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Chapter 2
Semiconductor Materials and Their Properties
In this chapter, we will cover the following topics
(1) Elemental and compound semiconductors (2) The valence bond model (3) The energy band theory (4) Concentration of electrons and holes including Fermi levels, energy distribution,
and temperature dependence
2.1 Elemental and Compound Semiconductors
Elemental semiconductors: two important ones: Si and Ge, both belong to group-IV
with 4 valence electrons in their outermost shell.
They crystallize in a diamond structure. Neighboring atoms are bound by covalent bonds.
Si is by far the widely used semiconductor for various device applications
Compound semiconductors: III-V and II-VI compounds.
III-Vs: GaN, GaP, GaAs, GaSb, InP, InAs, InSb
They crystallize in zinc blende structure. 8 valence electrons are shared between a pair of
nearest atoms. Therefore, the bonding has a covalent character. On the other hand, since
the group III elements are more electropositive and group V elements are more
electronegative. Hence, the bonding has a partial ionic character as well. But the covalent
nature is predominant.
II-VIs: ZnS, ZnSe, ZnTe, CdS, CdSe, CdTe
Crystal structures:
CdS and CdSe: wurtzite (two interpenetrating hexagonal close-packed lattices)
ZnTe and CdTe: zinc blende
ZnS and ZnSe: can be both (depeding on the substrate on which it is grown)
Bonding: mixture of covalent and ionic types. Stronger ionicity than III-Vs since group-
VI elements are considerably more electronegative than group II elements.
The ionic character has the effect of binding the valence electrons rather tightly to the
lattice atoms. Thus, the band gaps of these compounds are larger than those of the
covalent semiconductors of comparable atomic weights.
Ternary and quaternary compounds:
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Ga1-xAlxAs, ZnSxSe1-x, Zn1-xMgxSySe1-y, where 0 β€ π₯ β€ 1, 0 β€ π¦ β€ 1. The properties of the resulting compound vary gradually with the fraction yx, .
Advantages of compound semiconductors:
Wider choice of bandgap than elemental semiconductors: IR-visible-UV
Direct bandgap materials: optoelectronic applications, LEDs, lasers, sensors.
A major difficulty with compounds is that their preparation in single crystal form is more
difficult.
Two models can be used to study semiconductors
(1) Valence bond model which describes properties in domain of space and time. (2) Energy band model which describes properties in energy and momentum.
Energy band model is far more useful.
2.2 The Valence Bond Model
Use elemental semiconductors as example, Si and Se, they form diamond structure where
each atom bound to its four nearest neighbors by covalent bonds. These neighbors are all
equidistance from the central atom and lie at the four corners of the tetrahedron.
Fig.2.1 Illustration of covalent bond in tetrahedron.
Each bond has two electrons with opposite spin so that the central atom appears to have
eight electrons with opposite spins.
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Fig.2.2 An 2-D illustration of the diamond structure.
At π = 0K, all electrons are tightly bound in the bonds β a perfect insulator
At higher temperatures, lattice vibration occasionally shake loose some electrons so they
can move freely inside the crystal. The vacant side created by the broken bond has a net
positive charge known as hole (a particle of positive charge). Both electrons and holes are
responsible for semiconductor conductivity.
One way to visualize the movement of a hole is to consider the neighboring bond
electrons jumps over to the vacant site to create another vacant at the position from which
the electron came from. This is equivalent to hole moving from one site to another.
This picture has its limitation. It fails to explain the wave nature of the hole and the Hall
effect.
The above described generation of electrons and holes (both are called carriers) is due to
the thermal excitation. This is the case for intrinsic semiconductors where number of
electrons is equal to that of holes. The carrier concentration depends on temperature.
There is another way of introducing conduction carriers into the semiconductors β
dopping of impurity atoms intentionally.
For Si and Ge, elementals from group-III and V are commonly used as impurities.
The doped semiconductors are called extrinsic semiconductors.
n-type semiconductors
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A small amount of group V elements (As, P, Sb) is added into Si. These impurities
occupy lattice sites that are normally occupied by Si atoms β substitutional impurities
Fig.2.3 Illustration of group-V substitutional impurity in Si.
The 5th
electron is bound to the impurity atom only by weak electrostatic force.
Bohr theory of the hydrogen atom can be used to calculate the radius and the ionization
energy of its ground state
π0 = 0.53 ππ π0 π0ππβ Γ
πΈπΌ = 13.6 π0ππβ ππ π0
2
eV
Two modifications to the hydrogen formula:
(1) effective mass: ππβ < π0 (free electron mass)
(2) permittivity of the semiconductor: ππ > π0 (permittivity of the free space)
The ionization energy is typically small, therefore, at room temperature all of them
should be ionized to become conduction electrons β donors.
n-type semiconductor: majority carriers are electrons, minority carriers are holes.
p-type semiconductors
Group III elements occupy a substitutional position in the Si lattice (B, Al, In, Ga)
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Fig.2.4 Illustration of group-III substitutional impurity in Si.
Radius and ionization energy of the bound hole state can be calculated similarly.
These impurities cn contribute holes by accepting electrons β acceptors.
p-type semiconductor: majority carriers are holes, minority carriers are electrons.
If a semiconductor is doped with both donors and acceptors, the extra electrons attached
the donors fall into the incomplete bonds of the acceptors so that neither electrons nor
holes will be produced β compensation.
However, if ππ > ππ , we get n-type; if ππ < ππ , we get p-type.
Ambipolar semiconductors: can be doped to become both n-type and p-type. Elemental
and most of III-V compound semiconductors are ambipolar.
Unipolar semiconductors: can be doped either n-type or p-type, but not both. Many II-
VI compounds are such.
The unipolar behavior is due to the mechanism of self-compensation.
Consider ZnTe doped with iodine (I in group-VII):
Intention: I atoms replace Te atoms to make it n-type
Practice: For I atoms to replace Te atoms, temperature has to be raised, it will cause Zn
vacancies due to the evaporation of Zn atoms. Each Zn vacancy acts as a double acceptor.
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βπΈ Zn : energy required to create a Zn vacancy 2πΈπ : energy released when two donor electrons drop into the Zn vacancy.
If 2πΈπ > βπΈ Zn , the system energy is lowered, thus ZnTe will remain p-type.
2.3 Energy Band Model
The electronic states in a crystal is obviously different from that in an isolated atom. But
they are also related because a crystal is formed by binding atoms in a regular order.
Consider we have many isolated atoms (well separated), there are no interactions
between the atoms. A system of these atoms will have discrete energy levels and each
level is degenerate.
If we bring these atoms close to each other, the interaction between atoms will become
stronger. As a result, the wave functions of electrons in the outermost shell will begin to
overlap. The degeneracy of each energy level will be removed. The initially discrete level
will now split into many energy levels. The separation of these split energy levels
depends on the distance between atoms. The stronger the wave functions overlap, the
larger the energy splitting.
If we have π identical atoms bound to form a crystal, each energy level will split into π energy levels. Since π is usually very large, the density of these energy levels is high. It can be treated as they form a quasi-continuous energy band β allowed bands separated by
forbidden band β bandgap.
Fig.2.5 Illustration of energy band formation.
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The overlap of electron wave functions in inner shells is small, the interaction is weak,
therefore, the energy splitting can be neglected. The different properties of crystal and
constituent atoms are due to the different states of valence electrons. For example,
isolated neutral atoms (atom gas) do not conduct current. As they are bound together to
form a crystal, very different electronic properties can be determined (conductors,
semiconductors, insulators).
Another example, the optical spectrum associated with the transitions between different
levels in an isolated atom is discrete. However, the spectrum in a solid is continuous.
When π atoms form a crystal, one degenerate level splits into π energy levels which allow totally 2π electrons states according to Pauli principle that each energy level can be occupied by two electrons with opposite spin.
For atoms with one valence electron (Na, sodium and K, pottasium), the solids formed by
them have their energy band half filled, therefore, they are good conductors (metals). The
reason that only partially filled bands conduct current will be explained when we talk
about the energy band theory.
For atoms with two valence electrons (Mg), the outermost electrons are s2. Intuitively,
one would think that the 2π states will be completely filled therefore they are insulators. But the fact is these bands are overlapping each other with higher energy bands. As a
result both bands are partially filled, which leads to a good conductivity (metals).
For C(diamond), Si and Ge, the situation is more complicated. They all have 4 valence
electrons, s2 p
2. When they form a crystal, two energy bands should be produced. One
corresponds to s-state with 2π states. The other p-states with 6π states. It seems that the 6π band should be partially filled, therefore diamond, Si, and Ge are all good conductors. But they are not. Actually the orbit mixing between the s-state and p-state has led to a
new combination which result in two energy bands, each having 4π states. The lower one, called valence band, is then completely filled, leaving the upper one called
conduction band completely empty at low temperatures. Therefore, at low temperatures,
they act like insulators.
Electronic States in a Crystal
A detailed understanding of electronic states and the behavior of electrons in a crystal
requires calculations of quantum mechanics. The number of electrons involved in the
system is on the order of 1023
. This is a complex many body problem, an exact solution
of such a system is impossible.
The energy band theory is actually based on the single-eletron approximation. This
approximation takes into account the electron interaction by adding them into the
periodic potential field of the atoms. As a result, electrons can be treated independently,
while the interactions with other electrons are included in the potential field as a fixed
charge distribution.
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The Schrodinger equation becomes
βΡ2
2πββ2 + π π π· π = πΈπ· π
Where πβ is the effective mass which is different from the free electron mass due to the interaction with other electrons and π π is the periodic potential which includes the electron interaction with lattice atoms and other electrons.
Bloch Functions
F. Bloch proved that the solutions of the Schrodinger equation for a periodic potential
must be of a special form
π·π π = π’π π πππβπ
where π’π π has the period of the crystal lattice with π’π π+ π» = π’π π
This is a result of the Bloch theorem which states that the eigen functions of the wave
equation for a periodic potential are of the form of the product of a plane wave πππβπ and a function π’π π with the periodicity of the crystal lattice. The proof of this theorem can be found in Solid State Physics by Kittel.
Compared to free electron wave function πππβπ, Bloch function indicates that the probability of finding an electron in a lattice space is different from point to point within
one primitive unit cell, but the same between the corresponding points of different unit
cells.
In fact, π’π π describes the behavior of an electron around a lattice atom. πππβπ
demonstrates that the electron is not localized, but can propagate throughout crystal.
For a crystal with infinitely large volume, the value of π can vary continuously. Given a π, there exists a set of eigen energies πΈπ(π) and corresponding π·π π, π . The quantum number π indicates different energy band, intuitively can be considered as they are originated from different atomic energy levels. πΈπ(π) is therefore a continuous function of π. Due to the periodicity of a crystal, an electronic does not have a unique value of π. In fact, πβ² = π+π² can represent the same state as π does where π² is the so-called reciprocal lattice vector
π² = π1π1 + π2π2 + π3π3 Where π1 ,π2, π3 are integers and π1,π2,π3 are primitive vectors of the reciprocal lattice which can be constructed by the primitive vectors of the crystal lattice (π1,π2,π3)
π1 = 2ππ2 Γ π3
π1 β π2 Γ π3 ,π2 = 2π
π3 Γ π1π1 β π2 Γ π3
,π3 = 2ππ1 Γ π2
π1 β π2 Γ π3
It is easy to prove that
ππ β ππ = 2ππΏππ Actually, the Bloch wave function can be written as
π·π π, π = π’π ,π π πππβπ = π’π ,π π π
βππ²βπ πππβ² βπ = π’π ,πβ² π π
ππβ² βπ
where π’π ,πβ² π = π’π ,π π πβππ²βπ has the same periodicity as the lattice since
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πβππ²β π+π» = πβππ²βπ, π² β π» = 2ππ Thus, πβ² represents the same state as π does.
Unlike the situation for free electrons, the nonuniqueness of π suggests that strictly speaking, Ρπ will not carry the meaning of momentum. However, we will see later Ρπ can still be treated as if it is the momentum of an electron in a crystal.
Obviously, πΈπ(π) various with π periodically and for different n, πΈπ(π) varies within different energy intervals separated by energy gaps where electron states are forbidden.
Fig. 2.6 πΈπ(π) as a function of π.
If we take the complex conjugate of the Schrodinger equation, H remains unchanged,
therefore πΈπ(π) should have even symmetry with respect to π, i.e. πΈπ π = πΈπ(βπ) since
π·πβ π = π’π
β π πβππβπ is the eigen function of πΈπ(π) also.
Brillouin Zone
Since πβ² = π+π² represents the same electron state as π does with π² being the reciprocal lattice vector. We can actually limit π within a primitive unit cell in the reciprocal lattice, because π and πβ² point to the equivalent points within different primitive unit cell in the reciprocal lattice. Now for an fixed electron state, there exists a
unique corresponding π.
For example in 1-D, we can limit β π/2 < π < π/π.
A Brillouin zone is a special kind of primitive unit cell in a reciprocal lattice. It is defined
as a Wigner-Seitz cell in the reciprocal lattice, which is constructed by drawing
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perpendicular bisector planes in the reciprocal lattice from the chosen center to the
nearest equivalent reciprocal lattice sites.
The Wigner-Seitz cell is a primitive cell which maximally demonstrates the symmetry of
the crystal. For discussions on how to construct the Wigner-Seitz cell, one can refer to
Solid State Physics by Kittel.
State Density in k -Space
Consider a crystal with a dimension of πΏ1 Γ πΏ2 Γ πΏ3 with the dimension of a primitive cell of π1 Γ π2 Γ π3. For any state π = π1π + π2π + π3π , the periodic condition requires that
π·ππ 0,π¦, π§ = π·ππ πΏ1,π¦, π§ ,π·ππ π₯, 0, π§ = π·ππ π₯, πΏ2, π§ ,π·ππ π₯,π¦, 0 = π·ππ π₯,π¦, πΏ3 Thus
π1πΏ1 = π1 2π ,π2πΏ2 = π2 2π , π3πΏ3 = π3 2π .
Since, πΏ1 = π1π1, πΏ2 = π2π2, πΏ3 = π3π3, then
ππ =ππππ
2π
ππ, π = 1,2,3
with ππ β€ππ
2 since β π/2 < ππ < π/π. Therefore, βππ =
2π
πΏπ, then one state in π -space
takes a volume of
βπ1βπ2βπ3 =2π
πΏ1
2π
πΏ2
2π
πΏ3= 2π 3
π
where π is the volume of crystal.
Taking into account that each state can accommodate two electrons with opposite spin,
the density of states in π-space is then
ππ =2π
2π 3,
and within the Brillouin zone there are totally π = π1π2π3 allowed π states, which equals the number of primitive cells.
It can be shown that if the volume of a primitive cell in lattice is πΊ, the volume of a reciprocal primitive cell is 2π 3/πΊ.
Motion of Electrons in a Crystal
The time-dependent solution to the Schrodinger equation is
π·π π, π‘ = π’π π ππ πβπβππ‘
where π = πΈ/Ρ. Due to the linearity of the time-dependent Schrodinger equation, the superposition of the above wave function with different π
π· π, π‘ = πΆππ
π’ππ π ππ ππ βπβππ π‘
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will still be a solution to the Schrodinger equation. Since ππ βs are closely packed in π-space and can be treated as continuous, the superposition becomes an integral
π· π, π‘ = πΆ π π/2
βπ/2
π’π π ππ πβπβπ π π‘ π3π
Now at some instant of time, one tries to look at a wave packet that consists of Bloch
functions in the neighborhood of π, then the above integral can be rewritten as
π· π, π‘ = πΆ π+ πΌ βπ
ββπ
π’π+πΌ π ππ π+πΌ βπβπ π+πΌ π‘ π3πΌ
Since πΌ < βπ are small, then π’π+πΌ π β π’π π and
π π+ πΌ = π π + βππ π β πΌ +β― which is related to energy by π = πΈ/Ρ. Therefore
π· π, π‘ = π’π π ππ πβπβπ π π‘ πΆ π+ πΌ
βπ
ββπ
ππ πΌβ πββπππ‘ π3πΌ
The factor in front of the integral is a Bloch function. The integral represents a wave
packet with a center position π = βπππ‘ which moves with the velocity
π =ππ
ππ‘= βππ π =
1
ΡβππΈ π
which is the group velocity of electron in a crystal.
Now letβs consider a 1-D situation, with
πΆ π + π = πΆ, π < βπ
0, π > βπ
Then
π· π₯, π‘ = πΆπ’π π₯ ππ πβπ₯βπ π π‘ ππ π β π₯ββπππ‘ ππ
βπ
ββπ
= πΆπ’π π₯ ππ πβπ₯βπ π π‘
sin βπ π₯ β ππ/ππ π‘
βπ π₯ β ππ/ππ π‘ 2βπ
Hence, the wave packet probability
π· π₯, π‘ 2 β π’π π₯ 2
sin βπ π₯ β ππ/ππ π‘
βπ π₯ β ππ/ππ π‘
2
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Fig.2.7 1-D electron distribution described by a wave packet with a wave vector
distribution within π + βπ.
The center peak is bound by π₯ β ππ/ππ π‘ = Β±π/βπ, i.e. the half width βπ₯ is related to βπ by βπ₯βπ = π - exactly what is required by the uncertainty principle. For a wave packet with a space expansion of lattice constant π, we have βπ = π/π covers the entire Brillouin zone, i.e. π is completely uncertain, and vice versa. However, when we are dealing with the ranges of π and π that are much greater than the uncertainty βπ and βπ, we can consider both π and π have precise values β quasi classic.
Effective Mass
Suppose that an external field is applied, and the electron moves a distance ππ, the work done on the electron is
ππ€ = π β ππ = π β πππ‘ = π β βππππ‘ = π β βππΈππ‘/Ρ The work done on the electron causes its energy to change
ππΈ π = βππΈ β ππ = βππΈ β ππ/ππ‘ ππ‘ Since ππ€ = ππΈ, therefore π = Ρ ππ/ππ‘ = ππ/ππ‘ which is analogue to the Newtonβs law with π = Ρπ representing the momentum of the electron, similar to that of free electron.
The acceleration of an electron in a crystal is ππ
ππ‘=
1
Ρ
π
ππ‘ βππΈ =
1
Ρ ππ
ππ‘β βπ βππΈ =
1
Ρ2 π β βπ βππΈ
In tensor form, we can write ππ
ππ‘=
1
πβ π
where the inverse effective mass tensor
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1
πβ =
1
Ρ2
π2πΈ/πππ₯2 π2πΈ/πππ₯πππ¦ π
2πΈ/πππ₯πππ§
π2πΈ/πππ¦πππ₯ π2πΈ/πππ¦
2 π2πΈ/πππ¦πππ§
π2πΈ/πππ§πππ₯ π2πΈ/πππ§πππ¦ π
2πΈ/πππ§2
is a function of π. The tensor is symmetric, we can choose a proper coordinate system so that the tensor is diagonalized. Therefore, we can have for each direction
ππΌβππ£πΌππ‘
= πΉπΌ , πΌ = π₯,π¦, π§
which is analogue of Newtonβs 2nd
law.
Usually in semiconductors, we only deal with those states that are close to band edges (πΈπ and πΈπ£) because electrons are distributed a few ππ around the band edges.
For conduction band near πΈπ with π = 0, we can have
πΈπ π = πΈπ 0 +1
2 π2πΈππππ₯2
ππ₯2 +
π2πΈππππ¦2
ππ¦2 +
π2πΈππππ§2
ππ§2 +β―
= πΈπ 0 +Ρ2
2 ππ₯
2
ππ ,π₯β+
ππ¦2
ππ,π¦β+ππ§
2
ππ ,π§β +β―
Since πΈπ 0 is a minimum, ππΈπ
πππΌ= 0, and
π2πΈπ
πππ₯2 > 0, therefore, ππ ,πΌ
β > 0. Near the band
edge, πΈπ π has a parabolic relation with π.
For crystals with cubic symmetry, we have ππ ,π₯β = ππ,π¦
β = ππ ,π§β = ππ
β at π = 0. Then
πΈπ π = πΈπ 0 +Ρ2π2
2ππβ
and ππβππ
ππ‘= π, the same expressions as free electrons except that we have to use the
effective mass ππβ to replace the free electron mass.
For valence band near the top πΈπ£ with π = 0, we can have
πΈπ£ π = πΈπ£ 0 +1
2 π2πΈπ£πππ₯2
ππ₯2 +
π2πΈπ£πππ¦2
ππ¦2 +
π2πΈπ£πππ§2
ππ§2 +β―
= πΈπ£ 0 +Ρ2
2 ππ₯
2
ππ£,π₯β+
ππ¦2
ππ£,π¦β+ππ§
2
ππ£,π§β +β―
= πΈπ£ 0 +Ρ2π2
2ππ£β +β―
for cubic crystal. Since πΈπ£ 0 is a maximum, π2πΈπ£
πππ₯2 < 0 which will result in negative
electron effective mass (ππ£β < 0) near the top of the valence band. However, when we
look at the Newtonβs law expression of these electrons under an electric field π¬,
ππ£βππ
ππ‘= π = βππ¬
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the holes are introduced as having positive charge, therefore, the above Newtonβs law
should be modified
βππ£βππ
ππ‘= π
βππ
ππ‘= ππ¬
where the hole effective mass πβ = βππ£
β is actually positive at the top of the valence
band.
The effective mass is directly related with the curvature of the energy band, and therefore
related with the energy band width.
Thus, near the top of the valence band we can write
πΈπ£ π = πΈπ£ 0 βΡ2π2
2πβ
Fig.2.8 Illustration of the band curvature with the effective mass.
Energy Band Structures
The energy band structure is normally described as the relationship between energy πΈ
and π = ππ₯ ,ππ¦ ,ππ§ which is difficult to plot in 3-D. Typically, the relationship is
plotted along the principle directions of the crystal.
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Fig. 2.9 Band structure along the principle directions of the crystal (Si).
Direct band gap: the minimum of the conduction band is located at the same π as the maximum of the valence band.
Indirect band gap: the minimum of the conduction band and the maximum of the
valence band are located at the different πβs in π-space.
For direct band gap semiconductor, a photon with energy Ρπ = πΈπ = πΈπ β πΈπ£ can excite
an electron from the top of the valence band to the bottom of the conduction band. But
for indirect band gap semiconductor, such a transition requires that a photon with an
energy Ρπ > πΈπ = πΈπ β πΈπ£, because photons have very small momentum and the
absorption of a photon needs to be a vertical transition in π-space.
Electrons and holes are populated at the bottom of the conduction band and top of the
valence band, respectively. For direct band gap semiconductors, the probability of
electrons and holes recombine to emit photons is much higher than that of indirect
semiconductors. Therefore, direct band gap semiconductors have important applications
in optical devices.
Indirect: Si and Ge
Direct: most III-V and II-VI compounds
Constant energy surface is plotted as an surface area in π-space where each point on the surface has the same energy.
For a semiconductor with its conduction band minimum at π = 0 and isotropic effective mass, obviously the constant energy surface is spherical,
http://upload.wikimedia.org/wikipedia/commons/5/53/Fcc_brillouin.png
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πΈ = πΈπ +Ρ2π2
2ππβ
Fig. 2.10 Constant energy surface for conduction band minimum at π = 0 and isotropic effective mass.
If the tensor of the effective mass is anisotropic, the constant energy surface is an
ellipsoid.
For Ge and Si, the minimum of conduction band is not at π = 0, the center of the constant energy surface is not located at π = 0. Due to the crystal symmetry, there should be more than one constant energy surfaces, e.g. for Si there are 6 constant energy surfaces
along 6 equivalent principle axis 100 , the centers of the 6 ellipsoids are located at about ΒΎ of the distance from the Brillouin zone center.
πΈπ π = πΈπ +Ρ2
2 ππ₯ β ππ₯0
2
ππβ +
ππ¦2 + ππ§
2
ππ‘β
where ππβ and ππ‘
β are the longitudinal and transverse effective mass.
Fig.2.11 Constant energy surfaces in Si and Ge.
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For Ge, the centers of constant energy ellipsoids are along each of the 8 100 directions. The Brillouin zone boundary goes through the center of each ellipsoid. There are 8 half
ellipsoids within the 1st Brillouin zone, making 4 full ellipsoids within the 1
st Brillouin
zone.
The above description of energy band structure is detailed and emphasize on the πΈ β π relationship. There are times when the detail description is not necessary when we are
only interested in the band gap πΈπ and band alignment in real space.
At 0K the valence band is completely filled, the conduction band has no electrons. Under
this condition, the semiconductor behaves like an insulator.
This behavior is due to the fact that completely filled energy band as well as completely
empty band do not conduct electric current.
Fig. 2.12 Simplified band diagram.
Every moving electron contributes to electric current. But current is the total effect of all
electrons in a crystal. Within an energy band, if there is a state with momentum π and energy πΈ π , there must be another state with momentum βπ and πΈ βπ = πΈ π due to the symmetry of the crystal. Obviously, βππΈ π = ββππΈ βπ which implies that electron at π has same magnitude but opposite velocity as the electron at βπ. If the band is completely filled, then both states are occupied with electrons. The contribution to
current from the two electrons with π and βπ will be cancelled. Hence, the total current is zero.
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Fig. 2.13 Illustration of the symmetry of πΈ β π relationship and the corresponding electron velocity.
Under the condition of applied external field, the distribution of electrons within a
Brillouin zone is unchanged, even all electrons in k -space move according to
Ρππ
ππ‘= βππ¬
This is because some electrons will flow out of the Brillouin zone on the right side, more
electrons will flow into the Brillouin zone from the left side to fill up the empty states.
For partially filled energy band, it is easy to see why current is not zero under an electric
field.
Fig.2.14 Occupied πΈ β π states in partially filled band with zero and nonzero field.
Holes as empty states in valence band
The above analysis also provides an explanation for how an empty state in a valence band
acts as a hole β a conductive carrier.
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For completely filled band, the current density
π½ = βπ
π ππ
4π
π=1
= 0
where π is the volume since each band has 4π states. Now assume that some state π in the valence band is empty, we then can write the current density
π½π£ == βπ
π ππ
4π
π=1,πβ π
= βπ
π ππ
4π
π=1
β ππ =πππ
π
One empty state in the valence band acts as if it carries a positive charge π which conducts current.
Energy levels of impurities
Usually impurity levels lie in forbidden energy region. For intentionally doped donors
and acceptors, they are normally shallow levels in the range of πΈπΌ~0.001~0.01ππ. At room temperature, they are all ionized.
Fig.2.15 Impurity levels of donors and acceptors in forbidden band.
But some impurities tend to give deeper levels in the forbidden band, e.g. Au in Si; O in
Ge. They serve as recombination centers.
Compensation
Fig.2.16 Semiconductors doped with both donors and acceptors.
20
Electrons from the level πΈπ instead of going to the conduction band will drop into the acceptor states at πΈπ . Each such transition eliminates one electron-hole pair that would have been there if the transition could be prevented. Obviously, if ππ > ππ , p-type; and if ππ < ππ , n-type.
2.4 Equilibrium Carrier Concentration
In order to calculate the electron and hole concentrations in the conduction and valence
bands, we need to know the density of states and probability of occupancy of these states.
Density of States π πΈ
As we have learned the density of states in π-space is ππ =2π
2π 3, we need to convert this
into the density of states in energy π πΈ since the probability of occupancy described by the Fermi function is given in energy.
Assume that the constant energy surface is spherical so that the effective mass is a scalar
πΈ π = πΈπ +Ρ2π2
2ππβ
A constant energy surface of πΈ β πΈπ , the radius of the spherical surface is π =
2ππβ πΈ β πΈπ /Ρ. The volume encircled by this energy surface
4π
3ππ3 =
4π
3π 2ππ
β πΈ β πΈπ 3/2
Ρ3
The total number of states
2
2π 34π
3ππ3 =
2
2π 34π
3π 2ππ
β πΈ β πΈπ 3/2
Ρ3
=8π
3
2ππβ 3/2
3 πΈ β πΈπ
3/2
= ππ πΈ πΈβπΈπ
0
ππΈ
The density of state in energy
ππ πΈ =4π 2ππ
β 3/2
3 πΈ β πΈπ
1/2
for conduction band πΈ > πΈπ . Similarly
ππ£ πΈ =4π 2π
β 3/2
3 πΈπ£ β πΈ
1/2
for valence band πΈ < πΈπ£ .
21
Fermi-Dirac Distribution
The probability of occupancy of a state with energy πΈ by an electron
π πΈ =1
1 + exp πΈ β πΈπΉ /ππ
where πΈπΉ is the Fermi energy
Fig.2.17 Fermi-Dirac distribution
At π = 0πΎ,
π πΈ = 0, πΈ > πΈπΉ1, πΈ < πΈπΉ
,
in general,
π πΈ =
>
1
2, πΈ < πΈπΉ
=1
2, πΈ = πΈπΉ
<1
2, πΈ > πΈπΉ
In extreme cases when πΈβπΈπΉ
ππβ« 1, π πΈ = exp β πΈ β πΈπΉ /ππ - Maxwell-Boltzmann
Distribution, and when πΈβπΈπΉ
ππβͺ 1, π πΈ = 1.
Electron concentration
The number of electrons in an energy interval dE within the conduction band
ππ = π πΈ π πΈ ππΈ =4π 2ππ
β 3/2
3 πΈ β πΈπ
1/2 1 + exp πΈ β πΈπΉππ
β1
ππΈ
Total electron concentration distributed within the conduction band
22
π =4π 2ππ
β 3/2
3 πΈ β πΈπ
1/2πΈtop
πΈπ
1 + exp πΈ β πΈπΉππ
β1
ππΈ
This expression can be simplified if we assume that πΈπΉ lies below πΈπ by more than 3ππ so
π πΈ β exp β πΈ β πΈπΉ /ππ , an analytical expression can be obtained
π = ππ exp β πΈπ β πΈπΉ /ππ where
ππ = 2 2πππβππ/2 3/2
representing the density of states required at the conduction band edge πΈπwhich gives the concentration in the conduction band after multiplying with the probability of occupancy
at band edge. In reality, the density of states at πΈπ is zero as given by π πΈ β πΈ β πΈπ 1/2.
Hole concentration
The probability of a state not occupied by an electron
π πΈ = 1 β π πΈ =1
1 + exp πΈπΉ β πΈ /ππ
which can be reviewed as the probability of a state occupied by a hole. Similarly, the hole
concentration
π =4π 2π
β 3/2
3 πΈπ£ β πΈ
1/2πΈπ£
πΈbott om
1 + exp πΈπΉ β πΈ
ππ β1
ππΈ
Assuming that πΈπΉ lies above πΈπ£ by more than 3ππ so π πΈ β exp β πΈπΉ β πΈ /ππ ,
an analytical expression can be obtained
π = ππ£ exp β πΈπΉ β πΈπ£ /ππ
where
ππ£ = 2 2πππ£βππ/2 3/2
representing the density of states required at the valence band edge πΈπ£.
For semiconductors with anisotropic effective mass and multiple equivalent energy
minima in conduction band, all expressions concerning 2ππβ 3/2 should be modified as
π 8ππ₯βππ¦
βππ§β
1/2 or π 8ππ
βππ‘β2 1/2 where π is the number of equivalent energy valleys
(minima). We thus define a density-of-states effective mass with
2ππβ 3/2 = π 8ππ
βππ‘β2 1/2
such that ππβ = π 2ππ
βππ‘β2 1/3.
For intrinsic semiconductors (no doping, ππ = 0 and ππ = 0), we should have π = π =ππ , and we let πΈπ = πΈπΉ - the Fermi level of intrinsic semiconductor, then
ππ = ππ exp β πΈπ β πΈπ /ππ = ππ£ exp β πΈπ β πΈπ£ /ππ We can write in general
π = ππexp πΈπΉ β πΈπ /ππ , π = ππexp πΈπ β πΈπΉ /ππ
23
The electron-hole concentration product
ππ = ππ2 = ππ exp β πΈπ β πΈπ /ππ ππ£ exp β πΈπ β πΈπ£ /ππ
= ππππ£exp β πΈπ β πΈπ£ /ππ
= 32 π2ππ
βπβ
4
3/2
ππ 3exp βπΈπ/ππ
= π2 ππ 3exp βπΈπ/ππ
both ππβ and π
β are density-of-states effective masses.
The above expressions for electron and hole concentrations are valid for nondegenerate
semiconductors because the carrier concentrations are low so that the Fermi-Dirac (FD)
distribution function can be reduced to Maxwell-Boltzmann (MB) distribution.
For degenerate semiconductors, FD cannot be reduced to MB, the integral for n and p are
to be carried out numerically.
Temperature dependence of intrinsic carrier concentration
ππ = π ππ 3/2exp βπΈπ/2ππ
For most semiconductors, the band gap g
E decreases with the increase of temperature π,
πΈπ π = πΈπ0 β ππ accurate for 100πΎ < π < 400πΎ but inaccurate for low temperatures,
where πΈπ0 is the extrapolated value of the band gap at π = 0πΎ. The intrinsic concentration
ππ = π1 ππ 3/2exp βπΈπ0/2ππ
where π1 = πexp π/2π .
Carrier concentrations in extrinsic semiconductors
Ionization of impurities
As donors in semiconductors can lose one electron to become positively charged e.g.
As0 β π β As+ Let
ππ : total donor concentration ππ
0: neutral donor concentration
ππ+ : ionized donors
We obviously should have ππ = ππ0 + ππ
+. The occupation probability of impurity level
πΈπ is different from the regular FD distribution because of the spin degeneracy of the donor levels. When a donor is ionized, there are two possible quantum states
corresponding to each of the two allowed spins. An electron can occupy any one of these
states with the condition that as soon as one of them is occupied, the occupancy of the
other is prohibited.
Taking this into consideration, the FD distribution needs to be modified for impusities
π πΈπ =ππ
0
ππ= 1 +
1
2exp
πΈπ β πΈπΉππ
β1
24
The concentration of ionized donors
ππ+ = ππ β ππ
0 = ππ 1 +1
2exp
πΈπ β πΈπΉππ
β1
Similarly for acceptors with ionization process
Al0 + π β Alβ we should have ππ = ππ
0 + ππβ, occupation probability of the acceptor level πΈπ
π πΈπ =ππβ
ππ= 1 +
1
2exp
πΈπ β πΈπΉππ
β1
For n-type semiconductors, πΈπ β πΈπ is on the order of ππ at room temperature (π =300K), and for nondegenerate semiconductors, πΈπ β πΈπΉ > 3ππ. The ratio of electrons attached to the impurities to that in the conduction band
ππ0
π=ππ 1 +
12 exp
πΈπ β πΈπΉππ
β1
ππexp βπΈπ β πΈπΉππ
Under normal condition, πΈπ β πΈπΉ > 3ππ, then exp πΈπβπΈπΉ
ππ β« 1 and
ππ0
π=
2ππππ
exp πΈπ β πΈπΉππ
~ππππ
- on the same order, since πΈπ β πΈπ~ππ. For typical doping ππ < 1017/ππ3, most
impurities are ionized at room temperature with less than 1% remain unionized. We
should have π = ππ .
Consider a nondegenerate semiconductor to which impurities have been introduced. To
keep the discussion in general, we assume there are donors (ππ ) and acceptors (ππ ). Based on the condition of charge neutrality,
ππ+ + π = ππ
β + π which can be separated according to mobile and immobile charges as
π β π = ππ+ β ππ
β.
At π > 100πΎ, we should have practically all impurities ionized, ππ+ = ππ and ππ
β = ππ , thus
π β π = ππ β ππ . Since ππ = ππ
2, we arrive
π =1
2 ππ β ππ + ππ β ππ 2 + 4ππ
2
π =1
2 ππ βππ + ππ β ππ 2 + 4ππ
2
For intrinsic situation where ππ = ππ = 0 as well as for completely compensated doping where ππ = ππ , they reduces to π = π = ππ .
For net n-type doping where ππ β ππ β« 2ππ , we obtain
25
π β ππ β ππ β ππ if ππ β« ππ and π = ππ2/ππ (case of strongly extrinsic)
For net p-type doping where ππ β ππ β« 2ππ , we obtain π β ππ β ππ β ππ if ππ β« ππ and π = ππ
2/ππ (case of strongly extrinsic)
2.5 The Fermi level and Energy Distribution of Carriers
Intrinsic semiconductors (π = π), we have
ππexp βπΈπ β πΈπππ
= ππ£exp βπΈπ β πΈπ£ππ
Then
πΈπ =πΈπ + πΈπ£
2+
1
2ππln
ππππ£
=πΈπ + πΈπ£
2+
3
4ππln
πβ
ππβ
where πΈπ+πΈπ£
2 is the center of the band gap. Obviously, for π
β = ππβ , πΈπ is at the midpoint
exactly; for πβ > ππ
β , πΈπ is above the midpoint; for πβ < ππ
β , πΈπ is below the midpoint.
As a good approximation, 3
4ππln
πβ
ππβ is small for most semiconductors, πΈπ is roughly at
the midgap for intrinsic case.
Extrinsic semiconductors
n-type: π = ππ+, then
ππexp βπΈπ β πΈπΉππ
= ππ 1 + 2exp πΈπΉ β πΈπππ
β1
Introduce π = ππ
2ππ
1/2
exp βππ
2ππ where the ionization energy ππ = πΈπ β πΈπ . The
above equation can be written as 1
1 + 2exp πΈπΉ β πΈπππ
= 2π2exp
πΈπΉ β πΈπππ
Then
πΈπΉ = πΈπ + ππln 4 + π2 β π
4π
And
π = ππ π
2 4 + π2 β π
Now let us discuss the above expressions under different temperatures corresponding to:
(1) Weak ionization π βͺ 1 πΈπΉ = πΈπ + ππππ 1/2π (πΈπΉ above πΈπ ) and π = πππ (few donors are ionized) (2) Complete ionization π β« 1
26
πΈπΉ = πΈπ + ππππ 1/2π = πΈπ + ππππ ππ/ππ (πΈπΉ below πΈπ ) and π = ππ (all donors are ionized)
Since π is a function of temperature, given ππ and ππ (type of donors), we can determine a temperature which divides the situations of weak ionization and complete ionization
(transition temperature from weak to complete ionizations: π = 1). This temperature can be solved by
π = ππ
2ππ
1/2
exp βππ
2ππ = 1
And keep in mind that ππ = ππ π .
It can be seen that as T increases, π increases, and πΈπΉ moves from above πΈπ to below πΈπ . In fact, πΈπΉ decreases approximately linearly with the increase of temperature.
It is obvious that as the temperature continues to increase, πΈπΉ approaches πΈπ . At this point, the hole concentration approaches the electron concentration and cannot be
neglected any more, we must use
π = π +ππ With ππ = ππ
2, then
π =1
2ππ 1 + 1 +
4ππ2
ππ2
1/2
(1) ππ/ππ βͺ 1, π = ππ complete ionization (2) ππ/ππ β« 1, π = ππ intrinsic
The transition temperature from complete ionization to intrinsic can be determined by
setting π = ππ .
Temperature dependence of carrier concentration in n-type semiconductors
(1) At low temperatures, πΈπΉ > πΈπ , π < ππ , π βͺ π, weak ionization (2) At moderate temperatures, πΈπΉ < πΈπ , π = ππ , π βͺ π, complete ionization (3) At high temperatures, πΈπΉ β πΈπ , π β ππ β« ππ , π β π, intrinsic
27
Fig. 2.18 Temperature dependence of carrier concentration
Compensation
Consider semiconductors doped with both donors and acceptors, the temperature
dependence of the carrier concentration is different from that of doped with either donors
or accepters only at low temperatures.
Assume ππ > ππ , there will be ππelectrons make a transition from πΈπ to πΈπ . The remaining donors ππ β ππ can be excited into the conduction band.
The neutrality condition is then π + ππ = π + ππ
+
since all acceptors are ionized (occupied by electron) ππβ = ππ . Since donors are
partially ionized at least, πΈπΉ is in the neighborhood of πΈπ much closer to πΈπ than πΈπ£, thus, π βͺ π, or
π + ππ = ππ+ =
ππ
1 + 2exp πΈπΉ β πΈπππ
But exp πΈπΉβπΈπ
ππ =
π
ππexp
ππ
ππ where ππ = πΈπ β πΈπ is the ionization energy. Rearrange
the above equation
π π + ππ
ππ β ππ β π=ππ2
exp βππππ
Letβs assume light compensation, i.e. ππ β« ππ , we can examine two cases (1) π βͺ ππ (extremely low temperature)
28
π =ππ ππ β ππ
2ππexp β
ππππ
Since π = ππexp βπΈπβπΈπΉ
ππ , we have
πΈπΉ = πΈπ + ππlnππ β ππ
2ππ
πΈπΉ does not depend on the parameters of conduction band, only depends on π,ππ ,ππ , in this case. This is because the electrons are mainly distributed between the donor and
acceptor levels. The temperature dependence of π is then
lnπ = ππππ ππ β ππ
2ππβπππ
1
π
(2) π β« ππ (slightly higher temperature) We still have π βͺ ππ , then
π = ππππ
2
1/2
exp βππππ
This is the same expression for electron concentration at low temperature (weak
ionization) when the semiconductor was doped with donors only. This is because the
electrons are mainly distributed between the donor level and the conduction band, the
existence of small quantity of acceptors does not alter the distribution
πΈπΉ =πΈπ + πΈπ
2+
1
2ππππ
ππ2ππ
The temperature dependence of π is then
lnπ = ππ ππππ
2
1/2
βππ2π
1
π
Apparently, the slopes for lnπ~1
π are different between the two cases: a factor of Β½. The
slope at low temperature is twice as much as that at slightly higher temperature. At the
transition temperature, we should have π = ππ . Based on this behavior, one can tell whether the semiconductor is compensated or not.
29
Fig.2.19 Temperature dependence of n for compensated semiconductors.
As temperature increases beyond the weak ionization, we can have all donors ionized, but
not all electrons will be in the conduction band, actually, π = ππ β ππ (complete ionization). Further increase the temperature, the semiconductor reaches the intrinsic
region where π = π = ππ β« ππ β ππ . The major difference between a compensated and uncompensated semiconductor is the temperature dependence at the low temperature
region.
Energy Distribution of Carriers
Both π = ππexp βπΈπβπΈπΉ
ππ and π = ππ£exp β
πΈπΉβπΈπ£
ππ do not tell how electrons and holes
are distributed within the conduction and valence bands. If we need to know the
concentration as a function, we need π πΈ π πΈ where π πΈ is the density of states and π πΈ is the FD function.
Fig.2.20 Illustration of π πΈ , π πΈ , and π πΈ π πΈ .