solid state physics for nanotechnology i eee5425 introduction to nanotechnology1

EEE5425 Introduction to Nanotechnology 1

SOLID STATE PHYSICS FORNANOTECHNOLOGY I

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Structural Compositions of Solids

Periodic with long range order

Short range order

Small crystalline regions called grains.

May occur during thin film growth.

© Nezih Pala [email protected] EEE5425 Introduction to Nanotechnology

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MOSFET on Crystalline Si


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Crystalline Structures

Typical feature of crystalline solid is periodicity, which lead to long-range-order• Crystal structure= Lattice + basis• The whole crystal may be generated by repetition of unit cell• Size(s) of unit cell = lattice parameter


5

Lattice and Base

Lattice [lat-is]: A regular, periodic configuration of points, throughout an area or a space, especially in a crystalline solid.

Base: The same group of atoms positioned around each and every lattice point of a crystal.

The crystal consists of an atomic basis (or atomic cluster) attached to the lattice points. The “basis” can be a single atom or a group of atoms attached to each lattice point. Each lattice point receives an identical basis (or cluster). The (infinite) crystal consists of the collection of these regularly arranged clusters.

Crystal structure= Lattice + basis


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2D Lattice

Translation of a unit cell by vector r = 3a + 2b


7

Primitive cell vs Unit cellPrimitive cell: a minimum cell corresponding to a single lattice point of a structure with translational symmetry in 2D, 3D, or other dimensions. A lattice can be characterized by the geometry of its primitive cell.

Unit cell: small regions of space that, when duplicated, can be translated to fill the entire volume of the crystal.

Although there are an infinite number of ways to specify a unit cell, for each crystal structure there is a conventional unit cell, which is chosen to display the full symmetry of the crystal. However, the conventional unit cell is not always the smallest possible choice. A primitive unit cell of a particular crystal structure is the smallest possible volume one can construct with the arrangement of atoms in the crystal such that, when stacked, completely fills the space. This primitive unit cell does not always display all the symmetries inherent in the crystal. In a unit cell each atom has an identical environment when stacked in 3 dimensional space. In a primitive cell, each atom may not have the same environment.


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Bravais Lattices

August Bravais ( 1811 – 1863), a French physicist, was the first to count the categories different types of lattices correctly.

The 14 Bravais lattices are arrived at by combining one of the seven crystal systems (or axial systems) with one of the six lattice centerings.

From the point of view of symmetry, there are fourteen different kinds of Bravais lattices.


http://en.wikipedia.org/wiki/Image:Bravais2.gif

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Bravais Lattices

The six lattice centerings are:• Primitive centering (P): lattice points on the cell corners only • Body centered (I): one additional lattice point at the center of the cell • Face centered (F): one additional lattice point at center of each of the faces of

the cell • Centered on a single face (A, B or C centering): one additional lattice point at

the center of one of the cell faces.

The 14 Bravais lattices are arrived at by combining one of the seven crystal systems (or axial systems) with one of the six lattice centerings.


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Bravais Lattices

The seven crystal systems are:• Cubic • Tetragonal • Orthorhombic • Monoclinic• Triclinic • Trigonal • Hexagonal

Not all combinations of the crystal systems and lattice centerings are needed to describe the possible lattices. There are in total 7 × 6 = 42 combinations, but it can be shown that several of these are in fact equivalent to each other. For example, the monoclinic I lattice can be described by a monoclinic C lattice by different choice of crystal axes. Similarly, all A- or B-centered lattices can be described either by a C- or P-centering.


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Bravais Lattices


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Lattice Constant

Lattice constant –noun Crystallography. a measure of length that defines the size of the unit cell of a crystal lattice. Also called lattice parameter.


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Diamond Lattice

The diamond lattice is composed of two interpenetrating fcc lattices: one displaced 1/4 of a lattice constant in each direction from the other.© Nezih Pala [email protected] EEE5425 Introduction to Nanotechnology

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Diamond Lattice of Si

In the basic unit of a crystalline silicon solid, a silicon atom shares each of its four valence electrons with each of four neighboring atoms.


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Zincblende Lattice

The zincblende lattice represents the crystal structure of zincblende (ZnS), gallium arsenide (GaAs), indium phosphide (InP), cubic silicon carbide and cubic gallium nitride.


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Miller Indices

Crystal planes and directions are important to determine the physical and electrical properties. The notation system is a set of three integers and called Miller indices.

Miller Indices (h k l) are used to identify planes of atoms within a crystal structure


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Miller Indices

1. Find the intercepts of the plane with the crystal axes and express these intercepts as integral multiples of the basis vectors: (1,2,3)

2. Invert the three integers found in step 1: (1/1, 1/2, 1/3)

3. Using appropriate multiplier, convert the values found on step 2 to the smallest possible set of whole numbers h, k, and l: with 6 as multiplier we have (6,3,2)

4. Label the plane (hkl): (6 3 2)

If an intercept is negative the minus sign placed above Miller index (h k l)If a plane passes through the origin, it can be translated to a parallel position.


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Miller Indices

Step1: ( 2, 4, 1)

Step2: (1/2, 1/4, 1/1)

Step3: 4 x (1/2, 1/4, 1/1)

Step4: (2 1 4)


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Miller Indices

(001) (111) (221)? ? ?

? ? ? ? ? ?


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Miller Indices


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Miller Indices

Equivalence of the cube faces ( {100} planes ) by rotation of the unit cell within the cubic lattice.


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Miller Indices

Crystal directions in the cubic lattice.


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Miller Indices

Visualization and Miller indices of common crystalline directions


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Miller Indices

Convention Interpretation

(hkl) Crystal plane

{hkl} Set of equivalent planes

[hkl] Crystal direction

<hkl> Set of equivalent directions


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Comments on Crystal Planes

• mechanical, electrical, chemical and optical properties depend on crystal orientation.

• crystals can be cleaved along atomic planes very flat, smooth surfaces

• mirror facets for semiconductor lasers• wet chemical etchants etch along crystal planes


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Orientation-dependent Etching

KOHSi +OH- + 2H2O → SiO2(OH)2 2- + 2H2(g)Etch rate : {110} > {100} >> {111}

Etch rate of Si in KOH depending on crystallographic plane© Nezih Pala [email protected] EEE5425 Introduction to Nanotechnology

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Orientation-Dependent Mobility

Hole and electron mobility in CMOS devices fabricated on (100)-, (111)-, and (110)-orientated silicon substrates for HfO (lines) and oxynitride (dots) gate dielectrics. Nominal current flow on (100), (111), and (110) surfaces are in [110], [112] and [110] directions, respectively (solid lines and open dots).

Yang et al. IEEE Elect Dev Lett. 24 331 (2003)© Nezih Pala [email protected] EEE5425 Introduction to Nanotechnology

© Nezih Pala [email protected] EEE5425 Introduction to Nanotechnology 28

Electrons in Periodic Potential

||

))((

4

1)(

0 x

qqxU

When we wan to examine the properties of an electron in a periodic lattice, we need to consider Schrödinger's; equation such that the potential energy term V(r) reflects the fact that the electron sees a periodic potential.

In a one dimensional lattice, ionized atoms form a periodic potential distribution for electrons due to Coulomb potential.

xU(x)

Schrodinger’s equation can be written as

)()()(2

22

rErrVm

where the potential energy term is periodic V(r)=V(r+T) and where T is the crystal translation vector.


Bloch’s Theorem

Bloch’s theorem applies to waves in periodic structures in general. In a periodic potential distribution, wave function solutions of Schrodinger’s equation can be written as the product of a plane wave and a periodic function.

where k is the wavevector to be determined (called the Bloch wavevector) and where u is periodic

Thus

It is important to note that the Bloch theorem shows that the electronic can propagate through a perfect periodic medium without scattering(i.e. without hitting the atoms).

rkjerur .)()(

)()( Truru

TkjTkjrkjTrkj ereerueTruTr ...).( )()()()(


Kronig – Penny Model of Band Structure -1

As an approximate model for one dimensional crystal can be given as

0 ,

0 ,0)(

20

1

x -aV

axxV

where a=a1+a2 is the period of the lattice. This is known as Kronig-Penny model.

V=V0 V=V0V=V0

V=0 V=0-a2 a10

a

In the region -a2 ≤ x ≤ 0, the potential V=V0, and the solution of Schrödinger's equation is

xjxj BeAex )( 20 )(2

VEm

where

In the region 0 ≤ x ≤ a1, the potential V=0, and the solution of Schrödinger's equation is

xjxj FeDex )( 2

2

mE

where



Using Bloch theoremjkxexux )()( )()( axuxu and

therefore jkaaxjkaxjk exexueaxuax )()()()( )()(

Using these relationships the wavefunction in the period a1 ≤ x ≤ a1+a can be written as

aaxaeBeAe

axaeBeAexjkaaxjaxj

jkaaxjaxj

1)()(

1)()(

,

,)(

Enforcing the continuity of Ψ and Ψ’ at x=0 and x=a1 leads to the eignevalue equation, if 0 < E < V0

)sinh()sin(2

)cosh()cos(cos 21

22

21 aaaaka

)sin()sin(2

)cos()cos(cos 21

22

21 aaaaka

and if E > V0 20 )(2

EVm



In the preceding equations, the energy E is the only unknown parameter. For a solution to exist, we must have

1cos1 kaTherefore the right hand side of the last two equations denoted as r(E) must obey this condition. A typical plot of r(E) vs E is as follows:

The figure makes it that there are certain allowed values of energy, called allowed energy bands and certain unallowed values of energy , called band gaps. That is if E is in an allowed energy band, Schrodinger’s equation has solution, and if E is in an allowed energy band there is no solution. Within an allowed band, energy can take any value (i.e. it is not discretized). Note that as energy increases , the allowed energy bands increase in width, and so the forbidden bands decrease in width.



Since )(cos Erka we can generate an important figure called dispersion diagram, which is a diagram of energy versus wavenumber (E vs. k). To generate the dispersion diagram, start at E=0 and compute r(E). If |r(E)|=|cos ka|>1, we are at a forbidden energy (i.e. in a bandgap) and we need to increase E a bit and try again. If |r(E)|≤ 1, we are at an allowed energy (i.e. in an energy band) and in this case the corresponding wavenumber is

Since cosine is an even function –k will also be a solution. By increasing E by a small amount and checking the value of r(E) , we can generate the plot of allowed and unallowed energy bands. One form of the result will look like as follows. This depiction is known as the extended zone scheme.

))((cos1 1 Era

k



The various sections of wavenumber space are divided into what are called Brillouin zones with the range

Denoting the important first Brillouin zone. The second Brillouin zone is the range

and so on for higher zones.

ak

a

2

,2

ak

aak

a



Another depiction arises from noting that the energy bands in the higher Brillouin zones can be all translated to the first Brillouin zone by shifts of n2π/a . This results in what is called reduced zone scheme as shown below. This is the most common format for describing the band structure.


Effective Mass -1

An electron in empty space has a well defined constant mass. However, sometimes, it is useful to view mass merely a proportionality constant between force and acceleration, remembering the Newton’s second law F=ma. This is particularly appropriate when studying electrons in crystals where they appear to ac as if their mass is different from the free space value.

Quantum mechanically an electron can be represented by e wave packet with the electron velocity being its group velocity:

The influence of electron’s environment is contained in the energy relation E(k). For an electron in free space:

k

E

kvg

1

m

kVkE

2)(

22

0

However determining E(k) can not be easy for realistic crystals.


Effective Mass -2On the other hand, complete band structure of material does not need to be known, since only electrons in certain regions of band are of interest. We typically are more interested in the behavior of electrons near the band edge (conduction band or valence band) since they contribute to the conduction most. In this case only the local behavior of E-k curve is important.


Effective Mass -3

dt

kd

dt

dvmF g )(

*

From Newton’s law f motion

And from the group velocity definition

dt

kd

dk

Ed

dk

dE

dt

d

dt

dvg )(112

2

2

dt

kd

dt

kd

dk

Edm

)()(1*

2

2

2

2

2

2

*

dkEd

m

Hence

In a real three dimensional crystal, effective mass is

1

2,.

22

,, *

zyx

zyxk

Em


Effective Mass -4

Mathematically, the effective mass is inversely proportional to the curvature of an E vs k plot.

m* is positive near the bottoms of all bands.m* is negative near the tops of all bands.

A negative effective mass simply means that, in response to an applied force, the electron will accelerate in a direction opposite to that expected from purely classical considerations.

The negative effective mass helps to describe, conceptually and mathematically, charge transfer (and therefore conduction) in a partially filled band. We define an empty state left behind an electron as a positive charge and a positive effective mass and call it hole.

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Effective Mass –5

In any calculation involving the mass of the charge carriers, we must use effective mass values for the particular material involved.

mn*–the electron effective mass

mp*–the hole effective mass

The n subscript indicates the electron as a negative charge carrier, and the p subscript indicates the hole as a positive charge carrier.

Material mn*/m0 mp*/m0

Si 1.18 0.81

Ge 0.55 0.36

GaAs 0.066 0.52

m0–mass of electron in restm0= 9.10938188×10-31kg



Bloch Oscillations -1

Remembering the group velocity we can calculate and plot E vs k and E vs vg

k

Evg

1

When a DC bias is applied, electron will accelerate and its k will increase towards the value k=π/2a.

tq

ktk

)0()(

When a DC bias is applied, the electron will accelerate and its k will increase towards the value k=π/2a.

At k=π/2a there is an inflection point at E-k curve and electron velocity reaches to maximum. With further increase of k, electron decelerates and finally as k reaches the Brillouin zone boundary at π/a. Electron velocity goes to 0 indicating that the electron wavefunction is a standing wave rather than travelling wave.


Bloch Oscillations -2

As a result, when we apply DC field the electron, due to the band structure, oscillates back and forth. These are called Bloch Oscillations. In practice this behavior is not seen. Electrons are scattered before they reach to the amplitude of the Bloch oscillations.

Bloch oscillations, however, can be observed in superlattices where band structure can be precisely controlled .


Band Theory of Solids -1We concluded that, due to the periodic potential associated with the crystalline lattice, there are allowed and disallowed energy bands. Let us look at how carrier transport is affected if a band is filled with electrons or not.

1. If an allowed band is completely empty of electrons, obviously there are no electrons in the band to participate in electrical conduction. This can happen, for example, in a high-energy band where the energies of the band are above the energies of the systems electrons.

2. Similarly, and surprisingly, if an allowed band is completely filled with electrons, those electrons can not contributive to electric conduction either.

3. Only electrons in a partially filled energy band can contribute to conduction.


Band Theory of Solids -2

There is another way to view band structures that is particularly helpful in understanding how two systems interact when brought together.

It turns out that if a quantum system has energy levels E1, E2, E3, then if two such identical systems (e.g. two atoms) are brought together, it can be shown that each energy level ill split into two levels.

Where En± is an energy level slightly above or below the energy value En of the isolated

system.

nbn EEE ,

E2

E1

System 1E

~

E1

System 2E2

x

E2+

E2-

System 1E

~

E1

System 2

E2

x

E1+

E1-



The splitting is due to the overlap of each system's wavefunctions (orbitals). For example, in the case of two atoms that can come together to form a molecule, the atomic orbitals associated with each atom begin to overlap as atoms are brought together. This can be seen by considering a simplified liner model of forming a lithium (Li) molecule. Lithium has the electronic configuration 1s2 2s1 and in forming the molecule Li2 the s shell atomic orbitals form antibonding and bonding molecular orbitals as depicted . In the ground configuration the bonding molecular state is filled with two 2s1 electrons (one from each atom) and the antibonding state is empty.

If N identical atoms are brought together, each energy level of an isolated atom, E1, E2, E3 will split into N levels forming quasi-continous bands.

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When isolated atoms are brought together to form a solid, various interactions occur between neighboring atoms.

The forces of attraction and repulsion between atoms will find a balance at the proper interatomic spacing for the crystal.

In this process, important changes occur in the electron energy level configurations, and these changes result in the varied electrical properties of solids.


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•when two atoms are completely isolated from each other, they can have identical electronic structures

•as the spacing between the two atoms becomes smaller, electron wave functions begin to overlap. The Pauli exclusion principle dictates that no two electrons in a given interacting system may have the same quantum state; thus there must be a splitting of the discrete energy levels of the isolated atoms into new levels belonging to the pair rather than to individual atoms.


48



49


Linear combinations of atomic orbitals (LCAO): The LCO when 2 atoms are brought together leads to 2 distinct “normal modes – a higher energy antibonding orbital and a lower energy bonding orbital . Note that the electron probability density is high in the region between the ion cores (covalent “bond”), leading to lowering of the bonding energy level and the cohesion of the crystal. If instead of 2 atoms, one brings together N atoms, there will be N distinct LCAO, and N closely spaced energy levels in a band.


50



51


Energy levels in Si as a function of atomic spacing. The core levels (n=1,2) in Si are completely filled with electrons. At the actual atomic spacing of the crystal, the 2N electrons in the 3s subshell and the 2N electrons in the 3p subshell undergo sp3 hybridization, and all end up in the lower 4N states (valence band) while the higher lying 4N states (conduction band) are empty separated by a band gap.

1s2 2N states⇨ 2p6 6N states ⇨ 3p6 6N states⇨2s2 2N states⇨ 3s2 2N states⇨


52


Electrons must occupy different energies due to Pauli Exclusion principle.

Two atoms Six atoms Solid of N atoms


53

Intrinsic Materials –1

A perfect semiconductor crystal with no impurities or lattice defects is called an intrinsic semiconductor.

In such material there are no free charge carriers at T= 0 K, since the valence band is filled with electrons and the conduction band is empty. At higher temperatures electron-hole pairs are generated as valence band electrons are excited thermally across the band gap to the conduction band.

These EHPs are the only charge carriers in intrinsic material.

If one of the Si valence electrons is broken away from its position in the bonding structure such that it becomes free to move about in the lattice, a conduction electron is created and a broken bond (hole) is left behind.

The energy required to break the bond is the band gap energy Eg.


54


Since the electrons and holes are created in pairs, the conduction band electron concentration n (electrons per cm3) is equal to the concentration of holes in the valence band p (holes per cm3). Thus for intrinsic material

n= p= ni

where ni is concentration of EHPs in intrinsic material or intrinsic concentration. ni depends on temperature (!)

Obviously, if a steady state carrier concentration is maintained, there must be recombination of EHPs at the same rate at which they are generated.

Recombination occurs when an electron in the conduction band makes a transition (direct or indirect) to an empty state (hole) in the valence band, thus annihilating the pair.


55


Band diagram for intrinsic semiconductor


56

Extrinsic Materials –1

In addition to the intrinsic carriers generated thermally, it is possible to create carriers in semiconductors by purposely (controllably) introducing impurities into the crystal.

This process is called doping.

By doping, a crystal can be altered so that it has a predominance of either electrons or holes. There are two types of doped semiconductors:

•p-type (mostly holes)•n-type (mostly electrons)


57


When a crystal is doped such that the equilibrium carrier concentrations n0 and p0 are different from the intrinsic carrier concentration ni, the material is said to be extrinsic.

DonorsDopants increasing electron concentration

AcceptorsDopants increasing electron concentration

P (phosphorus) B (Boron)

As (Arsenic) Ga (Gallium)

Sb (Antimony) In (Indium)

Al (Aluminum)


58



59


When impurities or lattice defects are introduced into an otherwise perfect crystal, additional levels are created in the energy band structure, usually within the band gap.

n-type semiconductors

An impurity from V-column of the periodic table (P, As, and Sb) introduces an energy level very near the conduction band in Si or Ge. This level is filled with electrons at T= 0 K, and very little thermal energy is required to excite these electrons to the conduction band.

At T about 50–100 K virtually all of the electrons in the impurity level are "donated" to the conduction band. Such an impurity level is called a donor level. Thus semiconductors doped with a significant number of donor atoms will have n0>> ni or p0 at room temperature. This is n-type material.


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Semiconductor Si (Z= 14): 1s2 2s2 2p6 3s2 3p2

Dopant (donor) As (Z= 33): 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3

n-type


61


Band diagram for n-type semiconductor


62


Atoms from III-column (B, Al, Ga, and In) introduce impurity levels in Ge or Si near the valence band. These levels are empty of electrons at 0 K.

At low temperatures, enough thermal energy is available to excite electrons from the valence band into the impurity level, leaving behind holes in the valence band.

Since this type of impurity level "accepts" electrons from the valence band, it is called an acceptor level, and the column III impurities are acceptor impurities in Ge and Si. Doping with acceptor impurities can create a semiconductor with a hole concentration p0much greater than the conduction band electron concentration n0 or ni( this type is p-type material)


63


Semiconductor Si (Z= 14): 1s2 2s2 2p6 3s2 3p2

Dopant (acceptor) B (Z= 5): 1s2 2s2 2p1

p-type


64


Band diagram for p-type semiconductor


65

Extrinsic Materials –10Example:Calculate the approximate energy required to excite the extra electron of As donor atom into the conduction band of Si (the donor binding energy).

Solution:Let’s assume for rough calculations that the As (1s2 2s2 2p2 3s2 3p6 3d10 4s2 4p3) atom has its four covalent bonding electrons rather tightly bound and the fifth “extra” electron loosely bound to the atom.

We can approximate this situation by using the Bohr model results, considering the loosely bound electron as ranging about the tightly bound "core" electrons in a hydrogen-like orbit.

We have to find energy necessary to remove that “extra” electron from As atom.


66


We can approximate As dopant atom in Si lattice by using the Bohr model:the loosely bound “fifth” electron is ranging about the tightly bound "core" electrons in a hydrogen-like orbit.

The magnitude of the ground-state energy (n= 1) of such an electron in the Bohr model is

Constant K in this case is K=4πεrε0 where εr is relative dielectric constant of Si

Approximation of As dopant atom in Si lattice.

22

4

1 2 K

mqE


67


20

4*

220

4*

)(8)4(2 h

qmqmE

r

n

r

n

eV 0.1 J 101.6 1eV

(J) 10837.11010835.3

1045.70

10)63.6()8.11()85.8(8

)10()6.1(11.918.1

)1063.68.111085.8(8

)106.1)(1011.9(18.1

19

20926

107

92222

1074

23412

41931

E

E

Besides relative dielectric constant of Si, we have to use conductivity effective mass of electron mn* in Si in the formula for energy E:


68


Generally, the column-V donor levels lie approximately 0.01 eV below the conduction band in Ge, and the column-III acceptor levels lie about 0.01 eV above the valence band.

In Si the usual donor and acceptor levels lie about 0.03-0.06 eV from a band edge.

When a semiconductor is doped n-type or p-type, one type of carrier dominates.

For example, when we introduce donors, the number of electrons in conduction band is much higher than number of the holes in the valence band.

In n-type material:holes –minority carrierselectrons –majority carriers

In p-type material:holes –majority carrierselectrons –minority carriers



Excitons -1

We have considered band transitions in semiconductors and implicitly assumed that process of absorption (of a photon, thermal energy etc.) created a free electron and a free hole, each of which can contribute conduction. There is another effect that is worth mentioning , primarily because its importance in quantum-confined structures such as carbon nanotubes and quantum dots. The basic idea is that after an electron transition , it is possible for the electron and the created hole to be bound together by their mutual Coulomb attraction, forming a quasi particle called an exciton.


Excitons -2

The two-particle electron-hole exciton can be modeled like the two-particle hydrogen atom. However unlike the hydrogen atom, which consists of one proton and one electron (having different masses) in empty space, here the bound electron-hole pair moves through a material characterized by relative permittivity. Considering the formulas for energy and Bohr radius of the Hydrogen atom and using the reduced mass

)/( heher mmmmm

Then the binding energy and radius of the ground sate exciton are given by

6.138 22222

02

4

re

rY

re

r

r

r

m

mR

m

m

nh

qmE

)53.0(4

02

20

r

er

r

er

r

rex m

ma

m

m

qmr

Where RY is the Rydeberg energy and a0 is the Bohr radius.


Excitons -3

As an example, for GaAs(εr=13.3) using an average of the heavy and light hole masses mr=0.0502me, we find that

E=-3.86 meVrex=265a0=14nm

The binding energy of the pair, E, can be easily overcome by thermal effects (e.g. kT=25meV at room temp.) , thus breaking the exciton into free electrons and holes. Therefore in bulk materials exciton effects are usually only observed at very low temperatures and for relatively pure samples since impurities such as dopants tend to screen the Coulomb interaction much as occurs in conductors.


Excitons -4

To demonstrate the concept of optical absorption by bandgap transition and excitons the absorption coefficient versus incident photon energy for GaAs is shown n below for various temperatures. At room temperature one can observe the lack of absorption below the bandgap (Eg=1.43 eV). At low temperatures, the peaks at the onset of absorption are due to the creation of excitons.


Charge Carrier Statistics -1

In previous quantum well problems the allowed energy states were obtained, but here was no way to say which states would actually be “filled” by electrons. Here we will examine how many allowed states are near an energy of interest, and the probability that those states will actually be filled with electrons. Density of states and particle statistics concepts are indispensible in study of bulk materials as well as small material systems.

The density of states is required as the first step in determining the carrier concentrations and energy distributions of carriers within a semiconductor. Integrating the density of states function g(E) between two energies E1 and E2 tells us the number of allowed states available to electrons in the cited energy range per unit volume of the crystal. In principle, the density of states could be determined from band theory calculations for a given material. Such calculations however, would be rather involved and impractical. Fortunately, an excellent approximation for the density of states near the band edges can be obtained a simple and familiar approach of “particle in a box” problem.



zkykxkLLL

r zyxzyx

sinsinsin8

Remember the solution of Schrodinger’s equation for particle in a 3D box

...3,2,1n,n,n xxx z

zz

y

yy

x

xx L

nk

L

nk

L

nk

where

Each solution can be uniquely associated with a k-space vector

k = (nxπ/Lx)ax+(nyπ/Ly) ay+(nzπ/z) az

where ax ,ay ,az are unit vectors directed along k-space coordinate axes. In the figure, each point represents one solution of Schrodinger’s equation.



Taking note of lattice arrangements of the solution dots, we can deduce that a k-space “unit cell” of volume (π/Lx)(π/Ly) (π/Lz) contain one allowed solution. And therefore:

3space-k of eUnit volum

Solutions

zyx LLL

Considering that there is no physical difference between wavefunction solutions which differ only in sign, the total number should be divided to 8. On the other hand, for electrons, two allowed spin states (spin up and spin down ) must be associated with each independent solution. We therefore obtain

34space-k of eUnit volum

Solutions

zyx LLL



The next step is to determine the number of states with a k-value between arbitrarily chosen k and k+dk. This is equivalent to adding up the states lying between the two k-space spheres shown in the figure. Considering that the large dimensions of the system and close-packed density of k-space state s, the desired result is simply obtained by multiplying the k-space volume between the two spheres, 4πk2dk, times the last equation for the allowed states per unit k-space volume.

dkkLLL zyx 23

44dkk andk between

k with statesEnergy

E

dEmdk

m

kdkdE

mEk

m

kE

2

1 or

2 or

22

22

22



Therefore

dEmEmLLL zyx 2

dEE and Ebetween

E with statesEnergy 32

Then by definition

VdEEN /dEE and Ebetween

E with statesEnergy )(

where V is the volume of the crystal and N(E) is the density of states. Thus, finally

32

2)(

mEm

EN



To obtain the conduction and valence band densities of states near the band edges in real materials , the mass m of the particle in the forgoing derivation is replaced by the appropriate carrier effective mass. Also if EC is taken to be the minimum electron energy in the conduction band and EV the maximum hole energy in the valence band the E in the last equation must be replaced by E-EC in treating conduction band states and by EV-E in treating valence band states. Introducing the subscripts c and v to identify the conduction and valance band densities of states, respectively, we can then write in general

V

Vpp

V

CCnn

C

EEEEmm

EN

EEEEmm

EN

)(2

)(

)(2

)(

32

**

32

**



To gain appreciation of the last equations assuming m*=m we have

30

21 1/eVcm )(108.6)( VEEN eV

For example if E=0.1 eV and V0=0 then N(E)=2.15x1021 1/eVcm3

Thinking physically, if there are enough electrons to fill the various states, then the density of states N(E) is the density of electrons having energy E.

However, this is not case in reality and therefore we should find the probability for electrons to have the energy E.

80

Statistical Distributions –1

Given:System of N particles (air molecules, for example) in thermal equilibrium at temperature T.

Question:How is the total energy E distributed over the particles?

or:How many particles have the energy E1, E2, etc.?

How can we answer these questions?


81

Fermi-Diracstatistics

•identical particles•odd half-integral spin (fermions)•close together (overlapping ψ)

Statistical Distributions – 2

kT

E

AeEf

)(1

1)(

kT

E

Ae

Ef

Maxwell-Boltzmann statistics

•identical particles•“far” apart (no overlap of ψ)

Bose-Einstein statistics

•identical particles•integral spin (bosons)•close together (overlapping ψ))

Distinguishable particles (e.g. molecules in a gas)

Indistinguishable particles - Bosons (e.g. photons)

Indistinguishable particles - Fermions (e.g. electrons)

1

1)(

kT

E

Ae

Ef


82

Statistical Distributions – 3

kT

E

AeEf

)(1

1)(

kT

E

Ae

Ef

1

1)(

kT

E

Ae

Ef

Maxwell-Boltzmann statistics

Bose-Einstein statistics

Fermi-Diracstatistics


83

The Fermi Level –2

kT

EE F

e

Ef

1

1)(

The function f(E),-the Fermi-Dirac distribution function,-reflects probability that an available energy state at energy level E will be occupied by an electron at temperature T.

The quantity EF is called the Fermi level

Important property of the Fermi level:for an energy E equal to the Fermi level energy EF ,the occupation probability is

Fermi level is an energy state which has always a probability of ½ of being occupied by an electron.

2

1

11

1

1

1)(

kT

EEF FF

Ae

Ef


84

For any energy value E> EF:


101

1

1

1

1

1)(

0

ee

EfFEEF

Let’s take a look on what happens with Fermi-Dirac distribution function when temperature changes.

At T= 0 K, every energy state up to the Fermi level EF is filled with electrons and all states above EF are empty.

1. T = 0

For any energy value E< EF:

At T = 0 K the Fermi-Dirac distribution function f(E) takes the simple rectangular form what means:

01

1

1

1

1

1)(

0

ee

EfFEEF


85


2. T > 0

There is some probability for states above the Fermi level to be filled.

•At T >0,for E >EF probability that energy states above EF are filled f (E) ≠0 (> 0).

•At T >0,there is a corresponding probability [1-f (E)]≠0 that states below EF (E > EF) are empty.

Fermi function f (E) is symmetrical about the Fermi level EF for all temperatures:

the probability f (EF+ΔE) that a state ΔE above EF is filled is the same as the probability [1-f(EF+ΔE)] that a state ΔE below EF is empty.


86


The symmetry of the distribution of empty and filled states about EF makes the Fermi level a natural reference point in calculations of electron and hole concentrations in semiconductors.

The Fermi-Dirac distribution function


87


The symmetry of the distribution of empty and filled states about EF makes the Fermi level a natural reference point in calculations of electron and hole concentrations in semiconductors.

For intrinsic material we know:

the concentration of holes in the valence band is equal to the concentration of electrons in the conduction band.

⇓the Fermi level EF must lie at the middle of the band gap in

intrinsic material.

Since f (E) is symmetrical about EF, the electron probability "tail" of f (E) extending into the conduction band is symmetrical with the hole probability tail [1 –f (E)] in the valence band. The Fermi level in intrinsic material is located at energy Ei

Fermi level in intrinsic material


88


For intrinsic material at T > 0 K


89


In n-type material there is a high concentration of electrons in the conduction band compared with the hole concentration in the valence band.

⇓The distribution function f (E) lies above its intrinsic position on the energy scale.

n-type material has larger concentration of electrons at Ec and correspondingly smaller hole concentration at Ev, than intrinsic material.

Energy difference (Ec-EF) gives a measure of concentration of electrons in the conduction band.

Fermi level in n-type material


90


For n-type material at T > 0 K


91


In n-type material there is a high concentration of holes in the valence band as compared to the electron concentration in the conduction band.

⇓The distribution function f (E) lies below its intrinsic position on the energy scale.

The [1 –f (E)] tail below Ev is larger than the f (E) tail above Ec in p-type material.

The value of (EF - Ev) indicates how strongly p-type the material is.

Fermi level in p-type material


92


For p-type material at T > 0 K



Temperature Dependence of Carrier Concentrations

Typical temperature dependence of the majority carrier concentration in n-type semiconductor.

(Phosphorus doped Si, ND= 1015 1/cm2)

94

Charge Carrier Concentrations at Equilibrium

Our goal is to find carrier concentration n0 and p0 in semiconductor.

We can find concentration of carriers if we know:

1. the distribution function f (E) (probability of carriers to occupy energy state)

2. the densities of states N (E) in the valence and conduction bands.

Then concentration of electrons in the conduction band at equilibrium:

where f (E) is Fermi distribution function; N (E) is the density of states; N(E)dE is the density of states (cm-3) in the energy range dE.

CE

dEENEfn )()(0

The subscript “0”used with the electron and hole concentration symbols (n0, p0) will indicate equilibrium conditions.


95

Concentrations at Equilibrium –2

32

2/3* 2)()(

Cn EEm

EN

CE

dEENEfn )()(0

Using quantum mechanics approach, it can be shown that density of states N(E) in the conduction band is proportional to √E:

C

F

C

F

E kT

EECnCn

E kT

EE dE

e

EEmndE

EEm

e

n

1

2)(

2)(

1

132

2/3*

032

2/3*

0

Then concentration of carriers in the conduction band at equilibrium can be calculated as:

(we calculated this before!)


96


kTEEkTEE

F

Fe

e/)(

/)(1

1

dEeEEm

dE

e

EEmn

C

F

C

F

E

kT

EE

Cn

E kT

EECn

32

2/3*

32

2/3*

0

2)(

1

2)(

At room temperature kT is 0.026 eV → (Ec-EF) >>kT → Fermi distribution function f (E) can be simplified:


The last integral may be solved more easily by making a change of variable. If we let

kT

EE C



Then equation becomes

0

2/132

2/3*

0

2)(

dee

kTmn kT

EEn

FC

The integral is the gamma function with a value of

2

1

0

2/1

de

Then the equation for n0 becomes

kT

EEn

CF

eh

kTmn

2/3

2

*

0

22

which is the concentration of carriers in the conduction band at equilibrium

98


kT

EEn

CF

eh

kTmn

2/3

2

*

0

22

-concentration of carriers in the conduction band at equilibrium.

or

kT

EE

C

CF

eNn

0

2/3

2

*22

h

kTmN n

C

where

NC is the effective density of states located at the bottom of the conduction band Ec

We have found concentration of electrons in the conduction band at equilibrium.


99


The concentration of holes in the valence band at equilibrium is

where constant NV is the effective density of states in the valence band:

Thus, the concentration of holes in the valence band is

)](1[0 VV EfNp

2/3

2

*22

h

kTmN p

V

kT

EE

V

FV

eNp

0


100




Position of Fermi Level

For intrinsic semiconductors n0=p0=ni which leads to

kT

EEhkT

EEe

FVCF

eh

kTmpe

h

kTmn

2/3

2

*

0

2/3

2

*

0

22

22

which leads to

*

*

ln4

3

2 e

hVCFi m

mkTEEE

Therefore the intrinsic Fermi level is near the middle of the bandgap. If the electron effective mass and hole effective mass are equal the intrinsic Fermi level lies precisely in the middle of the bandgap.


Position of Fermi Level

For extrinsic n-type semiconductor, if we assume that the dopant concentration ND is much higher than the intrinsic carrier concentration ni and all the dopant atoms are ionized (the usual situation ) then,

DNn

Substituting this into kT

EE

C

CF

eNn

we find

D

CCF N

NkTEE ln

In a similar manner, for a p-type semiconductor with an acceptor doping density NA>>pi

A

VVF N

NkTEE ln


Exercise -1

The probability that a state is filled at the conduction band edge (EC) is precisely equal to the probability that a state is empty at the valence band edge (EV). Show mathematically where the Fermi level is located?


Exercise -2

The probability of a state being filled at Ei+kT is equal to the probability a state is empty at Ei+kT.

• Where is the Fermi level located? • What type of semiconductor is it in terms of

carrier type?

solid state physics for nanotechnology i eee5425 introduction to nanotechnology1

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