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Chapter 3. Introduction to The pQuantum Theory of Solids
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3.1 Allowed and Forbidden Energy gyBands
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Allowed and Forbidden Energy Bandsgy
Formation of Energy Bands
The wave functions of the two atom electrons overlap, which means that the two electrons will interact.
Th b l h d d This interaction or perturbation results in the discrete quantized energy
level splitting into two discrete energy levels.
The splitting of the discrete state into t o states is consistent ith the Pauli The splitting of the discrete state into two states is consistent with the Pauli
exclusion principle.
Fi r 3 1 ( ) P b bilit d it f ti f i l t d h d t
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Figure 3.1 (a) Probability density function of an isolated hydrogen atom. (b) Overlapping probability density functions of two adjacent hydrogen atoms. (c) The splitting of the n = 1 state.
Formation of Energy Bandsgy
If we start with a regular periodic arrangement of hydrogen-type atoms
that are initially very far apart, and begin pushing the atoms together, the initial quantized energy level will split into a band of discrete energy
l llevels.
This effect is shown schematically in Figure 3.2, where the parameter r0
represents the equilibrium interatomic distance in the crystalrepresents the equilibrium interatomic distance in the crystal.
Figure 3.2 The splitting of an energy state into a band of allowed energies.
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Formation of Energy Bandsgy
At the equilibrium, interatomic distance, there is a band of allowed
energies, but within the allowed band, the energies are at discrete 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 regardless of size.
H i l h h b h However since no two electrons can have the same quantum number, the discrete energy must split into a band of energies in order that each electron can occupy a distinct quantum state.can occupy a distinct quantum state.
At any energy level, the number of allowed quantum states is relatively small. In order to accommodate all of the electrons in a crystal, we must have many y yenergy levels within the allowed band.
The energy difference between two energy levels is extremely small, so
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that for all practical purposes, we have a quasi-continuous energy
distribution through the allowed energy band.
Example 3.1p
Objective: To calculate the change in kinetic energy of an electron when the velocity changes by a small value. Consider an electron traveling at a velocity of 107 cm/s. Assume the velocity
i b l f 1 / Th i i ki i i i bincreases by a value of 1 cm/s. The increase in kinetic energy is given by
Let v2 = v1+∆v. Then
2 2 2 22 1 2 1
1 1 1 ( )2 2 2
E mv mv m v v
Let v2 v1+∆v. Then
But ∆v << v1 , so we have that
2 2 2 22 1 2 1( ) 2 ( )v v v v v v v
Solution1 1
1 (2 )2
E m v v mv v
Substituting the number into this equation, we obtain
which may be converted to units of electron volts as
3 5 28(9.11 10 )(10 )(0.01) 9.11 10 JE
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which may be converted to units of electron volts as 28
-919
9.11 10 =5.7 10 eV1.6 10
E
Comment
A change in velocity of 1 cm/s compared with 107 cm/s results in a change in every of 5.7×10-9 eV, which is orders of magnitude larger than the h i f 5 7×10 19 V b i h ll dchange in energy of 5.7×10-19 eV between energy states in the allowed
energy band. This example serves to demonstrate that a difference in adjacent energy states of 10-9 eV is indeed very small, so that the discreteadjacent energy states of 10 eV is indeed very small, so that the discrete energies within an allowed band may be treated as a quasi-continuous distribution.
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The case for containing more than one electronone electron
Suppose the atom contains electrons up through n = 3 energy level.
If the aqtoms are initially very far apart the electrons in adjacent atoms will If the aqtoms are initially very far apart, the electrons in adjacent atoms will not interact and will occupy the discrete energy levels.
If these atoms brought closer together the outermost electrons in the n = 3 If these atoms brought closer together, the outermost electrons in the n 3 energy shell will begin to interact initially, so that this energy level will
split into a band of allowed energies.
If the atoms continue to mover closer together, the electrons in the n = 2 shell and n = 1 shell may begin to interact and also split into a band of
allowed energies.
Finally, if the atoms become sufficiently close together, the innermost l i h 1 l l i h hi l l l li ielectrons in the n = 1 level may interact, so that this level may also split into
a band of allowed energies. (Fig.3.3)
If the equilibrium interatomic distance is r then we have bands of allowed
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If the equilibrium interatomic distance is r0, then we have bands of allowed energies that the electrons may occupy separated by bands of forbidden energies.
The case for containing more than one electronone electron
This energy-band splitting and the formation of allowed and forbidden bands is the energy-band theory of single crystal materials
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Figure 3.3 Schematic showing the splitting of three energy states into allowed bands of energies.
Band Splitting in Silicon Atomp g
The actual band splitting in a crystal is much more complicated than indicated in Figure 3.3. A schematic representation of an isolated silicon atom is shown in Figure 3.4a.
Ten of the fourteen silicon atom electrons occupy deep-lying energy levels close to the nucleus. The four remaining valence electrons are relatively weakly bound and are the electrons involved in chemical reactionsweakly bound and are the electrons involved in chemical reactions.
Figure 3.4b shows the band splitting of silicon. We need only consider n = 3 level for the valence electrons, since the first two energy shells are3 level for the valence electrons, since the first two energy shells are completely full and are tightly bound to the nucleus.
The 3s state corresponds to n = 3 and l = 0 and contains two quantum p qstates per atom. This state will contain two electrons at T = 0 K. The 3p state corresponds to n = 3 and l = 1 and contains six quantum states per
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atom.
As the interatomic distance decreases, the 3s and 3p states interact and overlap. At the equilibrium interatomic distance, the bands have again split, but now four quantum states per atom are in the lower band and four
i h b dquantum states per atom are in the upper band.
At absolute zero degrees, electrons are in the lowest energy state, so that all states in the lower band (the valence band) will be full and all states in thestates in the lower band (the valence band) will be full and all states in the upper band (the conduction band) will be empty.
The bandgap energy E between the top of the valence band and theThe bandgap energy Eg between the top of the valence band and the bottom of the conduction band is the width of the forbidden energy band.
We have discussed qualitatively how and why bands of allowed and q y yforbidden energies are formed in a crystal. The formation of these energy bands is directly related to the electrical characteristics of the crystal, as we will see later in our discussion.
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Band Splitting in Silicon Atomp g
Fi r 3 4 ( ) S h ti f i l t d ili t (b) Th plitti f th
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Figure 3.4 (a) Schematic of an isolated silicon atom. (b) The splitting of the 3s and 3p states of silicon into the allowed and forbidden energy bands.
The Kronig-Penny Modelg y
Figure 3.5a – The potential function of a single, noninteracting, one-electron atomFi 3 5b S l i lFigure 3.5b – Several atoms are in close and the potential functions overlap.Figure 3.5c – The net potential functionFigure 3.5c The net potential function for one-dimensional single-crystal material
Kronig-Penny Model : idealized periodic potential representing a one-dimensional single crystal
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Figure 3.6 The one-dimensional periodic potential function of the Kronig-Penny model.
Figure 3.5
To obtain the solution to Schroedinger equation, we make use of a mathematical theorem by Bloch. The theorem states that all one-electron wave functions for problems involving periodically varying potential energy functions wave function must be of the formkenergy functions, wave function must be of the form
: a constant of motion, u(x): periodic function with period (a+b)
( ) ( ) jkxx u x e (3.1)
k
From Chapter 2, (total solution) = (time-independent solution) ⅹ (time dependent solution)
( )
(The traveling-wave solution)
If id i Ⅰi Fi 3 6 (0 ) i hi h V( ) 0 k h
( / ) ( ( / ) )( , ) ( ) ( ) ( ) ( )jkx j E t j kx E tx t x t u x e e u x e (3.2)
(3.3)
If we consider regionⅠin Figure 3.6 (0<x<a) in which V(x) = 0, take the second derivative of Eq.(3.1), and substitute this into the Eq.(2.13),
22 21 1( ) ( )2 ( ) ( ) 0d u x du xjk k u x (3 4)
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12 2 ( ) ( ) 0jk k u xdx dx
22
2mE
(3.4)
(3.5)
In region Ⅱ ( )00, ( )b x V x V
If d fi
22 2 02 2
22 2
2( ) ( )2 ( ) ( ) 0mVd u x du xjk k u xdx dx
(3.6)
If we define
Eq.(3.6) can be written as
2 2002 2
22 ( ) mVm E V
(3.7)
Eq.(3.6) can be written as2
2 22 222
( ) ( )2 ( ) ( ) 0d u x du xjk k u xdx dx
(3.8)
The solution to Eq.(3.4), for region Ⅰ( ) ( )
1( ) (0 )j k x j k xu x Ae Be x a (3.9)
The solution to Eq.(3.8), for region Ⅱ( ) ( )
2 ( ) ( 0)j k x j k xu x Ce De b x (3.10)
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Since V(x) is everywhere finite, both the ψ(x) and ∂ψ(x)/∂x must be continuous.
At the boundary (x = 0), the continuity condition 1 2(0) (0)u u1 2(0) (0)u u
0A B C D
1 2du du
(3.11)(3.12)
(3 13)1 2
0 0x x
du dudx dx
( ) ( ) ( ) ( ) 0k A k B k C k D
(3.13)
(3.14)
The periodicity and the continuity condition1 2( ) ( )u a u b
( ) ( ) ( ) ( )j k j k j k b j k b
(3.15)
( )( ) ( ) ( ) ( ) 0j k a j k a j k b j k bAe Be Ce De
1 2du dudx dx
(3.16)
(3.17)
The result is
x a x bdx dx
(3.18)( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 0j k a j k a j k b j k bk Ae k Be k Ce k De
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The result is2 2( ) (sin )(sin ) (cos )(cos ) cos ( )
2a b a b k a b
(3.19)
When E < V0, the electron bound within the crystal. β is then imaginary.
Eq.(3.19) can be written in terms of γj (3.20)
Let the barrier width b → 0 and the barrier height V →∞ but such that the
2 2
(sin )(sinh ) (cos )(cosh ) cos ( )2
a b a b k a b
(3.21)
Let the barrier width b → 0 and the barrier height V0→∞, but such that the product bV0 remains finite. Eq.(3.21) then reduces to
sinmV ba a
We may define a parameter P’ as
02
sin cos cosmV ba a a kaa
(3.22)
' 02
mV baP (3.23)
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Finally, we have the relation' sin cos cosaP a ka
a
(3.24)
The k-Space Diagramp g
For the case of V0 = 0, from Eq.(3.24) For the case of V0 0, from Eq.(3.24)
cos cosa ka (3.25)
k (3.26)
Since the potential is zero, the total energy is equal to the kinetic energy.21
22 ( )2 m mvmE p k (3.27)
P is the particle momentum. Therefore parameter k is related to the particle f h f l d i l f d b
2 2 k
(3.27)
momentum for the free electron, and is also referred to as a wave number.2 2 2
2 2p kEm m
(3.28)
Figure 3.7 The parabolic E versus k curve for the
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E versus k curve for the free electron.
The E-k Relation
As the P’ increases, the particle becomes more tightly bound to the potential well or atom. We may define the left side of Eq.(3.24) as
' sin( ) cosaf a P a (3 29)
Now, Eq.(3.24) becomes
( ) cosf a P aa
(3.29)
( )f k (3 30)
f(αa) must be bounded between +1 and -1.
The parameter α is related to the total energy E of the particle
( ) cosf a ka (3.30)
The parameter α is related to the total energy E of the particle.
A plot of the energy E of the particle as a function of the wave number can be generated from Figure 3.8c.can be generated from Figure 3.8c.
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Figure 3.8 A plot of (a) the first term in Eq.(3.29), (b) the second term
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in Eq.(3.29), and (c) the entire f(αa) function. The shaded areas show the allowed values of (αa) corresponding to real values of k.
Figure 3 8 The E versus k diagram generated
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Figure 3.8 The E versus k diagram generated from Figure 3.8. The allowed energy bands and forbidden energy bandgaps are indicated.
Example 3.2p To determine the lowest allowed energy bandwidth.
Assume that the coefficient P’ = 10 and that the potential width a = 5 Å.p Solution
To find the lowest allowed energy bandwidth, we need to find the diff r i l k h fr m 0 t (S Fi r 3 8 ) F r kdifference in αa values as ka changes from 0 to π (See Figure 3.8c). For ka= 0, Eq.(3.29) becomes
i l d fi d 6 d h f k
sin1 10 cosa aa
By trial and error, we find αa = 2.628 rad. We see that for ka = π, αa = π.For αa = π, we have
22mE a or 2 a
2 2 2 34 2
192 2 31 10 2
(1.054 10 ) 2.407 10 J 1.50 eV2 2(9.11 10 )(5 10 )
Ema
For αa = 2.628, we find that E1 =1.68×10-19 J =1.053 eV. The allowed energy bandwidth is then
1 50 1 053 0 447 VE E E
22 Comment ; We see from Figure 3.8c that, as the energy increases, the
width of the allowed bands increase from this Kronig-Penny model.
2 1 1.50 1.053 0.447 eVE E E
The cosine function is periodic.(3 31)
Therefore, we can displace portions of the curve by 2π. And the entire Er k pl t i nt in d ithin / <k< /
(3.31)cos cos( 2 ) cos( 2 )ka ka n ka n
versus k plot is contained within –π/a<k<π/a.
Fi 3 10 Th E k di
2
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Figure 3.10 The E versus k diagram showing 2π displacements of several sections of allowed energy bands.
Figure 3.11 The E versus k diagram in the reduced-zone representation.
Nearly Free Electron Model (Kittel, Chap 6)Ne y ee ec o ode ( e , C p 6)
The free particle Schrodinger equation in one-dimension is2 2
If the electrons are confined to a linear solid of length L, the wave function
2 2
2
( ) ( )2
x E xm x
(6)
is the standing wave
( ) sin( / )n x A nx L (7)
where n is a positive integer.
It is convenient to introduce wave functions that satisfy periodic boundary
conditions. We require the wave functions to be periodic in x with period L. Thus
( ) ( )x x L (8)
Wave functions satisfying the free-particle Schrodinger equations and the periodicity condition are of the form of a traveling plane wave:
( ) ( )x x L (8)
periodicity condition are of the form of a traveling plane wave:
24( ) exp( )x jkx (9)
Figure 2 6 Particle in an infinite potential well: (a) Four lowest discrete energy levels (b)Figure 2.6 Particle in an infinite potential well: (a) Four lowest discrete energy levels. (b) Corresponding wave functions. (c) Corresponding probability functions.
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provided that the component of the wavevector k satisfy
Any component of k is of the form 2nπ/L, where n is a positive or
2 40; ; kL L
(10)
negative integer. The components of k are the quantum numbers of the
problem. We confirm that these values of k satisfy (8), for
( ) exp[ ( )] exp[ 2 ( ) / ] exp( 2 / )exp( 2 )exp( 2 / ) exp( ) ( )x L ik x L j n x L L j n x L j n
j n x L jkx x
(11)
On substituting (9) in (6) we have the energy Ek of the orbital with wavevector k:
22E k
(12)
The magnitude of the wavevector is related to the wavelength by k = 2π/ . 2kE km
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Nearly Free Electron Model (Kittel, Chap7)Nearly Free Electron Model (Kittel, Chap7)
On the free electron model, the allowed energy values are distributed essentially continuously from zero to infinity. We saw in Chapter 6 that
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2kE km
(1)
where, for periodic boundary conditions over a linear solid of length L,2 40; ; kL L
(2)
The free electron wave functions are of the formL L
( ) exp( )x jkx (3)
They represent running waves and carry momentum p = ħk.
The band structure of a crystal can often be explained by the nearly
free electron model for which the band electrons are treated as perturbed only weakly by the periodic potential of the ion cores.
Thi d l l ll h li i i b h This model answers almost all the qualitative questions about the behavior of electrons in metals.
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We know that Bragg reflection is a characteristic feature of wave propagation in crystals. Bragg reflection of electron waves in crystals is the cause of
energy gaps. (At Bragg reflection, wavelike solutions of the Schrodinger i d i i Fi 2 ) Th f d i iequation do not exist, as in Fig. 2.) These energy gaps are of decisive
significance in determining whether a solid is an insulator or a conductor. We explain physically the origin of energy gaps in the simple problem of a linear We explain physically the origin of energy gaps in the simple problem of a linear
solid of lattice constant a. The low energy portions of the band structure are shown qualitatively in Fig. 2, in (a) for entirely free electrons and in (b) for electrons that are nearly free, but with an energy gap at k = ±π/a. The Bragg condition (k+G)² = k² for diffraction of a wave of wavevector k becomes in one dimension
1 nk G (4)
where G = 2πn/a is a reciprocal lattice vector and n is an integer. The first
reflections and the first energy gap occur at k = ±π/a. The region in k-space
2k G
a (4)
between -π/a and π/a is the first Brillouin zone of this lattice. Other energy gaps occur for other values of the integer n. 28
Brillouin ZonesBrillouin Zones
A linear lattice in one dimensions in Fig.11.
The zone boundaries of the linear lattice are at k = -π/a, where a is the primitive axis of the crystal lattice.
Figure 11 Crystal and reciprocal lattices in one dimension. The basis vectors in the reciprocal lattice is b, of length equal to 2π/a. The shortest reciprocal l i f h i i b d b h di l bi f
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lattice vectors from the origin are b and –b. the perpendicular bisectors of these vectors form the boundaries of the first Brillouin zone. The boundaries are at k = +π/a.
The wave functions at k = ±π/a are not the traveling waves exp(jπx/a) or exp(-jπx/a) of free electrons. At these special values of k, the wave functions are made up of
equal parts of waves traveling to the right and to the left.
When the Bragg reflection condition k = ±π/a is satisfied by the wave vector a When the Bragg reflection condition k = ±π/a is satisfied by the wave vector, a wave traveling to the right is Bragg-reflected to travel to the left, and vice versa. Each subsequent Bragg reflection will reverse the direction of travel of the wave. A wave that travels neither to the right nor to the left is a standing wave: it doesn’t go anywhere.
The time-independent state is represented by standing waves. We can form two
different standing waves from the two traveling waves exp(±jπx/a) namelydifferent standing waves from the two traveling waves exp(±jπx/a), namely
( ) exp( / ) exp( / ) 2cos( / );( ) exp( / ) exp( / ) 2 sin( / ).
j x a j x a x aj x a j x a j x a
(5)
The standing waves are labeled (+) or (-) according to whether or not they change sign when –x is substituted for x. Both standing waves are composed of equal parts of i h d l f di d liright- and left-directed traveling waves.
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Figure 2 (a) Plot of energy ε versus wave vector k for a free electron. (b) Plot of energy versus wave vector for an electron in a monatomic linear lattice of lattice constant a. The energy gap Eg shown is associated with the first Bragg reflection at k = ±π/a; other gaps are gg g pfound at ±nπ/a, for integral values of n.
Origin of The Energy GapOrigin of The Energy Gap
The two standing waves ψ(+) and ψ(-) pile up electrons at different regions, and therefore the two waves have different values of the potential energy. This is the
origin of the energy gap. The probability density ρ of a particle is ψ*ψ=|ψ|².
For a pure traveling wave exp(jkx) we have ρ = exp( jkx)exp(jkx) = 1 so that the For a pure traveling wave exp(jkx), we have ρ = exp(-jkx)exp(jkx) = 1, so that the charge density is constant. The charge density is not constant for linear combinations of plane waves. Consider the standing wave ψ(+) in (5); for this we have
The function piles up electrons (negative charge) on the positive ions centered at x = 0, 2 i Fi 3 h th p t nti l n r i l t
2 2( ) ( ) cos /x a
a, 2a, . . . in Fig. 3, where the potential energy is lowest.
Figure 3a pictures the variation of the electrostatic potential energy of a conduction electron in the field of the positive ion cores. The ion cores bear a net positive charge p p gbecause the atoms are ionized in the metal, with the valence electrons taken off to form the conduction band. The potential energy of an electron in the field of a positive ion is negative so that the force between them is attractiveis negative, so that the force between them is attractive.
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Figure 3 (a) Variation of potential energy of a conduction electron in the field of the ion cores of a linear lattice. (b) Distribution of probability density ρ in the lattice for |ψ(-)|²∝i ² / |ψ(+)|²∝ ² / d f t li Th f ti ψ(+) ilsin² πx/a; |ψ(+)|²∝ cos² πx/a; and for a traveling wave. The wave function ψ(+) piles up
electronic charge on the cores of the positive ions, thereby lowering the potential energy in comparison with the average potential energy seen by a traveling wave. The wave function ψ( ) piles up charge in the region between the ions thereby raising the potential energy inψ(-) piles up charge in the region between the ions, thereby raising the potential energy in comparison with that seen by a traveling wave. This figure is the key to understanding the origin of the energy gap.
For the other standing wave ψ(-), the probability density is
which concentrates electrons away from the ion cores.
2 2( ) ( ) sin /x a
In Fig. 3b we show the electron concentration for the standing waves
ψ(+), ψ(-), and for a traveling wave.
When we calculate the average or expectation values of the potential
energy over these three charge distributions, we find that the potential
f (+) i l h h f h li h henergy of ρ(+) is lower than that of the traveling wave, whereas the potential energy of ρ(-) is higher than the traveling wave. We have an energy gap of width Eg if the energies of ρ(-) and ρ(+) differ by E .energy gap of width Eg if the energies of ρ( ) and ρ( ) differ by Eg.
Just below the energy gap at points A in Fig. 2 the wave function is ψ(+), and just above the gap at points B, the wave function is ψ(-).j g p p ψ( )
34
Magnitude of The Energy GapMagnitude of The Energy Gap
The wave functions at the Brillouin zone boundary k = π/a are √2 cos πx/a and √2 sin πx/a, normalized over unit length of line. We write the potential energy of an electron in the crystal at point x as
( ) cos 2 /U x U x a (6)
The first-order energy difference between the two standing wave states is
( ) cos 2 /U x U x a
1 2 2( )[ ( ) ( ) ]E d U
(6)
0
2 2
( )[ ( ) ( ) ]
2 cos(2 / )(cos / sin / )
gE dxU x
dxU x a x a x a U
We see that the gap is equal to the Fourier component of the crystal potential.
35
3.2 Electrical Conduction in Solids
36
The energy band and the bond modelThe energy band and the bond model
At T = 0 K, each silicon atom is surrounded by eight valence electrons that are in their lowest energy state and are directly involved in the covalent bonding
and all of the valence electrons are in the valence band.
Th b d h d i b d i l l The upper energy band, the conduction band, is completely empty.
The 4N states in the lower band, the valence band, are filled with the valence electronselectrons.
Figure 3.12 Two-dimensional representation of the covalent bonding in a semiconductor at T = 0 K
37
Band Splitting in Silicon Atomp g
Fi r 3 4 ( ) S h ti f i l t d ili t (b) Th plitti f th
38
Figure 3.4 (a) Schematic of an isolated silicon atom. (b) The splitting of the 3s and 3p states of silicon into the allowed and forbidden energy bands.
The energy band and the bond modelThe energy band and the bond model
As the temperature increases above 0 K, a few valence band electrons may gain enough thermal energy to break the covalent bond and jump into the conduction band.
Figure 3.13 (a) Two-dimensional representation of the breaking of a covalent bond.
39
g ( ) p g(b) Corresponding line representation of the energy band and the generation of a negative and positive charge with the breaking of a covalent bond.
The energy band and the bond modelThe energy band and the bond model
The semiconductor is neutrally charged. This means that, as the negatively y g g ycharged electron breaks away from its covalent bonding position, a positive
charged “empty state” is created in the original covalent bonding position in the valence bond.
As the temperature further increases, more covalent bonds are broken, l t j t th d ti b d d iti “ tmore electrons jump to the conduction band, and more positive “empty
states” are created in the valence band.
We can also relate this bond breaking to the E versus k energy bands We can also relate this bond breaking to the E versus k energy bands.
Figure 3.14a shows the E versus k diagram of the conduction and valence bands at T = 0 K. The energy states in the valence band are completely full gy p yand the states in the conduction band are empty.
40
Figure 3.14b shows these same bands for T > 0 K, in which some electrons
have gained enough energy to jump to the conduction band and have left empty states in the valence band.
We are assuming at this point that no external forces are applied so the
electron and “empty state” distribution are symmetrical with k.
Fi 3 14 Th E kFigure 3.14 The E versus kdiagram of the conduction and valence bands of a semiconductor at (a) T = 0 Ksemiconductor at (a) T = 0 K and (b) T > 0 K.
41
Drift Current
Current is due to the net flow of charge. If we had a collection of
positively charged ions with a volume density N (cm-3) and an average
drift velocity vd (cm/s), then the drift current density is 2A/J N (3 32)
If, instead of considering the average drift velocity, we considered the individual ion velocities, then we could write the drift current density as
2 A/cmdJ qNv (3.32)
individual ion velocities, then we could write the drift current density as
where vi is the velocity of the ith ion. The summation in Eq.(3.33) is taken 1
N
ii
J q v
(3.33)
i y q ( )over a unit volume so that the current density J is still in units of A/cm2.
Since electrons are charged particles, a net drift of electrons in the conduction band will give rise to a current. The electron distribution in the conduction band, as shown in Fig. 3.14a, is an even function of when no
t l f i li d
42
external force is applied.
Recall that k for a free electron is related to momentum so that, since there are as many electrons with a +|k| value as there are with a -|k| value, the net drift current density due to these electrons is zero.
This result is certainly expected since there is no externally applied force.
If a force is applied to a particle and the particle moves, it must gain energy. Thi ff i dThis effect is expressed as
where F is force, dx is the differential distance the particle moves, v is thedE Fdx Fvdt (3.34)
where F is force, dx is the differential distance the particle moves, v is the velocity, and dE is the increase in energy.
If an external force is applied to the electrons in the conduction band, there ppare empty energy states into which the electrons can move; therefore, because of the external force, electrons can gain energy and a net
momentum.
43
The electron distribution in the conduction band may look like that shown in Fig. 3.15, which implies that the electrons have gained a net momentum.
We may write the drift current density due to the motion of electrons asn
(3 35)
where e is the magnitude of the electronic charge and n is the number of l i l i h d i b d A i h i i
1i
iJ e v
(3.35)
electrons per unit volume in the conduction band. Again, the summation is taken over a unit volume so the current density is A/cm2.
Figure 3.15 The asymmetric g ydistribution of electrons in the E-k diagram when an external force is applied.
44
pp
Electron Effective Mass The movement of an electron in a lattice will, in general, be different from
that of an electron in free space In addition to an externally applied forcethat of an electron in free space. In addition to an externally applied force there are internal force in the crystal due to positively charged ions and negatively charged electrons which will influence the motion of electrons g y g
in the lattice. We can write
total ext intF F F ma (3.36)
where Ftotal, Fext, and Fint are the total force, the externally applied force, and the internal forces, respectively, acting on a particle in a crystal. The
i h l i d i h f h i lparameter a is the acceleration and m is the rest mass of the particle.
Since it is difficult to take into account all of the internal forces, we will write the equationwrite the equation
where the acceleration a is now directly related to the external force. The
*extF m a (3.37)
45
where the acceleration a is now directly related to the external force. The parameter m* is called the effective mass, takes into account the particle
mass and also takes into account the effect of the internal forces.
We can also relate the effective mass of an electron in a crystal to the E-k curves, such as was shown in Fig. 3.11.
In a semiconductor material, we will be dealing with allowed energy bands
th t r lm t mpt f l tr n nd th r n r b nd r f ll f l tr nthat are almost empty of electrons and other energy bands are full of electrons.
To begin, consider the case of a free electron whose E – k curve was shown in Fig 3 7 Recalling the Eq (3 28) the energy and momentum are related by EFig. 3.7. Recalling the Eq. (3.28), the energy and momentum are related by E = p2/2m = ħ2k2/2m, where m is the mass of the electron.
The momentum and wave number k are related by p =ħk. If we take the y pderivative of Eq. (3.28) with respect to k, we obtain
2dE k pdk
(3.38)
Relating momentum to velocity, Eq.(3.38) can be written asdk m m
1 dE p vdk m
(3.39)
where v is the velocity of the particle. The first derivative of E with respect to k is related to the velocity of the particle. 46
dk m
If we now take the second derivative of E with respect to k we have,
We may rewrite Eq. (3.40) as
2 2
2
d Edk m
(3.40)
The second derivative of E with respect to k is inversely proportional to the
2
2 2
1 1d Edk m
(3.41)
mass of the particle.
For the case of a free electron, the mass is a constant, so the second
derivative function is a constant.
We may also note from Fig. 3.7 that d2E/dk2 is a positive quantity, which i li h h f h l i l i i iimplies that the mass of the electron is also a positive quantity.
If we apply an electric field to the electron and use Newton’s classical equation of motion we can write
47
of motion, we can writeF ma eE (3.42)
Solving for the acceleration, we haveeE (3 43)
The motion of the electron is in the opposite direction to the applied electric field because of the negative charge
eEam
(3.43)
field because of the negative charge.
We may now apply the results to the electron in the bottom of an allowed energy band Consider the allowed energy band in Fig 3 16aenergy band. Consider the allowed energy band in Fig. 3.16a.
48Figure 3.16 (a) The conduction band in reduced k space, and the parabolic approximation. (b) The valence band in reduced k space, and the parabolic approximation.
The energy near the bottom of this energy band may be approximated by a parabola, just as that of a free particle. We may write
Th E i h h b f h b d Si E > E h
21( )cE E C k (3.44)
The energy Ec is the energy at the bottom of the band. Since E > Ec , the parameter C1 is a positive quantity.
Taking the second derivative of E with respect to k from Eq (3 44) we obtain Taking the second derivative of E with respect to k from Eq.(3.44), we obtain2
12 2d E Cdk
(3.45)
We may put Eq.(3.45) in the form2
12 2 2
21 Cd Edk
(3.46)
Comparing Eq.(3.46) with Eq.(3.41), we may equate ħ2/2C1 to the mass of the particle However the curvature of the curve in Fig 3 16a will not in general
2 2 2dk
49
particle. However, the curvature of the curve in Fig. 3.16a will not, in general, be the same as the curvature of the free-particle curve. We may write
21
2 2 2 *
21 1Cd E (3.47)
The effective mass relates the quantum mechanical results to the classical force equations. If we apply an electric field to the electron in the bottom
2 2 2 *dk m (3.47)
q w pp y d bof the allowed energy band, we may write the acceleration as
*
eEam
(3.48)
where m*n is the effective mass of the electron. The effective mass m*
n
of the electron near the bottom of the conduction band is a constant.
nm
50
Concept of The Holep
When a valence electron is elevated into the conduction band, a positively
charged “empty state” is created.
For T > 0 K, all valence electrons may gain thermal energy; if a valence electron gains a small amount of thermal energy, it may hop into the empty state.
Th f l l i h i i l The movement of a valence electron into the empty state is equivalent
to the movement of the positively charged empty state itself.
Figure 3 17 shows the movement of valence electrons in the crystal Figure 3.17 shows the movement of valence electrons in the crystal alternately filling one empty state and creating a new empty state, a motion equivalent to a positive charge moving in the valence band.q p g g
This charge carrier is called a hole and, as we will see, can also be thought of as a classical particle whose motion can be modeled using Newtonian
51
mechanics.
The drift current density due to electrons in the valence band, such as shown in Fig. 3.14b, can be written as
(filled)i
iJ e v (3.49)
where the summation extends over all filled state. This summation is inconvenient since it extends over a nearly full valence band and takes into
nt r l r n mb r f t t W m r rit Eq (3 49) in th f rmaccount a very large number of states. We may rewrite Eq. (3.49) in the form
( total) (empty) (total) (empty)
( ) ( )i i i ii i i i
J e v e v e v e v (3.50)
52Figure 3.17 Visualization of the movement of a hole in a semiconductor.
If we consider a band that is totally full, all available states are occupied by electrons. The individual electrons can be thought of as moving with a velocity as given by Eq. (3.39):
1 1dE p dE
The band is symmetric in k and each state is occupied so that, for every l i h l i h i di l i h l i
1 1 ( ) ( )( )dE p dEv v Edk m dk
electron with velocity v, there is a corresponding electron with a velocity -v. Since the band is full, the distribution of electrons with respect to k cannot be changed with an externally applied force. The net drift current densitybe changed with an externally applied force. The net drift current density
from a completely full band, then, is zero, or
( t t l)0i
ie v (3.51)
We can write the drift current density from Eq.(3.50) for an almost full
band as
( total)i
(empty)i
i
J e v (3.52)
53
where the vi in the summation is the associated with the empty state.
(empty)i 1( ) ( )( )dEv Edk
Eq.(3.52) is entirely equivalent to placing a positively charged particle in the empty states and assuming all other states in the band are empty, or neutrally charged. This concept is shown in Fig 3.18. Fig. 3.18a shows the valence band with conventional electron-filled states and empty states while Figband with conventional electron filled states and empty states, while Fig. 3.18b shows the new concept of positive charges occupying the original empty states. This concept is consistent with the discussion of the positively
charged “empty state” in the valence band as shown in Fig. 3.17.
54
Figure 3.18 (a) Valence band with conventional electron-filled states and empty states. (b) Concept of positive charges occupying the original empty states.
The vi in the summation of Eq. (3.52) is related to how well this positively charged particle moves in the semiconductor.
Now consider an electron near the top of the allowed energy band shown in Fig. 3.16b. The energy near the top of the allowed energy band may again be approximated by a parabola so that we may write.
2( ) ( )E E C k (3 53)
The energy Ev is the energy at the top of the energy band. Since E < Ev for electrons in this band then the parameter C mist be a positive quantity
22( ) ( )vE E C k (3.53)
electrons in this band, then the parameter C2 mist be a positive quantity.
Taking the second derivative of E with respect to k from Eq.(3.53), we obtainobtain
We may rearrange this equation so that
2
22 2d E Cdk
(3.54)
55
We may rearrange this equation so that2
22 2 2
21 Cd Edk
(3.55)
Comparing Eq.(3.55) with Eq.(3.41), we may write
where m* is again an effective mass. We have argued that C2 is a positive
(3.56)2
22 2 2 *
21 1Cd Edk m
quantity, which now implies that m* is a negative quantity.
If we again consider an electron near the top of an allowed energy band
and use Newton’s force equation for an applied electric field, we will have
H * i i i i
*F m a eE (3.57)
However, m* is now a negative quantity, so we may write
A l i h f ll d b d i h* *| | | |
eE eEam m
(3.58)
An electron moving near the top of an allowed energy band moves in the same direction as the applied electric field.
56
The net motion of electrons in a nearly full band can be described by considering just empty states, provided that a positive electronic charge is associated with each state and that the negative of m* from Eq. (3.56) is
i d i h hassociated with each state.
We now can model this band as having particles with a positive electronic
charge and a positive effective masscharge and a positive effective mass.
The density of these particles in the valence band is the same as the density of empty electronic energy states.of empty electronic energy states.
This new particle is the hole.
The hole, then, has a positive effective mass (denoted by ) and a*pmThe hole, then, has a positive effective mass (denoted by ) and a
positive electronic charge so it will move in the same direction as an applied field.
p
57
Holes (Kittel, Chap. 8)Holes (Kittel, Chap. 8)
The properties of vacant orbitals in an otherwise filled band are important in semiconductor physics and in solid state electronics. Vacant orbitals in a band are commonly called holes. A hole acts in applied electric and magnetic fields as if it has a positive charge +e. The reason is given in five steps in the boxes that follow.a positive charge e. The reason is given in five steps in the boxes that follow.
1. kh = - ke . (17)
The total wavevector of the electrons in a filled band is zero: Σk = 0. This result follows from the geometrical symmetry of the Brillouin zone: every fundamental lattice type has symmetry under the inversion operation r → – r about any lattice point; it follows thatsymmetry under the inversion operation r → r about any lattice point; it follows that the Brillouin zone of the lattice also has inversion symmetry. If the band is filled all pairs of orbitals k and –k are filled, and the total wavevector is zero.
If an electron is missing from an orbital of wavevector ke , the total wavevector of the a e ect o s ss g o a o b ta o wavevecto ke , t e tota wavevecto o t esystem is – ke and is attributed to the hole. This result is surprising: the electron is missing from ke and the position of the hole is usually indicated graphically as situated at ke , as in Fig. 7. But the true wavevector kh of the hole is – ke, which is the wavevector of
58
e , g h e,the point G if the hole is at E. The wavevector – ke enters into selection rules for photon absorption.
Fi 7 Ab i f h f ħ d li ibl k lFigure 7. Absorption of a photon of energy ħω and negligible wavevector takes an electron from E in the filled valence band to Q in the conduction band. If ke was the wavevector of the electron at E, it becomes the wavevector of the electron at Q. The total wavevector of the valence band after the absorption is k and this is the wavevector we must ascribe tothe valence band after the absorption is - ke, and this is the wavevector we must ascribe to the hole if we describe the valence band as occupied by one hole. Thus kh = - ke ; the wavevector of the hole is the same as the wavevector of the electron which remains at C. For the entire system the total wavevector after the absorption of the photon is k + k = 0
59
For the entire system the total wavevector after the absorption of the photon is ke + kh = 0, so that the total wavevector is unchanged by the absorption of the photon and the creation of a free electron and free hole.
The hole is an alternate description of a band with one missing electron, and we either say that the hole has wavevector – ke or that the band with one missing electron has total wavevector khas total wavevector – ke.
2. Eh(kh) = – Ee(ke) . (18)
Let the zero of energy of the valence band be at the top of the band. The lower in the band the missing electron lies, the higher the energy of the system. The energy of the hole is opposite in sign to the energy of the missing electron, because it takes morethe hole is opposite in sign to the energy of the missing electron, because it takes more work to remove an electron from a low orbital than from a high orbital. Thus if the band is symmetric, Ee(ke)=Ee (- ke)=-Eh (- ke)=-Eh (kh). We construct in Fig. 8 a band scheme to represent the properties of hole. This hole band is a helpful representation p p p p pbecause it appears right side up.
60
Th h lf f h f h h h l b d h l hFigure 8 The upper half of the figure shows the hole band that simulates the dynamics of a hole, constructed by inversion of the valence band in the origin. The wavevector and energy of the hole are equal, but opposite in sign, to the
d f h l bi l i h l b d W d
61
wavevector and energy of the empty electron orbital in the valence band. We do not show the disposition of the electron removed from the valence band at ke .
¹Bands are always symmetric under the inversion k→ - k if the spin-orbit interaction is neglected. Even with spin-orbit interaction, bands are always symmetric if the crystal structure permits the inversion operation. Without a center of symmetry, but with spin-orbit interaction, the bands are symmetric if weof symmetry, but with spin orbit interaction, the bands are symmetric if we compare subbands for which the spin direction is reversed: E (k,↑)=E (-k,↓).
3 v = v (19)3. vh = ve . (19)
The velocity of the hole is equal to the velocity of the missing electron. From Fig. 9 we see that∇E (k )=∇E (k ) so that v (k )= v (k )we see that ∇Eh (kh)=∇Ee(ke), so that vh(kh)= ve(ke).
4. mh = – me . (20)
We show below that the effective mass is inversely proportional to the curvature d²E/dk², and for the hole band this has the opposite sign to that for an electron in the valence band. Near the top of the valence band me is negative, so that mh is positive.
62
p g , h p
Figure 9. (a) At t = 0 all states are filled except F at the top of the band, the velocity vx , is t F b dE/dk 0 (b) A l t i fi ld E i li d i th ± di ti Thzero at F because dE/dkx = 0 (b) An electric field E is applied in the ±x direction. The
force on the electrons is in the –kx direction and all electrons make transitions together in the -kx directtion, moving the hole to the state E. (c) After a further interval the electrons move farther along in k space and the hole is now at D
63
move farther along in k space and the hole is now at D.
1dk5. (21)
This comes from the equation of motion
1( )hh
d edt c
k E v B
q
(CGS) (22)1( )e
ed edt c
k E v B
that applies to the missing electron when we substitute –kh for ke and vh for ve. The equation of motion for a hole is that of a particle of positive charge e. The positive charge is consistent with the electric current carried by the valence band of Fig. 9: the current is carried by the unpaired electron in the orbital G:
J = (-e)v(G) = (-e)[-v(E)] = ev (E) , (23)
which is just the current of a positive charge moving with the velocity ascribed to the missing electron at E. The current is shown in Fig. 10.
64
Fi 10 M i f l i h d i b d d h l i hFigure 10. Motion of electrons in the conduction band and holes in the valence band in the electric field E. The hole and electron drift velocities are in opposite directions, but their electric currents arc in the same direction the direction of the electric field
65
direction, the direction of the electric field.
Metals, Insulators, and SemiconductorsMetals, Insulators, and Semiconductors
Each crystal has its own energy-band structure.
Insulator : energy bands are ether completely empty or completely full. The bandgap energy Eg is usually on the order of 3.5 to 6 eV or larger, so that at room temperature, there are very few thermally generated electrons and holes in an insulator.
S i d h l i l f l h l h b Semiconductor : has relatively few electrons or holes near the bottom or top of the bands. The bandgap energy is usually on the order of 1 eV. If an electric field is applied, the electrons or holes move through the crystal. The resistivityfield is applied, the electrons or holes move through the crystal. The resistivity can be controlled and varied over may orders.
Metal : partially full band in which there are many electrons available for p y yconduction or conduction and valence bands overlap at the equilibrium interatomic distance. There are large numbers of electrons as well as large
66
numbers of empty energy states into which the electron can move, so the material can exhibit a very high electrical conductivity.
Figure 3.19 Allowed energy bands showing (a) an empty band (b) a completely full
Figure 3.20 Allowed energy bands showing (a) an almost empty band (b) a almost full
67
(a) an empty band, (b) a completely full band, and (c) the bandgap energy between the two allowed bands.
(a) an almost empty band, (b) a almost full band, and (c) the bandgap energy between the two allowed bands.
Figure 3.21 Two possible energy bands of a metal showing (a) a partially filled band and (b) overlapping allowed
68
energy bands.
3.3 Extension to Three Dimensions
69
Extension to Three Dimensions
The basic concept of allowed and forbidden energy bands and the basic concept of effective mass have been developed in the last sections.
In this section, we will extend these concepts to three dimensions and to real crystals.
We will qualitatively consider particular characteristics of the three-
di i l l i f h E k l b d ddimensional crystal in terms of the E-k plots, bandgap energy, and effective mass.
We must emphasize that we will only briefly touch on the basic three We must emphasize that we will only briefly touch on the basic three-
dimensional concepts; therefore, many details will not be considered.
One problem encountered in extending the potential function to a three-One problem encountered in extending the potential function to a three
dimensional crystal is that the distance between atoms varies as the
direction through the crystal changes.
70
Extension to Three Dimensions
Figure 3.22 shows a face-centered cubic structure with the [100] and [110]
directions indicated.
Electrons travelling in the different directions encounter different potential
patterns and therefore different k-space boundaries.
The E versus k diagrams are in general a function of the k-space direction
i lin a crystal.
Figure 3.22 Two (100) plane of a face-centered cubic crystal showingface centered cubic crystal showing the [100] and [110] directions.
71
The k-Space Diagrams of Si and GaAsThe k Space Diagrams of Si and GaAs
Figure 3.23 shows an E-k diagram of gallium arsenide and of silicon.
These simplified diagrams show the basic properties considered in this text, but do not show many of the details more appropriate for advanced-level courses.
Note that in place of the usual positive and negative k axes, we now show diff l di itwo different crystal directions.
The E-k diagram for the one-dimensional model was symmetric in k so that no new information is obtained by displaying the negative axisthat no new information is obtained by displaying the negative axis.
It is normal practice to plot the [100] direction along the normal +k axis and to plot the [111] portion of the diagram so the +k point to the left.and to plot the [111] portion of the diagram so the k point to the left.
In the case of diamond or zincblende lattices, the maxima in the valence band energy and minima in the conduction band energy occur at k = 0 or
72
gy gyalong one of these two directions.
The k-Space Diagrams of Si and GaAsThe k Space Diagrams of Si and GaAs
73Figure 3.23 Energy band structures of (a) GaAs and (b) Si.
The k-Space Diagrams of Si and GaAsThe k Space Diagrams of Si and GaAs
Figure 3.23a shows the E-k diagram for GaAs.
The valence band maximum and the conduction band minimum both occur at k = 0. The electrons in the conduction band tend to settle at the minimum conduction band energy which is at k = 0.
Similarly, holes in the valence band tend to congregate at uppermost valence b dband energy.
In GaAs, the minimum conduction band energy and maximum
valence band energy occur at the same k value As semiconductor withvalence band energy occur at the same k value. As semiconductor with this property is said to be direct bandgap semiconductor.
Transitions between the two allowed bands can take place with no change in p gcrystal momentum.
This direct nature has significant effect on the optical properties of the
74
material. GaAs and other direct bandgap materials are ideally suited for use in semiconductor lasers and other optical devices.
The k-Space Diagrams of Si and GaAsThe k Space Diagrams of Si and GaAs
The E-k diagram for silicon is shown in Fig. 3.23b.
The maximum in the valence band energy occurs at k = 0 as before.
The minimum in the conduction band energy occurs not at k = 0, but along the [100] direction.
The difference between the minimum conduction band energy and the maximum valence band energy is still defined as the energy bandgap Eg.
A semiconductor whose maximum valence band energy and
i i d i b d d h k lminimum conduction band energy do not occur at the same k value
is called an indirect bandgap semiconductor.
When electrons make a transition between the conduction and valence When electrons make a transition between the conduction and valence bands, we must invoke the law of conservation of momentum.
A transition in an indirect bandgap material must necessarily include an
75
A transition in an indirect bandgap material must necessarily include an interaction with the crystal so that crystal momentum is conserved.
Additional Effective Mass ConceptsAdditional Effective Mass Concepts
The curvature of E-k diagrams near the minimum of the conduction band i l d h ff i f h lenergy is related to the effective mass of the electron.
We may note from Fig. 3.23, that the curvature of conduction band at its minim m val e for GaAs is larger than that of silicon so the effective mass ofminimum value for GaAs is larger than that of silicon, so the effective mass of an electron in the conduction band of GaAs will be smaller than in silicon.
For the one-dimensional E versus k diagram the effective mass was defined For the one dimensional E versus k diagram, the effective mass was defined by Eq.(3.41) as
2
2 2
1 1 d Em dk
A complication occurs in the effective mass concept in a real crystal.
A three-dimensional crystal can be described by three k vectors.
m dk
y y
The curvature of E-k diagram at the conduction band minimum may not be the same in the three k directions.
76
In later sections and chapters, the effective mass parameters used in calculations will be a kind of statistical average that is adequate for most device calculations.
A Linear Lattice in One Dimensione ce O e e s o
L = 6a
L is the length of a linear lattice and a is the lattice constant of the lattice.L = 6a
a
k = 2 / 0k 2 / 0
k = 2 / 2 /6a /3a = 2 /Lψ(k) = expjkx : travelling wave
k = 2 / 4 /6a 2 /3a = 4 /Lψ(k) = expjkx : travelling wave
k = 2 / 6 /6a /a = 6 /L
ψ(k) = 2coskx : standing wave
k = 2 / 8 /6a 4 /3a = 8 /L
k = 2 / 6 /6a /a = 6 /Lψ(k) = 2jsinkx : standing wave
k = 2 / 10 /6a 5 /3a = 10 /L
k = 2 / 12 /6a 2 /a = 12 /Lψ(k) = 2coskx : standing wave
∆k = 2 /3a - /3a = /3a = 2 /6a = 2 /L
If L = Na and N = 107, ∆k = 2 /L = 2 /Na is very small. → The k is quasi continuous,
The total number of wave vector k in an allowed band is N.
E-k Diagram in The Reduced-Zone RepresentationRepresentation
3.4 Density of States Functiony
We wish to describe the current–voltage (J-V) characteristics of semiconductor devices.
Since the current is due to the flow of charge, an important step in the i t d t i th b f l t d h l th t ill bprocess is to determine the number of electrons and holes that will be
available for conduction.
The number of carriers that can contribute to the conduction process is a p
function of the number of available energy or quantum states since, by the Pauli exclusion principle, only one electron can occupy a given quantum
statestate.
When we discussed the splitting of energy levels into bands of allowed and
forbidden energies, we indicated that the band of allowed energies was g g
actually made up of discrete energy levels.
We must determine the density of these allowed energy states as a
f i f i d l l h l d h l
81
function of energy in order to calculate the electron and hole
concentrations.
3.4.1 mathematical Derivation
We will consider a free electron confined to a three-dimensional infinite potential well, where the potential well represents the crystal.
The potential of the infinite potential well is define as
( ) 0 for 0V L ( , , ) 0 for 0 0 0
V x y z x Ly Lz L
(3.59)
where the crystal is assumed to be a cube with length L.
( , , ) elsewhereV x y z
y g
Schrodinger’s wave equation in three dimensions can be solved using the separation of variables technique.
2 2
where nx, ny, and nz are integers and n2 = nx2+ny
2+nz2.
2 2
2 2 2 2 2 2 2 22
2 2 2x y z x y z
mE k k k k n n n nL L
(3.60)
82
x y z g x y z
A Cubic Lattice in Three DimensionsA Cubic Lattice in Three Dimensions The energy of electrons in a three dimensional lattice can be written by
2 2 2 2 2 2 2 2 2( / 2 ) ( / 2 )( ) ( / 2 )(2 / )E m k m k k k m L n (3 60)where nx, ny, and nz are integers and n2 = nx
2+ ny 2+ nz
2. The energy E and the values of allowed k
x y z x y zn n n n n n
( / 2 ) ( / 2 )( ) ( / 2 )(2 / )x y zE m k m k k k m L n (3.60)
2 20, 0 0 0 2, n n
0
1 1 0E = 0, the total number of quantum states (k) = 1 -1 1 0 1 -1 0
2 1, 1 0 0 -1 -1 0 -1 0 0 n
1 0 1 0 1 0 -1 0 1 0 -1 0 1 0 -1
0 0 1 -1 0 -1 0 0 -1 0 1 1
22 21
2E = / 2 ( ) ; the total number of quantum statesmL = 6 0 -1 1
0 1 -10 -1 -1
83
0 1 1
2 22
2 E = / ( ) ;
the total number of quantum states = 12
mL
Figure 3.24 (a) A two-dimensional array of allowed quantum states in k
84
space. (b) The positive one-eight of the spherical k space
The distance between two quantum states in the k direction for example is given The distance between two quantum states in the kx direction, for example, is given by
12 2 2( 1)k k n n (3 61)
Generalizing this result to three dimensions, the volume Vk of a single quantum
1 ( 1)x x x xk k n nL L L
(3.61)
state is32
kVL
(3.62)
A differential volume in k space is shown in Figure 3.24b and is given by 4 k2dk, so the differential density of quantum state in k space can be written as
L
2
34( ) 22 /T
k dkg k dkL
(3.63)
85
2 / L
Th f 4 k2dk i i h diff i l l d h f (2 /L)3 i h l The factor 4 k2dk is again the differential volume and the factor (2 /L)3 is the volume of one quantum state. Equation (3.63) may be simplified to
2k dk23
3( )Tk dkg k dk L
(3.64)
Equation (3.64) gives the density of quantum states as a function of momentum, through the parameter k. We can now determine the density of quantum states as a
function of energy E For a free electron the parameters E and k are related byfunction of energy E. For a free electron, the parameters E and k are related by
22
2mEk
(3.65a)
or
1 2k mE (3.65b)
86
2k mE
(3.65b)
The differential is
12mdk dEE
(3.66)
Then, substituting the expressions for k2 and dk into Equation (3.64), the number
of energy states between E and E +dE is given by
3
3 2
2 1( )2T
L mE mg E dE dEE
(3.67)
Since ħ = h/2 , Equation (3.67) becomes34 L
Equation (3.68) gives the total number of quantum states between the energy E
3/ 23
4( ) (2 )TLg E dE m E dE
h
(3.68)
87
q ( ) g q gyand E +dE in the crystal space volume of L3.
If we divided by the volume L3, then we will obtain the density of quantum
states per unit volume of the crystal. Equation (3.68) then becomes
3/ 2
3
4 (2 )( ) mg E Eh
(3.69)
The density of quantum states is a function of energy E. As the energy of this free electron becomes small, the number of available quantum states decreases. This density function is really a double density, in that the units are given in terms of states per unit energy per unit volume.
88
3.4.2 Extension to Semiconductors3.4. e s o o Se co duc o s
The parabolic relationship between energy and momentum of a free electron is given as E = p2/2m = ħ2k2/2m. The E versus k curve near
k = 0 at the bottom of the conduction band can be approximated as a parabola, so we may writewe may write
2 2
*2ckE Em
(3.70)
where Ec is the bottom edge of the conduction band and is electron effective
mass. Equation (3.70) may be rewritten to give
2 nm*nm
Th l f f h E k l ti f l t i th b tt f
2 2 2 2
* ( . )2 2c
n
k kE E cf Em m
(3.71)(3.28)
The general form of the E versus k relation for an electron in the bottom of a
conduction band is the same as the free electron, except the mass is replaced by the effective mass. We can then think of the electron in the bottom of the conduction
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band as being a “free” electron with its own particle mass. “free” conduction
electron model
++
The density of allowed electronic energy states in the conduction band
* 3/ 2
3
4 (2 )( ) nc c
mg E E Eh
(3.72)
is valid for E ≥ Ec . As the energy of electron on the conduction band decreases, the number of available quantum states also decreases.
The density of quantum state in the valence band can be obtained by using the same The density of quantum state in the valence band can be obtained by using the same infinite potential well model, since the hole is also confined in the semiconductor crystal and can be treated as a free particle.
We may also approximate the E versus k curve near k = 0 by a parabola for a “free” hole, so that
2 2k (3 73)
Equation (3.73) may be rewritten to give
*2 p
kE Em (3.73)
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q ( ) y g2 2
*2 p
kE Em (3.74)
Equation (3.74) is of the same form used in the general derivation of the density of state function. We may the generalize the density of states function from Equation (3.69) to apply to the valence band, so thato Equat o (3.69) to app y to t e v e ce b d, so t at
* 3/ 2
3
4 (2 )( ) pm
g E E E
(3.75)
is valid for E ≤ Ev .
3( )gh ( )
Quantum states do not exist within the forbidden energy band, so g(E) = 0 for E ≤ E ≤ E Figure 3 25 shows the plot of the density of quantumg(E) = 0 for Ev ≤ E ≤ Ec . Figure 3.25 shows the plot of the density of quantum states as a function of energy.
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Figure 3.25 The density of energy states in the conduction band and valence band as a function of energy
3.5 Statistical Mechanics3.5 S s c ec cs
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.
In a crystal, the electrical characteristics will be determined by the statistical
b h i f l b f lbehavior of a large number of electrons.
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3.5.1 Statistical Laws In determining the statistical behavior of particles, we must consider the laws that
the particles obey There are three distribution laws determining the distributionthe particles obey. There are three distribution laws determining the distribution of particles among available energy states.
Maxwell-Boltzmann probability function
Distinguishable
No limit to the number of particles allowed in each energy state
E ) Gas molec les in a container at fairl lo press re Ex) Gas molecules in a container at fairly low pressure
Bose-Einstein function
Indistinguishable
No limit to the number of particles allowed in each quantum state
Ex) Photons or black body radiation
b b l f Fermi-Dirac probability function
Indistinguishable
One particle in one quantum state
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p q
Ex) Electrons in a crystal
In each case, the particles are assumed to be noninteracting.
A Cubic Lattice in Three DimensionsA Cubic Lattice in Three Dimensions
The energy of electrons in a three dimensional lattice can be written by2 2 2 2 2 2 2 2 2( / 2 ) ( / 2 )( ) ( / 2 )(2 / )E k k k k L
where nx, ny, and nz are integers and n2 = nx 2+ ny
2+ nz 2.
The energy E and the values of allowed k
2 2 2 2 2 2 2 2 2( / 2 ) ( / 2 )( ) ( / 2 )(2 / )x y zE m k m k k k m L n (3.60)
2 2
0
0, 0 0 0 2, 1 1 0E = 0, the total number of k
x y z x y zn n n n n n
n n = 1 -1 1 0
2
1 -1 01, 1 0 0 n -1 -1 0
-1 0 0 1 0 1 0 1 0 -1 0 1 0 -1 0 1 0 -1 0 0 1 -1 0 -1
2 21
0 0 -1 0 1 12E = / 2 ( ) ; the total number of k = 6 0 -1 1mL
0 1 1
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0 1 -1 0 -1 -1
2 22
2 E = / ( ) ; the total number of k = 12mL
3.5.2 The Fermi-Dirac Probability Function3.5.2 The Fermi Dirac Probability Function
A maximum of one particle is allowed in each quantum state by the Pauli
exclusion principle.
The total number of ways of arranging Ni particles in the i th energy level (where Ni ≤ gi ) isNi ≤ gi ) is
However, since the particles are indistinguishable, the Ni ! number of
!( ) ( 1) ( ( 1))( )!
ii i i i
i i
gg g g Ng N
(3.76)
p g i
permutations that the particles have among themselves in any given arrangement do not count as separate arrangements. Therefore, the actual number of independent ways of realizing a distribution of N particles in the i th level isindependent ways of realizing a distribution of Ni particles in the i th level is
!W!( )!
ii
i i i
gN g N
(3.77)!( )!i i iN g N
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Eq. (3.77) gives the number of independent ways of realizing a distribution of Ni
particles in the i th level.
The total number of ways of arranging (N1, N2, N3, · · · , Nn) indistinguishable particles among n energy levels is the product of all distributions, or
!n g1
!W!( )!
ni
i i i i
gN g N
(3.78)
The parameter W is the total number of ways in which N electrons can be arranged in this system, where is the total number electrons in the system.
1
n
ii
N N
The maximum W is found by varying Ni among the Ei levels, which varies the distribution, but at the same time, we will keep the total number of particles
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, , p p
and total energy constant.
We may write the most probable distribution function as
( ) 1= ( )FN E f E
(3 79)( )
( ) 1 expF
F
fE Eg E
kT
(3.79)
where EF is called the Fermi energy.
The number density N(E) is the number of particles per unit volume per unit energy.
(E) i th b f t t t it l it g(E) is the number of quantum states per unit volume per unit energy.
fF(E) is called the Fermi-Dirac distribution or probability function.
It gives the probability that a quantum state at the energy E will be occupied It gives the probability that a quantum state at the energy E will be occupied
by an electron.
fF(E) is the ratio of filled to total quantum states at any energy E.
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3.5.3 The Distribution Function and The Fermi EnergyThe Fermi Energy
1( )f 1( )1 exp( )
Ff
f E E EkT
(3.79)
kT
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100
1 1 1( )1 exp(0) 1 1 2F Ff E E 1 exp(0) 1 1 2
Figure 3.31 The Fermi probability function versus
f diffenergy for different temperatures.
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Figure 3.32 | The probability of a state being occupied, and the fF(E) probability of a state being empty,1 f (E)1 - fF(E).
Figure 3.33 The Fermi-Dirac probability function and theMaxwell-Boltzmann
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approximation.
Consider the case when E – EF >> kT, where the exponential term in the denominator of Equation(3.79) is much greater than unity. We may neglect the 1 in the denominator, so the Fermi-Dirac distribution function becomes
( )( ) exp FF
E Ef EkT
(3.80)
Equation (3.80) is known as the Maxwell-boltzmann approximation, or simply the Boltzmann approximation, to the Fermi-Dirac distribution function. Figure 3.33 shows the Fermi-Dirac probability function and the Boltzmann
approximation. This figure gives an indication of the range of energies over which the approximation is valid.the approximation is valid.
The actual Boltzmann approximation is valid when exp[(E-EF)/kT]>>1 .
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