chapter 2 semiconductor materials and their properties...compound semiconductors: iii-v and ii-vi...

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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 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 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 𝑇 = 0K, all electrons are tightly bound in the bonds – 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 –

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 – 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

𝑟0 = 0.53 𝜀𝑠𝜀0 𝑚0𝑚𝑒∗ Å

𝐸𝐼 = 13.6 𝑚0𝑚𝑒∗ 𝜀𝑠𝜀0

2

eV

Two modifications to the hydrogen formula:

(1) effective mass: 𝑚𝑒∗ < 𝑚0 (free electron mass)

(2) permittivity of the semiconductor: 𝜀𝑠 > 𝜀0 (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 – 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 – 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 – compensation.

However, if 𝑁𝑑 > 𝑁𝑎 , we get n-type; if 𝑁𝑑 < 𝑁𝑎 , 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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∆𝐸 Zn : energy required to create a Zn vacancy 2𝐸𝑔 : energy released when two donor electrons drop into the Zn vacancy.

If 2𝐸𝑔 > ∆𝐸 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 𝑁 identical atoms bound to form a crystal, each energy level will split into 𝑁 energy levels. Since 𝑁 is usually very large, the density of these energy levels is high. It can be treated as they form a quasi-continuous energy band – allowed bands separated by

forbidden band – 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 𝑁 atoms form a crystal, one degenerate level splits into 𝑁 energy levels which allow totally 2𝑁 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 2𝑁 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 2𝑁 states. The other p-states with 6𝑁 states. It seems that the 6𝑁 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 4𝑁 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

−ћ2

2𝑚∗∇2 + 𝑈 𝒓 𝛷 𝒓 = 𝐸𝛷 𝒓

Where 𝑚∗ is the effective mass which is different from the free electron mass due to the interaction with other electrons and 𝑈 𝒓 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

𝛷𝒌 𝒓 = 𝑢𝒌 𝒓 𝑒𝑖𝒌∙𝒓

where 𝑢𝒌 𝒓 has the period of the crystal lattice with 𝑢𝒌 𝒓+ 𝑻 = 𝑢𝒌 𝒓

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 𝑒𝑖𝒌∙𝒓 and a function 𝑢𝒌 𝒓 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 𝑒𝑖𝒌∙𝒓, 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, 𝑢𝒌 𝒓 describes the behavior of an electron around a lattice atom. 𝑒𝑖𝒌∙𝒓

demonstrates that the electron is not localized, but can propagate throughout crystal.

For a crystal with infinitely large volume, the value of 𝒌 can vary continuously. Given a 𝒌, there exists a set of eigen energies 𝐸𝑛(𝒌) and corresponding 𝛷𝒏 𝒌, 𝒓 . The quantum number 𝑛 indicates different energy band, intuitively can be considered as they are originated from different atomic energy levels. 𝐸𝑛(𝒌) is therefore a continuous function of 𝒌. Due to the periodicity of a crystal, an electronic does not have a unique value of 𝒌. In fact, 𝒌′ = 𝒌+𝑲 can represent the same state as 𝒌 does where 𝑲 is the so-called reciprocal lattice vector

𝑲 = 𝑛1𝒃1 + 𝑛2𝒃2 + 𝑛3𝒃3 Where 𝑛1 ,𝑛2, 𝑛3 are integers and 𝒃1,𝒃2,𝒃3 are primitive vectors of the reciprocal lattice which can be constructed by the primitive vectors of the crystal lattice (𝒂1,𝒂2,𝒂3)

𝒃1 = 2𝜋𝒂2 × 𝒂3

𝒂1 ∙ 𝒂2 × 𝒂3 ,𝒃2 = 2𝜋

𝒂3 × 𝒂1𝒂1 ∙ 𝒂2 × 𝒂3

,𝒃3 = 2𝜋𝒂1 × 𝒂2

𝒂1 ∙ 𝒂2 × 𝒂3

It is easy to prove that

𝒂𝑖 ∙ 𝒃𝑗 = 2𝜋𝛿𝑖𝑗 Actually, the Bloch wave function can be written as

𝛷𝒏 𝒌, 𝒓 = 𝑢𝑛 ,𝒌 𝒓 𝑒𝑖𝒌∙𝒓 = 𝑢𝑛 ,𝒌 𝒓 𝑒

−𝑖𝑲∙𝒓 𝑒𝑖𝒌′ ∙𝒓 = 𝑢𝑛 ,𝒌′ 𝒓 𝑒

𝑖𝒌′ ∙𝒓

where 𝑢𝑛 ,𝒌′ 𝒓 = 𝑢𝑛 ,𝒌 𝒓 𝑒−𝑖𝑲∙𝒓 has the same periodicity as the lattice since

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𝑒−𝑖𝑲∙ 𝒓+𝑻 = 𝑒−𝑖𝑲∙𝒓, 𝑲 ∙ 𝑻 = 2𝑛𝜋 Thus, 𝒌′ represents the same state as 𝒌 does.

Unlike the situation for free electrons, the nonuniqueness of 𝒌 suggests that strictly speaking, ћ𝒌 will not carry the meaning of momentum. However, we will see later ћ𝒌 can still be treated as if it is the momentum of an electron in a crystal.

Obviously, 𝐸𝑛(𝒌) various with 𝒌 periodically and for different n, 𝐸𝑛(𝒌) varies within different energy intervals separated by energy gaps where electron states are forbidden.

Fig. 2.6 𝐸𝑛(𝒌) as a function of 𝒌.

If we take the complex conjugate of the Schrodinger equation, H remains unchanged,

therefore 𝐸𝑛(𝒌) should have even symmetry with respect to 𝒌, i.e. 𝐸𝑛 𝒌 = 𝐸𝑛(−𝒌) since

𝛷𝒌∗ 𝒓 = 𝑢𝒌

∗ 𝒓 𝑒−𝑖𝒌∙𝒓 is the eigen function of 𝐸𝑛(𝒌) also.

Brillouin Zone

Since 𝒌′ = 𝒌+𝑲 represents the same electron state as 𝒌 does with 𝑲 being the reciprocal lattice vector. We can actually limit 𝒌 within a primitive unit cell in the reciprocal lattice, because 𝒌 and 𝒌′ 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 𝒌.

For example in 1-D, we can limit – 𝜋/2 < 𝑘 < 𝜋/𝑎.

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 𝐿1 × 𝐿2 × 𝐿3 with the dimension of a primitive cell of 𝑎1 × 𝑎2 × 𝑎3. For any state 𝒌 = 𝑘1𝒙 + 𝑘2𝒚 + 𝑘3𝒛 , the periodic condition requires that

𝛷𝑛𝒌 0,𝑦, 𝑧 = 𝛷𝑛𝒌 𝐿1,𝑦, 𝑧 ,𝛷𝑛𝒌 𝑥, 0, 𝑧 = 𝛷𝑛𝒌 𝑥, 𝐿2, 𝑧 ,𝛷𝑛𝒌 𝑥,𝑦, 0 = 𝛷𝑛𝒌 𝑥,𝑦, 𝐿3 Thus

𝑘1𝐿1 = 𝑛1 2𝜋 ,𝑘2𝐿2 = 𝑛2 2𝜋 , 𝑘3𝐿3 = 𝑛3 2𝜋 .

Since, 𝐿1 = 𝑁1𝑎1, 𝐿2 = 𝑁2𝑎2, 𝐿3 = 𝑁3𝑎3, then

𝑘𝑖 =𝑛𝑖𝑁𝑖

2𝜋

𝑎𝑖, 𝑖 = 1,2,3

with 𝑛𝑖 ≤𝑁𝑖

2 since – 𝜋/2 < 𝑘𝑖 < 𝜋/𝑎. Therefore, ∆𝑘𝑖 =

2𝜋

𝐿𝑖, then one state in 𝒌 -space

takes a volume of

∆𝑘1∆𝑘2∆𝑘3 =2𝜋

𝐿1

2𝜋

𝐿2

2𝜋

𝐿3= 2𝜋 3

𝑉

where 𝑉 is the volume of crystal.

Taking into account that each state can accommodate two electrons with opposite spin,

the density of states in 𝒌-space is then

𝑔𝑘 =2𝑉

2𝜋 3,

and within the Brillouin zone there are totally 𝑁 = 𝑁1𝑁2𝑁3 allowed 𝒌 states, which equals the number of primitive cells.

It can be shown that if the volume of a primitive cell in lattice is 𝛺, the volume of a reciprocal primitive cell is 2𝜋 3/𝛺.

Motion of Electrons in a Crystal

The time-dependent solution to the Schrodinger equation is

𝛷𝒌 𝒓, 𝑡 = 𝑢𝒌 𝒓 𝑒𝑖 𝒌∙𝒓−𝜔𝑡

where 𝜔 = 𝐸/ћ. Due to the linearity of the time-dependent Schrodinger equation, the superposition of the above wave function with different 𝒌

𝛷 𝒓, 𝑡 = 𝐶𝑚𝑚

𝑢𝒌𝑚 𝒓 𝑒𝑖 𝒌𝑚 ∙𝒓−𝜔𝑚 𝑡

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will still be a solution to the Schrodinger equation. Since 𝒌𝑚 ’s are closely packed in 𝒌-space and can be treated as continuous, the superposition becomes an integral

𝛷 𝒓, 𝑡 = 𝐶 𝒌 𝜋/2

−𝜋/2

𝑢𝒌 𝒓 𝑒𝑖 𝒌∙𝒓−𝜔 𝒌 𝑡 𝑑3𝒌

Now at some instant of time, one tries to look at a wave packet that consists of Bloch

functions in the neighborhood of 𝒌, then the above integral can be rewritten as

𝛷 𝒓, 𝑡 = 𝐶 𝒌+ 𝜼 ∆𝒌

−∆𝒌

𝑢𝒌+𝜼 𝒓 𝑒𝑖 𝒌+𝜼 ∙𝒓−𝜔 𝒌+𝜼 𝑡 𝑑3𝜼

Since 𝜼 < ∆𝒌 are small, then 𝑢𝒌+𝜼 𝒓 ≈ 𝑢𝒌 𝒓 and

𝜔 𝒌+ 𝜼 = 𝜔 𝒌 + ∇𝒌𝜔 𝒌 ∙ 𝜼 +⋯ which is related to energy by 𝜔 = 𝐸/ћ. Therefore

𝛷 𝒓, 𝑡 = 𝑢𝒌 𝒓 𝑒𝑖 𝒌∙𝒓−𝜔 𝒌 𝑡 𝐶 𝒌+ 𝜼

∆𝒌

−∆𝒌

𝑒𝑖 𝜼∙ 𝒓−∇𝒌𝜔𝑡 𝑑3𝜼

The factor in front of the integral is a Bloch function. The integral represents a wave

packet with a center position 𝒓 = ∇𝒌𝜔𝑡 which moves with the velocity

𝒗 =𝑑𝒓

𝑑𝑡= ∇𝒌𝜔 𝒌 =

1

ћ∇𝒌𝐸 𝒌

which is the group velocity of electron in a crystal.

Now let’s consider a 1-D situation, with

𝐶 𝑘 + 𝜂 = 𝐶, 𝜂 < ∆𝑘

0, 𝜂 > ∆𝑘

Then

𝛷 𝑥, 𝑡 = 𝐶𝑢𝑘 𝑥 𝑒𝑖 𝑘∙𝑥−𝜔 𝑘 𝑡 𝑒𝑖 𝜂 ∙ 𝑥−∇𝑘𝜔𝑡 𝑑𝜂

∆𝑘

−∆𝑘

= 𝐶𝑢𝑘 𝑥 𝑒𝑖 𝑘∙𝑥−𝜔 𝑘 𝑡

sin ∆𝑘 𝑥 − 𝑑𝜔/𝑑𝑘 𝑡

∆𝑘 𝑥 − 𝑑𝜔/𝑑𝑘 𝑡 2∆𝑘

Hence, the wave packet probability

𝛷 𝑥, 𝑡 2 ∝ 𝑢𝑘 𝑥 2

sin ∆𝑘 𝑥 − 𝑑𝜔/𝑑𝑘 𝑡

∆𝑘 𝑥 − 𝑑𝜔/𝑑𝑘 𝑡

2

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Fig.2.7 1-D electron distribution described by a wave packet with a wave vector

distribution within 𝑘 + ∆𝑘.

The center peak is bound by 𝑥 − 𝑑𝜔/𝑑𝑘 𝑡 = ±𝜋/∆𝑘, i.e. the half width ∆𝑥 is related to ∆𝑘 by ∆𝑥∆𝑘 = 𝜋 - exactly what is required by the uncertainty principle. For a wave packet with a space expansion of lattice constant 𝑎, we have ∆𝑘 = 𝜋/𝑎 covers the entire Brillouin zone, i.e. 𝑘 is completely uncertain, and vice versa. However, when we are dealing with the ranges of 𝒌 and 𝒓 that are much greater than the uncertainty ∆𝒌 and ∆𝒓, we can consider both 𝒌 and 𝒓 have precise values – quasi classic.

Effective Mass

Suppose that an external field is applied, and the electron moves a distance 𝑑𝒓, the work done on the electron is

𝑑𝑤 = 𝑭 ∙ 𝑑𝒓 = 𝑭 ∙ 𝒗𝑑𝑡 = 𝑭 ∙ ∇𝒌𝜔𝑑𝑡 = 𝑭 ∙ ∇𝒌𝐸𝑑𝑡/ћ The work done on the electron causes its energy to change

𝑑𝐸 𝒌 = ∇𝒌𝐸 ∙ 𝑑𝒌 = ∇𝒌𝐸 ∙ 𝑑𝒌/𝑑𝑡 𝑑𝑡 Since 𝑑𝑤 = 𝑑𝐸, therefore 𝑭 = ћ 𝑑𝒌/𝑑𝑡 = 𝑑𝒑/𝑑𝑡 which is analogue to the Newton’s law with 𝒑 = ћ𝒌 representing the momentum of the electron, similar to that of free electron.

The acceleration of an electron in a crystal is 𝑑𝒗

𝑑𝑡=

1

ћ

𝑑

𝑑𝑡 ∇𝒌𝐸 =

1

ћ 𝑑𝒌

𝑑𝑡∙ ∇𝒌 ∇𝒌𝐸 =

1

ћ2 𝑭 ∙ ∇𝒌 ∇𝒌𝐸

In tensor form, we can write 𝑑𝒗

𝑑𝑡=

1

𝑚∗ 𝑭

where the inverse effective mass tensor

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1

𝑚∗ =

1

ћ2

𝜕2𝐸/𝜕𝑘𝑥2 𝜕2𝐸/𝜕𝑘𝑥𝜕𝑘𝑦 𝜕

2𝐸/𝜕𝑘𝑥𝜕𝑘𝑧

𝜕2𝐸/𝜕𝑘𝑦𝜕𝑘𝑥 𝜕2𝐸/𝜕𝑘𝑦

2 𝜕2𝐸/𝜕𝑘𝑦𝜕𝑘𝑧

𝜕2𝐸/𝜕𝑘𝑧𝜕𝑘𝑥 𝜕2𝐸/𝜕𝑘𝑧𝜕𝑘𝑦 𝜕

2𝐸/𝜕𝑘𝑧2

is a function of 𝒌. The tensor is symmetric, we can choose a proper coordinate system so that the tensor is diagonalized. Therefore, we can have for each direction

𝑚𝛼∗𝑑𝑣𝛼𝑑𝑡

= 𝐹𝛼 , 𝛼 = 𝑥,𝑦, 𝑧

which is analogue of Newton’s 2nd

law.

Usually in semiconductors, we only deal with those states that are close to band edges (𝐸𝑐 and 𝐸𝑣) because electrons are distributed a few 𝑘𝑇 around the band edges.

For conduction band near 𝐸𝑐 with 𝒌 = 0, we can have

𝐸𝑐 𝒌 = 𝐸𝑐 0 +1

2 𝜕2𝐸𝑐𝜕𝑘𝑥2

𝑘𝑥2 +

𝜕2𝐸𝑐𝜕𝑘𝑦2

𝑘𝑦2 +

𝜕2𝐸𝑐𝜕𝑘𝑧2

𝑘𝑧2 +⋯

= 𝐸𝑐 0 +ћ2

2 𝑘𝑥

2

𝑚𝑐 ,𝑥∗+

𝑘𝑦2

𝑚𝑐,𝑦∗+𝑘𝑧

2

𝑚𝑐 ,𝑧∗ +⋯

Since 𝐸𝑐 0 is a minimum, 𝜕𝐸𝑐

𝜕𝑘𝛼= 0, and

𝜕2𝐸𝑐

𝜕𝑘𝑥2 > 0, therefore, 𝑚𝑐 ,𝛼

∗ > 0. Near the band

edge, 𝐸𝑐 𝒌 has a parabolic relation with 𝒌.

For crystals with cubic symmetry, we have 𝑚𝑐 ,𝑥∗ = 𝑚𝑐,𝑦

∗ = 𝑚𝑐 ,𝑧∗ = 𝑚𝑒

∗ at 𝒌 = 0. Then

𝐸𝑐 𝒌 = 𝐸𝑐 0 +ћ2𝑘2

2𝑚𝑒∗

and 𝑚𝑒∗𝑑𝒗

𝑑𝑡= 𝑭, the same expressions as free electrons except that we have to use the

effective mass 𝑚𝑒∗ to replace the free electron mass.

For valence band near the top 𝐸𝑣 with 𝒌 = 0, we can have

𝐸𝑣 𝒌 = 𝐸𝑣 0 +1

2 𝜕2𝐸𝑣𝜕𝑘𝑥2

𝑘𝑥2 +

𝜕2𝐸𝑣𝜕𝑘𝑦2

𝑘𝑦2 +

𝜕2𝐸𝑣𝜕𝑘𝑧2

𝑘𝑧2 +⋯

= 𝐸𝑣 0 +ћ2

2 𝑘𝑥

2

𝑚𝑣,𝑥∗+

𝑘𝑦2

𝑚𝑣,𝑦∗+𝑘𝑧

2

𝑚𝑣,𝑧∗ +⋯

= 𝐸𝑣 0 +ћ2𝑘2

2𝑚𝑣∗ +⋯

for cubic crystal. Since 𝐸𝑣 0 is a maximum, 𝜕2𝐸𝑣

𝜕𝑘𝑥2 < 0 which will result in negative

electron effective mass (𝑚𝑣∗ < 0) near the top of the valence band. However, when we

look at the Newton’s law expression of these electrons under an electric field 𝑬,

𝑚𝑣∗𝑑𝒗

𝑑𝑡= 𝑭 = −𝑒𝑬

14

the holes are introduced as having positive charge, therefore, the above Newton’s law

should be modified

−𝑚𝑣∗𝑑𝒗

𝑑𝑡= 𝑚

∗𝑑𝒗

𝑑𝑡= 𝑒𝑬

where the hole effective mass 𝑚∗ = −𝑚𝑣

∗ 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

𝐸𝑣 𝒌 = 𝐸𝑣 0 −ћ2𝑘2

2𝑚∗

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 𝐸

and 𝒌 = 𝑘𝑥 ,𝑘𝑦 ,𝑘𝑧 which is difficult to plot in 3-D. Typically, the relationship is

plotted along the principle directions of the crystal.

15

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 𝒌 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 𝒌’s in 𝒌-space.

For direct band gap semiconductor, a photon with energy ћ𝜔 = 𝐸𝑔 = 𝐸𝑐 − 𝐸𝑣 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 ћ𝜔 > 𝐸𝑔 = 𝐸𝑐 − 𝐸𝑣, because photons have very small momentum and the

absorption of a photon needs to be a vertical transition in 𝒌-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 𝒌-space where each point on the surface has the same energy.

For a semiconductor with its conduction band minimum at 𝒌 = 0 and isotropic effective mass, obviously the constant energy surface is spherical,

http://upload.wikimedia.org/wikipedia/commons/5/53/Fcc_brillouin.png

16

𝐸 = 𝐸𝑐 +ћ2𝑘2

2𝑚𝑒∗

Fig. 2.10 Constant energy surface for conduction band minimum at 𝒌 = 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 𝒌 = 0, the center of the constant energy surface is not located at 𝒌 = 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 ¾ of the distance from the Brillouin zone center.

𝐸𝑐 𝒌 = 𝐸𝑐 +ћ2

2 𝑘𝑥 − 𝑘𝑥0

2

𝑚𝑙∗ +

𝑘𝑦2 + 𝑘𝑧

2

𝑚𝑡∗

where 𝑚𝑙∗ and 𝑚𝑡

∗ are the longitudinal and transverse effective mass.

Fig.2.11 Constant energy surfaces in Si and Ge.

17

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 𝐸 − 𝒌 relationship. There are times when the detail description is not necessary when we are

only interested in the band gap 𝐸𝑔 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 𝒌 and energy 𝐸 𝒌 , there must be another state with momentum −𝒌 and 𝐸 −𝒌 = 𝐸 𝒌 due to the symmetry of the crystal. Obviously, ∇𝑘𝐸 𝒌 = −∇𝑘𝐸 −𝒌 which implies that electron at 𝒌 has same magnitude but opposite velocity as the electron at −𝒌. If the band is completely filled, then both states are occupied with electrons. The contribution to

current from the two electrons with 𝒌 and −𝒌 will be cancelled. Hence, the total current is zero.

18

Fig. 2.13 Illustration of the symmetry of 𝐸 − 𝑘 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

ћ𝑑𝒌

𝑑𝑡= −𝑒𝑬

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 𝐸 − 𝑘 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 – a conductive carrier.

19

For completely filled band, the current density

𝐽 = −𝑒

𝑉 𝒗𝑖

4𝑁

𝑖=1

= 0

where 𝑉 is the volume since each band has 4𝑁 states. Now assume that some state 𝑗 in the valence band is empty, we then can write the current density

𝐽𝑣 == −𝑒

𝑉 𝒗𝑖

4𝑁

𝑖=1,𝑖≠𝑗

= −𝑒

𝑉 𝒗𝑖

4𝑁

𝑖=1

− 𝒗𝑗 =𝑒𝒗𝑗

𝑉

One empty state in the valence band acts as if it carries a positive charge 𝑒 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 𝐸𝐼~0.001~0.01𝑒𝑉. 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 𝐸𝑑 instead of going to the conduction band will drop into the acceptor states at 𝐸𝑎 . Each such transition eliminates one electron-hole pair that would have been there if the transition could be prevented. Obviously, if 𝑁𝑎 > 𝑁𝑑 , p-type; and if 𝑁𝑎 < 𝑁𝑑 , 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 𝑁 𝐸

As we have learned the density of states in 𝒌-space is 𝑔𝑘 =2𝑉

2𝜋 3, we need to convert this

into the density of states in energy 𝑁 𝐸 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

𝐸 𝑘 = 𝐸𝑐 +ћ2𝑘2

2𝑚𝑒∗

A constant energy surface of 𝐸 − 𝐸𝑐 , the radius of the spherical surface is 𝑘 =

2𝑚𝑒∗ 𝐸 − 𝐸𝑐 /ћ. The volume encircled by this energy surface

4𝜋

3𝜋𝑘3 =

4𝜋

3𝜋 2𝑚𝑒

∗ 𝐸 − 𝐸𝑐 3/2

ћ3

The total number of states

2

2𝜋 34𝜋

3𝜋𝑘3 =

2

2𝜋 34𝜋

3𝜋 2𝑚𝑒

∗ 𝐸 − 𝐸𝑐 3/2

ћ3

=8𝜋

3

2𝑚𝑒∗ 3/2

3 𝐸 − 𝐸𝑐

3/2

= 𝑁𝑐 𝐸 𝐸−𝐸𝑐

0

𝑑𝐸

The density of state in energy

𝑁𝑐 𝐸 =4𝜋 2𝑚𝑒

∗ 3/2

3 𝐸 − 𝐸𝑐

1/2

for conduction band 𝐸 > 𝐸𝑐 . Similarly

𝑁𝑣 𝐸 =4𝜋 2𝑚

∗ 3/2

3 𝐸𝑣 − 𝐸

1/2

for valence band 𝐸 < 𝐸𝑣 .

21

Fermi-Dirac Distribution

The probability of occupancy of a state with energy 𝐸 by an electron

𝑓 𝐸 =1

1 + exp 𝐸 − 𝐸𝐹 /𝑘𝑇

where 𝐸𝐹 is the Fermi energy

Fig.2.17 Fermi-Dirac distribution

At 𝑇 = 0𝐾,

𝑓 𝐸 = 0, 𝐸 > 𝐸𝐹1, 𝐸 < 𝐸𝐹

,

in general,

𝑓 𝐸 =

>

1

2, 𝐸 < 𝐸𝐹

=1

2, 𝐸 = 𝐸𝐹

<1

2, 𝐸 > 𝐸𝐹

In extreme cases when 𝐸−𝐸𝐹

𝑘𝑇≫ 1, 𝑓 𝐸 = exp − 𝐸 − 𝐸𝐹 /𝑘𝑇 - Maxwell-Boltzmann

Distribution, and when 𝐸−𝐸𝐹

𝑘𝑇≪ 1, 𝑓 𝐸 = 1.

Electron concentration

The number of electrons in an energy interval dE within the conduction band

𝑑𝑛 = 𝑁 𝐸 𝑓 𝐸 𝑑𝐸 =4𝜋 2𝑚𝑒

∗ 3/2

3 𝐸 − 𝐸𝑐

1/2 1 + exp 𝐸 − 𝐸𝐹𝑘𝑇

−1

𝑑𝐸

Total electron concentration distributed within the conduction band

22

𝑛 =4𝜋 2𝑚𝑒

∗ 3/2

3 𝐸 − 𝐸𝑐

1/2𝐸top

𝐸𝑐

1 + exp 𝐸 − 𝐸𝐹𝑘𝑇

−1

𝑑𝐸

This expression can be simplified if we assume that 𝐸𝐹 lies below 𝐸𝑐 by more than 3𝑘𝑇 so

𝑓 𝐸 ≈ exp − 𝐸 − 𝐸𝐹 /𝑘𝑇 , an analytical expression can be obtained

𝑛 = 𝑁𝑐 exp − 𝐸𝑐 − 𝐸𝐹 /𝑘𝑇 where

𝑁𝑐 = 2 2𝜋𝑚𝑒∗𝑘𝑇/2 3/2

representing the density of states required at the conduction band edge 𝐸𝑐which gives the concentration in the conduction band after multiplying with the probability of occupancy

at band edge. In reality, the density of states at 𝐸𝑐 is zero as given by 𝑁 𝐸 ∝ 𝐸 − 𝐸𝑐 1/2.

Hole concentration

The probability of a state not occupied by an electron

𝑓 𝐸 = 1 − 𝑓 𝐸 =1

1 + exp 𝐸𝐹 − 𝐸 /𝑘𝑇

which can be reviewed as the probability of a state occupied by a hole. Similarly, the hole

concentration

𝑝 =4𝜋 2𝑚

∗ 3/2

3 𝐸𝑣 − 𝐸

1/2𝐸𝑣

𝐸bott om

1 + exp 𝐸𝐹 − 𝐸

𝑘𝑇 −1

𝑑𝐸

Assuming that 𝐸𝐹 lies above 𝐸𝑣 by more than 3𝑘𝑇 so 𝑓 𝐸 ≈ exp − 𝐸𝐹 − 𝐸 /𝑘𝑇 ,

an analytical expression can be obtained

𝑝 = 𝑁𝑣 exp − 𝐸𝐹 − 𝐸𝑣 /𝑘𝑇

where

𝑁𝑣 = 2 2𝜋𝑚𝑣∗𝑘𝑇/2 3/2

representing the density of states required at the valence band edge 𝐸𝑣.

For semiconductors with anisotropic effective mass and multiple equivalent energy

minima in conduction band, all expressions concerning 2𝑚𝑒∗ 3/2 should be modified as

𝑠 8𝑚𝑥∗𝑚𝑦

∗𝑚𝑧∗

1/2 or 𝑠 8𝑚𝑙

∗𝑚𝑡∗2 1/2 where 𝑠 is the number of equivalent energy valleys

(minima). We thus define a density-of-states effective mass with

2𝑚𝑒∗ 3/2 = 𝑠 8𝑚𝑙

∗𝑚𝑡∗2 1/2

such that 𝑚𝑒∗ = 𝑠2𝑚𝑙

∗𝑚𝑡∗2 1/3.

For intrinsic semiconductors (no doping, 𝑁𝑑 = 0 and 𝑁𝑎 = 0), we should have 𝑛 = 𝑝 =𝑛𝑖 , and we let 𝐸𝑖 = 𝐸𝐹 - the Fermi level of intrinsic semiconductor, then

𝑛𝑖 = 𝑁𝑐 exp − 𝐸𝑐 − 𝐸𝑖 /𝑘𝑇 = 𝑁𝑣 exp − 𝐸𝑖 − 𝐸𝑣 /𝑘𝑇 We can write in general

𝑛 = 𝑛𝑖exp 𝐸𝐹 − 𝐸𝑖 /𝑘𝑇 , 𝑝 = 𝑛𝑖exp 𝐸𝑖 − 𝐸𝐹 /𝑘𝑇

23

The electron-hole concentration product

𝑝𝑛 = 𝑛𝑖2 = 𝑁𝑐 exp − 𝐸𝑐 − 𝐸𝑖 /𝑘𝑇 𝑁𝑣 exp − 𝐸𝑖 − 𝐸𝑣 /𝑘𝑇

= 𝑁𝑐𝑁𝑣exp − 𝐸𝑐 − 𝐸𝑣 /𝑘𝑇

= 32 𝜋2𝑚𝑒

∗𝑚∗

4

3/2

𝑘𝑇 3exp −𝐸𝑔/𝑘𝑇

= 𝑏2 𝑘𝑇 3exp −𝐸𝑔/𝑘𝑇

both 𝑚𝑒∗ and 𝑚

∗ 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

𝑛𝑖 = 𝑏 𝑘𝑇 3/2exp −𝐸𝑔/2𝑘𝑇

For most semiconductors, the band gap g

E decreases with the increase of temperature 𝑇,

𝐸𝑔 𝑇 = 𝐸𝑔0 − 𝑐𝑇 accurate for 100𝐾 < 𝑇 < 400𝐾 but inaccurate for low temperatures,

where 𝐸𝑔0 is the extrapolated value of the band gap at 𝑇 = 0𝐾. The intrinsic concentration

𝑛𝑖 = 𝑏1 𝑘𝑇 3/2exp −𝐸𝑔0/2𝑘𝑇

where 𝑏1 = 𝑏exp 𝑐/2𝑘 .

Carrier concentrations in extrinsic semiconductors

Ionization of impurities

As donors in semiconductors can lose one electron to become positively charged e.g.

As0 − 𝑒 ↔ As+ Let

𝑁𝑑 : total donor concentration 𝑁𝑑

0: neutral donor concentration

𝑁𝑑+ : ionized donors

We obviously should have 𝑁𝑑 = 𝑁𝑑0 + 𝑁𝑑

+. The occupation probability of impurity level

𝐸𝑑 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

𝑓 𝐸𝑑 =𝑁𝑑

0

𝑁𝑑= 1 +

1

2exp

𝐸𝑑 − 𝐸𝐹𝑘𝑇

−1

24

The concentration of ionized donors

𝑁𝑑+ = 𝑁𝑑 − 𝑁𝑑

0 = 𝑁𝑑 1 +1

2exp

𝐸𝑑 − 𝐸𝐹𝑘𝑇

−1

Similarly for acceptors with ionization process

Al0 + 𝑒 ↔ Al− we should have 𝑁𝑎 = 𝑁𝑎

0 + 𝑁𝑎−, occupation probability of the acceptor level 𝐸𝑎

𝑓 𝐸𝑎 =𝑁𝑎−

𝑁𝑎= 1 +

1

2exp

𝐸𝑎 − 𝐸𝐹𝑘𝑇

−1

For n-type semiconductors, 𝐸𝑐 − 𝐸𝑑 is on the order of 𝑘𝑇 at room temperature (𝑇 =300K), and for nondegenerate semiconductors, 𝐸𝑐 − 𝐸𝐹 > 3𝑘𝑇. The ratio of electrons attached to the impurities to that in the conduction band

𝑁𝑑0

𝑛=𝑁𝑑 1 +

12 exp

𝐸𝑑 − 𝐸𝐹𝑘𝑇

−1

𝑁𝑐exp −𝐸𝑐 − 𝐸𝐹𝑘𝑇

Under normal condition, 𝐸𝑑 − 𝐸𝐹 > 3𝑘𝑇, then exp 𝐸𝑑−𝐸𝐹

𝑘𝑇 ≫ 1 and

𝑁𝑑0

𝑛=

2𝑁𝑑𝑁𝑐

exp 𝐸𝑐 − 𝐸𝐹𝑘𝑇

~𝑁𝑑𝑁𝑐

- on the same order, since 𝐸𝑐 − 𝐸𝑑~𝑘𝑇. For typical doping 𝑁𝑑 < 1017/𝑐𝑚3, most

impurities are ionized at room temperature with less than 1% remain unionized. We

should have 𝑛 = 𝑁𝑑 .

Consider a nondegenerate semiconductor to which impurities have been introduced. To

keep the discussion in general, we assume there are donors (𝑁𝑑 ) and acceptors (𝑁𝑎 ). Based on the condition of charge neutrality,

𝑁𝑑+ + 𝑝 = 𝑁𝑎

− + 𝑛 which can be separated according to mobile and immobile charges as

𝑛 − 𝑝 = 𝑁𝑑+ − 𝑁𝑎

−.

At 𝑇 > 100𝐾, we should have practically all impurities ionized, 𝑁𝑑+ = 𝑁𝑑 and 𝑁𝑎

− = 𝑁𝑎 , thus

𝑛 − 𝑝 = 𝑁𝑑 − 𝑁𝑎 . Since 𝑝𝑛 = 𝑛𝑖

2, we arrive

𝑛 =1

2 𝑁𝑑 − 𝑁𝑎 + 𝑁𝑑 − 𝑁𝑎 2 + 4𝑛𝑖

2

𝑝 =1

2 𝑁𝑎 −𝑁𝑑 + 𝑁𝑎 − 𝑁𝑑 2 + 4𝑛𝑖

2

For intrinsic situation where 𝑁𝑎 = 𝑁𝑑 = 0 as well as for completely compensated doping where 𝑁𝑎 = 𝑁𝑑 , they reduces to 𝑛 = 𝑝 = 𝑛𝑖 .

For net n-type doping where 𝑁𝑑 − 𝑁𝑎 ≫ 2𝑛𝑖 , we obtain

25

𝑛 ≈ 𝑁𝑑 − 𝑁𝑎 ≈ 𝑁𝑑 if 𝑁𝑑 ≫ 𝑁𝑎 and 𝑝 = 𝑛𝑖2/𝑁𝑑 (case of strongly extrinsic)

For net p-type doping where 𝑁𝑎 − 𝑁𝑑 ≫ 2𝑛𝑖 , we obtain 𝑝 ≈ 𝑁𝑎 − 𝑁𝑑 ≈ 𝑁𝑎 if 𝑁𝑎 ≫ 𝑁𝑑 and 𝑛 = 𝑛𝑖

2/𝑁𝑎 (case of strongly extrinsic)

2.5 The Fermi level and Energy Distribution of Carriers

Intrinsic semiconductors (𝑛 = 𝑝), we have

𝑁𝑐exp −𝐸𝑐 − 𝐸𝑖𝑘𝑇

= 𝑁𝑣exp −𝐸𝑖 − 𝐸𝑣𝑘𝑇

Then

𝐸𝑖 =𝐸𝑐 + 𝐸𝑣

2+

1

2𝑘𝑇ln

𝑁𝑐𝑁𝑣

=𝐸𝑐 + 𝐸𝑣

2+

3

4𝑘𝑇ln

𝑚∗

𝑚𝑒∗

where 𝐸𝑐+𝐸𝑣

2 is the center of the band gap. Obviously, for 𝑚

∗ = 𝑚𝑒∗ , 𝐸𝑖 is at the midpoint

exactly; for 𝑚∗ > 𝑚𝑒

∗ , 𝐸𝑖 is above the midpoint; for 𝑚∗ < 𝑚𝑒

∗ , 𝐸𝑖 is below the midpoint.

As a good approximation, 3

4𝑘𝑇ln

𝑚∗

𝑚𝑒∗ is small for most semiconductors, 𝐸𝑖 is roughly at

the midgap for intrinsic case.

Extrinsic semiconductors

n-type: 𝑛 = 𝑁𝑑+, then

𝑁𝑐exp −𝐸𝑐 − 𝐸𝐹𝑘𝑇

= 𝑁𝑑 1 + 2exp 𝐸𝐹 − 𝐸𝑑𝑘𝑇

−1

Introduce 𝜒 = 𝑁𝑐

2𝑁𝑑

1/2

exp −𝜖𝑑

2𝑘𝑇 where the ionization energy 𝜖𝑑 = 𝐸𝑐 − 𝐸𝑑 . The

above equation can be written as 1

1 + 2exp 𝐸𝐹 − 𝐸𝑑𝑘𝑇

= 2𝜒2exp

𝐸𝐹 − 𝐸𝑑𝑘𝑇

Then

𝐸𝐹 = 𝐸𝑑 + 𝑘𝑇ln 4 + 𝜒2 − 𝜒

4𝜒

And

𝑛 = 𝑁𝑑 𝜒

2 4 + 𝜒2 − 𝜒

Now let us discuss the above expressions under different temperatures corresponding to:

(1) Weak ionization 𝜒 ≪ 1 𝐸𝐹 = 𝐸𝑑 + 𝑘𝑇𝑙𝑛 1/2𝜒 (𝐸𝐹 above 𝐸𝑑 ) and 𝑛 = 𝜒𝑁𝑑 (few donors are ionized) (2) Complete ionization 𝜒 ≫ 1

26

𝐸𝐹 = 𝐸𝑑 + 𝑘𝑇𝑙𝑛 1/2𝜒 = 𝐸𝑐 + 𝑘𝑇𝑙𝑛 𝑁𝑑/𝑁𝑐 (𝐸𝐹 below 𝐸𝑑 ) and 𝑛 = 𝑁𝑑 (all donors are ionized)

Since 𝜒 is a function of temperature, given 𝑁𝑑 and 𝜖𝑑 (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: 𝜒 = 1). This temperature can be solved by

𝜒 = 𝑁𝑐

2𝑁𝑑

1/2

exp −𝜖𝑑

2𝑘𝑇 = 1

And keep in mind that 𝑁𝑐 = 𝑁𝑐 𝑇 .

It can be seen that as T increases, 𝜒 increases, and 𝐸𝐹 moves from above 𝐸𝑑 to below 𝐸𝑑 . In fact, 𝐸𝐹 decreases approximately linearly with the increase of temperature.

It is obvious that as the temperature continues to increase, 𝐸𝐹 approaches 𝐸𝑖 . At this point, the hole concentration approaches the electron concentration and cannot be

neglected any more, we must use

𝑛 = 𝑝 +𝑁𝑑 With 𝑛𝑝 = 𝑛𝑖

2, then

𝑛 =1

2𝑁𝑑 1 + 1 +

4𝑛𝑖2

𝑁𝑑2

1/2

(1) 𝑛𝑖/𝑁𝑑 ≪ 1, 𝑛 = 𝑁𝑑 complete ionization (2) 𝑛𝑖/𝑁𝑑 ≫ 1, 𝑛 = 𝑛𝑖 intrinsic

The transition temperature from complete ionization to intrinsic can be determined by

setting 𝑛 = 𝑁𝑑 .

Temperature dependence of carrier concentration in n-type semiconductors

(1) At low temperatures, 𝐸𝐹 > 𝐸𝑑 , 𝑛 < 𝑁𝑑 , 𝑝 ≪ 𝑛, weak ionization (2) At moderate temperatures, 𝐸𝐹 < 𝐸𝑑 , 𝑛 = 𝑁𝑑 , 𝑝 ≪ 𝑛, complete ionization (3) At high temperatures, 𝐸𝐹 → 𝐸𝑖 , 𝑛 ≈ 𝑛𝑖 ≫ 𝑁𝑑 , 𝑝 ≈ 𝑛, 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 𝑁𝑑 > 𝑁𝑎 , there will be 𝑁𝑎electrons make a transition from 𝐸𝑑 to 𝐸𝑎 . The remaining donors 𝑁𝑑 − 𝑁𝑎 can be excited into the conduction band.

The neutrality condition is then 𝑛 + 𝑁𝑎 = 𝑝 + 𝑁𝑑

+

since all acceptors are ionized (occupied by electron) 𝑁𝑎− = 𝑁𝑎 . Since donors are

partially ionized at least, 𝐸𝐹 is in the neighborhood of 𝐸𝑑 much closer to 𝐸𝑐 than 𝐸𝑣, thus, 𝑝 ≪ 𝑛, or

𝑛 + 𝑁𝑎 = 𝑁𝑑+ =

𝑁𝑑

1 + 2exp 𝐸𝐹 − 𝐸𝑑𝑘𝑇

But exp 𝐸𝐹−𝐸𝑑

𝑘𝑇 =

𝑛

𝑁𝑐exp

𝜖𝑑

𝑘𝑇 where 𝜖𝑑 = 𝐸𝑐 − 𝐸𝑑 is the ionization energy. Rearrange

the above equation

𝑛 𝑛 + 𝑁𝑎

𝑁𝑑 − 𝑁𝑎 − 𝑛=𝑁𝑐2

exp −𝜖𝑑𝑘𝑇

Let’s assume light compensation, i.e. 𝑁𝑑 ≫ 𝑁𝑎 , we can examine two cases (1) 𝑛 ≪ 𝑁𝑎 (extremely low temperature)

28

𝑛 =𝑁𝑐 𝑁𝑑 − 𝑁𝑎

2𝑁𝑎exp −

𝜖𝑑𝑘𝑇

Since 𝑛 = 𝑁𝑐exp −𝐸𝑐−𝐸𝐹

𝑘𝑇 , we have

𝐸𝐹 = 𝐸𝑑 + 𝑘𝑇ln𝑁𝑑 − 𝑁𝑎

2𝑁𝑎

𝐸𝐹 does not depend on the parameters of conduction band, only depends on 𝑇,𝑁𝑎 ,𝑁𝑑 , in this case. This is because the electrons are mainly distributed between the donor and

acceptor levels. The temperature dependence of 𝑛 is then

ln𝑛 = 𝑙𝑛𝑁𝑐 𝑁𝑑 − 𝑁𝑎

2𝑁𝑎−𝜖𝑑𝑘

1

𝑇

(2) 𝑛 ≫ 𝑁𝑎 (slightly higher temperature) We still have 𝑛 ≪ 𝑁𝑑 , then

𝑛 = 𝑁𝑐𝑁𝑑

2

1/2

exp −𝜖𝑑𝑘𝑇

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

𝐸𝐹 =𝐸𝑐 + 𝐸𝑑

2+

1

2𝑘𝑇𝑙𝑛

𝑁𝑑2𝑁𝑐

The temperature dependence of 𝑛 is then

ln𝑛 = 𝑙𝑛 𝑁𝑐𝑁𝑑

2

1/2

−𝜖𝑑2𝑘

1

𝑇

Apparently, the slopes for ln𝑛~1

𝑇 are different between the two cases: a factor of ½. The

slope at low temperature is twice as much as that at slightly higher temperature. At the

transition temperature, we should have 𝑛 = 𝑁𝑎 . 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, 𝑛 = 𝑁𝑑 − 𝑁𝑎 (complete ionization). Further increase the temperature, the semiconductor reaches the intrinsic

region where 𝑛 = 𝑝 = 𝑛𝑖 ≫ 𝑁𝑑 − 𝑁𝑎 . The major difference between a compensated and uncompensated semiconductor is the temperature dependence at the low temperature

region.

Energy Distribution of Carriers

Both 𝑛 = 𝑁𝑐exp −𝐸𝑐−𝐸𝐹

𝑘𝑇 and 𝑝 = 𝑁𝑣exp −

𝐸𝐹−𝐸𝑣

𝑘𝑇 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 𝑁 𝐸 𝑓 𝐸 where 𝑁 𝐸 is the density of states and 𝑓 𝐸 is the FD function.

Fig.2.20 Illustration of 𝑁 𝐸 , 𝑓 𝐸 , and 𝑁 𝐸 𝑓 𝐸 .