MLE2105 Electronic Properties of Materials

Workload: 26 lecture hours + 6 tutorials + 6 assignmentsg

Dr Chen JingshengDr. Chen Jingsheng

Tel: 65167574 (O) 98223576 (Hp)Office: E3A 04-13E il j@ d

1

E-mail: msecj@nus.edu.sg

Main Text Book:

(1) S.O. Kasap, Principles of Electronic Materials and Devices, McGraw-Hill,(2006), TK453kas2006.

Reference Books:

(1) William D. Callister, Jr., Materials Science and Engineering An Introduction,John Wiley & Sons, (2003), TA403 cal 2003.

(2) D. Halliday, R. Resnick and J. Walker, Fundamentals of Physics, JohnWiley(2) D. Halliday, R. Resnick and J. Walker, Fundamentals of Physics, JohnWiley& Sons, (2001), QC 21.2 Hal 2001.

2

Percentages

Final examination 60%

Experimental report and VIVA 15%

Mid-term test 10%Mid term test 10%

Assignment (including one term paper) 15%

Mid-term test

Early of March

Biweekly Tutorials. Three slots available: Wed, 14:00-15:00, E1-06-05Thursday: 13:00-14:00 14:00-15:00 E1-06-03

Early of March

Thursday: 13:00 14:00, 14:00 15:00, E1 06 03.

3

switch Position sensor IC circuit

4Light emitting diodequantum dots with several different colors

Honda‘s two seated Dream car is powered by photovoltaics

Chapter I: Electrical Conduction in Solids

Chapter II: Quantum Theories of solids

Ch t III S i d tChapter III: Semiconductors

Chapter IV: Semiconductor Devicesp

5

Chapter I Electrical conduction in solids

magnetic field effectHall effect and Hall deviceConductivity (σ) in

classic model-the Drude model

Application in IC-interconnect

Temperature dependence of σ -pure metal σ of polycrystalline

thin film

Electrical Conductionin metal

Doping effects on σ :Matthiessen’s rule σ of heterogeneous

Structure: mixture rule

in metal

σ of solid solution:Nordheim’s ruleElectrical

Conductionin non metal

6

in non-metal

Characteristics of metal

Cu: 1s22s22p63s23p63p104s1

Valence electron is detached

electrons wander in the whole latticeelectrons wander in the whole lattice

Interaction between electron and ionsis the origin of existence of metal solidg

Kinetic energy of gas atom=3/2 kTElectron speed in metal is insensitive temperature -almost constant Electrons is not really like gas.

7

one assumptions in the classic mode:Valence electrons are free to move in metal (free electron gas model) although the speed of electron is constant.

Electric current density J : the net amount of charge flowing across a unit areaper unit time, Electric current is the net amount of charge flowing per unit time,which are,

No electric field

x

ΔtΔqI

tAqJ =ΔΔ

= and

Cu: 1s22s22p63s23p63p104s1

M f l i l lu

Movement of electrons in metal crystal:

Collide with vibrating atomsVibrating Cu+ ions

Move randomly

Cross its initial x plane position No net displacementin any one direction

8

Common sense tells us: No current.

With an applied electric field

Ex AΔ A

Jx

Δxvdx

Δx

VV

The electron experiences an acceleration (force eEx) in the x direction inaddition to its random motion. There is a net displacement of electrons in xdi ti Th l t t d ift l it ( ) i di tidirection. The electrons get a drift velocity (vdx) in x direction.

Current > 0

9Current: net movement of electrical charges in one direction.

The relationship between electric current density J and the applied electric

1. The Drude modelThe relationship between electric current density J and the applied electricfield:

tAqJΔΔ

= 1-1AeNJΔ

=tAΔ 1 1

AΔxΔt

tAΔ

1

Jx

vdx ]...[1321 xNxxxdx vvvv

Nv ++++= 1-2

eNnVN = n is the number of electrons per unit volume (n = N/V).

tAeNJ x Δ

=

tAvV dxΔ= 1-3dx

dxx env

AtenAv

AeN

AqJ =

ΔΔ

=Δ

=ΔΔ

=

10

dxx tAtAtA ΔΔΔ

1-4)()( tenvtJ dxx =

How to calculate the average electron velocity (vdx) under an electric field?

Ex First consider the velocity vxi of the ithelectron in the x direction at time t.

Δx

Suppose its last collision was at time ti, itsvelocity is uxi (initial velocity).

In the presence of an electric field, fortime (t-ti), the electron is accelerated freeof collisions and the acceleration is

x

V

of collisions, and the acceleration iseEx/me.

Last collision

Present time t

vxi

timeti t

uxi

11

i

So, the velocity vxi in the x direction at time t will be:

)(xeENewton’s second law: )( ie

xxixi tt

muv −+= 1-5Newton’s second law:

F = ma, v = v0 + at

We assume that immediately after a collision with a vibrating ion, the electron maymove in any random direction; that is, it can just as likely move along the negativeor positive x, so that uxi averaged over many electrons is zero. Thus,

1 eE )(]...[1321 i

e

xxNxxxdx tt

meEvvvv

Nv −=++++= 1-6

12

is the average free time for N electrons between collisions, or the meantime between collisions (also known as mean scattering time and relaxation

)( itt −

time), which is denoted as τ, so eqn 1-6 can be written as,

xdx Emev τ

= 1-7em

The constant eτ/me is called drift mobility μd, so,

xddx Ev μ= 1-8

So, the current density Jx is,

xdx EenJ μ= 1-9

Reciprocal 1/τ : the mean probability per unit time that the electron will be

xx EJ σ=denμσ =

13

p p y pscattered.

Consider an infinitesimally small time interval dt at time t. Let N be the numberunscattered electrons at time t The probability of scattering during dt is (1/τ)dt andunscattered electrons at time t. The probability of scattering during dt is (1/τ)dt andthe number of scattered electrons during dt is N(1/τ)dt . The change dN in N is thus:

dtNdN )1(τ

−=

0 at time electrons dunscattere ofnumber total theis

),exp(

0

0

=

−=

tN

tNNτ

The mean free time can be calculated from the mathematical definition of .t

τ==∫∫∞

∞

0

Ndt

tNdtt

14

∫0 Ndt

( ) ( ) ( )αααα α Γ=+Γ=Γ ∫∞ −− 1 ;

0

1dtte t

2. Temperature dependence of resistivity: pure metals

Relate the temperature to resistivity (the electron is only scattered by thermalvibration in pure metals):

T → σ=1/ρ denμσρ

==1

T → μd

eτμ =

T → τe

d mμ =

T → Scattering

15The mean speed u of conduction electrons in a metal can be shown to be onlyslightly temperature dependent.

τ is inversely proportional toπS = a2

y p pthe area πa2 that scatters theelectron.

Th ki i f

a

u

l=uτ

The average kinetic energy ofthe oscillations is 1/4Ma2ω2,where ω is the oscillationfrequency. From the kinetic

u

q ytheory of matter, this averagekinetic energy must be on theorder of 1/2kT. Therefore,

Electron

Fi 1 3

kTMa21

41 22 ≈ω

Figure 1-3

Ta ∝2

16Ta11

2 ∝∝π

τTC

=τ 1-10So,

eCd =μ 1 11d

eτμ =Tme

dμ 1-11e

d mμ

C 12

TmnCeen

edT

12

== μσ

ATnCeTm

ene

dTT ==== 2

11μσ

ρ 1-12

Where A is a temperature-independent constant. We term this conductivity lattice-scattering-limited conductivity.g y

17

3. Matthiessen’s rule

(1) Matthiessen’s rule and the temperature coefficient resistivity (α)

For metallic alloys, their resistivities are only weakly temperature dependent. Why?y , y y p p y

τI

τΤ

Figure 1-4

The impurity atom results in a local distortion of the crystal lattice, and then the

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p y ylocal potential energy is changed. It will be effective in scattering.

Two types of mean free times between collisions: τT, for scattering from thermalvibrations only and τI, for scattering from impurities only.vibrations only and τI, for scattering from impurities only.

IT τττ111

+= 1-14IT

μd = eτ/me μμμ111

+= 1-15ILd μμμ

Where, μL = eτT/me μI = eτI/me.

The effective resistivity ρ of the material is simply 1/enμd, so,

ITILd enenen

ρρμμμ

ρ +=+==111

1-16Matthiessen’s rule

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τI depends on the separation between the impurity atoms and therefore on the concentration of those atoms.

There may also be electrons scattering from other crystal defects. All of thesescattering processes add to the resistivity of a metal, just as the scattering processg p y j g pfrom impurities, so,

IRT ρρρρ ++= 1-17

Where ρR is called the residual resistivity and is due to the scattering of all thecrystal defects. The residual resistivity shows very little temperature dependence,whereas ρT = AT, so,w e e s ρT , so,

BAT +≈ρ 1-18

Where A and B are temperature independent constantsWhere A and B are temperature-independent constants.

20

The temperature coefficient of resistivity (TCR) α0 is defined as the fractionalchange in the resistivity per unit temperature increase at the reference temperaturechange in the resistivity per unit temperature increase at the reference temperatureT0, that is,

⎤⎡

000

1

TTT =⎥⎦⎤

⎢⎣⎡=δδρ

ρα 1-19

⎥⎦

⎤⎢⎣

⎡ −= 0

01

TTρρ

ρα

)](1[ TT

⎦⎣ − 00 TTρ

)](1[ 000 TT −+= αρρ 1-20

For pure metal, ρ = AT, is substituted in equation 1-20, then

21α0 = T0

-1

Metal ρ0 (nΩ m) α0 (1 /K) n Comment

Table 1.1 Resistivity, thermal coefficient of resistivity α0 at 273K (0 °C) for various metals. The resistivity index n in ρ ∝ Tn or some of the metals is also shown.

Aluminum, Al 25.0 1/233

Antimony, Sb 38 1/196

Copper, Cu 15.7 1/232 1.15

Gold, Au 22.8 1/251

Indium, In 78.0 1/196

Platinum, Pt 98 1/255 0.94

Silver, Ag 14.6 1/244 1.11

Tantalum Ta 117 1/294 0 93 Tantalum, Ta 117 1/294 0.93

Tin, Sn 110 1/217 1.11

Tungsten, W 50 1/202 1.20

Iron, Fe 84.0 1/152 1.80 Magnetic metal; 273 < T < 1043K

Nickel, Ni 59.0 1/125 1.72 Magnetic metal; 273 < T < 627K

22•For some magnetic metals, the conduction electron is not scattered simply by atomicvibrations, but is affected by its magnetic interaction with the ions in the lattice. This leadsto a complicated T dependence.

10

100ρ∝T

Figure 1-5

1

10

ρ(nΩm)n Ωm )

•As the T decreases, typically below~ 100 K for many metals, our simpleassumption that all the atoms areib ti ith t t f

0.01

0.1

22.5

33.5

ρ∝Tρ∝T5

ρ(nΩm)

Resis

ti vit y

(n Ω vibrating with a constant frequencyfails. The mean free time τ becomeslonger and strongly T dependent,leading to a smaller resistivity than

0.0001

0.001

00.5

11.5

0 20 40 60 80 100

ρ∝T5g y

the ρ∝ T. For some metals, such ascopper, ρ∝ T5 (Figure 1-5).

0.000011 10 100 1000 10000

Temperature (K)

T (K)

The resistivity versus temperature behavior can be empirically described bya power law of the form:

n⎤⎡

23TT⎥⎦

⎤⎢⎣

⎡=

00ρρ 1-13

(2) Solid solutions and Nordheim’s rule

F i h ll f t t lFor an iso-morphous alloy of two metals,

ρ = ρT + ρR + ρI

How to work out ρI?

Concentration of impurity atoms related

dII enμ

ρ 1= A semi-empirical equation.

dIμ

Ieτμ = lI=τ

24e

dI mμ =

uI =τ

600

C u-N i A lloys

300

400

500

(nΩ

m)

100

200

300

Res

isti

vity

100% C u at.% N i

00 20 40 60 80 100

100% N i

(b )(b )

The resistivity of the Cu-Ni alloy as a function of Ni content (at.%) at roomtemperature.

25

Nordheim’s rule: relates the impurity resistivity ρI to the atomic fraction X ofsolute atoms in a solid solution, as follows.

)1( XCXI −=ρ 1-21

Where C is the constant termed the Nordheim coefficient, which represents theeffectiveness of the solute atom in increasing the resistivity. Obviously, for smallamounts of impurity, X << 1 and ρI = CX.

26

Resistivity ρ=1/(enμd), change in n also affects the resistivity

Nordheim’s rule: the alloying does not significantly vary the number ofconduction electrons per atom on the alloy.

Alloy from same column i th P i d t bl

Alloy of different valency such as Cu-Zn

Cu-Ni alloyin the Period table-same valence electrons such as Cu-Au and Ag-Au

such as Cu Zn

Other scattering h i

Nordheim’s rule is valid for any concentration

Resistivity of Cu-Zn at high Zn content predicted by Eq.1-21 is greater than actual value.

mechanism-magnetic interaction

y

Nordheim’s rule is only valid at low concentration

Resistivity at high Ni concentration by Eq.1-21 is smaller than actual

27

is smaller than actual value.

With Nordheim’s rule in eqn 1-21, the resistivity of an alloy of composition X is,

)1( XCXmatrix −+= ρρ 1-22

Where ρmatrix = ρT + ρR is the resistivity of the matrix due to scattering fromthermal vibrations and from other defects, in the absence of alloying elements. Toreiterate the value of C depends on the alloying element and the matrix Forreiterate, the value of C depends on the alloying element and the matrix. For,example, C for gold in copper would be different from C for copper in gold.

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4. Heterogeneous mixture rules

Nordheim’s rule only applies to solid solutions that are single-phase solids. In otherwords, it is valid for homogeneous mixtures in which the atoms are mixed at thegatomic level throughout the solid.

How about multiphase solids?How about multiphase solids?

29Heterogeneous mixture rules.

Three kinds of mixtures of two phases

L LA

Dispersed phaseContinuous phase

y

Jx

A A Ax

Jx

α β L

(a) Resistivity of materials with layer structure

(b) Resistivity of materials with layer structure

(c) Resistivity of materials with a dispersed phase inlayer structure

along a direction perpendicular to the layers

yalong a direction parallel to the layers

dispersed phase in a continuous matrix

30

AL The effective resistance Reff for the whole material is,

Jx

A

AL

ALReff

ββαα ρρ+= 1-23

α β

AL

R effeff

ρ= 1-24

di l

Where ρeff is the effective resistivity, Lα , Lβ are the total length of α and β,respectively, and L = Lα + Lβ. Using volume fraction χα = Lα/L and χβ = Lβ/L in

perpendicular

eqn 1-23, we find,

ββαα ρχρχρ +=eff 1-25

Which is called the resistivity-mixture rule or series rule of mixtures.

31

L

A

The effective resistance Reff for the whole material is,

A

β

β

α

α

ρρρ LA

LA

LA

R effeff

+==1

parallelL

AL

AL

A eff ββαα σσσ+=

parallel

Where σeff is the effective conductivity and A = Aα + Aβ. Using χα = Aα/A andχβ = Aβ/A, we find,χβ β , ,

ββαα σχσχσ +=eff 1-26

Which is called the conductivity-mixture rule or parallel rule of mixtures.

32

A

Dispersed phaseContinuous phase

y

x

1) the resistivities of the two phases are not verydifferent,

xJx

2) The resistivities of the two phases are very different?L

RandomRandom

33

1) the resistivities of the two phases are not very different.

Divide the solid into a bundle of N parallel fibers of

L

A

Divide the solid into a bundle of N parallel fibers of length L and cross-sectional area A/N

In the fiber α and β phases are in series volume

β

α(a)

βα

In the fiber, α and β phases are in series, volume fractions χα=Vα/V, χβ=Vβ/V, total length of α and βregion are χαL and χβL. The resistivity of the fiber is

βA/N

α

L(b)

(a) A two phase solid. (b) A thin fiber cut out from the solid.

( )( )

( )( )NA

LNA

LRfiber //ββαα χρχρ

+=

Two types of resistivities ρα of αand ρβ of β

The resistance of the solid is made up of N such fibers in parallel, that is

ρβ β

( ) ( )A

LA

LN

RR fiber

solidββαα χρχρ

+==Rsolid= ρeffL/A

( ) ( )ρ LLL

34

( ) ( )

ββαα

ββαα

χρχρρ

χρχρρ

+=

+=

eff

eff

AL

AL

AL

Assume that ρ and ρ are the resistivities of the continuous and dispersed phases (ρ

2) the resistivities of the two phases are very different.

Assume that ρc and ρd are the resistivities of the continuous and dispersed phases (ρcand ρd are very different), and χc and χd are their volume fractions.

If the dispersed phase is much more resistive with respect to the matrix, that is, ρd >

)211( dχ+ 1-27

10ρc, then,

)1(

)2

(

d

d

ceff χ

χρρ

−=

On the other hand, if ρd < (ρc/10), then,

)1( dff

χρρ −= 1 28)21( d

ceff χρρ

+ 1-28

We therefore have at least four mixture rules, the uses of which depends on the

35

mixture geometry and the resistivities of the various phases.

5. Hall EffectThe force F (Lorentz force) acting on a charge q

Jy=0

The force F (Lorentz force) acting on a charge qmoving with a velocity v in a magnetic B is given through the vector product:

Jx

VVH

xz

yBz

Jx E

+ + ++ + BvF ×= q 1-29

Jxvdx

A

Jx Ex

BqvF = Di i ?Bz

V

A

+

zdx BqvF = Direction?

+

36

BvF ×= q

vq=-e

Direction of v x B: Downward (-y).B

v v is swept into B through a smaller angle

F = qv´Bq is -e, so F is still in –y direction.

VVH

Jy=0

yBz

+ + ++ +

Jxvdx

VVHeEH xzJx

EH

Ex

+ + ++ +

evdxBz

Bz

A

EH

37V+

The accumulation of electrons near the bottom results in an internal electric field EHin the –y direction. This is called Hall field and gives rise to Hall voltage VHbetween the top and the bottom.

Two forces will be balanced at last:

zdxH BeveE = 1-30

However, Jx = envdx. Therefore,

⎞⎛zxH BJ

enE ⎟

⎠⎞

⎜⎝⎛=

1 1-31

A useful parameter called the Hall coefficient RH is defined as,

yER = 1 32

38

zxH BJ

R = 1-32

The quantity RH measures the resulting Hall field, along +y, per unit transverseapplied current and magnetic field. The larger RH, the greater Ey for a given Jx andB Th f R i f h i d f h h ll ff A i f 1Bz. Therefore, RH is a gauge of the magnitude of the hall effect. A comparison of 1-31 and 1-32 shows that for metal,

E

zx

yH BJ

ER = 1-32

yH EE −=

zx

H

BJenR

⎟⎠⎞

⎜⎝⎛−

=

1zxH BJ

enE ⎟

⎠⎞

⎜⎝⎛=

1

1

zxH BJ

R⎠⎝

enRH

1−= 1-33

39

40

Application of Hall effect

Wattmeter - measure the power dissipated in the load

Bz∝ IL

IX= VL/R

Hall voltage VH = Ehw=wRHJxBz∝I B ∝V I∝IxBz ∝VLIL

41

6. The conduction of thin metal film

Criteria for defining bulk metal material and thin metal film

Mean free path for electron scattering

Bulk metal material Thin metal film

Any dimension of the specimen is much larger than th f th

The thickness or average grain size is comparable with the mean free path for electronthe mean free path mean free path for electron scattering

Thermal scattering and Thermal scattering and impurity scattering

42

gimpurity scattering

Thermal scattering and impurity scatteringGrain boundary scattering and surface scattering

Grain boundary scattering Assumption: after one grain boundary scattering, electrons are randomized.

Average grain size -d corresponds to the mean free path-uτ. According to Matthiessen’s rule

dl111

+=λ

Where λ is the mean free path of the conductionWhere λ is the mean free path of the conduction electron in the single crystal.

ρ 11∝∝l dlfilm

111+=∝

λρ

λτρ ∝∝crystal dlfilm λ

Elastic scattering +=film ;3311 βρ

)(1dcrystal

film λρρ

+=Elastic scattering from boundary

⎟⎠⎞

⎜⎝⎛−

=

+=

RR

d

crystal

1

;33.11

λβ

βρ

1-34

43

⎠⎝

Where R is probability of elastic scattering

1 34

1-35

Surface scatteringAssumption: the scattering from surface is inelastic- nonspecular.

The mean free path lsurf of the electrons depends on its direction after the scattering

inelastic nonspecular.

on its direction after the scattering.

( )θcos/Dlsurf =

Since the electrons cannot escape from the thin film, it therefore take two collisions to randomize the velocity.

As a result, the effective mean free path is twice as long that islong, that is

( )θcos/2Dlsurf =

44

According to Mathiessen’s rule, the overall mean free path-l, λ is the mean free path of bulk crystal (no surface scattering)

Dll surf 2cos1111 θ

λλ+=+=

Average for all possible θ values per scattering, θis from -π/2 to π/2

π

⎟⎞

⎜⎛∫

Dd

dDDlsurf π

θ

θθ

θ π

π

π

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=⎟⎠⎞

⎜⎝⎛=

∫

∫

−

−

2

2

2

cos/2

cos2

⎟⎠

⎜⎝∫−

2

+=λλ 1

Elastic scattering (specular scattering)

⎟⎠⎞

⎜⎝⎛+=

D

Dlsurf

λρ

π

11

(specular scattering)

( ) ,3.0 ,11381 >−+=

Dp

Dbulk

λπ

λρρ

1-37

45

⎟⎠

⎜⎝ Dbulk πρ where p is fraction of elastic scattering from

surface1-36

35As deposited

300(a) (b)

Examples

30Annealed at 100 CAnnealed at 150 C

As deposited

100

(a) (b)

20

25 50

0 0.05 0.01 0.015 0.02 0.02515 50 100 500105

10ρbulk = 16.7 n m

Film thickness (nm)1/d (1/micron) Film thickness (nm)

(a) ρfilm of Cu polycrystalline films versus reciprocal mean grain size. Film thi k D 250 900 d t ff t

(b) ρfilm of Cu polycrystalline films versus film thickness. Annealing the fil t d th l t lli dthickness D=250-900nm does not affect

the resistivity.

Interception with Y axis and slope can deduce the

film to reduce the polycrystalline does not affect the resistivity since ρfilm is controlled by surface scattering.

Interception with Y axis and slope can deduce the single crystal resistivity and mean free path of single crystal

7. Application of metal film: Interconnects in microelectronics

Why Cu interconnect has replaced Al interconnect with increase of the chip speed?

Lower resistivity

S i i t t

Heavier atom

47

Superior resistance to electromigrationLower power

consumption- I2RLower RC time constant-favoring high operation speed Any metal better than Cu?

RC time constant induced by interconnects

(a) A single line interconnect surrounded by(a) A single line interconnect surrounded by dielectric insulation.

(b) Interconnects crisscross each other. There are three levels of interconnect: M 1 M d M + 1

(c) An interconnect has vertical and horizontal capacitances Cv and CH.

Resistance of an interconnect is ( )TWLR /ρ=

M – 1, M, and M + 1

The interconnects couples with other interconnets around it

From the simple parallel plate capacitance formula,

HWL

dAC

XTL

dAC r

Vr

Hεεεεεε 00 ; ==== 1-38

The four capacitances are in parallel, the effective capacitance is

( ) ⎟⎠⎞

⎜⎝⎛ +=+=

HW

XTLCCC rVHeff εε 022 1-39⎠⎝ HX

⎟⎠⎞

⎜⎝⎛ +⎟⎟⎠

⎞⎜⎜⎝

⎛=

HW

XT

TWLRC r

2

02 ρεε 1-40⎠⎝⎠⎝

Three important factors affect the RC and thus the chip speed,

(1) Resistivity of interconnects

(2) Permeability of dielectronic material

(3) Geometry of the interconnects-architecture of the interconnects

49

Electromigration

(a) Electrons bombard the metal ions and force them to slowly migrate(b) i f id d hill k i l lli l i b h(b) Formation of voids and hillocks in a polycrystalline metal interconnect by theelectromigration of metal ions along grain boundaries and interfaces. (c) Acceler-ated tests on 3 mm CVD (chemical vapor deposited) Cu line. T = 200 oC, J = 6MA cm-2: void formation and fatal failure (break) and hillock formation

50

MA cm-2: void formation and fatal failure (break), and hillock formation.

|SOURCE: Courtesy of L. Arnaud et al, Microelectronics Reliability, 40, 86, 2000.

8. Electrical Conductivity of Nonmetals

Semiconductors ConductorsInsulators

Many ceramics

Diamond

Superconductors

Mica

Alumina

Inorganic GlassesMetals

AgGraphite NiCrTeIntrinsic Si

DegeneratelyDoped Si

SiO2

PETPVDF

Amorphous

Borosilicate Pure SnO2 Alloys

Intrinsic GaAs

Soda silica glassPolypropylene

10610310010-310-610-910-1210-1510-18 109 1012

Conductivity (Ωm)-1

AgGraphite NiCrTeSiO2 As2Se3

Intrinsic GaAs

Conductivity (Ωm) 1

51

(1) SemiconductorsE

e-hole

(b) (c)(a)

Figure 1-9

Suppose that n and p are the concentrations of electrons and holes in aSuppose that n and p are the concentrations of electrons and holes in asemiconductor crystal. If electrons and holes have drift mobilities of μe and μh,respectively, then the overall conductivity of the crystal is given by

52eh enep μμσ += 1-41

(2) Ionic crystals and glasses

S f bil h

EV id th diff i f iti i

Sources of mobile charges:

1. Crystal defects: vacancies and interstitialVacancyaids the diffusion of positive ion impurities which are often ionized or

charged.

2 Derivations from stoichiometry in2. Derivations from stoichiometry incompound solids: mobile electrons or holes.

I t titi l ti diffAnion vacancy

Interstitial cation diffusesacts as a donor

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Glasses

EE

Na+Na

Many glasses and polymers contain a certain concentration of mobile ionsin the structure.

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Conductivity σ of the material depends on all the conduction mechanisms withh i f h i ki t ib ti it i i beach species of charge carrier making a contribution, so it is given by

iiinq μσ ∑= 1-42

Where ni is the concentration, qi is the charge carried by the charge carrier speciesi , qi g y g pof type i (for electrons and holes qi = e), and μi is the drift mobility of these carriers.

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For many insulators, whether ceramic, glass or polymer, the conductivity followsan exponential or Arrhenius-type temperature dependence so that σ is thermallyactivated,

⎟⎠⎞

⎜⎝⎛−=

kTEσσσ exp0 1-43

⎠⎝

Where Eσ is the activation energy for conductivity.

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