α=
h2
h2
h2
h
h
⎧ p = hk
⎪
⇒⎨
p 2 h 2k 2
=
⎪E =
2m 2m
⎩
p: particle momentum
k: wave number
For the case of free particle, parabolic relation is
valid between energy E and momentum p.
p: momentum of wave motion
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19
Case 2: bound electrons in crystal:
bVo is finite value, but
b→0
Periodic δ function potential barrier
Vo→∞ ⇒ E<<Vo
V(x)
Periodic δ function potential barrier
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20
− (α 2 + β 2 )
(sin α a )(sin β b) + (cos α a )(cos β b) = cos k (a + b)
2αβ
α=
2mE
h2
Q E << Vo ⇒ β = jγ = j
2m(Vo − E )
, γ >0
h2
decaying wave at barrier region
(γ 2 − α 2 )
(sin α a)(sinh γ b) + (cos α a)(cosh γ b) = cos k (a + b)
2αγ
1
1
b→0 ⇒sinhγb≈ γb & coshγb→1
Q sin θ = [e jθ − e − jθ ] cos θ = [e jθ + e − jθ ]
2j
2
mVoba sin α a
+ (cos α a) = cos ka
h2
αa
sin α a
E - k equation P'
+ (cos α a) = cos k a
αa
energy term
momentum term
P' =
mVoba
h2
α=
2mE
h2
The above eq does not lend itself to an analytical solution, but must be solved
using numerical or graphical techniques to obtain the relation between k, E
and Vo.
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21
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Q sin θ =
e x + e− x
cosh x =
2
1
1 jθ
[e − e − jθ ] cos θ = [e jθ + e − jθ ]
2j
2
e x − e− x
sinh x =
2
e x + e− x
cosh x =
2
x
e x − e− x
sinh x =
2
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22
f (αa) = P'
sin αa
+ (cos αa)
αa
− 1 ≤ f (αa ) = cos ka ≤ +1
a is fixed, k varies
The parameter α is related to the total energy E of the particle
α=
2mE
h2
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Electrophysics, NCTU
23
W.K. Chen
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24
E-k diagram
„
„
„
The right figure shows the
energy E as a function of the
wave number k
The plot shows the concept of
allowed energy bands for
particle propagating in the
crystal lattice
Since the energy E is
discontinuities, we also have
the concept of forbidden
energy for particles in the
crystal
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25
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Example 3.2 bandwidth
Coefficient P’ =10, potential width a=5Å
f (αa) = P'
sin αa
+ (cos αa) = cos ka
αa
Solution
The lowest allowed band occurs as ka changes from 0 to π
For ka=0
P'
sin αa
+ (cos αa) = +1
αa
Using numerical technique (trial and error), we find
α=
Elower
W.K. Chen
αa = 2.628
2mEupper
2mE
⇒
=
α
a
a = 2.628
h2
h2
(2.628) 2 h 2 (2.628) 2 (1.054 ×10−34 ) 2
=
=
= 1.68 ×10−19 J = 1.053 eV
2
−31
−10 2
2ma
2(9.11×10 )(5 ×10 )
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26
For ka=π, as seen form the figure it happens αa=π
αa =
Eupper =
2mEupper
h2
π 2h 2
2ma2
=
a =π
π 2 (1.054 ×10−34 ) 2
2(9.11×10−31 )(5 ×10−10 ) 2
= 2.407 ×10−19 J = 1.50 eV
ΔE = Eupper − Elower = 1.50 − 1.053 = 0.447 eV
W.K. Chen
P'
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sin αa
+ (cos αa ) = cos ka
αa
energy term
27
cos ka = cos(ka + 2nπ ) = cos(ka − 2nπ )
momentum term
Displacement of portions of the curve in left figure by 2π still satisfy the E-k
equation
W.K. Chen
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28
„
„
The entire E versus k plot is contained within π/a<k<π/a
For free electron, the particle momentum and
the wave number k are related by
momentum p = hk (free electron)
„
Given the similarity between the free electron
solution and the results of the single crystal,
The parameter ħk in a single crystal is
referred to as the crystal momentum, which
is not the actual momentum of the electron in
the crystal , but is a constant of motion that
includes the crystal interaction
crystal momentum : hk
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29
3.2 Electrical conduction in solids
„
„
„
For covalent bonding of silicon crystal,
each silicon atom is surrounded by 8
valence electrons. 4 from itself, 4 from the
4 nearest Si neighbor
For N silicon atoms in the crystal, there are
4 N energy states in lower valence band
and 4N energy states in higher conduction
band. Each energy state allow only one
electron to reside
At T=0K, all valence electrons are lying in
their lowest energy states, valence band,
i.e. the valence band is fully occupied and
the conduction band is completely empty
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30
Covalent bond breaking
„
At T>0K, a few valence band (VB) electrons may gain enough thermal
energy to break the covalent band and jump into conduction band (CB)
‰
‰
„
VB: covalent bonding position
CB: elsewhere
Since the pure semiconductor is neutrally charged, as the negatively
charged electron breaks away from its covalent bonding position, a
positively charged “empty state” is created in the original covalent
bonding position in the VB
CB
VB
W.K. Chen
„
„
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31
At T=0K, the energy states in the VB in E-k diagram are completely full,
and the states in CB are empty (Fig. (a))
At T>0K, a few valence band (VB) electrons may gain enough thermal
energy to break the covalent band and jump into CB (Fig (b))
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32
3.2.3 Drift current
l
l
N: total number of flow charge
n: volume density of flow charge
A: cross-sectional area
υ: average drift velocity
l: traveling length of carrier per Δt
A
N nAl nA(υΔt )
#
= nAυ ( )
=
=
Δt
Δt
Δt
sec
Φ
#
)
Flux density φ = = nυ (
A
sec ⋅ cm 2
Flux Φ =
Drift current density J = qφ = qnυ d (
n
J = q ∑υi
A
)
cm 2
Ampere
υi: the velocity of the ith charged carrier
i =1
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33
E-k diagram under external bias
„
„
Under no external force, the electron
distribution in CB is an even function of k,
since the net momentum is zero under
thermal equilibrium.
Under applied force, electrons in CB can
gain energy and a net momentum
Work - Eenergy Theorem
dE = Fdx = Fυdt
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34
3.2.3 Electron effective mass
The movement of an electron in a lattice is different
from that of an electron in free space
r
Fext
Ftotal = Fext + Fint = ma
r
Fext
Fint: due to the interaction between the moving
electron and other charged particles, such as ions,
protons and else electrons in the lattice
Fext = m*a
Free space
r
Fext = ma
crystal
r
Fext = m*a
m*: effective mass
Which takes into account of the particle mass
and the effects of internal forces
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35
E-k diagram and particle mass in free space
(hk ) (parabolic curve)
1
p2
2
E = moυ =
=
2
2mo 2mo
2
( p = hk )
∂E 2h 2 k hp hmoυ
⇒
=
=
=
mo
∂k 2mo mo
1 ∂E
=
velocity
:
υ
(free electron)
⇒
h ∂k
h2
∂ 2 E ∂ h 2k
=
(
)=
mo
∂k 2 ∂k 2mo
1
1 ∂2E
=
⇒ mass :
mo h 2 ∂k 2
W.K. Chen
„
„
The first derivative of E with k is
related to the velocity of free
particle
The second derivative of E versus
k is inversely proportional to the
mass of free particle
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36
Electron effective mass & E-k diagram in crystal
Conduction band
The energy near the bottom of CB may
be approximated by a parabola, just as
that of free particle
E − Ec = C1k 2 , C1 > 0
1
1 ∂2E
=
mass :
mo h 2 ∂k 2
(free electron)
W.K. Chen
1
1 ∂2E
effective mass : * = 2
mn h ∂k 2
(Block electron)
37
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∂E
∂2E
= 2C1k (C1 > 0),
= 2C1
∂k
∂k 2
1
1 ∂ 2 E 2C1
⇒ * = 2
= 2
h
mn h ∂k 2
Thus acceleration of Block electron with external force is Fext = mn a
*
If we apply an electric field, the acceleration is
a=
Fext − eE
= *
mn*
mn
a
Fext
E: electric field
E: total energy
„
„
E
Effective mass parameter is used to relate quantum mechanics and
classical mechanics
The larger the bowing of E-k diagram, the small the effective mass is
W.K. Chen
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38
3.2.4 Concept of hole
„
„
„
„
The empty state is created when a covalent bond is broken
The movement of a valence electron into the empty state is equivalent to the
movement of a positively charged empty state itself
The above positive charge carrier is called “hole”
The crystal now has a second equally important charge carrier, hole, that
can give rise to a current
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39
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The drift current in VB
Under external biased voltage, it is the electrons that moving in the VB, not
the positively charged ions
The drift current due to electrons in VB
J = ( − e)
∑υ
n
Q J = q ∑υi , q = −e for electron
i
i ( filled )
J = ( − e) ∑ υ i + ( + e)
i ( full )
i =1
∑υ
q = +e for + ion
i
i ( empty )
=
+
filled emptyfilled
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40
(−e) ∑υi = ? (full valence band)
i ( full )
The velocity of individual electron is given
υe (E ) =
„
„
1 ∂E
h ∂k
The valence band is symmetric in k and every state in VB is occupied
so that every electron with a velocity lυ l, there is a corresponding
electron with a velocity -lυ l
Since the valence band is full, the distribution of electrons with respect
to k has no chance to be changed by an external applied. The net drift
current density generated from a completely full band electrons, then, is
zero
( − e) ∑ υ i = 0
i ( full )
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41
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The net drift current density due to electrons in VB
0
J = ( − e) ∑ υ i + ( + e)
i ( full )
„
„
∑υ
i
i ( empty )
= ( + e)
∑υ
i
i ( empty )
The net drift current due to electrons in VB is entirely equivalent to placing a
positively charged particle in the empty state (assigned as hole) and
assuming all other states in the band are neutrally charged
The associated hole velocity is exactly the velocity of the electron who
occupy the empty state
neutral
W.K. Chen
hole
neutral
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42
Effective mass & E-k diagram in crystal
Valence band
The energy near the bottom of VB may
be approximated by a parabola, just as
that of free particle
Eυ − E = C2 k 2
∂E
∂2E
= −2C2 k (C2 > 0),
= −2C2
∂k
∂k 2
1
1 ∂ 2 E − 2C 2
⇒ *= 2
=
<0
2
2
me h ∂k
h
„
An electron moving near the top of valence band behaves as if it has a
negative mass
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43
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Thus acceleration of Block electron near the top of VB with external force is
Fext = me*a = (−e) E
We may rewrite
me: negative quantity
F
F
a = ext* = ext* > 0
me
me
=
− eE + eE
= *
me*
mp
a
Fext
E
filled emptyfilled
m*p = − me*
1
1
1 ∂2E
=− * =− 2
m*p
me
h ∂k 2
1 ∂E
υ h ( E ) = −υ e ( E ) = −
h ∂k
W.K. Chen
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E
a
Fext
neutral hole neutral
44
„
Now we can model VB as having particles with a positive electronic charge
and a positive effective mass, so it will move in the same direction as the
direction of applied field
„
„
The new particle is the hole
The density of hole is the same as the density of empty states in VB
W.K. Chen
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45
3.2.5 Metals, insulators and semiconductors
„
Insulator:
‰
‰
‰
The allowed bands are either completely empty or completely full
Since empty band and full band contribute no current at all, the resistivity of an
insulator is very large
The bandgap of insulator is usually on the order of 3.5 to 6 eV or larger
W.K. Chen
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46
„
Semiconductor:
‰
‰
‰
‰
The allowed bands are at conditions either almost empty or almost full
Both the electrons in CB and holes in VB can contribute the current
The bandgap of semiconductor is on the order of 1 eV
The resistivity of semiconductor can be controlled and varied over many orders of
magnitude
W.K. Chen
„
Electrophysics, NCTU
47
Metal:
‰
‰
‰
The band diagram for a metal may be in one of two forms
partially full band
overlapped conduction and valence bands
In the case of partially full band, many electrons are available for conduction, so
that the material can exhibit a large electrical conductivity
For overlapped conduction and valence bands , there are large numbers of
electrons as well as large numbers of holes can move, so this material can also
exhibit a very high electrical conductivity
W.K. Chen
Electrophysics, NCTU
48
3.3 Extension to three dimensions
„
„
The basic concept of allowed and forbidden energy bands comes from the
electrons moving in periodic potential arrangement in crystal lattice
For three dimensional crystal, the distance between atoms varies as the
direction through the crystal changes
W.K. Chen
„
„
Electrophysics, NCTU
49
The E-k diagram is symmetric in k so that no new information is obtained by
displaying the negative axis
In following plot of E-k diagrams, E-k diagrams at two directions are plotted.
[100] portion of the diagram with +k to the right and [111] portion of the
diagram with +k to the left are plotted
W.K. Chen
Electrophysics, NCTU
50
„
„
Direct bandgap
the minimum conduction band is at k=0 so that the transitions for electron
between CB & VB can take place with no change in crystal momentum
Indirect bandgap
the minimum conduction band occurs not at k=0 so that the transitions for
electron between CB & VB includes an interaction with the crystal
W.K. Chen
Electrophysics, NCTU
51
Effective mass concept in 3-dim crystal
W.K. Chen
Electrophysics, NCTU
52
3.4.1 Density of states for free electron
„
„
To determine the density of allowed quantum states as a function of energy,
we need to consider an appropriate mathematical model
Consider a free electron confined to a three-dimensional cubic with length a
with infinite potential well
a
V ( x, y , z ) = 0
for 0 < x < a
for 0 < y < a
for 0 < z < a
a
V ( x, y, z ) = ∞ elsewhere
a
Using the separation of variables technique in 3 dimensional Schrodinger’s
wave equation, we can have
r
k
2mE
= k 2 = k x2 + k y2 + k z2
2
h
W.K. Chen
Electrophysics, NCTU
r
kx
r
kz
r
ky
53
Two-dimensional plot of allowed quantum states in k space
k x = nx
π
a
, k y = ny
π
a
, k z = nz
π
a
nx, ny, nz: positive integers
Positive and negative values of
kx,ky or kz have the same energy
and represent the same quantum
state
The volume Vk of a single quantum state in k-space is
π
Vk = ( ) 3
a
W.K. Chen
Electrophysics, NCTU
54
The density of quantum state in k-space for whole cubic box is
1 4πk 2 dk
gT (k )dk = 2( )
Vk
8
2: due to two spin states allowed for each quantum state
1/8: due to the positive values of kx,ky, and kz
gT (k )dk =
k 2 dk
π2
⋅ a3
W.K. Chen
55
Electrophysics, NCTU
Density of quantum states versus energy
g k (k )dk =
k 2 dk
π2
⋅a
4π (2m) 3 / 2
⇒ g ( E )dE =
h3
E dE
1
2mE
h
1
1 1
1 m
2m ⋅ ⋅
dk =
dE =
dE
2 E
h
h 2E
Qk2 =
2mE
,
h2
3
k=
The number of energy states for whole cubic box between E and E+dE, i.e.,
between k and k+dk is given by
g k (k )dk =
k 2 dk
π
2
⋅ a 3 = g ( E )dE
2mE 1 1 m
4πa 3
3
g ( E )dE = 2 ⋅ 2 ⋅
dE ⋅ a = 3 ⋅ (2m) 3 / 2 ⋅ E dE
h
π h 2E
h
W.K. Chen
Electrophysics, NCTU
56
The density of quantum states per unit energy per unit volume of the crystal
g ( E )dE =
gT ( E )dE gT ( E )dE
=
V
a3
E
4π (2m) 3 / 2
g ( E )dE =
h3
E dE
g(E)
„
„
The density of states is proportional to the energy E. As the energy of free
particle becomes large, the number of available quantum states increase
The density of states is dependent on the mass of the free particle
W.K. Chen
Electrophysics, NCTU
57
Example 3.3 Density of states
Calculate the total density of states for free electron per unit volume with
energies between 0 and 1 eV
4π (2m) 3 / 2 1eV
N =∫
g ( E )dE =
∫0 E dE
0
h3
4π (2m) 3 / 2 2 3 / 2
N=
⋅ ⋅E
h3
3
4π [2 ⋅ (9.11×10 −31 )]3 / 2 2
N=
⋅ ⋅ (1.6 ×10 −19 ) 3 / 2 = 4.5 ×10 27 m −3
−34 3
(6.625 ×10 )
3
1eV
N = 4.5 × 10 21 states/cm −3
„
„
The density of states is typically a large number ranging from 1018 to 1022
cm-3
The density of quantum states in semiconductor is usually less than the
density of atoms in semiconductor crystal
W.K. Chen
Electrophysics, NCTU
58
3.4.2 Density of states for semiconductor
The parabolic relationship between energy and momentum of a free electron
is
p 2 h 2 k2
=
E=
2m 2m
The E-k curve near k=0 at conduction band can be approximated as a
parabola, so we have
p 2 h 2 k2
E − Ec =
=
2m 2m
The density of states in CB is
4π (2mn* ) 3 / 2
g c ( E )dE =
h3
W.K. Chen
E − Ec dE
Electrophysics, NCTU
59
Similarly, the E-k parabolic relationship for a free hole in VB
h 2 k2
p2
=
Eυ − E =
2m*p 2m*p
The density of states for holes (electrons) in VB is
gυ ( E )dE =
W.K. Chen
4π (2m*p ) 3 / 2
h3
Eυ − E dE
Electrophysics, NCTU
60
3.5 Statistical mechanics
„
„
In dealing with large numbers of particles, we are interested only in the
statistical behavior of the group as a whole rather than in the behavior of
each individual particle
There are three distribution laws determining the distribution of particles
among available energy states
W.K. Chen
„
‰
‰
Particles are distinguishable
No limit to the number of particles allowed in each energy state
The exemplary case is low-pressure gas molecules in a container
Bose-Einstein distribution
‰
‰
‰
„
61
Maxwell-Boltzmann distribution
‰
„
Electrophysics, NCTU
Particles are indistinguishable
No limit to the number of particles allowed in each energy state
The exemplary case is photon
Fermi-Dirac distribution
‰
‰
‰
Particles are indistinguishable
Only one particle is permitted in each energy state
The exemplary case is electrons in crystal
W.K. Chen
Electrophysics, NCTU
62
3.5.2 The Fermi-Dirac Probability Function
For ith energy level with gi quantum states
Only one particle is allowed in each state
First particle:
gi ways of choosing to place the particle
Second particle:
(gi -1) ways of choosing to place the particle
Ni particle:
[gi –(Ni-1)] ways of choosing to place the particle
Then the total number of ways of arranging Ni particles in the ith energy level
is
( g i )( g i − 1) L ( g i − ( N i − 1)) =
W.K. Chen
gi !
( g i − N i )!
Electrophysics, NCTU
63
Since the particles are indistinguishable, the interchange of any two electrons
does not produce a new arrangement
The actual number of independent ways to distribute the Ni particles in the ith
level is
Wi =
W.K. Chen
gi !
N i !( g i − N i )!
Electrophysics, NCTU
64
N ( E )dE
= f E (E) =
g ( E )dE
1
⎛ E − Ef
1 + exp⎜⎜
⎝ kT
⎞
⎟⎟
⎠
( N ( E ) < g ( E ))
N(E)dE (number density): Number of particles per unit volume per unit
energy (E to E+dE)
g(E)dE: number of quantum states per unit volume per unit energy
W.K. Chen
„
„
Electrophysics, NCTU
65
Meaning of Fermi energy
At T=0K,
‰
‰
The probability of a quantum state being occupied is unity for E<Ef
The probability of a quantum state being occupied is zero for E>Ef
f E (E) =
W.K. Chen
1
⎛ E − Ef
1 + exp⎜⎜
⎝ kT
⎞
⎟⎟
⎠
f E (E) = 0
for E > E f
f E (E) = 1
for E < E f
Electrophysics, NCTU
66
3.5.3 Features of Fermi-Dirac Function
„
For T>0K
‰
The probability of a state being occupied at E=Ef is always ½
f E (E) =
1
1
=
1 + exp(0 ) 2
W.K. Chen
Electrophysics, NCTU
67
Example 3.6 3kT above the Fermi-energy
fE (E) =
1
⎛ E − Ef
1 + exp⎜⎜
⎝ kT
⎞
⎟⎟
⎠
=
1
1
=
= 4.74%
⎛ 3kT ⎞ 1 + 20.09
1 + exp⎜
⎟
⎝ kT ⎠
At energy above Ef, the probability of a state being occupied by an electron
become significantly less than unity
W.K. Chen
Electrophysics, NCTU
68
Example 3.7 99% probability
Assume Fermi energy is 6.25 eV, Calculate the temperature at which there
is 1 % probability that a state 0.30 below the Fermi energy level will not
contain an electron
1 − f E (E) = 1 −
1
⎛ E − Ef
1 + exp⎜⎜
⎝ kT
⎞
⎟⎟
⎠
1
⎛ − 0.30 ⎞
1 + exp⎜
⎟
⎝ kT ⎠
⇒ kT = 0.06529eV ⇒ T = 756K
0.01 = 1 −
k = 1.38 ×10 −23 J/K
= 8.62 ×10 −5 eV/K
W.K. Chen
69
Electrophysics, NCTU
Maxwell-Boltzmann Approximation
f E (E) =
1
⎛ E − Ef
1 + exp⎜⎜
⎝ kT
⎞
⎟⎟
⎠
When E-Ef>>kT, the exponential term in
the denominator is much greater than
the unity
⎡ E − Ef ⎤
f E ( E ) ≈ exp ⎢−
⎥ for E − E f > 3kT
kT
⎣
⎦
„
„
kT = 0.026 eV at T = 300K
The Fermi-Dirac probability for E-Ef> 3kT can be approximated by
Boltzmann distribution
For E-Ef>3kT, the Boltzmann approximation is slightly larger than the FermiDirac curve with inaccuracy less than 5%
W.K. Chen
Electrophysics, NCTU
70
Example 3.8 3kT
Calculate the energy at which the difference between Boltzmann
approximation and the Fermi-Dirac function is 5%
⎛ E − Ef
exp⎜⎜ −
kT
⎝
⎞
1
⎟⎟ −
⎠ 1 + exp⎛⎜ E − E f
⎜ kT
⎝
1
⎛ E − Ef ⎞
⎟⎟
1 + exp⎜⎜
kT
⎝
⎠
⎞
⎟⎟
⎠ = 0.05
⎛ E − Ef
exp⎜⎜ −
kT
⎝
⎞ ⎧
⎛ E − Ef
⎟⎟ ⋅ ⎨1 + exp⎜⎜
⎠ ⎩
⎝ kT
⎛ E − Ef
exp⎜⎜ −
kT
⎝
⎞
1
⎟⎟ = 0.05 ⇒ E − E f = kT ln(
) ≈ 3kT
0.05
⎠
W.K. Chen
⎞⎫
⎟⎟⎬ − 1 = 0.05
⎠⎭
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