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Crystal Structure Fermi-Dirac Statistics Intrinsic Semiconductors What are we learning today? What are we learning today?

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Page 1: Eng323 Lect01 My

✒ Crystal Structure

✒ Fermi-Dirac Statistics

✒ Intrinsic Semiconductors

What are we learning today?What are we learning today?

Page 2: Eng323 Lect01 My

Crystal StructuresCrystal Structures

Galena is lead sulfide, PbS, and has a cubic crystal structure|SOURCE: Photo by SOK

Cubic FeS2, iron sulfide, or pyrite, crystals. The crystals look brass-like yellow (“fool’s gold”).|SOURCE: Photo by SOK

Page 3: Eng323 Lect01 My

Unit CellUnit Cell• The unit cell is the most convenient small cell in the crystal

structure that carries the properties of the crystal.

• The repetition of the unit cell in three dimensions generates the whole crystal structure

Page 4: Eng323 Lect01 My

2R

a

a

a

a

(c)(b)

FCC Unit Cell(a)

(a) The crystal structure of copper is Face Centered Cubic (FCC). Theatoms are positioned at well defined sites arranged periodically and there isa long range order in the crystal.(b) An FCC unit cell with closed packed spheres.(c) Reduced sphere representation of the FCC unit cell. Examples: Ag, Al,Au, Ca, Cu, γ -Fe (>912°C), Ni, Pd, Pt, Rh

Face Centered Cubic StructureFace Centered Cubic Structure

Page 5: Eng323 Lect01 My

Body Centered Cubic StructureBody Centered Cubic Structure

a

a

b

Examples: Alkali metals (Li, Na, K, Rb), Cr,Mo, W, Mn, α -Fe (< 912 ° C), β -Ti (> 882 ° C).

Body centered cubic (BCC) crystal structure. (a) A BCC unit cellwith closely packed hard spheres representing the Fe atoms. (b) Areduced-sphere unit cell.

Body centered cubic (BCC) unit cell has an atom at each corner of the cube and one atom at the center of the cube.

Page 6: Eng323 Lect01 My

Layer ALayer BLayer A

(a) (b)

Layer A

Layer B

Layer A

c

a(c) (d)

Examples: Be, Mg, α -Ti ( < 882°C ), Cr, Co, Zn, Zr, Cd

The Hexagonal Close Packed (HCP) Crystal Structure. (a) The Hexagonal Close Packed (HCP) Structure. A collection of many Zn atoms. Color difference distinguishes layers (stacks).(b) The stacking sequence of closely packed layers is ABAB (c) A unit cell with reduced spheres (d) The smallest unit cell with reduced spheres.

Hexagonal Close Packed StructureHexagonal Close Packed Structure

Page 7: Eng323 Lect01 My

a

C

a

a

The diamond unit cell is cubic. The cell has eight atoms. Grey Sn(α -Sn) and the elemental semiconductors Ge and Si have thiscrystal structure.

Diamond StructureDiamond Structure

Page 8: Eng323 Lect01 My

Unit cell is repeated to construct the crystal.

Page 9: Eng323 Lect01 My

S

Zn

a

a

a

The Zinc blende (ZnS) cubic crystal structure. Many importantcompound crystals have the zinc blende structure. Examples: AlAs,GaAs, GaP, GaSb, InAs, InP, InSb, ZnS, ZnTe.

Zinc-Blende StructureZinc-Blende Structure

The diamond & the zinc blende crystal structures have similarities. Both are cubic crystals & both have 8 atoms in the unit cell.

Page 10: Eng323 Lect01 My

http://www.crystallographic-software.com/bravaislattice.htm

Page 11: Eng323 Lect01 My

Miller IndicesMiller Indices

• determine the origin• determine the intercepts on the axes in terms of the lattice

constant a, b & c • write down the intercepts in terms of the numbers of the lattice

constant • take the reciprocal of these numbers • multiply or divide the reciprocal to reduce them to the smallest

three integers • the integers are enclosed in parentheses (h k l) • negative indices are represented by a bar over the appropriate

index

Planes in a crystal are specified by Miller indices (h k l), determined with the following steps:

Page 12: Eng323 Lect01 My

x

y

z

o

Page 13: Eng323 Lect01 My

Diamond cubic crystal structure and planes.

Page 14: Eng323 Lect01 My

Planar ConcentrationPlanar Concentration

• Planar concentration is the number of atoms per unit area on a given plane in the crystal.

• It is the surface concentration of atoms on a given plane.

• To calculate the planar concentration n(hkl) on a given (h k l) plane we consider a bound area A. Only atoms whose centers line on A are involved in n(hkl).

• For each atom, we then evaluate what portion of the atomic cross section cut by the plane (h k l) is contained within A.

Page 15: Eng323 Lect01 My
Page 16: Eng323 Lect01 My

a

FCC Unit cell

z

y

x

z = 1/2ay = Ða

(012)

O′

(012)(a) plane

a a

a

(b) (100) plane (c) (110) plane

A = a2

A = a2√ 2

a√ 2

The (012) plane and planar concentrations in an FCC crystal.

Page 17: Eng323 Lect01 My

Fermi-Dirac StatisticsFermi-Dirac StatisticsThe Fermi-Dirac probability function for f(E), describes thestatistics of electrons in a solid where electrons can interact with each other and the environment.

EF is the reference energy called Fermi energy.

At energies where E is several kT above EF, E-EF >> kT

& the probability is much like the Boltzmann probability function, which describes the statistics of particles where there are many more available states than the number of particles.

Page 18: Eng323 Lect01 My
Page 19: Eng323 Lect01 My

Intrinsic SemiconductorsIntrinsic SemiconductorsAn intrinsic semiconductor is an ideal perfect crystal of material like Si, that has no impurities or crystal defects such as dislocations.

Such a crystal consists of, say, Si atoms perfectly bonded to each other in a diamond lattice.

Other materials – Ge, GaAs, InP, etc.

Page 20: Eng323 Lect01 My

SiliconSilicon

A silicon ingot is a single crystal of Si. Within the bulk of the crystal, the atoms are arranged on a well-defined periodical lattice. The crystal structure is that of diamond.

Page 21: Eng323 Lect01 My

Silicon is the most important semiconductor in today’s electronics.|SOURCE: Courtesy of IBM

Page 22: Eng323 Lect01 My

Properties of Intrinsic SiProperties of Intrinsic Si• In a Si lattice, the four outer electrons of each Si atom

forms a covalent bond with a corresponding electron of a neighboring Si.

• At a temperature T (> 0 K), Si atoms in the crystal lattice• vibrate, with a distribution in vibration energy.• Average energy of vibration < 3kT is insufficient to break

covalent bond.• Some lattice vibrations do however have sufficient energy

to break covalent bond.• When a Si-Si bond is broken, a free electron is created that

can wander around and contribute to electrical conduction.

Page 23: Eng323 Lect01 My

Si = 1s22s22p63s23p2

Si = [Ne]3s23p2

Electronic Configuration

Page 24: Eng323 Lect01 My

(a) A simplified two dimensional illustration of a Si atom with four hybridorbitals, ψ hyb. Each orbital has one electron. (b) A simplified two dimensionalview of a region of the Si crystal showing covalent bonds. (c) The energy banddiagram at absolute zero of temperature.

Electron energy

VALENCE BAND(VB)Full of electronsat 0 K.

Ec

Ev

0

Ec+χ

ψ B

Si crystal in 2-D

Bandgap = Eg

(c)(b)

CONDUCTIONBAND (CB)Empty of electronsat 0 K.

ψ hyb orbitals

Si ion core(+4e)

Valenceelectron

(a)

Page 25: Eng323 Lect01 My

A two dimensional pictorial view of the Si crystal showing covalentbonds as two lines where each line is a valence electron.

Page 26: Eng323 Lect01 My

Energy Band Diagram of SiEnergy Band Diagram of Si

• We first start with the electronic structure of Si

• Each arrow represents an electron.

• The Si atom has 14 electrons.

Page 27: Eng323 Lect01 My

Formation of sp3 hybridized orbitals

Page 28: Eng323 Lect01 My

• Each Si atom forms four sp3 hybrid orbitals with one electron each, directed towards four neighboring Si atoms in a tetrahedral formation.

• This means that a pair of two neighboring Si atoms can have their hybrid orbitals overlap to form bonds.

• Thus one Si atom bonds with four other Si atoms in tetrahedral directions.

• Once a hybrid orbital (with one electron) of one Si atom overlaps with another hybrid orbital (with one electron) of a neighboring Si atom, they form two molecular orbitals.

• One adds in phase & another, out of phase, forming two types of molecular orbitals, bonding molecular orbital ψB (with two electrons & completely filled) & antibonding molecular orbital ψA (with no electrons & completely empty), with distinct energies EB & EA respectively.

Page 29: Eng323 Lect01 My
Page 30: Eng323 Lect01 My
Page 31: Eng323 Lect01 My

Electron energy

VALENCE BAND(VB)Full of electronsat 0 K.

Ec

Ev

0

Ec+χ

ψ B

Si crystal in 2-D

Bandgap = Eg

(c)(b)

CONDUCTIONBAND (CB)Empty of electronsat 0 K.

ψ hyb orbitals

Si ion core(+4e)

Valenceelectron

(a)

• Conduction band is separated from valence band by an energy band gap Eg. • Ev marks the top of the valence band, Ec marks the bottom of the conduction

band. • Conduction band contains electronic states that are at high energies, formed

by overlap of antibonding orbitals.• χ refers to electron affinity, the energy distance from the• bottom of conduction band to the vacuum level.

Page 32: Eng323 Lect01 My

Thermal Generation of CarriersThermal Generation of Carriers

• The broken bond has a missing electron that causes the region to be positively charged.

• The vacancy left behind by a missing electron in the bonding orbital is called a “hole”

• Hole movement takes place by a neighboring electron tunneling through to fill up the vacancy.

• This can happen under the influence of an external force, when both electrons & holes take part in contributing to electrical conduction through the semiconductor.

Page 33: Eng323 Lect01 My

h+ EgVB

CBh+(a)

h+h+

h+h+

h+h+ (d)

h+ eÐ

h+

(f)

(b)

(c)

(e)

e-

e-

e-

An illustration of a hole in the valence band wandering around the crystal due to the tunneling of electrons from neighboring bonds.

Page 34: Eng323 Lect01 My

Conduction in SemiconductorsConduction in Semiconductors

• Total electron energy E = KE + PE

• When an electric field εx is applied, the electrostatic potential is given by

• Electrostatic PE = - e.V(x)

• The electron potential energy and hence total energy is an increasing function of distance x.

Page 35: Eng323 Lect01 My

VB

CB

VB

CB

xx = 0 x = L

V(x) Electrostatic PE(x)

(a) (b)

Ex

Ex

Holeenergy

ElectronEnergy

When an electric field is applied, electrons in the CB and holes in theVB can drift and contribute to the conductivity. (a) A simplifiedillustration of drift in Ex. (b) Applied field bends the energy bandssince the electrostatic PE of the electron is -eV(x) and V(x) decreases inthe direction of Ex whereas PE increases.

Page 36: Eng323 Lect01 My

• Electrons in conduction band gain energy from the field and move in a direction opposite to that of the field.

• They also collide due to thermal vibrations of lattice and lose energy.

• This results in electrons drifting as shown, with a drift velocity vde.

• Similar process results for holes in valence band, in an opposite direction, with a drift velocity vdh.

Page 37: Eng323 Lect01 My

conductivity σ = e · (n · µe + p · µh)

Drift Velocity, v = µ · E

Mobility, µ = e · ts /mWhere m = mass

Ts = mean free time between scattering

Current Density, J = e · (n · ve + p · vh)

Resistivity = 1/σ

Page 38: Eng323 Lect01 My

• To determine the conductivity of the semiconductor, we have to know the electron concentration in the CB which is n and the hole concentration in the VB which is p.

• We derive expressions for n and p mathematically, using

• probability functions and statistical approach.

• gcb(E) = Density of states in CB (No. of energy states per unit energy interval per unit volume)

• This means is that if you consider a range between E and E+dE, the number of energy states per unit volume in the interval dE will be gcb(E).dE

• f(E) = Probability of finding an electron in a state with energy E.

• The probability of finding a hole in a state with energy E will be [1- f(E)]

Electron and Hole ConcentrationsElectron and Hole Concentrations

Page 39: Eng323 Lect01 My

(a) Energy band diagram. (b) Density of states (number of states per unit energyper unit volume). (c) Fermi-Dirac probability function (probability of occupancyof a state). (d) The product of g(E) and f(E) is the energy density of electrons inthe CB (number of electrons per unit energy per unit volume). The area undernE(E) vs. E is the electron concentration in the conduction band.

Ev

Ec

0

Ec+χ

EF

VB

CB

nE(E) or pE(E)

E

nE(E)

pE(E)

Area = p

Ec

Ev

f(E)

EF

E

For electrons

For holes

[1- f(E)]

(c)

E

g(E)(b)

g(E) (E-Ec)1/2

(a) (d)

Area = n

Page 40: Eng323 Lect01 My

• If nE(E) is the number of electrons per unit volume per unit energy interval, then the number of electrons in the energy range E to E+dE is

• Integrating this from Ec to Ec+χ will give us “n”.

• We will assume that (EC-EF) >> kT so that the Boltzman Probability equation can be used for f(E) instead of the Fermi Dirac function.

• There is no loss of generality if we replace EC+χ by ∞.

Page 41: Eng323 Lect01 My

• As for gcb(E), since almost all the electrons are concentrated in the vicinity above EC , we can use

• Substituting in the integral for n, the result is

• Here

is called the effective density of states at the CB edge.

Page 42: Eng323 Lect01 My

• A similar analysis can be carried out for holes in VB, to give the hole concentration

where

Note that the product

is a constant dependent only on temperature & bandgap energy

Page 43: Eng323 Lect01 My

• A special case is when the semiconductor is intrinsic, in which case electrons and holes are always generated in pairs, and hence n = p= ni.

• The parameter ni is called the intrinsic carrier concentration, same for both electrons in CB and holes in VB at any given temperature.

• At higher temperature, ni should be higher.

• For an intrinsic semiconductor,

• This is called the Law of Mass Action.• An intrinsic semiconductor is a pure crystal where n = p.

Page 44: Eng323 Lect01 My

• In a semiconductor, we not only have thermal generation of

• electron-hole pairs, but we also have recombination of electrons and holes (that results in a broken bond being “repaired”) that results in their annihilation.

• The rate of recombination R n & R p ∝ ∝

• This means that R np∝

• The rate of generation G will depend on how many electrons are available for excitation at Ev, that is , Nv and how many empty states are available at Ec, that is , Nc.

• It is also dependent on the probability that the electron will make the transition, that is exp(-Eg/kT). Thus

Page 45: Eng323 Lect01 My

• At thermal equilibrium, we know that n = p = ni is constant.

• Hence, G must be equal to R in an intrinsic semiconductor.

• Generation rate is exactly equal to Recombination rate.

• Unlike an intrinsic semiconductor, which is pure, it is possible to add some impurities and alter its properties.

• Such semiconductors are called extrinsic semiconductorsextrinsic semiconductors.

• In such semiconductors it is possible to have

n>>p [called n-type semiconductor], or

p>>n [ called p-type semiconductor]

Thermal EquilibriumThermal Equilibrium