chap3 1 semiconductor 1

7/30/2019 Chap3 1 Semiconductor 1

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Chapter 3: semiconductorscience and light emitting diodes

Of the 18 atoms shown in the figure, only 8 belong to the volume ao3.

Because the 8 corner atoms are each shared by 8 cubes, they contribute a

total of 1 atom; the 6 face atoms are each shared by 2 cubes and thuscontribute 3 atoms, and there are 4 atoms inside the cube.The atomic density is therefore 8/ao

3, which corresponds to 17.7, 5.00, and4.43 X 1022 cm-3, respectively.

Semiconductor Lattice Structures

Diamond Lattices

The diamond-crystal lattice characterized byfour covalently bonded atoms.

The lattice constant, denoted by ao, is 0.356,0.543 and 0.565 nm fordiamond, silicon, andgermanium, respectively.

Nearest neighbors are spaced units apart.)4/

3 oa

(After W. Shockley: Electrons and Holes in Semiconductors, Van Nostrand, Princeton, N.J., 1950.)

Semiconductor Lattice Structures

Diamond and Zincblende Lattices

Diamond latticeSi, Ge

Zincblende latticeGaAs, InP, ZnSe

Diamond lattice can be though of as an FCC structures with anextra atoms placed at a/4+b/4+c/4 from each of the FCC atoms

The Zincblende lattice consist of a face centered cubic Bravais point lattice which containstwo different atoms per lattice point. The distance between the two atoms equals one quarterof the body diagonal of the cube.

How Many Silicon Atoms per cm-3? Number of atoms in a unit cell:

4 atoms completely inside cell

Each of the 8 atoms on corners are shared among cells

count as 1 atom inside cell

Each of the 6 atoms on the faces are shared among 2

cells count as 3 atoms inside cell

Total number inside the cell = 4 + 1 + 3 = 8

Cell volume:

(.543 nm)3 = 1.6 x 10-22 cm3

Density of silicon atoms

= (8 atoms) / (cell volume) = 5 x 1022 atoms/cm3


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diamond cubic lattice

The Si Crystal

Each Si atom has 4 nea

rest neighbors

lattice constant

= 5.431

Semiconductor Materials forOptoelectronic Devices

400~450 450~470 470~557 557~567 567~572 572~585 585~605 605~630 630~700Pure Blue Blue Pure Green Green Yellow Green Yellow Amber Orange Red

Semiconductor Optical Sources

0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6

Blue

Green

Orange

Yellow

Red

1.7

I n f r a red (um)Violet

GaAs

GaAs

0.5

5P0.4

5

GaAs1-yPy

InP

In0.14

Ga0.8

6As

In1-xGaxAs1-yPy

AlxGa1-xAs

x = 0.43

GaP(

N)

GaSb

InGaN

SiC(A

l)

In0.7G

a0.3

As0

.66P0.3

4

In0.57

Ga0.4

3As0

.95P0.0

5

In0.49AlxGa0.51-xP

Semiconductor Materials forOptoelectronic Devices


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Quantization Concept

plank constant

Core electrons

Valence electrons

Periodic Table of the Elements

Group**

Period 1IA

1A

18VIIIA

8A

11H

1.008

2IIA

2A

13IIIA

3A

14IVA

4A

15VA

5A

16VIA

6A

17VIIA

7A

2He4.003

23

Li6.941

4Be9.012

5B

10.81

6C

12.01

7

N14.01

8O

16.00

9F

19.00

10Ne20.18

8 9 103

11Na22.99

12Mg24.31

3

IIIB3B

4

IVB4B

5

VB5B

6

VIB6B

7

VIIB7B

------- VIII -------

------- 8 -------

11

IB1B

12

IIB2B

13Al26.98

14Si

28.09

15

P30.97

16S

32.07

17Cl

35.45

18Ar39.95

419

K39.10

20Ca40.08

21Sc44.96

22Ti

47.88

23V

50.94

24Cr52.00

25Mn54.94

26Fe55.85

27Co58.47

28Ni58.69

29Cu63.55

30Zn65.39

31Ga69.72

32Ge72.59

33

As74.92

34Se78.96

35Br79.90

36Kr83.80

537

Rb85.47

38Sr

87.62

39

Y88.91

40Zr

91.22

41Nb92.91

42Mo95.94

43Tc(98)

44Ru101.1

45Rh102.9

46Pd106.4

47Ag107.9

48Cd112.4

49In

114.8

50Sn118.7

51

Sb121.8

52Te127.6

53I

126.9

54Xe131.3

655Cs132.9

56Ba137.3

57La*138.9

72

Hf178.5

73Ta180.9

74W

183.9

75Re186.2

76Os190.2

77Ir

190.2

78Pt

195.1

79Au197.0

80Hg200.5

81Tl

204.4

82Pb207.2

83

Bi209.0

84Po(210)

85At(210)

86Rn(222)

787Fr

(223)

88Ra(226)

89Ac~(227)

104

Rf(257)

105

Db(260)

106

Sg

(263)

107

Bh(262)

108

Hs(265)

109

Mt(266)

110---

()

111

---()

112

---()

114

---()

116

---()

118---

()

IV CompoundsSiC, SiGe

III-V Binary CompoundsAlP, AlAs, AlSb, GaN, GaP, GaAs,GaSb, InP, InAs, InSb

III-V Ternary CompoundsAlGaAs, InGaAs, AlGaP

III-V Quternary CompoundsAlGaAsP, InGaAsP

II-VI Binary CompoundsZnS, ZnSe, ZnTe, CdS, CdSe, CdTe

II-VI Ternary CompoundsHgCdTe

Semiconductor Materials Semiconductor Materials


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Atomic Bonding

a. Ionic bonding (such as NaCl)b. Metallic bonding (all metals)

c. Covalent bonding (typical Si)

d. Van der Waals bonding (water)

e. Mixed bonding (GaAs, ZnSe, ionic & covalent)

Bonding forces in Solids

Covalent Bonding


plank constant

Core electrons

Valence electrons

2s2p

1sK

L


The shell model of the atom in which the electrons are confinedto live within certain shells and in sub shells within shells.

The Shell Model

1s22s22p2 or [He]2s22p2

L shell withtwo sub shells

Nucleus


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Band theory of solids

Two atoms brought together to form moleculesplitting of energy levels for outer electron shells

Energy Band Formation (I)

=Splitting of energy states into allowed bandsseparated by a forbidden energy gap

as the atomic spacing decreases.The electrical properties of a crystalline material

correspond to specific allowed and forbidden energies

associated with an atomic separation related to the

lattice constant of the crystal.

Allowed energy levels ofan electron acted on by

the Coulomb potential ofan atomic nucleus.

Energy Band Formation (I)

Energy Band Formation (II)

Strongly bonded materials: small

interatomic distances. Thus, the strongly bonded materials can

have larger energy bandgaps than doweakly bonded materials.

Energy Bandgapwhere no states exist

As atoms are brought closer towards

one another and begin to bond

together, their energy levels must

split into bands of discrete levelsso closely spaced in energy, they

can be considered a continuum of

allowed energy.

Pauli Exclusion Principle

Only 2 electrons, of spin 1/2, canoccupy the same energy state at

the same point in space.

Energy Band Formation (III)

Conceptual development of the energy band model.

Electrone

nergy

E

lectronenergy

isolatedSi atoms

Si latticespacing

Decreasing atom spacing

s

p

sp n = 3

Nisolated Si-atoms

6N p-states total2N s-states total

(4Nelectrons total)

Electrone

nergy

Crystalline SiN-atoms

4Nallowed-states

(Conduction Band)

4Nallowed-states(Valance Band)

No states

4Nempty states

2N+2Nfilled states

Electrone

nergy

Mostlyempty

Mostlyfilled

Etop

EcEg

Ev

Ebottom


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Broadening of allowed energy levels into allowed energy bandsseparated by forbidden-energy gaps as more atoms influence eachelectron in a solid.

Energy Band Formation (IV)

One-dimensionalrepresentation

Two-dimensional diagram in whichenergy is plotted versus distance.

Nelectrons filling halfof the 2Nallowed states, as can occur in a metal.

Energy Band

Energy band diagrams.

A completely empty band separatedby an energy gapEg from a bandwhose 2Nstates are completely filled

by 2Nelectrons, representative of aninsulator.

2s Band

Overlapping

energy bands

Electrons2s2p

3s3p

1s 1s

SOLIDATOM

E= 0

Free electronElectron Energy, E

2p Band

3s BandVacuumlevel

In a metal the various energy bands overlap to give a single band of energiesthat is only partially full of electrons.

There are states with energies up to the vacuum level where the electron is free.

Typical band structures ofMetal

Metals, Semiconductors, and Insulators

Electron energy, E

ConductionBand(CB)Emptyofelectronsat 0K.

ValenceBand(VB)Fullofelectronsat 0K.

Ec

Ev

0

Ec+

Covalent bondSiioncore(+4e)

A simplified two dimensional

view of a region of theSicrystal showing covalentbonds.

The energy band diagram of

electrons in theSicrystal atabsolute zero of temperature.

Typical band structures ofSemiconductor


Band gap=Eg


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Carrier Flow for Metal


Carrier Flow for Semiconductors.mov

Carrier Flow for Semiconductor

Carrier Flow for Metals.mov


Insulator Semiconductor Metal

Typical band structures at 0 K.

10610310010-310-610-910-1210-1510-18 109

Semiconductors Conductors

1012

AgGraphite NiCrTeIntrinsic Si

Degeneratelydoped Si

Insulators

Diamond

SiO2

Superconductors

PETPVDF

Amorphous

As2Se

3

Mica

Alumina

Borosilicate Pure SnO2

Inorganic Glasses

Alloys

Intrinsic GaAs

Soda silicaglass

Manyceramics

Metals

Polypropylene


Range ofconductivities exhibited by various materials.

Conductivity (m)-1

Energy Band Diagram

Electrons within an infinite potential energy well of spatial widthL,its energy is quantized.

e

nn

m

kE

2

)( 2h=L

nkn

= ...3,2,1n

Energy increasesparabolically with thewavevector kn.

nkh : electron momentum

This description can be used to the behavior of electron in aMetalwithin which their average potential energy is V(x) 0 inside, and verylarge outside.

x

0+a/2-a/2

V(x)

V=0

m

infinite square potential

0 +a/2-a/2

0 +a/2-a/2

x

x

2

n=1

n=2

n=3

E1

E2

E3

Energy state Wavefunction Probability density


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r

PE(r)

x

V(x)

x = Lx = 0 a 2a 3a

0aa

Surface SurfaceCrystal

Potential Energyof the electronaround an isolated atom

When Natoms are arranged toform the crystal then there is anoverlap of individual electron PEfunctions.

PEof the electron, V(x), inside thecrystal is periodic with a period a.

The electron potential energy [PE, V(x)], inside the crystal is periodic with thesame periodicity as that of the crystal, a.

Far away outside the crystal, by choice, V= 0 (the electron is free and PE= 0).

3.1 Energy Band Diagram

E-kdiagram,Bloch function.

Moving through Lattice.mov

within the Crystal!


[ ] 0)(222

2 =xVEmdxd e

h

Schrdinger equation

)()( maxVxV +Periodic Potential

xkikk exUx )()( =

Periodic Wave function

Bloch Wavefunction

There are many Bloch wavefunction solutions to the one-dimensional crystal eachidentified with a particularkvalue, say kn which act as a kind of quantum number.

Each k(x) solution corresponds to a particularkn andrepresents a state with an energyEk.

E-kdiagram,Bloch function.

...3,2,1m

Ek

k

/a/a

Ec

Ev

Conduction

Band (CB)E

c

Ev

CB

The Energy Band

Diagram

Emptyk

Occupiedk

h+

e-

Eg

e-

h+

h

VB

h

Valence

Band (VB)

TheE-kcurve consists of many discrete points with each point corresponding to apossible state, wavefunction k(x), that is allowed to exist in the crystal.

The points are so close that we normally draw theE-krelationship as a continuouscurve. In the energy rangeEv toEc there are no points [no k(x) solutions].

3.1 Band Diagram

E-kdiagram of a direct bandgap semiconductor

E-kdiagram


The bottom axis describe different directions of the crystal.

Si Ge GaAs

The energy is plotted as a function of the wave number, k,along the main crystallographic directions in the crystal.


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E

CB

kk

Direct Bandgap

GaAs

E

CB

VB

Indirect Bandgap, Eg

kk

kcb

Si

E

kk

Phonon

Si witha recombinationcenter

Eg

Ec

Ev

Ec

Ev

kvb VB

CB

Er

Ec

Ev

Photon

VB

In GaAs the minimum ofthe CB is directly abovethe maximum of the VB.

direct bandgapsemiconductor.

InSi, the minimum of the CB isdisplaced from the maximum of

the VB.indirect bandgap semiconductor

Recombination of an electronand a hole inSiinvolves arecombination center.


E-kdiagram

3.1 Energy Band

A simplified energy band diagram with the highest almost-filledband and the lowest almost-empty band.

valence band edge

conduction band edge

vacuum level

: electron affinity

ehole

CB

VB

Ec

Ev

0

Ec

+

Eg

Free eh >EgHole h+

Electron energy, E

h

3. 1 Electrons and Holes

A photon with an energy greaterthenEgcan excitation an electron

from the VB to the CB.

Each line betweenSi-Siatoms is avalence electron in a bond.

When a photon breaks aSi-Sibond, afree electron and a hole in theSi-Sibond is created.

Generation of Electrons and Holes Electrons: Electrons in the conduction band that are free to move throughout the crystal.

Holes: Missing electrons normally found in the valence band(or empty states in the valence band that would normally be filled).

Electrons and Holes

These particles carry electricity.Thus, we call these carriers


10/67

3.1 Effective Mass (I)

An electron moving in respond to an applied electric field.

EE

within a Vacuum within a Semiconductor crystal

dtdmEqF v0 dtdmEqF n v=

It allow us to conceive of electron and holes as quasi-classical particlesand to employ classical particle relationships in semiconductor crystals orin most device analysis.

3.1 Carrier Movement Within the Crystal

Density of States Effective Masses at 300 K

Ge and GaAs have lighter electrons than Si which results in faster devices

3.1 Effective Mass (II)

Electrons are not free but interact with periodic potential of the lattice.

Wave-particle motion is not as same as in free space.

Curvature of the band determine m*.m* is positive in CB min., negative in VB max.

Moving through Lattice.mov


The bottom axis describe different directions of the crystal.

Si Ge GaAs

The energy is plotted as a function of the wave number, k,along the main crystallographic directions in the crystal.


11/67

The motion of electrons in a crystal can be visualized and describedin a quasi-classical manner.

In most instances

The electron can be thought of as a particle.

The electronic motion can be modeled using Newtonianmechanics.

The effect of crystalline forces and quantum mechanical propertiesare incorporated into the effective mass factor.

m* > 0 : near the bottoms of all bands

m* < 0 : near the tops of all bands

Carriers in a crystal with energies near the top or bottom of anenergy band typically exhibit a constant (energy-independent)effective mass.

` : near band edge

3.1 Mass Approximation

constant2

2 =

dk

Ed

Covalent Bonding

Covalent Bonding Band Occupation at Low Temperature


12/67

Band Occupation at High Temperature Band Occupation at High Temperature

Band Occupation at High Temperature Band Occupation at High Temperature


13/67

Band Occupation at High Temperature

Without help the total number of carriers (electrons andholes) is limited to 2ni.

For most materials, this is not that much, and leads to veryhigh resistance and few useful applications.

We need to add carriers by modifying the crystal.

This process is known as doping the crystal.

Impurity Doping

The need for more control over carrier concentration

RegardingDoping, ...

Concept of a Donor Adding extra Electrons Concept of a Donor Adding extra Electrons


14/67

Concept of a Donor Adding extra Electrons Concept of a Donor Adding extra Electrons

Band diagram equivalent view

eAs+

x

As+ As+ As+ As+

Ec

Ed

CB

Ev

~0.05 eV

Asatom sites every106 Siatoms

Distance into

crystal

Electron Energy

The four valence electrons ofAs allowit to bond just likeSibut the 5thelectron is left orbiting the As site.The energy required to release to freefifth- electron into the CB is verysmall.

Energy band diagram for an n-typeSidoped

with 1 ppmAs. There are donor energy levelsbelow Ecaround As+ sites.

Concept of a Donor Adding extra Electrons

n-type Impurity Doping ofSi

just

Energy band diagram of an n-typesemiconductor connected to avoltage supply ofVvolts.

The whole energy diagram tiltsbecause the electron now has anelectrostatic potential energy aswell.

Current flowing

V

n-Type Semiconductor

Ec

EF

eV

A

B

V(x), PE(x)

x

PE(x) = eV

E

Electron Energy

EceV

EveV

V(x)

EF

Ev

Concept of a Donor Adding extra Electrons

Energy Band Diagram in anApplied Field


15/67

Concept of a Acceptor Adding extra Holes

All regionsof

materialare neutrally

charged

One less bondmeans

the acceptor is

electrically

satisfied.

One less bond

meansthe neighboring

Silicon is left with

an empty state.

Hole Movement

Empty state is located next to the Acceptor

Hole Movement

Another valence electron can fill the empty state located next tothe Acceptor leaving behind a positively charged hole.

Hole Movement

The positively charged hole can move throughout the crystal.(Really it is the valance electrons jumping from atom to atom that creates the hole motion)


16/67

Hole Movement


Hole Movement


Regionaround thehole hasone lesselectronand thus ispositivelycharged.

Hole Movement


Regionaround theacceptorhasone extraelectronand thus isnegativelycharged.


Band diagram equivalent view


17/67

B

h+

x

B

Ev

Ea

B atom sites every106 Si atoms

Distance

into crystal

~0.05 eV

B B B

h+

VB

Ec

Electron energy

p-type Impurity Doping ofSi


Boron dopedSicrystal.B hasonly three valence electrons.When it substitute for aSiatomone of its bond has an electronmissing and therefore a hole.

Energy band diagram for ap-typeSicrystaldoped with 1 ppm B. There are acceptorenergy levels just aboveEv aroundB

- site.These acceptor levels accept electronsfrom the VB and therefore create holes inthe VB.

Ec

Ev

EFi

CB

EFp

EFn

Ec

Ev

Ec

Ev

VB

Intrinsicsemiconductors

In all cases, np=ni2

Note that donor and acceptor energy levels are not shown.

Intrinsic, n-Type, p-Type Semiconductors

Energy band diagrams

n-typesemiconductors

p-typesemiconductors

Impurity Doping Impurity Doping

Valence Band

Valence Band


18/67

Impurity Doping

Position ofenergy levels within the bandgap ofSi forcommon dopants.

Energy-band diagram for a semiconductorshowing the lower edge of theconduction bandEc, a donor levelEdwithin the forbidden band gap,and Fermi levelEf, an acceptor levelEa, and the top edge of the valencebandEv.

Energy Band

Energy band diagrams.

3.2B Semiconductor Statistics

dEEgc )(

The number of conduction band

states/cm3 lying in the energyrange betweenEandE+ dE

(ifE Ec).

The number of valence band

states/cm3 lying in the energy

range betweenEandE+ dE

(ifE Ev).

dEEgv )(

Density of States Concept

General energy dependence of

gc (E) and gv (E) near the band edges.


Density of States Concept

Quantum Mechanics tells us that the number of available states in a

cm3 per unit of energy, the density of states, is given by:

Density of States

in Conduction Band

Density of States

in Valence Band


19/67

3.2B Fermi- Dirac function

Probability of Occupation (Fermi Function) Concept

Now that we know the number of available states at each energy, then

how do the electrons occupy these states? We need to know how the electrons are distributed in energy.

Again, Quantum Mechanics tells us that the electrons follow the Fermi-distribution function.

Ef Fermi energy (average energy in the crystal)

k Boltzmann constant (k=8.61710-5eV/K)T Temperature in Kelvin (K)

f(E) is the probability that a state at energy Eis occupied.

1-f(E) is the probability that a state at energyEis unoccupied.

kTEE feEf

/)(1

1)( +

Fermi function applies only under equilibrium conditions, however, isuniversal in the sense that it applies with all materials-insulators,

semiconductors, and metals.

The Fermi function f(E) is aprobability

distribution function that tells one the ratio of

filled to total allowed states at a given energy

E

How do electrons and holes populate the bands?

Probability of Occupation (Fermi Function) Concept

Fermi-Dirac Distribution


Ef

Fermi Function

Probability that an available state at energyEis occupied:

EF is called the Fermi energy or the Fermi level

There is only one Fermi level in a system at equilibrium.

IfE>>EF :

IfE


20/67


Probability of Occupation (Fermi function) Concept

Maxwell Boltzmann Distribution Function


21/67

Boltzmann Approximation

Probability that a state is empty (occupied by a hole):

kTEE

FFeEfkTEE

/)()(,3If >

kTEE

FFeEfkTEE

/)(1)(,3If >

kTEEkTEE FF eeEf/)(/)(

)(1 =

TYU

Assume the Fermi level is 0.30eV below theconduction band energy (a) determine the pro

bability of a state being occupied by an electr

on at E=Ec+KT at room temperature (300K).

TYU

Determine the probability that an allowed ene

rgy state is empty of electron if the state is below the fermi level by (i) kT (ii) 3KT (iii)

6 KT


22/67


Example 2.2

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 bandedge (Ev).

Where is the Fermi energy locate?

Solution

The Fermi function, f(E), specifies the probability of electron occupyingstates at a given energy E.The probability that a state is empty (not filled) at a given energy E is equalto 1- f(E).

( ) ( )VC EfEf 1( ) ( ) kTEEC FCeEf /1 1+ ( ) ( ) ( ) kTEEkTEEV VFFV eeEf // 1 11 111 ++

kT

EE

kT

EE FVFC =2

VCF

EEE

+=

The density ofelectrons (or holes) occupying the statesin energy betweenEandE + dEis:


Probability of Occupation Concept

0 Otherwise

dEEfEgc )()(Electrons/cm3 in the conduction

band betweenEandE+ dE

(ifEEc).

Holes/cm3 in the conduction

band betweenEandE+ dE

(ifEEv).dEEfEgv )()(


Probability of Occupation Concept


23/67

Typical band structures of Semiconductor

Ev

Ec

0

Ec+

EF

VB

CB

E

g(E)

g(E) (EEc)1/2

fE)

EF

E

For

electrons

Forholes

[1f(E)]

Energy banddiagram

Density of states Fermi-Diracprobabilityfunction

probability ofoccupancy ofa state

nE

(E) orpE

(E)

E

nE

(E)

pE

(E)

Area =p

Area

Ec

Ev

ndEEnE = )(

g(E) Xf(E)Energy density of electrons in

the CB

number of electrons per unitenergy per unit volumeThe area undernE(E) vs.Eis theelectron concentration.

number ofstates per unitenergy per unitvolume


The Density of Electrons is:

Probability the state is filled

Number of states per cm-3 in energy range dE

Probability the state is empty

Number of states per cm-3 in energy range dE

units ofn andp are [ #/cm3]

The Density of Hole is:

Developing the Mathematical Modelfor Electrons and Holes concentrations

Electron Concentration (no)

TYU

Calculate the thermal equilibrium electron concen

tration in Si at T=300K for the case when the Fermi level is 0.25eV below the conduction band

.

EC

EV

EF0.25eV


24/67

Hole Concentration (no)

TYU

Calculate thermal equilibrium hole concentrati

on in Si at T=300k for the case when the Fermilevel is 0.20eV above the valance band energy

Ev.

EC

EV 0.20eV

EF


25/67

Degenerate and Nondegenerate Semiconductors

Nondegenerate Case

Useful approximations to the Fermi-Dirac integral:

( ) kTEEC

CfeNn=

( ) kTEEV

fVeNp=

Developing the Mathematical Modelfor Electrons and Holes

( ) kTEECi

CieNn =When n = ni, Ef= Ei (the intrinsic energy), thenor

and

( ) kTEEVi

iVeNn =( ) kTEE

iVVienN =or

( ) kTEEiC

iCenN =

The intrinsic carrier concentration

( ) kTEECo

CfeNn= ( ) kTEEV

fVo eNp

= Other useful relationships: np product:

( ) kTEECi

CieNn = and ( ) kTEEVi iVeNn =( ) kTE

VC

kTEE

VCi

gVC

eNNeNNn=2

kTE

VCigeNNn

2=

Semiconductor Statistics


26/67

TYU

Determine the intrinsic carrier concentration in

GaAs (a) at T=200k and (b) T=400K

2

ioo npn =

Law of mass Action

kTEE

ioifenn

=( ) kTEE

iofiepp

=andSince

It is one of the fundamental principles of semiconductorsin thermal equilibrium

Example

Law of mass action

An intrinsic Silicon wafer has 1x1010 cm-3 holes. When 1x1018

cm-3

donors are added, what is the new hole concentration?

2

ioo npn =


27/67

DNn DiNn

p

2

andAD NN iD nN andif

TYU

Find the hole concentration at 300K, if theelectron concentration is no=1 x 10

15 cm-3,

which carrier is majority carrier and which

carrier is minority carrier?

TYU

: The concentration of majority carrier

electron is no=1 x 1015

cm-3

at 300K. Determine the concentration of phosphorus th

at are to be added and determine the concentr

ation minority carriers holes.


28/67

Partial Ionization,

Intrinsic Energyand Parameter Relationships.

Energy band diagramshowing negativecharges

Energy band diagram

showing positivecharges

Ifexcess charge existed within the semiconductor, random motionof charge would imply net (AC) current flow.

Not possible! Thus, all charges within the semiconductor must cancel.

Charge Neutrality:

( ) ( )[ ]

( ) ( )[ ] 0=+

+=++

+

nNNpq

nNNp

dA

ad

Mobile+charge

Immobile-charge

Immobile+

charge

Mobile-charge

3.5 Carrier concentration-effects of doping

NA = Concentration of ionized acceptors = ~NA

ND+ = Concentration of ionized Donors = ~ND

Charge Neutrality: Total Ionization case

( ) ( ) 0+ nNNp dA

3.5 Developing the Mathematical Modelfor Electrons and Holes


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The intrinsic carrier concentration as a function of

temperature.

Electron concentration versus temperature for n-typeSemiconductor.

Carrier Concentration vs. Temper

ature

position of Fermi Energy level

( ) kTEEco

fceNn][ =

)/ln( occ nNkTEE F =

)/ln( dcFc NNkTEE =

Nd >> ni

Note: If we have a compensated semiconductor , then the Nd termin the above equation is simply replaced by Nd-Na.


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( ) kTEEVo

fVeNp=

)/ln(ovvF

pNkTEE =

)/ln( avvF NNkTEE =

Na >> ni


Note: If we have a compensated semiconductor , then the Na term

in the above equation is simply replaced by Na-Nd.

position of Fermi level as a function of carrier concentration

Where is Ei?

Extrinsic Material:

Note: The Fermi-level is pictured here for 2 separate cases: acceptor and donor doped.

TYU

Determine the position of the Fermi level with res

pect to the valence band energy in p-type GaAs atT=300K. The doping concentration are Na=5 x 1

016 cm-3 and Na=4 x 1015 cm-3.


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Extrinsic Material:

( ) kTEEio

fifenn = ( ) kTEEioffienp =

Solving for (Ef- Efi)

=

=

ii

fifn

pkT

n

nkTEE lnln

=

i

Dfif

n

NkTEE ln

=

i

Afif

n

NkTEE ln

AD NN iD nN andfor DA NN iA nN andfor

TYU 3.8

Calculate the position of the Fermi level in n-

type Si at T=300K with respect to the intrinsic Fermi energy level. The doping concentrati

on are Nd=2 x 1017 cm-3 and Na=3 x 10

16 cm-3

.


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EC

EV

EFi

EF

Mobile Charge Carriers in

Smiconductor devices

Three primary types of carrier action occur inside asemiconductor:

Drift: charged particle motion under the influence of an

electric field.

Diffusion: particle motion due to concentration gradient

or temperature gradient.

Recombination-generation (R-G)

Carrier Motion

Carrier Dynamics

Electron Drift

Hole Drift

Electron Diffusion

Hole Diffusion

Carrier Drift Direction of motion

Holes move in the direction of the electric field. (F\) Electrons move in the opposite direction of the electric field. (\F) Motion is highly non-directional on a local scale, but has a net direction

on a macroscopic scale.

Average net motion is described by the drift velocity, vd [cm/sec].

Net motion of charged particles gives rise to a current.

Instantaneous velocity is extremely fast

Describe the mechanism of the carrier drift and drift currentdue to an applied electric field.


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Drift

Drift of Carriers

Electric Field

Drift of electron in a solid

The ball rolling down the smooth hill speeds upcontinuously, but the ball rolling down thestairs moves with a constant average velocity.

[cm2/Vsec] : mobilityRandom thermal motion.

Combined motion due to random thermalmotion and an applied electric field.

Drift

Schematic path of an electron in a semiconductor.

EE

Drift

Random thermal motion.Combined motion due to

random thermal motion and an

applied electric field.

Drift

Conduction process in an n-type semiconductor

Thermal equilibrium Under a biasing condition


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Drift

Given current densityJ(I = JxArea ) flowing in a semiconductor blockwith face areaA under the influence ofelectric fieldE, is volumedensity, the component ofJdue to drift of carriers is:

Hole Drift Current Density

dp vpqJ

vJdrf

drf

d

=

=

Electron Drift Current Density

dn vneJdrf

=and

dp

drf

vpeJ

vJ

drf

d

=

=

Drift

At Low Electric Field Values,

EpeJ pDriftp= EneJ nDriftn = and

[cm2/Vsec] is the mobility of the semiconductor and measures theease with which carriers can move through the crystal.

The drift velocity increases with increasing applied electric field.:

EnpqJJJ npDriftnDriftpdrf +=+= )(

Electron and hole mobilities of selected

intrinsic semiconductors (T=300K)

Si Ge GaAs InAs

n (cm2/Vs) 1350 3900 8500 30000

(cm2/Vs) 480 1900 400 500

sV

cm

V/cm

cm/s 2

= has the dimensions ofv/ :

Electron and Hole Mobilities EX 4.1

Consider a GaAs sample at 300K with dopin

g concentration of Na=0 and Nd=1016

cm-3

.Assume electron and hole mobitities given in

table 4.1. Calculate the drift current density if

the applied electric filed is E=10V/cm.


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[cm2/Vsec] is the mobility of the semiconductor and measures the

ease with which carriers can move through the crystal.

Mobility

n ~ 1360 cm2/Vsec for Silicon @ 300K

p ~ 460 cm2/Vsec for Silicon @ 300K

n ~ 8000 cm2/Vsec for GaAs @ 300K

p ~ 400 cm2/Vsec for GaAs @ 300K

[ ]sec2*

,

, Vcmm

q

pn

pn

= is the average time between particle collisions in the

semiconductor.

Collisions can occur with lattice atoms, charged dopant atoms, or withother carriers.

Drift velocity vs. Electric field inSi.

Saturation velocity Saturation velocity

Drift velocity vs. Electric field

Designing devices to work atthe peak results in fasteroperation

1/2mvth2=3/2kT=3/2(0.0259)

=0.03885eV

Ohms law is valid only in the low-field region where drift velocity is independentof the applied electric field strength.

Saturation velocity is approximately equal to the thermal velocity (107 cm/s).

[ ]sec2*

,

, Vcmm

q

pn

pn

=


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Drift

Drift velocity vs. Electric field inSiand GaAs.

Note that forn-type GaAs,there is a region of negativedifferential mobility.

[ s2*

,

, Vcmm

q

pn

pn

=

Negative differential mobility

Electron distributions under various conditions of electricfields for a two-valley semiconductor.

m*n=0.067mom*n=0.55mo

Figure 3.24.

Velocity-Field characteristic of a Two-valley semiconductor.

Negative differential mobility

TYU

Silicon at T=300K is doped with impurity

concentration of Na=5 X 1016

cm-3

and Nd=2x 1016 cm-3. (a) what are the electron and hole

mobilities? (b) Determine the resistivity and

conductivity of the material.


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Mean Free Path

Average distance traveled between collisions

mpthvl =

EX 4.2Using figure 4.3 determine electron and hole nobilities.

EX 4.2Using figure 4.3 determine electron and hole mobilities in (a) Si for Nd=1017 cm-3,Na=5 x 1016 cm-3 and (b) GaAs for Na=Nd=1017cm-3

Ex 4.2


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Effect of Temperature on Mobility

Temperature dependence ofmobility with both lattice and impurity scattering.

A carrier moving through the latticeencounters atoms which are out oftheir normal lattice positions due tothe thermal vibrations.

The frequency of such scatteringincreases as temperature increases.

At low temp. lattice scattering is less important.

At low temperature, thermalmotion of the carriers is

slower, and ionized impurityscattering becomes dominant.

Since the slowing moving carrier islikely to be scattered more strongly byan interaction with charged ion.

Impurity scattering events cause adecrease in mobility with decreasingtemperature.

As doping concentration increase, impurityscattering increase, then mobility decrease.

Mobility versus temperatureMobility versus temperature

Effect of Temperature on Mobility

Electron mobility in silicon

versus temperature forvarious donor concentrations.

Insert shows the theoreticaltemperature dependence ofelectron mobility.

Electron and hole mobilities inSilicon as functions of the totaldopant concentration.

Effect of Doping concentration on Mobility

300 K

Resistivity and Conductivity

Ohms Law

Ohms Law]2cmAEEJ

=

Conductivity[ ]cmohm 1Resistivity[ ]cmohm


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semiconductor conductivity and resistivity

Adding the Electron and Hole Drift Currents (at low electric fields)

Drift CurrentEnpeJJJ npDriftnDriftpdrf +=+= )(

Conductivity)( npe np +=

Resistivity[ ])(11 pne pn

+==

But since n and p change very little and n andp change severalorders of magnitude:

for n-type with n>>p

pe

ne

p

n

for p-type with p>>n


40/67

[ ]sec2*

,

, Vcmm

q

pn

pn

=

Particles diffuse from regions of higher concentration

to regions of lower concentration region, due to

random thermal motion.

DiffusionDiffusion

Nature attempts to reduce concentration gradients to zero.Example: a bad odor in a room, a drop of ink in a cup of water.

In semiconductors, this flow of carriers from one region of higherconcentration to lower concentration results in a Diffusion Current.

Visualization of electron and hole diffusion on a macroscopic scale.

DiffusionpJ

DiffusionnJ

Diffuse Diffuse


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dx

dneDJ N=diffN,

dx

dpeDJ P=diffP,

D is the diffusion constant, ordiffusivity.

x x

Diffusion Current

Diffusion current density

Ficks law

Diffusion as the flux, F, (of particles in our case) is proportional tothe gradient in concentration.

DF : ConcentrationD : Diffusion Coefficient

For electrons and holes, the diffusion current density( Flux of particles times q )

nDqJ

pDqJ

nDiffusionn

pDiffusionp

The opposite sign for electrons and holes

JN = JN,drift + JN,diff= qnn+dxdnqDN

JP = JP,drift + JP,diff= qppdx

dpqDP

J = JN + JP

Total Current


42/67

Total Current

Total Current = Drift Current + Diffusion Current

nDqEnqJJJ

pDqEpqJJJ

nnDiffusionnDriftnn

ppDiffusionpDriftpp

np JJJ +

TYU

Consider a sample of Si at T=300K. Assume thatelectron concentration varies linearly with distance,

as shown in figure.The diffusion current density is

found to be Jn=0.19 A/ cm2. If the electron diffusio

n coefficient is Dn=25cm2/sec, determine the electr

on concentration at x=0.

dxdneDJ N=diffN,

dx

dpeDJ

P

=diffP,

Jp=0.270 A/cm2

Dp=12 cm2/secFind the hole concentration at x=50um


43/67

Graded impurity distribution

Energy band diagram of a semiconductor in thermal equilibrium

with a nonuniform donor impurity concentration

Carrier Generation

Generation and Recombination

Band-to-band generation

Generation Mechanism

Band-to-Band Generation

Thermal Energy

or

Light

Band-to-Band or direct (directly across the band) generation.

Does not have to be a direct bandgap material.

Mechanism that results in ni.

Basis forlight absorption devices such as semiconductorphotodetectors, solar cells, etc

Gno=Gpo


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Band-to-Band Recombination

Recombination Mechanism

Photon(single particle of light)

or

multiple phonons(single quantum of latticevibration - equivalent tosaying thermal energy)

Band to Band or direct (directly across the band) recombination.

Does not have to be a direct bandgap material, but is typicallyvery slow in indirect bandgap materials.

Basis for light emission devices such as semiconductor Lasers,LEDs, etc

Rno=Rpo

In thermal equilibrium: Gno=Gpo=Rno=Rpo

Low-Level-Injection implies

00 , nnnp

chap3 1 semiconductor 1

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