CHAPTER 10 Electrical Properties of Materials - … · CHAPTER 10 Electrical Properties of...

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CHAPTER 10 Electrical Properties of Materials (Callister Chap. 18 p 665 – 686, 702 –704, 711 ; Ashby p 313 - 325)

Transcript of CHAPTER 10 Electrical Properties of Materials - … · CHAPTER 10 Electrical Properties of...

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CHAPTER 10Electrical Properties of Materials

(Callister Chap. 18 p 665 – 686, 702 –704, 711 ; Ashby p 313 -

325)

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• Electrical Conduction in Metals

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• In the absence of an electric potential, the motion of the valence electrons is at random and there is no net current or current flow.

• When an electric potential is applied, the electrons attain a directed drift velocity, which results in directed flow of electrons (or current).

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Electrical Conduction

• Resistivity, ρ and Conductivity, σ:-- geometry-independent forms of Ohm's Law

E: electricfieldintensity

resistivity(Ohm-m)

J: current density

conductivity

-- Resistivity is a material property & is independent of sample

ρ=∆AI

LV

σ = 1

ρ• Resistance:

σ=ρ=

AL

AL

R

• Ohm's Law:∆V = I R

voltage drop (volts = J/C)C = Coulomb

resistance (Ohms)current (amps = C/s)

Ie-A

(cross sect. area) ∆V

L

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• Ohm's law can also be expressed in its microscopic form, which is independent of the shape of the conductor:

• J = E/ρρρρ or J = EσσσσJ = current density [A / m2 ]E = electrical field [V / m ]ρ = electrical resistivity [Ω.m]σ = electrical conductivity [Ω.m] -1

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• Room T values (Ohm-m)-1

Selected values from Tables 18.1, 18.3, and 18.4, Callister 7e.

Conductivity: Comparison

Silver 6.8 x 10 7

Copper 6.0 x 10 7

Iron 1.0 x 10 7

METALS conductors

Silicon 4 x 10-4

Germanium 2 x 10 0

GaAs 10-6

SEMICONDUCTORS

semiconductors

= (Ω - m)-1

Polystyrene <10-14

Polyethylene 10-15-10-17

Soda-lime glass 10Concrete 10-9

Aluminum oxide <10-13

CERAMICS

POLYMERS

insulators

-10-10-11

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• Factors affecting the resistivity and the conductivity of materials.

• Pure defect free metals at low temperatures have very low resistivities. For other conditions, the total resisitivity of a conductor is made up of a thermal (ρt), impurity (ρi) and deformation (ρd) resistivity component

• ρTot = ρt + ρi + ρd

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= ρT + ρr

Thus ρr = ρi + ρd

ρT = ρt = thermal component of resistivityρr = residual component of resistivity

Temperature ρt increases almost linearly with temperature. At higher temperatures the metal ions at the lattice positions vibrate more vigorously, by which the conduction (valence) electrons are increasingly scattered. Thus, the free path of electrons are decreased and resistivity is increased

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• The electrical resistivity of most metals can be approximated by the following equation

ρT = ρ0°C ( 1 + αT T) [Ω.m]ρ0°C = electrical resistivity at 0°CαT = temperature coefficient of resistivity

(αT is a function of the type of metal)T = temperature of conductor (°C)

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Metals: Resistivity vs T, Impurities• Imperfections increase resistivity

-- grain boundaries-- dislocations-- impurity atoms-- vacancies

These act to scatterelectrons so that theytake a less direct path.

• Resistivityincreases with:-- temperature-- wt% impurity-- %CW

Adapted from Fig. 18.8, Callister 7e. (Fig. 18.8 adapted from J.O. Linde, Ann. Physik 5, p. 219 (1932); and C.A. Wert and R.M. Thomson, Physics of Solids, 2nd ed., McGraw-Hill Book Company, New York, 1970.)

ρ = ρthermal

+ ρimpurity

+ ρdeformation

deformed Cu + 1.12 at%Ni

T (°C)-200 -100 0

Cu + 3.32 at%Ni

Cu + 2.16 at%Ni

123

456

Res

istiv

ity,

ρ(1

0-8

Ohm

-m)

0

Cu + 1.12 at%Ni

“Pure” Cu

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• Calculate the resistivity ρ for pure Al at 90 oC

• What is the conductivity σ of pure Al at 90 oC

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Alloying Elements When alloying elements are added to pure metals, solid solutions are formed, whereby the crystal lattice of the parent metal is deformed. This has the effect to cause additional scattering of the conduction electrons, with an increase of the resistivity

Crystal DefectsCrystal defects like dislocations and grain boundaries also cause additional scattering of the conducting electrons

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The Energy-band Model for Electrical Conduction

The energy-band model explains the ability of a material to conduct electrons

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• Sodium (Na) Atomic Number: 11Electron configuration = 1s2 2s2 2p6 3s1

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Band Structure• Valence band – filled – highest occupied energy

levels• Conduction band – empty – lowest unoccupied energy

levels

valence band

Conductionband

Adapted from Fig. 18.3, Callister 7e.

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Schematic energy band diagram for metallic conductors

(a) Sodium (b) Magnesium (c) Aluminium

a b cc

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• The energy-band model for insulatorsIn insulators all the electrons are tightly boundto their bonding atoms by covalent or ionic bonding. They are therefore not free to conduct electricity, unless highly energisedThe valence electron's energy band is therefore separated from the upper (empty) conduction band by a large energy gap (Eg)

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• Intrinsic semiconductors are pure semiconducting materials, like silicon and germanium

• Silicon and germanium are both covalent materials in which bonding is effected by four highly directed sp3 hybrid covalent bonds, to form a diamond cubic crystal structure. The four valence electrons from each atom are therefore tightly attached to the bonding atoms.

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• Electric conduction in these materials is extremely difficult and occurs because some of the valence electrons are exited from their bonding positions by thermal activation. The electrons jump from the valance band across the energy gap Eg to conduction band.

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• Conduction will occur under the applied electric field due to the movements of the positive "holes" to the negative pole and the negative electrons to positive hole

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• Quantitative relationships for conduction in intrinsic semiconductors

• There are two charge carriers contributing to electrical conduction in intrinsic semiconductors i.e. free electrons and positive holes

• The conduction could be expressed as:σ = nqµn + pqµp

n = number of conduction electrons / m3

p = number of conduction holes / m3

q=absolute value of electron or hole charge(q = 1,6 x 10 -19 coulombs)

µn = mobility of electrons [m2/V.s]µp = mobility of holes [m2/V.s]

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Conduction in Terms of Electron and Hole Migration

Adapted from Fig. 18.11, Callister 7e.

electric field electric field electric field

• Electrical Conductivity given by:

# electrons/m3 electron mobility

# holes/m 3

hole mobilityhe epen µ+µ=σ

• Concept of electrons and holes:

-

electron holepair creation

-

no applied applied

valence electron Si atom

applied

electron holepair migration

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But n = p = ni

ni = intrinsic carriers / m3

Thus: σ = ni q (µn + µp ) (Ωm)-1

• Effect of temperature on intrinsic semiconductivityAt zero Kelvin the valence bands of intrinsic semi-conductors such as Si and Ge are completely filled and their conduction bands are completely empty

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• In contrast to metals, the conductivity of semiconductors increase with increase in temperature.

• σ = σo e-Eg/2kT

with σo = temp. independent constantk = Boltzman constant = 8.620x10-5 (eV/K)Eg = Energy gap (eV)

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• Variation of charge carriers with concenteration (fig 18.12 Callister)

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• Extrinsic Semiconductors :Extrinsic semiconductors are very dilute substitutional solid solutions in which the solute impurity atoms (dopants) have different valence characteristics from the solvent (parent) lattice atoms

Two types are distinguishedn-Type (negative-type) extrinsic semiconductors

p-Type (positive-type) extrinsic semiconductors

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The n-Type extrinsic semiconductor If an atom from group 5A e.g. phosphorous, replaces a silicon atom, which is a Group 4A element, there will be one excess valence electron above the four needed for the covalent bonding in the silicon lattice.This extra electron is only loosely bonded and requires only 0,044 eV to be able to jump the energy gap. This is opposed to the 1,1 eV(Eg) which is required for the other valence electrons in the silicon crystal

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• Intrinsic:# electrons = # holes (n = p)--case for pure Si

• Extrinsic:--n p--occurs when impurities are added with a different

# valence electrons than the host (e.g., Si atoms)

Intrinsic vs Extrinsic Conduction

• n-type Extrinsic: (n >> p)

no applied electric field

5+

4+ 4+ 4+ 4+

4+

4+4+4+4+

4+ 4+

Phosphorus atom

valence electron

Si atom

conductionelectron

hole

een µ≈σ

• p-type Extrinsic: (p >> n)

no applied electric field

Boron atom

3+

4+ 4+ 4+ 4+

4+

4+4+4+4+

4+ 4+ hep µ≈σ

Adapted from Figs. 18.12(a) & 18.14(a), Callister 7e.

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• The influence of doping on carrier concentration in extrinsic semiconductors

• At constant temperature, the product of the negative free electron concentration (n) and the positive hole concentration (p) is a constant

np = ni2

Where ni is the intrinsic concentration of carriers in a semiconductor and is constant at a given temperarture.

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• Charge densities in extrinsic semiconductors

• The material must be electrically neutral• Charge carriers in extrinsic semiconductors are

donor ions (Nd), acceptor ions (Na), free electrons (n) and holes (p).

• In general then, sum of negative carriers = sum of positive carriers, or:

• Na + n = Nd + p

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• For n-type extrinsic semiconductors:

• Where nn = concentration (-) electrons in n-type semiconductor

pn = concentration (+) holes in n-type semiconductorNA =0 and nn>> pn

nn ≈≈≈≈ Nd

Nd = concenteration of dopant atoms like P,As and Sb

n p nn n i2=

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• for minority carriers :

• Conductivity (n-type): σσσσ = nn q µn + pn q µp

• ≈ nnqµn because nn>>pn

• Therefore resistivity =1/ σσσσ = 1/ nnqµn

pnn

nNn

i2

n

i2

d= ≈

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• For p-type extrinsic semiconductors• nppp =ni

2

• Where np = concentration (-) electrons in p-type semiconductor

pp = concentration (+) holes in p-type semiconductor

Nd =0 and pp >> np

pp≈ Na

(Na = concentration of dopant atoms, like B, Al and Ga.)

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• Conductivity: (p-type) σσσσ = np q µn + pp q µp

≈ pp q µp

• A silicon wafer is doped with 1021 phosphorous atoms / m3. Calculate:

(a) the majority carrier concentration.(b) the minority carrier concentration.

(c) the electrical resistivity of the doped silicon at 300 K. Data for silicon: ni = 1,5 x 1016 [carriers / m3]

µn = 0,135 [m2 / V s]

µp = 0,048 [m2 / V s]

nnp

nNp

i2

p

i2

a= ≈

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• Dielectric (insulation) BehaviourA dielectric material is one that is electically insulating and

exhibits or is made to exhibit electric dipole stuctureHigh and low voltage electrical insulators.Examples of ceramic materials used as electrical

insulators:# electrical porcelain , alumina (Al2O3) , steatiteExamples of polymer materials used as electrical

insulators:# phenol formaldehyde (bakelite) , nylon , polyethyleneInserts for capacitors. (A capacitor is an electrical device

which can store electric charge.)Examples of ceramic materials used in capacitors:# barium titanate (BaTiO3) . BaTiO3 mixed with other

ceramics

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• Piezoelectric ceramics• some ceramics which can convert

electrical pulses into mechanical vibrations and vice versa

• Utilized in transducers• Examples of piezoelectric ceramics:# barium titanate (BaTiO3)# mixtures of lead zirconate (PbZrO3) and lead titanate(PbTiO3)