university of technology materials engineering department ...€¦ · university of technology...

University of Technology

Materials Engineering Department

Lecturer Name : Assistant Lecture Sura Salim Ahmed

Subject Name : Electronic and Magnetic Materials

Brunch : Building and Ceramic Materials Engineering

Stage : 3rd

Semester : 1st

Semester Topic : Electrical Properties of Mateials

Topic-1 Outline

1 Introduction

2 Electrical Properties (Conductivity)

3 Resistivity in Metal and Alloy

4 Band Structures of Solids

5 Superconductivity

6 Applications of Superconductivity

7 Semiconductors

8 Semiconductors

9 Insulators and Dielectric Properties

10 Polarization in Dielectrics

11 Electrostriction, Piezoelectricity, and Ferroelectricity

Lecture 2: Resistivity and Band Structures Electrical Properties of Materials

References

1) Rolf E. Hummel, Understanding Materials Science – History * Properties *

Application, Second Edition, © Springer – Verlag New York, LLC, 2004.

2) Rolf E. Hummel, Eectronic Properties of Materials, Fourth Edition, © Springer

Science + Business Media, LLC, 2011.

3) William D. Callister and David G. Rethwisch, Materials Science and

Engineering: An Introduction, Seventh Edition, © John Wiley & Sons, Inc.,

2007.

4) Donald R. Askeland – Pradeep P. Phulé, The Science and Engineering of

Materials, Ch. 18: Electronic Materials, Fourth Edition, Power Point Lectures.

5) Brain S. Mitchell, An Introduction to Materials Engineering and Science,

© John Wiley & Sons, Inc., Hoboken, New Jersey, 2004.


3 Electrical Resistivity of Metals

As mentioned previously, most metals are extremely good conductors of

electricity because of the large numbers of free electrons. Thus n has a large value in the

conductivity expression, Eq. (6) in lecture – 1.

At this point it is convenient to discuss conduction in metals in terms of the

resistivity, the reciprocal of conductivity.

Since crystalline defects serve as scattering centers for conduction electrons in

metals, increasing their number raises the resistivity (or lowers the conductivity). The

concentration of these imperfections depends on temperature, composition, and the

degree of cold work of a metal specimen. In fact, it has been observed experimentally

that the total resistivity of a metal is the sum of the contributions from thermal

vibrations, impurities, and plastic deformation. This may be represented in

mathematical form as follows:

in which ρth, ρimp and ρdef represent the individual thermal, impurity, and deformation

resistivity contributions, respectively. Eq. (7) is sometimes known as Matthiessen’s

rule. The influence of each variable on the total resistivity is demonstrated in Figure (5),

a plot of resistivity versus temperature for copper and several copper – nickel alloys in

annealed and deformed states. The additive nature of the individual resistivity

contributions is demonstrated at – 100°C.

Figure (5): The electrical resistivity

versus temperature for copper and three

copper–nickel alloys, one of which has

been deformed. Thermal, impurity, and

deformation contributions to the resistivity

are indicated at [Adapted from J. O.

Linde, Ann. Physik, 5, 219 (1932); and C.

A. Wert and R. M. Thomson, Physics of Solids, 2

nd edition, McGraw-Hill Book

Company, New York, 1970.]


Influence of Temperature

When the temperature of a metal

increases, thermal energy causes the

atoms to vibrate, Figure (6). At any

instant, the atom may not be in its

equilibrium position, and it therefore

interacts with and scatters electrons.

The mean free path decreases, the

mobility of electrons is reduced, and

the resistivity increases. The change in

resistivity of pure metal as a function of

temperature, thus,

where ρth the resistivity at any temperature T, ρRT the resistivity at room temperature

(i.e., 25°C), ΔT =(T – TRT)is the difference between the temperature of interest and

room temperature, and α is the temperature resistivity coefficient. This dependence of

the thermal resistivity component on temperature is due to the increase with temperature

in thermal vibrations and other lattice irregularities (e.g., vacancies), which serve as

electron – scattering centers. So, the relationship between resistivity and temperature is

linear over a wide temperature range, Figure (7). Table (3), given some examples of the

temperature resistivity coefficient.

Figure (6): Movement of an electron through

(a) a perfect crystal, (b) a crystal heated to a high

temperature, and (c) a crystal containing atomic

level defects. Scattering of the electrons reduces

the mobility and conductivity.

Figure (7): The effect of temperature on the

electrical resistivity of a metal with a perfect

crystal structure. The slope of the curve is the

temperature resistivity coefficient.


Table (3): The temperature resistivity coefficient α for selected metals.

Example – 3: Resistivity of Pure Copper

Calculate the electrical conductivity of pure copper at (a) 400oC and (b) – 100

oC

Example – 3 Solutions

Since the conductivity of pure copper is 5.98 105 Ω

-1.cm

-1. the resistivity of

copper at room temperature is 1.67 10-6

ohm.cm. The temperature resistivity

coefficient is 0.0068 (°C)

-1.


Influence of Impurities

For additions of a single impurity that forms a solid solution, the impurity

resistivity ρimp is related to the impurity concentration cimp in terms of the atom fraction

as follows:

where A is the impurity resistivity coefficient. The influence of nickel impurity additions

on the room temperature resistivity of copper is demonstrated in Figure (8), up to 50

wt% Ni; over this composition range nickel is completely soluble in copper Figure (9).

Again, nickel atoms in copper act as scattering centers, and increasing the concentration

of nickel in copper; results in an enhancement of resistivity.

For a two-phase alloy consisting of α and β phases, a rule-of-mixtures expression

may be utilized to approximate the resistivity as follows:

Figure (8): Room temperature electrical

resistivity versus composition for copper–

nickel alloys.

Figure (9): The copper – nickel phase

diagram.


where the V’s and ρ’s represent volume fractions and individual resistivity for the

respective phases.

Example – 4: Estimate the electrical conductivity of a Cu-Ni alloy that has a yield

strength of 125 MPa.

Example – 4Solution:

Influence of processing and/or Plastic Deformation

Plastic deformation and/or metal processing techniques affect the electrical

properties of a metal in different ways, Table (4). Where, raising the electrical resistivity

as a result of increased numbers of electron-scattering dislocations. The effect of

deformation on resistivity is also represented in Figure (5). Furthermore, its influence is

much weaker than that of increasing temperature or the presence of impurities. For

example; solid solution strengthening is not a good way to obtain high strength in metals

intended to have high conductivities. The mean free paths are very short due to the

random distribution of the interstitial or substitution atoms. Figure (10) shows the defect

of zinc and other alloying elements on the conductivity of copper; as the amount of

alloying element increases, the conductivity decreases substantially.

Also, age hardening and dispersion strengthening reduce the conductivity to an

extent that is less than solid – solution strengthening, since there is a longer mean free


path between precipitates, as compared with the path between point defects. Strain

hardening and grain – size control has even less effect on conductivity, Figure (10) and

Table (4).

Table (4):

Figure (10): (a) the effect of solid-solution strengthening and cold working on the

electrical conductivity of copper, and (b) the effect of addition of selected elements

on the electrical conductivity of copper.


Example – 5: from Figure (10), Estimate the defect resistivity coefficient for tin

(Sn) in copper (Cu). Notice; the atomic weight of Sn = 118.69g/mol and the atomic

weight of Cu = 63.54 g/mol.

Example – 5 Solution:

1- Find the σ and ρ of pure Cu (from Table 1).

σ = 5.98 × 105 (Ω.cm)

-1 , → ρ = 1/σ = 0.167 × 10

-5 (Ω.cm)

2- Convert the weight percent to the atomic percent for impurities

For 0.2% wt% Sn in Cu

⁄

⁄ ⁄

⁄

⁄ ⁄

3- Find the cimp(1-cimp)

For Sn → cSn(1- cSn) = 0.00107(1- 0.00107) ≈ 0.00107

4- Find ρ for impurities

For 0.2% Sn, from Figure (10b) shows that the σSn is 92% that of pure Cu, or

σ0.2%Sn = (5.98 × 105)(0.92) = 5.50×10

5 (Ω.cm)

-1, → ρ0.2%Sn= 0.182 × 10

-5 (Ω.cm)

5- Find Δρ =(ρimp – ρpure)

For 0.2% Sn → Δρ = 0.182 × 10-5

- 0.167 × 10-5

= 0.015 × 10-5

(Ω.cm)

6- Find the defect resistivity coefficient A

a) For one component (or percent), A = ρimp/cimp(1 – cimp)

b) For many compositions, tabulated the data as a table below, then find A from

the slope after plotted the data.


4 Energy Band Structures in Solids

According to Bohr atomic model, in which electrons are

assumed to revolve around the atomic nucleus in discrete

orbitals, and the position of any particular electron is more or

less well defined in terms of its orbital. This model of the atom

is represented in Figure (11). For each individual atom there exist discrete energy levels that

may be occupied by electrons, arranged into shells and

subshells. Shells are designated by integers (1, 2, 3, etc.), and

subshells by letters (s, p, d, and f ). For each of s, p, d, and f

subshells, there exist, respectively, one, three, five, and seven

states. The electrons in most atoms fill only the states having the lowest energies,

two electrons of opposite spin per state, in accordance with the Pauli Exclusion

Principle. The electron configuration of an isolated atom represents the arrangement

of the electrons within the allowed states.

Figure (11): Schematic

representation of the

Bohr atom.


A solid material consisting of a large number, say, N, of atoms initially separated

from one another, which are subsequently brought together and bonded to form the

ordered atomic arrangement found in the crystalline material, Figure (12).

When atoms close of one another, electrons are acted upon, or perturbed, by the

electrons and nuclei of adjacent atoms. This influence is may split each atomic state

into a series of closely spaced electron states in the solid, to form an electron energy

band. The extent of splitting depends on interatomic separation, Figure (13) and

begins with the outermost electron shells, since they are the first to be perturbed as

the atoms merge. Within each band, the energy states are discrete, yet the difference

between adjacent states is exceedingly small. At the equilibrium spacing, band

formation may not occur for the electron subshells nearest the nucleus, as illustrated

in Figure (14). Furthermore, gaps may exist between adjacent bands, as also

indicated in the figure; normally, energies lying within these band gaps are not

available for electron occupancy. The conventional way of representing electron

band structures in solids is shown in Figure (14).

Energy levels of

an isolated atom

Figure (12): The energy level into bands as the number of electrons grouped

together increase.


The electrical properties of a solid material are a consequence of its electron band

structure; the arrangement of the outermost electron bands and the way in which

they are filled with electrons.

Four different types of band structures are possible at 0 K, Figure (15):

Figure (13): Schematic plot of electron

energy versus interatomic separation

for an aggregate of 12 (N=12) atoms.

Upon close approach, each of the 1s

and 2s atomic states splits to form an

electron energy band consisting of 12

states.

Figure (14): (a) The conventional representation of

the electron energy band structure for a solid

material at the equilibrium interatomic separation.

(b) Electron energy versus interatomic separation

for an aggregate of atoms, illustrating how the

energy band structure at the equilibrium separation

in (a) is generated.

This energy band

structure is typified

by some metals, such

as copper, in which

electron states (Fermi

energy) are available

above and adjacent to

filled states, in the

same band metals at

0K.

The electron band

structure also of

metals, such as

magnesium, where

in there is an

overlap of filled

band and empty

band.

In these two band structures the filled valence band is

separated from the empty conduction band, and an

energy band gap lies between them. For very pure

materials, electrons may not have energies within this

gap. The difference between the two band structures

lies in the magnitude of the energy gap;

Insulators: have relatively

large band gap (>2 eV)

Semiconductors: have relatively

narrow band gap (<2 eV).

Note: The Fermi energy for these two band structures

lies within the band gap—near its center.

Figure (15):


5 Conduction and Electron Transport

Only electrons with energies greater than the Fermi energy Ef (i.e. free electrons)

may be move when the electric field is applied and also participate in conduction

process.

Holes have energies less than Ef and also participate in electronic conduction.

They found in semiconductors and insulators.

Thus, the electrical conductivity depends on the numbers of free electrons and

holes.

The distinction between conductors and nonconductors (insulators and

semiconductors) lies in the numbers of these free electron and hole charge

carriers.

Metals

Since electron is free, it must be excited into

one of the empty and available energy states

above Ef .

In both band structures of metals, there are

vacant energy states adjacent to the highest

filled state at Ef .

Thus, very little energy is required to excite

electrons into the low-lying empty states,

Figure (16).

Generally, the energy provided by an electric

field is sufficient to excite large numbers of

electrons into these conducting states.

Insulators and Semiconductors

The empty states adjacent to the top

of the filled valence band are not

available.

Electrons to become free must be

across the energy band gap and into

empty states at the bottom of the

conduction band.

This is possible only by supplying the

electron with energy, which is

approximately equal to the band gap

Figure (16): For a metal, occupancy

of electron states (a) before and (b)

after an electron excitation.

Figure (17): For an insulator or semiconductor,

occupancy of electron states (a) before and (b)

after an electron excitation from the valence

band into the conduction band, in which both a

free electron and a hole are generated.


energy. This excitation process is demonstrated in Figure (17).

For many materials this band gap is several electron volts wide. Most often the

excitation energy is from a nonelectrical source such as heat or light, usually the

former.

The number of electrons excited thermally (by heat energy) into the conduction

band depends on the energy band gap width as well as temperature.

Increasing the temperature of either a semiconductor or an insulator results in an

increase in the thermal energy that is available for electron excitation.

Thus, more electrons are transport (move) into the conduction band, which gives

rise to an enhanced conductivity.

For electrically insulating materials, interatomic bonding is ionic or strongly

covalent. Thus, the valence electrons are tightly bound to or shared with the

individual atoms. In other words, these electrons are highly localized and are not

be free to move throughout the crystal.

The bonding in semiconductors is covalent (or predominantly covalent) and

relatively weak, which means that the valence electrons are not as strongly bound

to the atoms. Consequently, these electrons are more easily removed by thermal

excitation than they are for insulators.

Homework:

7) Calculate the electrical conductivity of nickel at - 50°C and at +500°C. Hence the

resistivity at room temperature 6.84*10-6

Ω.cm and α = 0.0069 Ω.cm/°C.

8) The electrical resistivity of pure chromium is found to be 18*10-6

Ω.cm. estimate the

temperature at which the resistivity measurement was made, where the resistivity at

room temperature 12.9*10-6

Ω.cm and α = 0.0030 Ω.cm/°C.

9) The electrical resistivity of a beryllium alloy containing 5 at% of an alloying element

is found to be 50 × 10-6

ohm.cm at 400°C. Determine the contributions to resistivity due

to temperature and due to impurities by finding the expected resistivity of pure

beryllium at 400°C, the resistivity due to impurities, and the defect resistivity

coefficient. What would be the electrical resistivity if the beryllium contained 10 at% of

the alloying element at 200°C?

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