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Page 1: 1 Chapter 18 Electronic Materials. 2 Objectives of Chapter 18  To study electronic materials – insulators, dielectrics, conductors, semiconductors, and

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Chapter 18 Electronic Materials

Page 2: 1 Chapter 18 Electronic Materials. 2 Objectives of Chapter 18  To study electronic materials – insulators, dielectrics, conductors, semiconductors, and

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Objectives of Chapter 18 To study electronic materials – insulators,

dielectrics, conductors, semiconductors, and superconductors.

To study conductivity in electronic materials.

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18.1 Ohm’s Law and Electrical Conductivity

18.2 Band Structures of Solids 18.3 Conductivity of Metals and Alloys 18.4 Superconductivity 18.5 Conductivity in Other Materials 18.6 Semiconductors 18.7 Applications of Semiconductors 18.8 Insulators and Dielectric Properties 18.9 Polarization in Dielectrics 18.10 Electrostriction, Piezoelectricity,

Pyroelectricity, and Ferroelectricity

Chapter Outline

Page 4: 1 Chapter 18 Electronic Materials. 2 Objectives of Chapter 18  To study electronic materials – insulators, dielectrics, conductors, semiconductors, and

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.1 Classification of technologically useful electronic materials.

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Current density - The current flowing through per unit cross-sectional area.

Electric field - The voltage gradient or volts per unit length.

Drift velocity - The average rate at which electrons or other charge carriers move through a material under the influence of an electric or magnetic field.

Mobility - The ease with which a charge carrier moves through a material.

Dielectric constant - The ratio of the permittivity of a material to the permittivity of a vacuum, thus describing the relative ability of a material to polarize and store a charge; the same as relative permittivity.

Section 18.1 Ohm’s Law and Electrical Conductivity

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Figure 18.2 (a) Charge carriers, such as electrons, are deflected by atoms or defects and take an irregular path through a conductor. The average rate at which the carriers move is the drift velocity v. (b) Valence electrons in the metallic bond move easily. (c) Covalent bonds must be broken in semiconductors and insulators for an electron to be able to move. (d) Entire ions must diffuse to carry charge in many ionically bonded materials.

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Page 8: 1 Chapter 18 Electronic Materials. 2 Objectives of Chapter 18  To study electronic materials – insulators, dielectrics, conductors, semiconductors, and

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TABLE 18-1 Electrical conductivity of selected materials at TF300 K* (Continued)

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Page 10: 1 Chapter 18 Electronic Materials. 2 Objectives of Chapter 18  To study electronic materials – insulators, dielectrics, conductors, semiconductors, and

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Design an electrical transmission line 1500 m long that will carry a current of 50 A with no more than 5 105 W loss in power. The electrical conductivity of several materials is included in Table 18-1.

Example 18.1 SOLUTION

Electrical power is given by the product of the voltage and current or:

Example 18.1 Design of a Transmission Line

From Equation 18-2:

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Example 18.1 SOLUTION (Continued)

Let’s consider three metals—aluminum, copper, and silver—that have excellent electrical conductivity. The table below includes appropriate data and some characteristics of the transmission line for each metal.

Any of the three metals will work, but cost is a factor as well. Aluminum will likely be the most economical choice, even though the wire has the largest diameter. However , other factors, such as whether the wire can support itself between transmission poles, also contribute to the final choice.

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Some manufacturers of automotives are considering using a 42-volt battery system instead of the standard 12/14-volt battery system. Why do you think this change is being considered?

Example 18.2 SOLUTION

There is a need for increased overall power as newer cars come equipped with many electronic features such as power windows, power doors, power liftgates, and advanced lighting, braking, and steering systems.

Increasing the voltage means less current is needed (V = IR), so the total weight of the copper wires can be reduced. Less weight in both the wiring and the motors contributes to fuel efficiency.

Example 18.2 A 42-volt Battery System for Cars

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Example 18.3 Drift Velocity of Electrons in Copper

Assuming that all of the valence electrons contribute to current flow, (a) calculate the mobility of an electron in copper and (b) calculate the average drift velocity for electrons in a 100-cm copper wire when 10 V are applied.

Example 18.3 SOLUTION

1. The valence of copper is one: therefore, the number of valence electrons equals the number of copper atoms in the material.The lattice parameter of copper is 3.6151 10-8 cm and, since copper is FCC, there are 4 atoms/unit cell. From Table 18-1

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Example 18.3 SOLUTION (Continued)

2.The electric field is:

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Valence band - The energy levels filled by electrons in their lowest energy states.

Conduction band - The unfilled energy levels into which electrons can be excited to provide conductivity.

Holes - Unfilled energy levels in the valence band.Because electrons move to fill these holes, the holes move and produce a current.

Hybridization - When valence and conduction bands are separated by an energy gap, leading to the semiconductive behavior of silicon and germanium.

Energy gap (Bandgap) - The energy between the top of the valence band and the bottom of the conduction band that a charge carrier must obtain before it can transfer a charge.

Section 18.2 Band Structures of Solids

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.3 The energy levels broaden into bands as the number of electrons grouped together increases.

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Figure 18.4 The simplified band structure for sodium. The energy levels broaden into bands. The 3s band, which is only half filled with electrons, is responsible for conduction in sodium.

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.5 (a) At absolute zero, all of the electrons in the outer energy level have the lowest possible energy. (b) When the temperature is increased, some electrons are excited into unfilled levels. Note that the Fermi energy is unchanged.

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Figure 18.6 The band structure of carbon in the diamond form. The 2s and 2p levels combine to form two hybrid bands separated by an energy gap, Eg.

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.7 Schematic of band structures for (a) metals, (b) semiconductors, and (c) dielectrics or insulators. (Temperature is 0 K.)

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Mean free path - The average distance that electrons can move without being scattered by other atoms.

Temperature Effect - When the temperature of a metal increases, thermal energy causes the atoms to vibrate

Effect of Atomic Level Defects - Imperfections in crystal structures scatter electrons, reducing the mobility and conductivity of the metal

Matthiessen’s rule - The resistivity of a metallic material is given by the addition of a base resistivity that accounts for the effect of temperature (ρT), and a temperature independent term that reflects the effect of atomic level defects, including impurities forming solid solutions (ρd).

Effect of Processing and Strengthening

Section 18.3 Conductivity of Metals and Alloys

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.8 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.

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.9 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.

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

Example 18.4 Resistivity of Pure Copper

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Example 18.4 SOLUTION

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 ohm . cm/oC.

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Figure 18.10 The electrical resistivity of a metal is composed of a constant defect contribution ρd and a variable temperature contribution ρT.

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.11 (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.

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Intrinsic semiconductor - A semiconductor in which properties are controlled by the element or compound that makes the semiconductor and not by dopants or impurities.

Extrinsic semiconductor - A semiconductor prepared by adding dopants, which determine the number and type of charge carriers.

Doping - Deliberate addition of controlled amounts of other elements to increase the number of charge carriers in a semiconductor.

Thermistor - A semiconductor device that is particularly sensitive to changes in temperature, permitting it to serve as an accurate measure of temperature.

Radiative recombination - Recombination of holes and electrons that leads to emission of light; this occurs in direct bandgap materials.

Section 18.6 Semiconductors

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.16 When a voltage is applied to a semiconductor, the electrons move through the conduction band, which the electron holes move through the valence band in the opposite direction.

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.17 the distribution of electrons and holes in the valence and conduction bands (a) at absolute zero and (b) at an elevated temperature.

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Figure 18.18 The electrical conductivity versus temperature for intrinsic semiconductors compared with metals. Note the break in y-axis scale.

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For germanium at 25oC, estimate (a) the number of charge carriers, (b) the fraction of the total electrons in the valence band that are excited into the conduction band, and (c) the constant n0.

Example 18.7 Carrier Concentrations in Intrinsic Ge

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Example 18.7 SOLUTION

1. From Equation 18-12:

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2. The lattice parameter of diamond cubic germanium is 5.6575 10-8 cm. The total number of electrons in the valence band of germanium is:

3. From Equation 18-14(a):

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.19 When a dopant atom with a valence greater than four is added to silicon, an extra electron is introduced and a donor energy state is created. Now electrons are more easily excited into the conduction band.

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.20 The effect of temperature on the carrier concentration of an n-type semiconductor. At low temperatures, the donor or acceptor atoms are not ionized. As temperature increases, the ionization process is complete and the carrier concentration increases to a level that is dictated by the level of doping. The conductivity then essentially remains unchanged, until the temperature becomes too high and the thermally generated carriers begin to dominate. The effect of dopants is lost at very high temperatures and the semi-conductor essentially shows “intrinsic” behavior.

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.21 When a dopant atom with a valence of less than four is substituted into the silicon structure, a hole is created in the structure and an acceptor energy level is created just above the valence band. Little energy is required to excite the holes into motion.

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Design a p-type semiconductor based on silicon, which provides a constant conductivity of 100 ohm-1. cm-1 over a range of temperatures. Comment on the level of purity needed.

Example 18.8 SOLUTION

In order to obtain the desired conductivity, we must dope the silicon with atoms having a valence of +3, adding enough dopant to provide the required number of charge carriers. If we assume that the number of intrinsic carriers is small, then

Example 18.8 Design of a Semiconductor

The number of charge carriers required is:

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Example 18.8 SOLUTION (Continued)

Let’s compare the concentration of donor atoms in Si with the concentration of Si atoms. Assume that the lattice constant of Si remains unchanged as a result, of doping:

Possible dopants include boron, aluminum, gallium, and indium. High-purity chemicals and clean room conditions are essential for processing since we need 26 dopant atoms (apples) in a million silicon atoms (oranges).

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A light-emitting diode display made using a GaAs-GaP solid solution of composition 0.4 GaP-0.6 GaAs has a direct bandgap of 1.9 eV. What will be the color this LED display?

Example 18.9 SOLUTION

The wavelength of the light emitted is related to the bandgap by:

Example 18.9 Creating the Color of a Light-Emitting

Diode Display

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Example 18.9 SOLUTION (Continued)

You can show that if the bandgap is in eV and λ is to be in μm, then:

Therefore, λ = 0.652 μm or 652 nm. This is the wavelength of red light, and, therefore, the LED display would be red in color.

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Diodes, transistors, lasers, and LEDs are made using semiconductors. Silicon is the workhorse of very large scale integrated (VLSI) circuits.

Forward bias - Connecting a p-n junction device so that the p-side is connected to positive. Enhanced diffusion occurs as the energy barrier is lowered, permitting a considerable amount of current can flow under forward bias.

Reverse bias - Connecting a junction device so that the p-side is connected to a negative terminal; very little current flows through a p-n junction under reverse bias.

Avalanche breakdown - The reverse-bias voltage that causes a large current flow in a lightly doped p-n junction.

Transistor - A semiconductor device that can be used to amplify electrical signals.

Section 18.7 Applications of Semiconductors

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Figure 18.22 Behavior of a p-n junction device: (a) When no bias is applied electron and hole currents due to drift and diffusion cancel out and there is no net current and a built-in potential V0 exists. (b) Under a forward bias causes the potential barrier to be reduced to Vf - V0 and the depletion region becomes smaller causing a current to flow. (c) Under reverse bias, the potential barrier increases to Vr + V0 and very little current flows. The internal electric field that develops is shown as E. (Source: From Solid State Electronic Devices, Third Edition, by. B.G. Streetman, Fig. 5-10. Copyright © 2000 Prentice Hall. Reprinted by permission Pearson Education, Inc., Upper Saddle River, NJ.)

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Figure 18.23 (a) The current-voltage characteristic for a p-n junction. Note the different scales in the first and third quadrants. (b) If an alternating signal is applied, rectification occurs and only half of the input signal passes the rectifier. (Source: From Electronic Devices, Sixth Edition, by T.L. Floyd, Fig. 1-30. Copyright © 2002 Prentice Hall. Reprinted by permission Pearson Education, Inc., Upper Saddle River, NJ.)

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.24 (a) A circuit for an n-p-n bipolar junction transistor. The input creates a forward and reverse bias that causes electrons to move from the emitter, through the base, and into the collector, creating an amplified output. (b) Sketch of the cross-section of the transistor.

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Figure 18.25 Operation of a n-p-n transistor. (a) Schematic of a n-p-n transistor. The dashes indicate electrons, the + sign indicates holes. Electrons in source and gate regions are not shown. (b) The gate has a positive voltage. Electrons in the p-type substrate move to the region between the source and drain. (c) When positive voltage is applied to a drain, electrons from the source move to the drain via the channel region between the source and drain (created by a positive gate voltage). The transistor is now ‘‘on.’’ (d) When a gate voltage is removed. The electron channel between the source is broken and the transistor is turned ‘‘off.’’ (Source: www.intel.com.)

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Figure 18.25 (c) When positive voltage is applied to a drain, electrons from the source move to the drain via the channel region between the source and drain (created by a positive gate voltage). The transistor is now ‘‘on.’’ (d) When a gate voltage is removed. The electron channel between the source is broken and the transistor is turned ‘‘off.’’ (Source: www.intel.com.)

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Figure 18.26 (a) Czochralski growth technique for growing single crystals of silicon. (Source: From Microchip Fabrication, Third Edition, by P. VanZant, Fig. 3-7. Copyright © 1997 The McGraw-Hill Companies. Reprinted with permission The McGraw-Hill Companies.) (b) Overall steps encountered in the processing of semiconductors. (Source: From Fundamentals of Modern Manufacturing, by M.P. Groover, p. 849, Fig. 34-3. Copyright © 1996 Prentice Hall.)

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Figure 18.26 (c) Production of an FET semiconductor device: (i) A p-type silicon substrate is oxidized. (ii) Photolithography, ultraviolet radiation passing through a photomask, exposes a portion of the photoresist layer. (iii) The exposed photoresist is dissolved. (iv) The exposed silica is removed by etching. (v) An n-type dopant is introduced to produce the source and drain. (vi) The silicon is again oxidized. (vii) Photolithography is repeated to introduce other components, including electrical leads, for the device.

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Superconductivity - Flow of current through a material that has no resistance to that flow.

Applications of Superconductors - Electronic circuits have also been built using superconductors and powerful superconducting electromagnets are used in magnetic resonance imaging (MRI). Also, very low electrical-loss components, known as filters, based on ceramic superconductors have been developed for wireless communications.

Section 18.4 Superconductivity

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Kamerlingh Onnes, 1911

Superconductivity Theoretical predictions of resistivity of metals at low T

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Kamerlingh Onnes, 1911

perfect conductivity Superconductivity

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Experiment Theoretical predictions

R(T) at low temperatures

1 )perfect conductivity

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Superconductivity Transition Temperatures and Critical Fields

Li ...

Be 0.026

Transition temperature in Kelvin B ...

C ...

N ...

O ...

F ...

Ne ...

Na ...

Mg ...

Al

1.140 Si* 7

P* 5

S* ...

Cl ...

Ar ...

K ...

Ca ...

Sc ...

Ti 0.39

V 5.38

Cr* ...

Mn ...

Fe ...

Co ...

Ni ...

Cu ...

Zn 0.875

Ga 1.091

Ge* 5

As* 0.5

Se* 7

Br ...

Kr ...

Rb ...

Sr ...

Y* ...

Zr 0.546

Nb 9.50

Mo 0.90

Tc 7.77

Ru 0.51

Rh 0.0003

Pd ...

Ag ...

Cd 0.56

In 3.4035

Sn 3.722

Sb* 3.5

Te* 4

I ...

Xe ...

Cs* 1.5

Ba* 5

La(fcc) 6.00

Hf 0.12

Ta 4.483

W 0.012

Re 1.4

Os 0.655

Ir 0.14

Pt ...

Au ...

Hg 4.153

Tl 2.39

Pb 7.193

Bi* 8

Po ...

At ...

Rn ...

Fr ...

Ra ...

Ac ...

... Ce*

2 Pr ...

Nd ...

Pm ...

Sm ...

Eu ...

Gd ...

Tb ...

Dy ...

Ho ...

Er ...

Tm ...

Yb ...

Lu 0.1

...

... Th

1.368 Pa 1.4

U* 2

Np ...

Pu ...

Am ...

Cm ...

Bk ...

Cf ...

Es ...

Fm ...

Md ...

No ...

Lr ...

...

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Isotope Natural abundance (atom %)

196Hg 0.15 198Hg 9.97 199Hg 16.87 200Hg 23.10 201Hg 13.18 202Hg 29.86 204Hg 6.87

Isotope effect

Tc M -

M

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Isotope Natural abundance (atom %)

196Hg 0.15 198Hg 9.97 199Hg 16.87 200Hg 23.10 201Hg 13.18 202Hg 29.86 204Hg 6.87

Isotope effect

Tc M -

M

2 )Lattice is important

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Perfect diamagnetismMeissner and Ochsenfeld, 1933

perfect conductor = 0 => E =0 => dB/dt = 0constant flux

B = 0

zero fluxMeissner effect

superconductor

Meissner effect

3 )Perfect diamagnetism

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Magnetization of superconductors

B

H

B = H B = H

normal metal

diamagneticB < H

paramagneticB > H

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B

H

superconductor (type I)

B = H

HcB=0

MeissnerB=H

normal

B = 0

Magnetization of superconductors

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B

H

superconductor (type I)

B = H

HcB=0

Meissner

B=H

normal

B = 0

4MH

Hc

Meissner normal

4M

4M = B - H

Magnetization of superconductors

4M = - H M = 0

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4M

HHc

Meissner

B=0

normal

B=H

Type I

4M

Magnetization of superconductors

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4M

HHc

Meissner

B=0

normal

B=H

Type I

4M

HHc1 Hc2

normal

B=HMeissner

B=0

mixed

B<H

Type II

4 )Magnetic field suppresses superconductivity

Magnetization of superconductors

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4M

HHc

Meissner

B=0

normal

B=H

Type I

4M

HHc1 Hc2

normal

B=HMeissner

B=0

mixed

B<H

Type II

Magnetization of superconductors

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Figure 18.13 (a) The effect of a magnetic field on the temperature below which superconductivity occurs for a Type I superconductor. (b) The critical surface of Nb3Sn, a Type II superconductor. The superconducting envelope or surface showing the combined effects of temperature, magnetic field, and current density on a Nb3Sn superconductor. Conditions within the envelope produce superconductivity.

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Figure 18.13 (c) The effect of a magnetic field on the temperature below which superconductivity occurs for a Type II superconductor. (d) Critical-current density as a function of magnetic-flux density (in Tesla) created by the applied magnetic fields (for different superconductors).

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Phase diagram

Type II

H

T

Normal

metal

Meissner state

Hc2(T)

Hc1(T)

Tc

mixed state

H

T

Normal

metal

Meissner state

Hc(T)

Tc

Type I

Hc

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Nobel 1973

thininsulator

normal

SC

Tunneling experiments

Al

Al - Al2O3 Al - Al2O3 - Pb

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thininsulator

normal

SC

Al

PbI

V

V

dI/dVPb normal

Pb superconducting

Tunneling experiments

5 )Energy gap in density of states

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2

3.5kTc (meV)

6 )Tc and energy gap are related

Superconducting energy gap

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76

H

R

time

Little-Parks experiment

VI+

-

H

0 = h/2e = 2.07x10-7 G

cm2

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77

VI+

-

H

0 = h/eAu

h/2e

7 )Quantization in units of h/2e

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Critical temperature of superconductors

Tc < 30 K

conventional superconductors

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Critical temperature of superconductors

High-temperature superconductors (HTS)Tc > 140 K

8) Superconductivity mechanism in HTS is different from LTS

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80

Cu

O

Sr

Bi

Ca

High-temperature superconductors

Bi2Sr2CaCu2O8

CuO2 double layer

J. Georg BednorzK. Alexander Müller

Nobel Prize in Physics 1987

Bi2Sr2CaCu2O8

Bi2Sr2Ca2Cu3O10

(La1.85Ba.15)CuO4

YBa2Cu3O7

Hg0.8Tl0.2Ba2Ca2Cu3O8.33

Tl2Ba2Ca2Cu3O10

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Superconductivity phenomena

• Perfect conductivity

• Dynamics of the lattice is important Tc M-

• Perfect diamagnetism B = 0

• Magnetic field suppresses superconductivity Hc(T), Hc1(T), Hc2(T)

• Energy gap 2

• Tc and energy gap are related

• Magnetic flux is quantized in units of h/2e

• Superconductivity mechanism in HTS is different from LTS

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Design the limiting magnetic field that will permit niobium to serve as a superconductor at liquid helium temperatures. Assume H0 = 1970 oersteds.

Example 18.5 Design of a Superconductor System

Example 18.5 SOLUTION

From Table 18-5, Tc = 9.25 K. We are given the value of and H0 = 1970 oersted. Since the operating temperature will be 4 K, we need to determine the maximum permissible magnetic field, using Equation 18-9:

Therefore, the magnetic field (H) must remain below 1602 oersted in order for the niobium to remain superconductive.

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Conduction in Ionic Materials - Conduction in ionic materials often occurs by movement of entire ions, since the energy gap is too large for electrons to enter the conduction band. Therefore, most ionic materials behave as insulators.

Conduction in Polymers - Because their valence electrons are involved in covalent bonding, polymers have a band structure with a large energy gap, leading to low-electrical conductivity.

Section 18.5 Conductivity in Other Materials

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Suppose that the electrical conductivity of MgO is determined primarily by the diffusion of the Mg2+ ions. Estimate the mobility of the Mg2+ ions and calculate the electrical conductivity of MgO at 1800oC.

Example 18.6 SOLUTION

From Figure (5.18), the diffusion coefficient of Mg2+ ions in MgO at 1800oC is 10-10 cm2/s. For MgO, Z = 2/ion, q = 1.6 10-19 C, kB = 1.38 10-23 J/K, and T = 2073 K:

Example 18.6 Ionic Conduction in MgO

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Example 18.6 SOLUTION (Continued)

Since the charge in coulomb is equivalent to Ampere . second, and Joules is equivalent to Amp . sec . volts:

MgO has the NaCl structure, with four magnesium ions per unit cell. The lattice parameter is 3.96 10-8 cm, so the number of Mg2+ ions per cubic centimeter is:

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©20

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Figure 18.15 Effect of carbon fibers on the electrical resistivity of nylon.

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Materials used to insulate an electric field from its surroundings are required in a large number of electrical and electronic applications.

Electrical insulators obviously must have a very low conductivity, or high resistivity, to prevent the flow of current.

Porcelain, alumina, cordierite, mica, and some glasses and plastics are used as insulators.

Section 18.8 Insulators and Dielectric Properties

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Capacitor - A microelectronic device, constructed from alternating layers of a dielectric and a conductor, that is capable of storing a charge.These can be single layer or multi-layer devices.

Permittivity - The ability of a material to polarize and store a charge within it.

Linear dielectrics - Materials in which the dielectric polarization is linearly related to the electric field; the dielectric constant is not dependent on the electric field.

Dielectric strength - The maximum electric field that can be maintained between two conductor plates without causing a breakdown.

Section 18.9 Polarization in Dielectrics

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Figure 18.27 Polarization mechanisms in materials: (a) electronic, (b) atomic or ionic, (c) high-frequency dipolar or orientation (present in ferroelectrics), (d) low-frequency dipolar (present in linear dielectrics and glasses), (e) interfacial-space charge at electrodes, and (f ) interfacial-space charge at heterogeneities such as grain boundaries. (Source: From Principles of Electronic Ceramics, L.L. Hench and J.K. West, p. 188, Fig. 5-2. Copyright © 1990 Wiley Interscience. Reprinted by permission. This material is used by permission of John Wiley & Sons, Inc.)

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Suppose that the average displacement of the electrons relative to the nucleus in a copper atom is 1 10-8 Å when an electric field is imposed on a copperplate. Calculate the electronic polarization.

Example 18.10 SOLUTION

The atomic number of copper is 29, so there are 29 electrons in each copper atom. The lattice parameter of copper is 3.6151 Å . Thus

Example 18.10 Electronic Polarization in Copper

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©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.28 (a) The oxygen ions are at face centers, Ba+2 ions are at cube corners and Ti+4 is at cube center in cubic BaTi03. (b) In tetragonal BaTi03 ,the Ti+4 is off-center and the unit cell has a net polarization.

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Figure 18.29 A charge can be stored at the conductor plates in a vacuum (a). However, when a dielectric is placed between the plates (b), the dielectric polarizes and additional charge is stored.

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Figure 18.30 Frequency dependence of polarization mechanisms. On top is the change in the dielectric constant with increasing frequency, and the bottom curve represents the dielectric loss. (Source: From Electroceramics: Material, Properties, Applications, by A.J. Moulson and J.M. Herbert, p. 68, Fig. 2-38. Copyright © 1990 Chapman and Hall. Reprinted with kind permission of Kluwer Academic Publishers and the author.)

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When any material undergoes polarization, its ions and electronic clouds are displaced, causing the development of a mechanical strain in the material. This effect is seen in all materials subjected to an electric field and is known as the electrostriction.

Piezoelectrics - Materials that develop voltage upon the application of a stress and develop strain when an electric field is applied.

Pyroelectric - The ability of a material to spontaneously polarize and produce a voltage due to changes in temperature.

Ferroelectric - A material that shows spontaneous and reversible dielectric polarization.

Section 18.10 Electrostriction, Piezoelectricity,

Pyroelectricity, and Ferroelectricity

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Figure 18.31 The (a) direct and (b) converse piezoelectric effect. In the direct piezoelectric effect, applied stress causes a voltage to appear. In the converse effect (b), an applied voltage leads to development of strain.

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Example 18.11 Design of a Spark Igniter

A PZT spark igniter is made using a disk that has a 5-mm diameter and 20-mm height.Calculate the voltage generated if

the g coefficient for PZT used is . Assume that

a compressive force of 10,000 N is applied on the circular face.

Example 18.11 SOLUTION

The g coefficient can be defined as:

N

mV 31035

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Example 18.11 SOLUTION (Continued)

In this equation, X is the applied stress, E is the electric field generated.

This field appears across a height (H) of 20 mm. The voltage (V) generated will be:

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Figure 18.32 Examples of ceramic capacitors. (a) Single-layer ceramic capacitor (disk capacitors). (b) Multilayer ceramic capacitor (stacked ceramic layers). (Source: From Principles of Electrical Engineering Materials and Devices, by S.O. Kasap, p. 559, Fig. 7-29. Copyright © 1997 Irwin. Reprinted by permission of The McGraw-Hill Companies.)

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Figure 18.33 (a) Different polymorphs of BaTiO3 and accompanying changes in lattice constants and dielectric constants. (Source: From Principles of Electronic Ceramics, L.L. Hench and J.K. West, p. 247, Fig. 6-5. Copyright © 1990 Wiley Interscience. This material is used by permission of John Wiley & Sons, Inc.). The ferroelectric hysteresis loops for (b) single crystal. (Source: From Electroceramics: Material, Properties, Applications, by A.J. Moulson and J.M. Herbert, p. 76, Fig. 2-46. Copyright © 1990 Chapman and Hall. Reprinted with kind permission of Kluwer Academic Publishers and the author.)

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Figure 18.33 (b) single crystal. (Source: From Electroceramics: Material, Properties, Applications, by A.J. Moulson and J.M. Herbert, p. 76, Fig. 2-46. © 1990 Chapman and Hall. Reprinted with kind permission of Kluwer Academic Publishers and the author.) (c) Polycrystalline BaTiO3 showing the influence of the electric field on polarization. (Source: From Principles of Electrical Engineering Materials and Devices, by S.O. Kasap, p. 565, Fig. 7-35. Copyright © 1997 Irwin. Reprinted by permission of The McGraw-Hill Companies.)

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Figure 18.34 (a) The effect of temperature and grain size on the dielectric constant of barium titanate. Above the Curie temperature, the spontaneous polarization is lost due to a change in crystal structure and barium titanate is in the paraelectric state. The grain size dependence shows that similar to yield-strength dielectric constant is a microstructure sensitive property. (Source: From Electroceramics: Material, Properties, Applications, by A.J. Moulson and J.M. Herbert, p. 78, Fig. 2-48. Copyright © 1990 Chapman and Hall. Reprinted with kind permission of Kluwer Academic Publishers and the author.) (b) Ferroelectric domains can be seen in the microstructure of polycrystalline BaTiO3. (Courtesy of Dr. Rodney Roseman, University of Cincinnati.)

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Calculate the maximum polarization per cubic centimeter and the maximum charge that can be stored per square centimeter for barium titanate.

Example 18.12 Polarization in Barium Titanate

©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 18.28 (a) The oxygen ions are at face centers, Ba+2 ions are at cube corners and Ti+4 is at cube center in cubic BaTi03. (b) In tetragonal BaTi03 ,the Ti+4 is off-center and the unit cell has a net polarization.

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Example 18.12 SOLUTION

In BaTiO3, the separations are the distances that the Ti4+ and O2- ions are displaced from the normal lattice points (Figure 18.28). The charge on each ion is the product of q and the number of excess or missing electrons. Thus, the dipole moments are:

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Example 18.12 SOLUTION (Continued)

Each oxygen ion is shared with another unit cell, so the total dipole moment in the unit cell is:

The polarization per cubic centimeter is:

The total charge on a BaTiO3 crystal 1 cm 1 cm is:

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A multi-layer capacitor is to be designed using a BaTiO3-based formulation containing SrTiO3. The dielectric constant of the material is 3000. (a) Calculate the capacitance of a multi-layer capacitor consisting of 100 layers connected in parallel using Ni electrodes. The sides of the capacitor are 10 5 mm and the thickness of each layer is 10 µm. (b) What is the role of SrTiO3? (c) What processing technique will be used to make these?

Example 18.13 SOLUTION

(a) The capacitance of a parallel plate capacitor is given by:

Example 18.13 Design of a Multi-layer Capacitor

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Example 18.13 SOLUTION (Continued)

The permittivity of free space is ε0, 8.85 10-12 F/m.The relative permittivity of BaTiO3 formulation εr or dielectric constant (k) is 3000.

Capacitance per layer will be:

Note the conversion to SI units.

We have 100 layers connected in parallel. Capacitances add up in this arrangement. All layers have same geometric dimensions in this case.

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Example 18.13 SOLUTION (Continued)

(b) The role of SrTiO3 addition to the formulation is to shift the Curie temperature below room temperature. In this way, the capacitor dielectric is in the high dielectric constant, but paraelectric state. Thus, if it is subjected to stress due to vibrations, it will not generate any spurious voltages due to the piezoelectric effect present in the tetragonal ferroelectric state.

(c) Tape casting will be the best way to make such thin-interconnected layers.

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Figure 18.11 (Repeated for Problems 18.15 and 18.17) (b) the effect of selected elements on the electrical conductivity of copper.

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Figure 18.35 Ferroelectric hysteresis loops (for Problems 18.64 and 18.67).