351-2: Physics of Materials II - Northwestern Universitymsecore.northwestern.edu/351-2/351-2text.pdf · or equivalent; PHYSICS 135 2,3. 2 Course Outcomes 3 351-2: Solid State Physics

351-2: Physics of Materials II

Bruce Wessels and Peter GirouardDepartment of Materials Science and Engineering

Northwestern University

October 1, 2019

Contents

1 Catalog Description (351-1,2) 6

2 Course Outcomes 6

3 351-2: Solid State Physics 6

4 Semiconductor Devices 74.1 Law of Mass Action . . . . . . . . . . . . . . . . . . . . . . . . . . 74.2 Chemistry and Bonding . . . . . . . . . . . . . . . . . . . . . . . 84.3 Reciprocal Lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.4 Nearly Free Electron Model . . . . . . . . . . . . . . . . . . . . . 114.5 Two Level Model . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

5 Band Diagrams 125.1 P- and N-Type Semiconductors . . . . . . . . . . . . . . . . . . . 135.2 P-N Junctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135.3 Boltzmann Statistics: Review . . . . . . . . . . . . . . . . . . . . 135.4 P-N Junction Equilibrium . . . . . . . . . . . . . . . . . . . . . . 145.5 Charge Profile of the p-n Junction . . . . . . . . . . . . . . . . . . 145.6 Calculation of the Electric Field . . . . . . . . . . . . . . . . . . . 155.7 Calculation of the Electric Potential . . . . . . . . . . . . . . . . . 165.8 Junction Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . 175.9 Rectification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185.10 Junction Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . 19

6 Transistors 196.1 pnp Device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206.2 pnp Device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216.3 Amplifier Gain . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

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CONTENTS CONTENTS

6.4 Circuit Configurations . . . . . . . . . . . . . . . . . . . . . . . . 226.5 MOSFETs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.6 Oxide-Semiconductor Interface . . . . . . . . . . . . . . . . . . . 236.7 Depletion and Inversion . . . . . . . . . . . . . . . . . . . . . . . 246.8 Channel Pinch-Off . . . . . . . . . . . . . . . . . . . . . . . . . . 246.9 Tunnel Diodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256.10 Gunn Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266.11 Gunn Diodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

7 Heterojunctions 267.1 Quantum Wells . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277.2 Heterostructures and Heterojunctions . . . . . . . . . . . . . . . 287.3 Layered Structures: Quantum Wells . . . . . . . . . . . . . . . . 287.4 Transitions between Minibands . . . . . . . . . . . . . . . . . . . 297.5 Quantum Dots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297.6 Quantum Dot Example: Biosensor . . . . . . . . . . . . . . . . . 307.7 Quantum Cascade Devices . . . . . . . . . . . . . . . . . . . . . . 30

8 Optoelectronics 318.1 I-V Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . 328.2 Power Generation . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

8.2.1 Solar Cell Efficiency . . . . . . . . . . . . . . . . . . . . . 338.2.2 Other Types of Solar Cells . . . . . . . . . . . . . . . . . . 338.2.3 Metal-Semiconductor Solar Cells . . . . . . . . . . . . . . 348.2.4 Schottky Barrier and Photo Effects . . . . . . . . . . . . . 348.2.5 Dependence of φB on Work Function . . . . . . . . . . . 358.2.6 Pinned Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 368.2.7 Light Emitting Diodes (LEDs) . . . . . . . . . . . . . . . . 378.2.8 Light Emitting Diodes (LEDs) . . . . . . . . . . . . . . . . 378.2.9 Solid Solution Alloys . . . . . . . . . . . . . . . . . . . . . 378.2.10 LED Efficiency . . . . . . . . . . . . . . . . . . . . . . . . 38

9 Lasers 389.1 Emission Rate and Laser Intensity . . . . . . . . . . . . . . . . . 399.2 Two Level System . . . . . . . . . . . . . . . . . . . . . . . . . . . 399.3 Planck Distribution Law . . . . . . . . . . . . . . . . . . . . . . . 409.4 Transition Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409.5 Two Level System . . . . . . . . . . . . . . . . . . . . . . . . . . . 419.6 Three Level System . . . . . . . . . . . . . . . . . . . . . . . . . . 419.7 Lasing Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459.8 Laser Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

9.8.1 Ruby . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459.8.2 Others Examples . . . . . . . . . . . . . . . . . . . . . . . 46

9.9 Threshold Current Density . . . . . . . . . . . . . . . . . . . . . . 469.10 Comparison of Emission Types . . . . . . . . . . . . . . . . . . . 48

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CONTENTS CONTENTS

9.11 Cavities and Modes . . . . . . . . . . . . . . . . . . . . . . . . . . 489.11.1 Longitudinal Laser Modes . . . . . . . . . . . . . . . . . . 499.11.2 Transverse Laser Modes . . . . . . . . . . . . . . . . . . . 49

9.12 Semiconductor Lasers . . . . . . . . . . . . . . . . . . . . . . . . 499.13 Photonic Bandgap Materials . . . . . . . . . . . . . . . . . . . . . 50

10 Band Diagrams 5010.1 Band Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5110.2 Heterojunctions and the Anderson Model . . . . . . . . . . . . . 5110.3 Band Bending at p-n Junctions . . . . . . . . . . . . . . . . . . . 51

10.3.1 Calculate the Conduction Band Discontinuity . . . . . . 5210.3.2 Calculate the Valence Band Discontinuity . . . . . . . . . 5310.3.3 Example: p-GaAs/n-Ge . . . . . . . . . . . . . . . . . . . 54

11 Dielectric Materials 5411.1 Macroscopic Dielectric Theory . . . . . . . . . . . . . . . . . . . . 5511.2 Microscopic Structure . . . . . . . . . . . . . . . . . . . . . . . . . 56

11.2.1 Relation of Macroscopic to Microscopic . . . . . . . . . . 5711.3 Polarizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5711.4 Polarization in Solids . . . . . . . . . . . . . . . . . . . . . . . . . 5711.5 Dipole Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5811.6 Polarizability of Solids . . . . . . . . . . . . . . . . . . . . . . . . 5911.7 Dielectric Constant . . . . . . . . . . . . . . . . . . . . . . . . . . 5911.8 Claussius Mossotti Relation . . . . . . . . . . . . . . . . . . . . . 6011.9 Frequency Dependence . . . . . . . . . . . . . . . . . . . . . . . . 6011.10Quantum Theory of Polarizability . . . . . . . . . . . . . . . . . 6211.11Ferroelectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

12 Phase Transitions 6312.1 Lattice Instabilities . . . . . . . . . . . . . . . . . . . . . . . . . . 6312.2 Curie Weiss Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6312.3 Ferroelectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

12.3.1 First versus Second Order Transitions . . . . . . . . . . . 6612.3.2 Ferroelectric Example: BaTiO3 . . . . . . . . . . . . . . . 66

12.4 Other Instabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . 6712.5 Piezoelectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

13 Diamagnetism and Paramagnetism 6813.1 Diamagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6913.2 Paramagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

13.2.1 Calculation of Susceptibility . . . . . . . . . . . . . . . . . 7213.2.2 Calculation of Total Angular Momentum J . . . . . . . . 7313.2.3 Spectroscopic Notation . . . . . . . . . . . . . . . . . . . . 7313.2.4 Paramagnetic Susceptibility . . . . . . . . . . . . . . . . . 7313.2.5 Calculation of J . . . . . . . . . . . . . . . . . . . . . . . . 7313.2.6 Spin Orbit Interactions . . . . . . . . . . . . . . . . . . . . 74

3

CONTENTS CONTENTS

13.2.7 Effective Magnetic Number . . . . . . . . . . . . . . . . . 7413.2.8 Paramagnetic Properties of Metals . . . . . . . . . . . . . 7513.2.9 Band Model . . . . . . . . . . . . . . . . . . . . . . . . . . 7513.2.10 Multivalent Effects . . . . . . . . . . . . . . . . . . . . . . 76

14 Ferromagnetism 7714.1 Ferromagnetic Phase Transition . . . . . . . . . . . . . . . . . . . 7714.2 Molecular Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

14.2.1 Prediction of TC . . . . . . . . . . . . . . . . . . . . . . . . 7914.2.2 Temperature Dependence of M(T ) for Ferromagnetism . 8014.2.3 Ferromagnetic-Paramegnetic Transition . . . . . . . . . . 8014.2.4 Ferromagnetic-Paramagnetic Transition . . . . . . . . . . 8014.2.5 Ferromagnetic-Paramagnetic Transition . . . . . . . . . . 8114.2.6 Ferromagnetic-Paramagnetic Transition . . . . . . . . . . 8114.2.7 Ferromagnetism of Alloys . . . . . . . . . . . . . . . . . . 8214.2.8 Transition Metals . . . . . . . . . . . . . . . . . . . . . . . 8214.2.9 Ni Alloys . . . . . . . . . . . . . . . . . . . . . . . . . . . 8214.2.10 Band Model . . . . . . . . . . . . . . . . . . . . . . . . . . 8314.2.11 Spin Waves . . . . . . . . . . . . . . . . . . . . . . . . . . 8314.2.12 Magnon Dispersion . . . . . . . . . . . . . . . . . . . . . 84

14.3 Ferrimagnetic Order . . . . . . . . . . . . . . . . . . . . . . . . . 8414.3.1 Magnetic Oxides . . . . . . . . . . . . . . . . . . . . . . . 8514.3.2 Magnetization and Hysteresis . . . . . . . . . . . . . . . . 86

14.4 Domains and Walls . . . . . . . . . . . . . . . . . . . . . . . . . . 8614.4.1 Energy of Bloch Domain Walls . . . . . . . . . . . . . . . 87

14.5 Anisotropy of Magnetization . . . . . . . . . . . . . . . . . . . . 8714.5.1 Anisotropy of Magnetization . . . . . . . . . . . . . . . . 88

14.6 Ferrimagnetic Ordering . . . . . . . . . . . . . . . . . . . . . . . 8814.6.1 Exchange Terms and Susceptibility . . . . . . . . . . . . . 8914.6.2 Structure Dependence of Jex . . . . . . . . . . . . . . . . 89

15 Optical Materials 8915.1 Frequency Doubling . . . . . . . . . . . . . . . . . . . . . . . . . 9015.2 Nonlinearity in Refractive Index . . . . . . . . . . . . . . . . . . 9015.3 Electro-Optic Modulators . . . . . . . . . . . . . . . . . . . . . . 9115.4 Optical Memory Devices . . . . . . . . . . . . . . . . . . . . . . . 9115.5 Volume Holography . . . . . . . . . . . . . . . . . . . . . . . . . 9215.6 Photorefractive Crystals . . . . . . . . . . . . . . . . . . . . . . . 9215.7 Photorefractive Crystals . . . . . . . . . . . . . . . . . . . . . . . 9315.8 All-Optical Switching . . . . . . . . . . . . . . . . . . . . . . . . . 9315.9 Acousto-Optic Modulators . . . . . . . . . . . . . . . . . . . . . . 9415.10Integrated Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . 9415.11Dielectric Waveguides . . . . . . . . . . . . . . . . . . . . . . . . 9515.12Phase Shifter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9515.13LiNbO Phase Shifter . . . . . . . . . . . . . . . . . . . . . . . . . 96

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CONTENTS CONTENTS

16 Superconductivity 9616.1 BCS Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9616.2 Density of States and Energy Gap . . . . . . . . . . . . . . . . . . 9716.3 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9716.4 Superconductor Junctions . . . . . . . . . . . . . . . . . . . . . . 9816.5 Junctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9816.6 Type I and Type II Superconductors . . . . . . . . . . . . . . . . 9816.7 High TC Superconductors . . . . . . . . . . . . . . . . . . . . . . 9916.8 Cuprate Superconductors . . . . . . . . . . . . . . . . . . . . . . 99

17 351-2 Problems 100

18 351-2 Laboratories 10118.1 Laboratory 1: Measurement of Charge Carrier Transport Param-

eters Using the Hall Effect . . . . . . . . . . . . . . . . . . . . . . 10118.1.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10118.1.2 Outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . 10218.1.3 References . . . . . . . . . . . . . . . . . . . . . . . . . . . 10218.1.4 Pre-Lab Questions . . . . . . . . . . . . . . . . . . . . . . 10218.1.5 Experimental Details . . . . . . . . . . . . . . . . . . . . . 10318.1.6 Instructions/Methods . . . . . . . . . . . . . . . . . . . . 10418.1.7 Link to Google Form for Data Entry . . . . . . . . . . . . 10418.1.8 Lab Report Template . . . . . . . . . . . . . . . . . . . . . 10418.1.9 Hints for derivation . . . . . . . . . . . . . . . . . . . . . 105

18.2 Laboratory 2: Diodes . . . . . . . . . . . . . . . . . . . . . . . . . 10618.2.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10618.2.2 Outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . 10618.2.3 Pre-lab Questions . . . . . . . . . . . . . . . . . . . . . . . 10618.2.4 References . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

18.3 Laboratory 3: Transistors . . . . . . . . . . . . . . . . . . . . . . . 11018.3.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11018.3.2 Outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . 11018.3.3 Pre-lab questions . . . . . . . . . . . . . . . . . . . . . . . 11018.3.4 Experimental Details . . . . . . . . . . . . . . . . . . . . . 111

18.4 Laboratory 4: Dielectric Materials . . . . . . . . . . . . . . . . . . 11318.4.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11318.4.2 Outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . 11318.4.3 Pre-lab questions . . . . . . . . . . . . . . . . . . . . . . . 11318.4.4 Experimental Details . . . . . . . . . . . . . . . . . . . . . 11418.4.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . 11518.4.6 Instructions/Methods . . . . . . . . . . . . . . . . . . . . 11518.4.7 Lab Report Template . . . . . . . . . . . . . . . . . . . . . 115

18.5 Laboratory 5: Magnetic Properties . . . . . . . . . . . . . . . . . 118

5

3 351-2: SOLID STATE PHYSICS

1 Catalog Description (351-1,2)

Quantum mechanics; applications to materials and engineering. Band struc-tures and cohesive energy; thermal behavior; electrical conduction; semicon-ductors; amorphous semiconductors; magnetic behavior of materials; liquidcrystals. Lectures, laboratory, problem solving. Prerequisites: GEN ENG 205 4or equivalent; PHYSICS 135 2,3.

2 Course Outcomes

3 351-2: Solid State Physics

At the conclusion of 351-2 students will be able to:

1. Given basic information about a semiconductor including bandgap anddoping level, calculate the magnitudes of currents that result from theapplication of electric fields and optical excitation, distinguishing be-tween drift and diffusion transport mechanisms.

2. Explain how dopant gradients, dopant homojunctions, semiconductor-semiconductor hetero junctions, and semiconductor-metal junctions per-turb the carrier concentrations in adjacent materials or regions, identifythe charge transport processes at the interfaces, and describe how theapplication of an electric field affects the band profiles and carrier con-centrations.

3. Represent the microscopic response of dielectrics to electric fields withsimple physical models and use the models to predict the macroscopicpolarization and the resulting frequency dependence of the real andimaginary components of the permittivity.

4. Given the permittivity, calculate the index of refraction, and describehow macroscopic phenomena of propagation, absorption, reflection andtransmission of plane waves are affected by the real and imaginary com-ponents of the index of refraction.

5. Identify the microscopic interactions that lead to magnetic order in ma-terials, describe the classes of magnetism that result from these interac-tions, and describe the temperature and field dependence of the macro-scopic magnetization of bulk crystalline diamagnets, paramagnets, andferromagnets.

6. Specify a material and microstructure that will produce desired magneticproperties illustrated in hysteresis loops including coercivity, remnantmagnetization, and saturation magnetization.

6

4 SEMICONDUCTOR DEVICES

7. Describe the output characteristics of p-n and Schottky junctions in thedark and under illumination and describe their utility in transistors, lightemitting diodes, and solar cells.

8. For technologies such as cell phones and hybrid electric vehicles, iden-tify key electronic materials and devices used in the technologies, specifybasic performance metrics, and relate these metrics to fundamental ma-terials properties.

4 Principles of Semiconductor Devices

Recall that the conductivity of semiconductors is given by

σ = neµ+ peµ (4.1)

where n is the number of electrons, e is the electronic charge, µ is the mobilityin cm2/V/s. The following trends for conductivity versus temperature arenoted for metals, semiconductors, and insulators:

1. Metals: n and p are constant with temperature. Mobility is relatedto temperature as µ ∝ T−a, where T is temperature and a is a con-stant. Conductivity decreases with increasing temperature. The resis-tivity ρ (ρ = 1/σ) can be written as a sum of contributing factors usingMatthiessen’s rule as

ρ = ρ0 + ρ(T )

where ρ0 is a constant and ρ(T ) is the temperature dependent resistivity.A typical carrier concentration for a metal is 1023cm−3. Metals do nothave a gap between the conduction and valence bands.

2. Semiconductors: n and p are not constant with temperature but are ther-mally activated. Mobility is related to temperature as µ ∝ T−a. The typi-cal range of carrier concentrations for semiconductors is 1014−1020cm−3.The range of bandgaps for semiconductors is typically 0.1− 3.0 eV.

3. Insulators: n and p are much lower than they are in metals and semicon-ductors. A typical carrier concentration for an insulator is < 107cm−3.The conductivity is generally a function of temperature. The bandgapfor insulators is > 3.0 eV.

4.1 Law of Mass Action

The equilibrium concentration of electrons and holes can be determined bytreating them as chemical species. At equilibrium,

[np] [n] + [p] (4.2)

7

4.2 Chemistry and Bonding 4 SEMICONDUCTOR DEVICES

The rate constant is given by

[n][p]

[np]= K(T ) (4.3)

where

K(T ) = [F (T )]2 exp[−∆E/kT ] (4.4)

Consider the intrinsic case, that is, when the semiconductor is not doped withchemical impurities. For this case,

[n] = [p] = [ni] = A exp[−∆E/2kT ] (4.5)

where ni is the intrinsic carrier concentration. For an intrinsic semiconductor,the conductivity is

σ = 2nieµ ∼= σ0 exp[−∆E/2kT ] (4.6)

A log σ versus 1/T plot gives a straight line as shown in Fig. 4.1a.

IOS

IOS

K1/T(a)

1/T(b)

Figure 4.1: (a) Log versus 1/T plot of conductivity and resistivity in semicon-ductors. (b) Log of conductivity versus 1/T plot for a semiconductor spanningthe regimes of extrinsic and intrinsic conduction.

At lower temperatures where ∆Eg > kT , the conductivity is dominated bycarriers contributed from ionized impurities. For temperatures where ∆Eg <kT , thermal energy is sufficient to excite electrons from the valence to conduc-tion band. In this regime, the intrinsic carriers dominate the conduction. Thetwo regimes of conduction are illustrated in Fig. 4.1b.

4.2 Chemistry and Bonding

From the periodic table, we know the valence, atomic weight, and atomic size.A portion of the periodic table with groups IIIA-VIA is shown in Fig. 4.2.On the left side (group IIIA) are metals. The right hand side (group VIA) con-tains nonmetals. In between these two groups are elemental semiconductorsand elements that form compound semiconductors. At the bottom of the table(In, Sn, and Sb) are more metallic elements. Group IVA consists of the covalentsemiconductors Si, Ge, and gray Sn.

8


B C N O

Al Si P

Ga Ge As

In Sn Sb

IIIA IVA VA VIA

more metallic

morenon-metallic

Figure 4.2: Portion of the periodic table showing elements in typical elementaland compound semiconductors.

Electronic bands are made from an assembly of atoms with individual quan-tum states. The spectroscopic notation for quantum states is given by

1s22s22p63s23p63d104s2... (4.7)

By the Pauli exclusion principle, no two electrons can share the same quantumstate. This results in a formation of electronic bands when atoms are broughttogether in a periodic lattice. The atomic bonding energy as a function of dis-tance between atoms is shown in Fig. 4.3. In this figure, deq is the equilibrium

E

d

Figure 4.3: Potential energy versus distance between atoms. The equilibriumdistance, deq corresponds to the minimum in the energy curve. The differentenergies corresponding to deq form discrete levels in an electronic band.

distance between atoms with the lowest potential energy and is equal to thelattice constant. The energies correponding to deq form the allowed energiesin electronic bands. A diagram showing the energy levels in electronic bandsis shown in Fig. 4.4.Core electrons do not contribute to conduction. Conduction of charge carri-ers occurs in only in partially filled bands. A diagram showing the filling ofenergy states in the valence band is given in Fig. 4.5. Note that two electronsoccupy each energy state having different values of the spin quantum number,that is, “spin up” or “spin down.”Recall that semiconductors have band gap energies of 0.1−3.0 eV that separatethe conduction from the valence band. An electron can be promoted from thevalence to conduction band if it is supplied with energy greater than or equalto the band gap energy. The source of this energy may be thermal, optical, or

9


forbiddenenergyband

statesin aperiodiclattice

E

coreelectrons

1s

2s

2p etc anti-bonding

single atomvalence elctrons bonding

Figure 4.4: Energy diagram showing the electronic bands formed in a crys-talline solid.

spin upspin down valence

filledbond

(a)

partially filled metallic

(b)

Figure 4.5: (a) Diagram of energy states in a filled valence band. Up arrowsindicate the “spin up” state and down arrows indicate the “spin down” state..(b) Filling of states in a partially filled band. Since the band is partially filled,the electrons can contribute to conduction.

electrical in nature. A simple two-level band model for a semiconductor show-ing the conduction and valence band levels is shown in Fig. 4.6a Also shown(Fig. 4.6b) is a diagram illustrating the process of promoting an electron fromthe valence to conduction band. When an electron is excited to the conduc-tion band, it leaves behind a “hole” in the valence band, which represents anunfilled energy state. As discussed earlier with regards to the conductivity insemiconductors, both electrons and holes contribute to conduction (see Eqn.4.1).

h (p)

e (n)

(a) (b)

Figure 4.6: (a) Energy band diagram for a simple two-level system. (b) Dia-gram illustrating the generation of an electron and hole pair.

10

4.3 Reciprocal Lattice 4 SEMICONDUCTOR DEVICES

4.3 Reciprocal Lattice and the Brillouin Zone

• Reciprocal space: k-space, momentum space

~p = ~~k

• ~k is the reciprocal lattice vector with units 1/cm

Reciprocal 2D Square Lattice

IrreducibleBrillouinZone

4.4 Nearly Free Electron Model

• Fermi wavevector: kF = (3π2n)1/3

• Fermi energy: EF =~2k2F2m

• Conductivity is proportional to the Fermi surface area, SF . σ ∝ SF

First Brillouin Zone

Fermi surface of second BZ

Constant energy surface for free electrons

4.5 Two Level Model

11

5 BAND DIAGRAMS

E

K

• Parabolic bands: E = ~2k2/2m∗.�� m∗ ≡ effective mass

• Origin? Solution to the Schrodinger Equation, HΨ = EΨ for free elec-trons

5 Semiconductor Band Diagrams

Direct Bandgap

E

K

Indirect Bandgap

12

5.1 P- and N-Type Semiconductors 5 BAND DIAGRAMS

E

K

• Direct bandgap semiconductors: III-V’s

• Indirect bandgap semiconductors: Si, Ge

5.1 P- and N-Type Semiconductors

E

X

Intrinsic

n-type

p-type

5.2 P-N Junctions

n-type

p-typetransition region W (space change region)

n-type

p-type

p-n Junction

5.3 Boltzmann Statistics: Review

For Ev as the reference energy (EV = 0),

n = Ns exp

[−Ec − EF

kT

](5.1)

13

5.4 P-N Junction Equilibrium 5 BAND DIAGRAMS

p = Ns exp

[−EF − EV

kT

]= NS exp

(−EFkT

)(5.2)

For fully ionized acceptors, p ∼= NA. For fully ionized donors, n ∼= ND.�� Ns = Density of states

The Fermi Level for electrons and holes are calculated from Eq.5.1 and 5.2,respectively:

Electrons: EF = Ec − kT ln(NS/n)

Holes: EF = EV + kT ln(NS/p)

Note: these equations are true for non-generate semiconductors.

5.4 P-N Junction Equilibrium

Poisson’s equation - describes potential distribution φ(x):

d2φ(x)

dx2= −ρ(x)

ε�

�

�

�φ(x) ≡ Potential

ε ≡ Dielectric Constant of Materialρ(x) ≡ Charge Density

Assume that the donors and acceptors are fully ionized

p-type: NA → N−A + p+

n-type: ND → N+D + e−

Note: the space charge region is positive on the n-side and negative on thep-side.

5.5 Charge Profile of the p-n Junction

+-

neutral

W

14

5.6 Calculation of the Electric Field 5 BAND DIAGRAMS

• Total charge: ρ(x) = e [p(x) +ND(x)− n(x)−NA(x)]

• Space charge width: w = xn + xp

• Charge balance: NAxp = NDxn

• Note that x = 0 at the junction interface.

5.6 Calculation of the Electric Fieldd2φ

dx2= −ρ

ε

dφ

dx= −E

⇒ dEdx

=ρ

ε

E(x) =

∫ρ

εdx

Separate the integral for the n- and p-regions:

E(x) =

∫ 0

−xp

ρ(x)

εdx︸︷︷︸

p-region

+

∫ xn

0

ρ(x)

εdx︸︷︷︸

n-region

Important assumptions made:

1. ρ ≈ eND in the n region, and ρ ≈ −eNA in the p region.

2. ρ is constant in the separate n and p regions.

Boundary conditions:

1. E(x = −xp) = E(x = xn) = 0

2. E is continuous at the junction

Resulting electric field across the junction:

E(x) =

{− eNA

ε (x+ xp), −xp ≤ x ≤ 0eND

ε (x− xn), 0 ≤ x ≤ xn

E(x) =

{− eNA

ε (x+ xp), −xp ≤ x ≤ 0eND

ε (x− xn), 0 ≤ x ≤ xn

Electric Field in the Junction

15

5.7 Calculation of the Electric Potential 5 BAND DIAGRAMS

5.7 Calculation of the Electric Potential

Recall

φ = −∫Edx

Boundary conditions:

1. φ is continuous at the junction boundary

2. φ = 0 at x = −xp (chosen arbitrarily)

Resulting potential across the junction:

φ(x) =

{eNA

ε

(12x

2 + x · xp + 12x

2p

), −xp ≤ x ≤ 0

eND

ε

(NA

2NDx2p + x · xn − 1

2x2), 0 ≤ x ≤ xn

Electric Potential in the Junction

16

5.8 Junction Capacitance 5 BAND DIAGRAMS

5.8 Junction Capacitance

• Space charge width:

w = xp + xn =

[2εφ0

e(NA +ND)

]1/2 [(NAND

)1/2

+

(NDNA

)1/2]

φ0 ≡ Built-in potential

• Note that, from charge balance, NAxp = NDxn

• Junction capacitance:

C =ε

wF/cm2

• One-sided abrupt junction: heavily doped p or n

• Consider the case of a p+-n junction. For NA � ND,

w =

(2εφ0eND

)1/2

C =

(eεND2φ0

)1/2

17

5.9 Rectification 5 BAND DIAGRAMS

5.9 Rectification

transition region

V

I

Reverse Bias

Forward Bias

Band Diagram Current-Voltage Characteristic

• Larger Eg , larger φ0

• ID = I0D exp(−eφ0/kT )

Eg ni (cm−3 × 1014)Ge 0.66 0.24Si 1.08 0.00015

GaAs 1.4 -GaP 2.25 -GaN 3.4 -

• Under a forward bias +φ1, the built in potential is reduced by φ1 andthere is a higher probability of electrons going over the barrier.

• Barrier height under forward bias: e(φ0 − φ)

• Current increases exponentially with forward bias voltage.

Forward Bias

Reverse Bias

p n

p n

• Under a reverse bias +φ1, the built in potential is increased by φ1 andthere is a lower probability of electrons going over the barrier.

18

5.10 Junction Capacitance 6 TRANSISTORS

5.10 Junction Capacitance

Definition of capacitance:

C ≡ ∂Q

∂φ

For a one-sided abrupt junction, recall that Q = eNDxn, where

xn =

[2ε(φ0 + φ1)NAeND(ND +NA)

]1/2The total charge Q for a one sided junction is then

Q =

[2eε(φ0 + φ1)

NANDNA +ND

]1/2Calculate the capacitance as a function of bias voltage φ1:

C =∂Q

∂φ1=

[eε

2(φ0 + φ1)

NANDNA +ND

]1/2A plot of 1/C2 versus φ1 is called a Mott Schottky plot and yields NA or ND.

U

6 Transistors: Semiconductor Amplifiers

• Three terminal device

19

6.1 pnp Device 6 TRANSISTORS

• Bipolar Junction Transistor (BJT): consists of two p-n junctions back-to-back

• The three seminconductor regions are the emitter, base, and collector

• pnp and npn devices

p n p+

emitter base collector

outputinput

6.1 pnp Device

• Emitter-Base junction is forward biased, and the Collector-Base junctionis reverse biased

• Little recombination in the Base region

• Holes are injected into the Base from the Emitter and collected at theCollector.

Device Diagram

emitter (p)

forwardbiased junction

reversedbiased junction

collector (p)

base (n)

+ +

Circuit Diagram

20

6.2 pnp Device 6 TRANSISTORS

6.2 pnp Device

Device Diagram

emitter (p)

forwardbiased junction

reversedbiased junction

collector (p)

base (n)

+ +

Band Diagram

emitter collector

base

6.3 Amplifier Gain

Voltage across the junctions:

icφr � ieφfφrφf≈ 10 V/V

Nodal analysis at the base for pnp and npn devices:

21

6.4 Circuit Configurations 6 TRANSISTORS

b

e

c

npn

ic ≈ ie and ic = αie

From nodal analysis at the base, ic = ie − ib.

⇒ ic =αib

1− α= hfeib with hfe =

α

1− α

hfe is the current gain parameter. α ∼ 0.9.

6.4 Circuit Configurations

• Circuit configurations include the common-base, common-collector, andcommon-emitter.

• The term following “common” indicates which terminal is common tothe input and output.

Common-Emitter Diagram

+

+np

n

Common-Base I-V Characteristic

22

6.5 MOSFETs 6 TRANSISTORS

2mA3mA

= 4mA

6.5 MOSFETs

• MOSFET: Metal Oxide Semiconductor Field Effect Transistor

• An applied electric field induces a channel between the source and drainthrough which current conducts.

n

source gate drain

body

conductionchannel

oxide

6.6 Oxide-Semiconductor Interface

Changing the gate bias causes the following:

1. Band bending at the oxide-semiconductor interface.

2. Accumulation of charge at the interface

3. An increase in channel conductance.

23

6.7 Depletion and Inversion 6 TRANSISTORS

oxide

filled states(conducting electrons)

semiconductor

6.7 Depletion and Inversion

• Band bending in the opposite sense leads to depletion of charge carriers.

• Inversion occurs when the Fermi level of an n (p) type semiconductorintercepts the valence (conduction) band due to band bending.

• The channel in a MOSFET is an inversion layer.

filled states(conducting holes)

Depletion Inversion

6.8 Channel Pinch-Off

• A threshold voltage VT is required for channel inversion.

• For a positive gate voltage VG, the potential across the oxide-semiconductor junction is VG − VT .

• For a positive voltage between the drain and source (VDS), the potentialnear the drain is VG − VT − VDS .

• For VDS ≥ VG − VT , the channel is depleted near the drain, resulting in“pinch-off.”

p

source gate drain

pinched-offchannelchannel

24

6.9 Tunnel Diodes 6 TRANSISTORS

• In saturation mode (VDS ≥ VG − VT ), ID ∝ V 2GS

• A small change in VDS causes a large change in ID.

IncreasingGate Bias

�

�

�

�G = GateS = SourceD = Drain

6.9 Tunnel Diodes

• Consists of a heavily doped p+-n+ structure

• Tunneling current: I ∝ Neffe−x/a

• Electrons tunnel from filled to empty states across the junction

insulating region (barrier)

• Electrons can go over the barrier or through the barrier (tunneling)

negative resistance region

I

V V

I

V

I

Rectifying Tunneling Thin Junction

25

6.10 Gunn Effect 7 HETEROJUNCTIONS

6.10 Gunn Effect

• Excitation of electrons under a high field to a higher energy conductionband with larger effective mass

σ = neµ and µ =eτ

m∗, m∗ =

1

~2∂2E

∂k2︸︷︷︸band curvature

−1

GaAsband diagram

light electron

heavy electron

K

E

6.11 Gunn Diodes

• Negative resistance due to increased effective mass

J = σE =ne2τ

m∗E

• Transfer of electrons from one valley to another (Transferred ElectronDevice, TED)

I

0 V

BApositiveresistance

negativeresistance

7 Heterojunctions and Quantum Wells

• Composed of semiconductor super-lattices (alternating layers of differ-ent semiconductors)

• Multilayers grown by Molecular Beam Epitaxy (MBE) and Metal-Organic Chemical Vapor Deposition (MOCVD)

26

7.1 Quantum Wells 7 HETEROJUNCTIONS

• Used in the construction of lasers, detectors, and modulators

• Utilizes band offset, can be used to confine carriers

GaAlAs GaAs

7.1 Quantum Wells

• From the particle in a box solution, minibands are formed within the well

GaAlAs GaAs GaAlAs

miniband

HΨ = EΨ; Ψ = A exp(−knx)

kn =nπ

Lx

Ex =~2k2

2m∗=

~2

2m∗

(π2n2xL2x

)

I II

27

7.2 Heterostructures and Heterojunctions 7 HETEROJUNCTIONS

• Smaller Lx, larger energy of miniband

• Lx can be tuned to engineer the band gap

7.2 Heterostructures and Heterojunctions

• Devices that use heterojunctions and heterostructures: lasers,modulation-doped field effect transistors (MODFETs)

• Types of heterostructures: quantum wells (2D), wires (1D), and dots (0D)

2D(confinement in

1 direction)

1D(confinement in

2 directions)

0D(confinement in

3 directions)

Well Wire Dot

7.3 Layered Structures: Quantum Wells

• “Cladding” layer can confine carriers due to difference in bandgap andlight due to difference in refractive index.

• Examples: GaAs/GaAlAs, InP/InGaAsP, GaN/InGaN

• Semiconductor 1 is “lattice matched” (same lattice constant as substrate)

• Semiconductor 2 is the strained layer (different lattice constant)

semiconductor 2

semiconductor 1

semiconductor 2Z

28

7.4 Transitions between Minibands 7 HETEROJUNCTIONS

7.4 Transitions between Minibands

• Allowed initial and final states are governed by “selection rules.”

• Will have conduction in minibands

• Energy of states:

E =~2k2

2m∗=

~2

2m∗

(nπ

Lx

)2

GaAlAs GaAs GaAlAs

miniband

n=2

n=1

n=1

n=2

7.5 Quantum Dots

Quantum Dots

Substrate

29

7.6 Quantum Dot Example: Biosensor 7 HETEROJUNCTIONS

• Electrons are confined in all dimensions to “dots.”

• Quantum Dots: clusters of atoms 3-10 nm in diameter.

• QDs can be created by colloidal chemical synthesis or island growth epi-taxy.

• Modeled after the hydrogen atom with radius R. Quantized energies:

Ei =~2

2mi

(αiR

)2αi ≡ quantum number

7.6 Quantum Dot Example: Biosensor

CdTe

bioactivematerial

ZnSe

• Solution synthesized colloidal quantum dots (CQD)

• Fluorescence energy depends on surface and size of dot

• Binding of molecules to the surface of the QD quenches the photolumi-nescent intensity.

7.7 Quantum Cascade Devices

• Consist of multiple quantum wells

• Band bending occurs with bias

30

8 OPTOELECTRONICS

8 Optoelectronic Devices: Photodetectors and So-lar Cells

• Light incident on a p-n junction generates electron-hole pairs which areseparated by the built in potential.

• Separated carriers contribute to the photocurrent.

+

p n

I

V

with light

h

e

31

8.1 I-V Characteristics 8 OPTOELECTRONICS

8.1 I-V Characteristics

• Diode equation: I = I0(eqV/kT − 1

)• Short circuit current: Isc = Aq(Le + Lh)G�

�

�

�Le ≡ Electron diffusion lengthLh ≡ Hole diffusion lengthG ≡ Carrier generation rate (depends on light intensity)

• Current under illumination: I = I0(eqV/kT − 1

)− Isc

• Open circuit voltage: V (I = 0) = Voc =kT

qln

(IscI0

+ 1

)

8.2 Power Generation

• Quadrant IV is the power quadrant.

• Theoretical maximum power: Pmax, theory = VocIsc

• Maximum obtainable power: Pmax = ff(IscVoc)�� Fill Factor: ff ≡ ImVmIscVoc

I

V

32

8.2 Power Generation 8 OPTOELECTRONICS

8.2.1 Solar Cell Efficiency

• Efficiency is governed by the fill factor and bandgap

• Want larger fill factors and absorption that matches the solar spectrum

Intensity

a-Si

Band Gap (eV)1.1 1.30 1.40 1.70

19

23

Si

GaAs

8.2.2 Other Types of Solar Cells

1. Multi-junction solar cells: GaAlAs/GaAs. More efficient absorption.

2. Thin films: amorphous silicon (a-Si), CdTe. Inexpensive.

3. Metal-semiconductor solar cells

GaAlAs

GaAs

Heterojunction Cell

33


8.2.3 Metal-Semiconductor Solar Cells

• Energy barrier φB at metal-semiconductor junction

• Current density, J:

J = J00 exp

(−φBkT

)[exp

(qV

kT

)− 1

]

J ≡ currentarea

=I

A

Slope =

Intercept =

E

metal

8.2.4 Schottky Barrier and Photo Effects

• Barrier height is obtained from log(J) versus 1/T plot

• Electrons can tunnel across the barrier

log J

1/T

slope =

34


E

metal

(Barrier height)tunnel

semiconductor

• Photo-response is characterized by two regions: metal photo-emission,and band-to-band excitation.

• Photo-response ∝√φB

Photoemissionfrom metal

overbarrier

Response

band-to-band

Photon Energy (eV)

Photon Energy

Resp

onse

1/2

(μ

A1

/2)

8.2.5 Dependence of φB on Work Function�

�φ ≡ energy required to take an electron from the Fermi Level

to the Vacuum Level

φB = φm − φs

35


unpinned

nearly metalindependent

Al As Pt Au

GaAs

Mg

InAu

Electronegativity Difference

ZnS

metal semiconductor

8.2.6 Pinned Surfaces

• Surface states can influence the amount of band bending at ametal/semiconductor interface

• For “pinned” surfaces, the barrier height is nearly independent of themetal work function.

• Total amount of band bending is equal to the band gap energy:

φB(n) + φB(p) = Eg

• Example, InP (n): φB(n) ∼ 0.5 eV φB(p) ∼ 0.8 eV Eg ∼ 1.27 eV

• For making an Ohmic contact to an n-type semiconductor, φB = φm −φs < 0

electronicsurface states

metal semiconductor

36


8.2.7 Light Emitting Diodes (LEDs)

• Forward biasing a p-n junction results in electron-hole recombination atthe junction

• Recombination results in light emission. The wavelength depends on Eg

hν = Eg

p-n Junction under Forward Bias

8.2.8 Light Emitting Diodes (LEDs)

Material Eg at 300K (eV) CommentsGaP 2.25

GaAsP 1.7-2.25 AlloyGaAs 1.43

InGaAsP 0.8-1.4ZnSe 2.58 BlueGaN 3.4 Ultraviolet

AlGaInN 4.2

8.2.9 Solid Solution Alloys

• The bandgap can be engineered by alloying

• Vegard’s law can be used to approximate the bandgap

Eg ≈ xEg1 + (1− x)Eg2

1.43

GaAs X GaP

Eg (eV)

Indirect

Direct

2.25

37

9 LASERS

8.2.10 LED Efficiency

• LED efficiency is the ratio of the radiative recombination probability(WR) to the total recombination probability (Wtotal)

η =WR

Wtotal=

WR

WR +WNR

• WNR: the probability of non-radiative recombination

• Non-radiative processes limit the efficiency

• Recombination probability is related to the lifetime (τ )

W ∝ 1

τ[s−1]

1

τtotal=

1

τR+

1

τNR

• Efficiency in terms of probability (W ) and lifetime (τ )

η =WR

Wtotal=

WR

WR +WNR

η =1/τR

1/τtotal=

1

1 + τR/τNR

• For τR � τNR, τNR determines the radiation efficiency ∴ want as fewnon-radiative defects (dislocations, impurities,etc.) as possible

η =τNRτR

• GaN example: 107 dislocations per cm2

9 Lasers

• LASER: Light Amplification through Stimulated Emission of Radiation

• Laser light has narrow divergence and consists of photons of the samefrequency and phase

• Photons of a certain frequency and phase stimulate emission of photonswith the same frequency and phase

same wavelength same phase

"coherent"

Input Intensity

Outp

ut

Inte

nsi

ty

38

9.1 Emission Rate and Laser Intensity 9 LASERS

9.1 Emission Rate and Laser Intensity

• Calculate emission rate and laser intensity by accounting for both emis-sion and absorption processes using Boltzmann Statistics

• Calculate the transition rate between states and Einstein A and B coeffi-cients

• Consider 2 and 3 level systems

Two Level System Three Level System

9.2 Two Level System

• At equilibrium, the population of levels is determined by the Boltzmanndistribution

N2 ∝ Ne−E2/kT orN2

N=e−E2/kT

Z

Z ≡ sum over states

• Ratio of populations in a two level system:

N2

N1= e−(E2−E1)/kT

At equilibrium, fewer electrons are in level 2

Triggers of Stimulated Emission

• Internal or external radiation can cause the (stimulated) transition to oc-cur

• Internal sources are described by “black body” radiation.

int. ext.

stimulatedemission

39

9.3 Planck Distribution Law 9 LASERS

9.3 Planck Distribution Law

• Number of photons inside the cavity is determined by Planck’s Distribu-tion Law

• Spectral density of photons of frequency ν in linewidth dν

ρ(ν)dν =8πn3hν3

c3︸︷︷︸Density of states

1

exp(hν/kT )− 1︸︷︷︸Planck distribution

dν

9.4 Transition Rates

• Transition rate Rij : rate at which electrons transition from state i to statej.

• Absorption (“up”) rate R12:

R12 = N1B12ρ(ν21)dν

N1 ≡ Number of photons in lower state

ρ(ν21) ≡ number of photons available having transition energy hν

• Emission (“down”) rate R21:

R21 = N2[A21︸︷︷︸Spont.

+B21ρ(ν21)]︸︷︷︸Stim.

dν

• A21 and B21 are the Einstein A and B coefficients for spontaneous andstimulated transitions, respectively, between states 2 and 1.

40

9.5 Two Level System 9 LASERS

9.5 Two Level System

• At equilibrium (steady state):

R12 = R21

N1B12ρ(ν21)dν = N2 [A21 +B21ρ(ν21)] dν

• Substitute in N2

N1= e−(E2−E1)/kT , simplifying and solving for ρ(ν21) :

ρ(ν21)dν =A21dν

B12 exp [(hν21)/kT ]−B21

• Note: same form as Planck Distribution Law with B12 = B21

• In a two level system, the stimulated emission and absorption rates areequal, i.e., B12 = B21

• Comparing to the Planck Distribution Law,

A21 = B218πn3hν3

c3

• The coefficient of spontaneous emission A21 is related to the stimulatedemission rate

Two Level System - Summary

• Number of photons inside the cavity is determined by Planck’s Distribu-tion Law

• The stimulated emission and absorption rates are equal

• The coefficient of spontaneous emission A21 is related to the stimulatedemission rate

• A population inversion is required for lasing

9.6 Three Level System

• A three level system consists of a ground state (E1), excited state (E2),and intermediate state (E3).

• Electrons are pumped from E1 to E3 by a pump source with energy ≥hν31; N2 population is unaffected

• More electrons are in the upper state E3 than the lower (metastable) stateE2⇒ a population inversion is achieved

41

9.6 Three Level System 9 LASERS

Carrier Population

• The carrier population is given by the Boltzmann distribution

N(E) = N0 exp(−E/kT )

logN = −E/kT + logN0

• Lower energy states have higher carrier populations at equilibrium

log N

Energy

thermalequilibrium

non-equilibrium(excited)

Spontaneous Emission

• A31: spontaneous emission rate; measure of the spontaneous depopula-tion

• Assuming first order kinetics,

dN3

dt= −A31N3

A31 =1

tspont

• Solution for initial population N3,i: N3(t) = N3,i exp (−A31t)

slope = 1/tspont

t

log N

3

finitelifetime

42


Populations under High Pumping

• Under high pumping at steady state, the populations in levels 1 and 3are

N∗1 = N∗3 =N1 +N3

2

• Equal probabilities of emission and absorption: material is “transparent”

• From energy balance,

B23N2ρ(ν)︸︷︷︸Stim. absorption

= A32N3︸︷︷︸Spont. emission

+B32N3ρ(ν)︸︷︷︸Stim. emission

Laser Gain

• The net number of photons gained per second per unit volume is

NW12 = (N2 −N1)W12

W12 ≡ Transition rate from state 1 to 2

• For N > 0: population inversion, material can act as an amplifier

• For N = 0: the material is transparent

• For N < 0: light is attenuated

For amplification, a population inversion must exist.

Derivation of Laser Gain Factor

• Stimulated-emission photons travel in the same direction (z) as the inci-dent photons and further stimulate emission

• ⇒ photon flux and intensity increase in z direction

mirror

lasing medium

• Increase in photon intensity through a volume of thickness dz:

43


dI = (N2 −N1)W12hν12dz

∴dI

dz= (N2 −N1)W12hν12

• Transition probability is defined as

W12 =I

hν

c2g(ν)

8πn2ν2tspont

• g(ν) is the lineshape factor

• Laser gain factor γ(ν) is defined by the equation

dI

dz= γ(ν)I

∴ γ(ν) = (N2 −N1)c2g(ν)

8πn2ν2tspont

• The change in laser intensity with distance z is

dI

dz= γ(ν)I

• Solution for I(z):

I(z) = I(0) exp [γ(ν)z]

• Recall laser gain factor from previous slide:

γ(ν) = (N2 −N1)c2g(ν)

8πn2ν2tspont

• For amplification, I(z)/I(0) > 1 is required ∴ N2 > N1. A populationversion is required for amplification

• Laser threshold: operation at which I(z)/I(0) = 1

44

9.7 Lasing Modes 9 LASERS

9.7 Lasing Modes

• Lasing frequencies are chosen by confining the stimulated-emission pho-tons inside a cavity called a Fabry Perot resonator

• Standing waves of certain discrete frequencies are established inside thecavity

mirror

lasing medium

partially transmissive mirror

• Laser intensity after a round trip in the cavity:

I = I0R1R2 exp [(γ − α)2l]

α ≡ absorption and scattering lossesγ ≡ laser gain factor

• Criteria for net gain: gain ≥ sum of losses

R1R2 exp [(γ − α)2l] ≥ 1

• Modes that are in phase after a round trip are stabilized

9.8 Laser Examples

9.8.1 Ruby

• Transition metal ion (Cr) doped alumina (Al2O3)

• 3 level system - first demonstrated laser

• Optically pumped by a “discharge lamp”

laser transition

pump levels 3

2

1

45

9.9 Threshold Current Density 9 LASERS

9.8.2 Others Examples

Semiconductor:

• Emission from carrier recombination in a forward biased p-n junction

• Heterostructures form both p-n junction and waveguiding region (cav-ity)

Others:

• Ceramic: Nd^{3+} :YAG, 4-level system

• Glass fiber laser Er3+ :SiO2 Erbium Doped Fiber Amplifier (EDFA), basisof modern optical communication

p n

V

active region thickness

9.9 Threshold Current Density

• Number of electrons injected into lasing region per unit volume: Ne

• Recombination rate:

Nelwd

trec=Iiη

e�

�

�

�η ≡ EfficencyIi ≡ Injection currente ≡ Fundamental charge

• Recall the gain coefficient, γ(ν)

γ(ν) = (N3 −N2)c2g(ν)

8πn2ν2tspont

trec ∼= tspont

Ne = N3 −N2

• Substituting in N3 −N2 = Ne = Iiηetreclwd and tspont ∼= trec,

46

9.9 Threshold Current Density 9 LASERS

γ(ν) =c2g(ν)η

8πn2ν2elwdIi (cm−1)

Relates gain coefficient to Ii

• The linewidth factor is g(ν) ∼= 1/∆ν

• Let the mirror reflectivity be R1R2 = R. Recall the lasing threshold con-dition

R exp(γ − α)l′ = 1

(γ − α)l′ + lnR = 0

• Define threthold current density: jth = Ii,th/wl

• Substituting in for γ and solving for jth:

jth =8πn2ν2ed

c2ηg(ν)︸︷︷︸β−1

(α− 1

l′lnR

)[A/cm2]

• Lasing threshold current density:

jth =α

β− 1

βl′lnR (R < 1)

Lasing Threshold Condition:

jth =α

β− 1

βl′lnR (R < 1)

0100 200

47

9.10 Comparison of Emission Types 9 LASERS

Energy (eV)

Inten

sity

Lasing Spectrum

Inte

nsi

tyspontaneous stimulated

LaserLED

Lasing Threshold

9.10 Comparison of Emission Types

Emission Intensity Linewidth Directional CoherentSpontaneous Weak Broad No NoStimulated Intense Narrow Not No(Super-Radiant) NecessarilyStimulated Intense Very Yes YesLasing Narrow

9.11 Cavities and Modes

• Inside a cavity, standing waves are formed

• Light frequencies that are in phase after a round trip through the cavitywill lase. This condition is given by

Integer number of wavelengths = Optical path length

mλ = 2nl

mλ

2n= l m = 1, 2, 3...

• n is the refractive index of the cavity and l is the cavity length

these willbe amplifiedstanding waves are

formed

48

9.12 Semiconductor Lasers 9 LASERS

9.11.1 Longitudinal Laser Modes

• Modes whose condition for lasing depends on the length of the cavityare called longitudinal modes

• Wavelength spacing between modes:

∆λ =λ2

2l(n− λdndλ

)• dn/dλ is the dispersion of the refractive index

Imodes modulate

intensity

9.11.2 Transverse Laser Modes

• If the other facets are flat, then transverse modes are supported in thecavity.

• The effect of transverse modes is visible in the angular dependence ofthe laser output intensity

m = 1 m = 2

Transverse Modes

Emitted Intensity

0 20 40 degrees

m=1m=2

m=0

Output Intensity

9.12 Semiconductor Lasers

• Light is confined to regions of high refractive index

• Semiconductor heterostructure is designed to confine light in the junc-tion

49

9.13 Photonic Bandgap Materials 10 BAND DIAGRAMS

GaAs

Ga0.94Al0.06As

Ga0.98Al0.02As

Ga0.94Al0.06As

GaAs Substrate

index

GaAs Laser Example

9.13 Photonic Bandgap Materials

• Light interacts with periodic structures with periodicity on the order ofthe wavelength

• Photonic crystals: structures with spatial periodic variation in the refrac-tive index in 1, 2, or 3 dimensions. The periodicity is on the order ofoptical wavelengths.

• Photonic crystals can be engineered to slow down the propagation ofoptical pluses (“slow light”)

• The speed (vg) of a pulse is described by the group index (ng)

vg = c/ng as ng →∞, vg → 0

10 Band Diagrams

• Energy band levels are represented in real space

• Energy levels of interest: Conduction band edge, valence band edge, andFermi Level

• Energies of interest: work function and electron affinity

φ ≡Work Function. Energy required to move an electron from the fermi levelto the vacuum level.χ ≡Electron Affinity. Energy required to move an electron from the conductionband edge to the vacuum level.Ec ≡Conduction band energy.Ev ≡Valence band energy.EF ≡Fermi level.

50

10.1 Band Diagrams 10 BAND DIAGRAMS

10.1 Band Diagrams

Vacuum Level

Metal Band DiagramSemiconductor Band Diagram

Vacuum Level

10.2 Heterojunctions and the Anderson Model

• Heterojunction: metallurgical junction between two different types ofsemiconductors

• Anderson Model: junction between different semiconductors results indiscontinuities in the energy bands.

• Example: n-GaAs (Eg = 1.45 eV) and p-Ge (Eg = 0.7 eV)

n-GaAs p-Ge

10.3 Band Bending at p-n Junctions

Band diagrams for isolated p- and n-type semiconductors are shown in Fig.10.1below.When brought into contact without application of an external bias field, ther-mal equilibrium requires that the Fermi levels equilibrate, i.e.

EF,n = EF,p (10.1)

This results in a bending of the conduction and valence bands as shown inFig. 10.2. With the Fermi levels equilibrated, the following expression is trueconsidering the path A-B in Fig. 10.2:

EF,n−EF,p = 0 = (χGaAs+δGaAs+qVD,n+qVD,p)−(χGe+Eg,Ge−δGe) (10.2)

51

10.3 Band Bending at p-n Junctions 10 BAND DIAGRAMS

n-GaAs

1.45 eV

= 4.07

p-Ge

0.7 eV

= 4.13 eV

Vacuum Level

Figure 10.1: Band diagrams of isolated n-GaAs and p-Ge.

Rearranging Eqn. 10.2,

q(VD,n + VD,p) = (χGe − δGe + Eg,Ge)︸︷︷︸p

− (χGaAs + δGaAs)︸︷︷︸n

(10.3)

Substituting in known values,

q(VD,n + VD,p) = (4.13− 0.1 + 0.7)− (4.07 + 0.1) (10.4)=0.56 eV < Eg,Ge

where

VD,nVD,p

=NAεGeNDεGaAs

(10.5)

10.3.1 Calculate the Conduction Band Discontinuity

Let the Fermi level be zero on the energy scale.

52

10.3 Band Bending at p-n Junctions 10 BAND DIAGRAMS

A B

VacuumLevel

n-GaAs p-Ge

Figure 10.2: Band diagram of a n-GaAs/p-Ge heterojunction.

δGaAs + qVD,n = (Eg,Ge − δGe)− qVD,p + ∆Ec

∆Ec = δGaAs + qVD,n − (Eg,Ge − δGe) + qVD,p

On substituting in Eqn. 10.3, we get

∆Ec = χGe − χGaAs = +0.6

The positive value for ∆Ec indicates a spike in the conduction band.

10.3.2 Calculate the Valence Band Discontinuity

The valence band discontinuity ∆Ev is calculated in a similar manner. In thiscase, a negative value for ∆Ev indicates a spike in the valence band. ∆Ev iscalculated as follows:

∆Ev = ∆Eg −∆χ

∆Ev = (Eg,GaAs − Eg,Ge)− (χGe − χGaAs)

Substituting in known values,

∆Ev = (1.45− 0.7)− (4.13− 4.07)

∆Ev = +0.69

Since ∆Ev is positive, there is no spike in the valence band.

53

11 DIELECTRIC MATERIALS

p-GaAs

= 4.07 eV

= 1.45 eV

Vacuum Level

(a)

n-Ge

= 4.13 eV

= 0.7

Vacuum Level

(b)

Figure 10.3: Isolated band diagrams of (a) p-GaAs and (b) n-Ge.

10.3.3 Example: p-GaAs/n-Ge

The isolated band diagrams for p-GaAs and n-Ge are shown in Figs. 10.3aand 10.3b. Here, we calculate the conduction and valence band discontinuitiesusing the approaches outlined in Sections 10.3.1 and 10.3.2. For the conductionband discontinuity,

q(VD,n + VD,p) = (χGaAs + Eg,GaAs − δGaAs)︸︷︷︸p

− (χGe + δGe)︸︷︷︸n

= (4.07 + 1.45− 0.1)− (4.13 + 0.1)

= 1.19 < Eg,GaAs

For calculating the valence band discontinuity,

∆Ev = (Eg,Ge − Eg,GaAs)− (χGaAs − χGe)

∆Ev = −0.69⇒ spike (10.6)

The resulting band diagram is shown in Fig. 10.4.

11 Dielectric Materials

• Dielectrics are insulating materials which include glasses, ceramics,polymers, and ferroelectrics

• Dielectric constant (ε) describes how well a material stores charge

• ε is related to capacitance (C)

Parallel Plate Capacitance

C =εA

dd ≡ distance between platesA ≡ plate area

54

11.1 Macroscopic Dielectric Theory 11 DIELECTRIC MATERIALS

p-GaAs n-Ge

VacuumLevel

Figure 10.4: Band diagram of a p-GaAs/n-Ge heterojunction as predicted us-ing the approach in Section10.3.3. Note that there is a spike in the valence bandwhich impedes the conduction of holes.

+ + + +Dielectric+V

11.1 Macroscopic Dielectric Theory

• From Maxwell’s equations,

D = εE [C/m2]��

��

D ≡ Electric flux densityε ≡ Dielectric constantE ≡ Electric field

• Note: ε = εrε0��

�

εr ≡ Relative dielectric constant (unitless)ε0 ≡ Permittivity of free space [F/m]

• Total charge stored in a capacitor in vacuum:

Q = ε0V

d[C]

with

55

11.2 Microscopic Structure 11 DIELECTRIC MATERIALS

V ≡ Voltage [V]d ≡ Distance [m]

• Putting in a dielectric,

Q′ = ε0εrV

dand C = Q′/V

• An increase in charge is stored in the capacitor with insertion of the di-electric

• D (electric flux density) is equal to the surface charge

• P (polarization density) is the additional surface charge

P ≡ D − ε0EP ≡ (εrε0 − ε0)E = ε0 (εr − 1)︸︷︷︸

χ

E

P ≡ ε0χE

• χ = (εr − 1) is the dielectric susceptibility

11.2 Relation of ε to the Microscopic Structure

• Knowledge of the microscopic structure is needed to explain dielectricphenomena. Example: ferroelectric εr(T )

T

• In the microscopic approach, we consider 1) Atomic behavior and 2) De-formation of the atomic charge cloud (orbitals) when a field is applied.

With Applied FieldNo Applied Field

56

11.3 Polarizability 11 DIELECTRIC MATERIALS

11.2.1 Relation of Macroscopic to Microscopic

• Induced dipole moment due to an applied field: µ = qδ

• ForNm molecules per unit volume, the polarization density is P = Nmqδor

P = Nmµ

• For small fields, µ = αE ′ and P = NmαE ′

E ′ ≡ Local electric fieldα ≡ Polarizability of species (atoms, molecules)

Polarized DielectricSingle Dipole

11.3 Contributions to the Polarizability

1. Electronic: Deformation of the electronic cloud (orbitals) due to opticalfields.

2. Molecular: Bonds between atoms can stretch, bend, and rotate

3. Orientational: Polymers, liquids, and gases can be reoriented in an ap-plied electric field. Example: poled polymers, useful for electro-opticdevices.

11.4 Polarization in Solids

• In an applied field, some molecules will be aligned, others will not becompletely aligned. The orientation energy is:

E = −µE cos θ

θ ≡ Angle between µ and E

• The number of dipoles with a certain E will depend on the Boltzmannfactor exp(−E/kT ). Increasing temperature will favor more randomalignment.

57

11.5 Dipole Moment 11 DIELECTRIC MATERIALS

+

-can alignwith field

Dipoles in a Solid

11.5 Calculation of the Average Dipole Moment

• Average Dipole Moment (〈µ〉)

〈µ〉 =weighted average energy for each orientation

total energy for all orientations

• Consider the 3-dimensional case. In a sphere with radius r, the totalnumber of molecules with energy E in a volume element 2πr2 sin θdθ is

# of dipoles with energy E = 2πA sin θdθ exp(−E/kT )

A = Constant including radius

• Recall the orientation energy is given by E = −µE cos θ. Substituting infor E,

# of dipoles with energy E = 2πA sin θdθ exp

(µE cos θ

kT

)• The total energy for all orientations is then∫ π

0

A exp

(µE cos θ

kT

)2π sin θdθ

• The weighted average for each orientation is∫ π

0

A exp

(µE cos θ

kT

)(µ cos θ) 2π sin θdθ

• The average dipole moment is then

〈µ〉 =

∫ π0A exp

(µE cos θkT

)(µ cos θ) 2π sin θdθ∫ π

0A exp

(µE cos θkT

)2π sin θdθ

• Integrating,

〈µ〉µ

= L(a) = coth a− 1

a

with a =µEkT

58

11.6 Polarizability of Solids 11 DIELECTRIC MATERIALS

• For small a, coth a ≈ 1a + a

3 + · · ·

∴〈µ〉µ

=a

3for small a

〈µ〉 =µ2E3kT

or〈µ〉µ

=µE3kT

11.6 Polarizability of Solids

• The polarizability of the solid is

P = Nm〈µ〉

• P is proportional to the field E and inversely proportional to the temper-ature T

P = Nmµ2E3kT

• The definition for the electric susceptibility is χ = ∂P∂E ⇒ χ ∝ 1

T

11.7 Dielectric Constant for a Solid

• Under an applied field E0, induced dipoles in the solid create an oppos-ing field E1. The field inside the material is E = E0 + E1 .

++

++

++

_

__

_

___

_+++

++ _ _

__

• For a crystal, the polarizability will depend on the interaction of dipolesin the lattice

P =∑j

NjPj =∑j

NjαjEloc,j

59

11.8 Claussius Mossotti Relation 11 DIELECTRIC MATERIALS

• Eloc,j is the local field of atom j due to interactions with other dipoles.For cubic materials,

Eloc = E +P

3ε0

• The polarizability is then

P =

∑j

Njαj

(E +P

3ε0

)

11.8 Claussius Mossotti Relation

• From the relation for the macroscopic electric susceptibility (see slide11.1), χ = P/ε0E

χ =

∑j

Njαj

E + P3ε0

ε0E

• Recall that χ = εr − 1 and P = χε0E = ε0E(εr − 1)

εr − 1 =

∑j

Njαj

E + 13ε0ε0E(εr − 1)

ε0E

Simplifying and rearranging gives the Claussius Mossotti Relation:

εr − 1

εr + 2=

1

3ε0

∑j

Njαj (SI Units)

• The Claussius Mossotti relation relates the dielectric constant to theatomic polarizability

11.9 Frequency Dependence of the Polarizability

• Dipolar, ionic, and electronic contributions to the polarizability

α(ω) =∑

αi(ω)

60

11.9 Frequency Dependence 11 DIELECTRIC MATERIALS

• Classical approach: treat electrons as harmonic oscillators

• Apply Hooke’s Law

eEloc = Cx = mω20x

C = mω20 ≡ Force constant

Eloc =mω2

0x

e

• Note: ω0 is the natural or resonance frequency

• The electronic contribution to the polarizability is P = NmαelEloc. Recallthat P = Nmqδ = Nex

αel =P

NEloc=

Nex

NEloc=

e2

mω20

• For an applied field with frequency ω, the equation of motion for theelectron response is

md2x

dt2+mω2

0x = −eEloc sinωt

• The solution is x = x0 sinωt. Substituting the solution into the equationof motion,

m(−ω2 + ω20)x0 = −eEloc

• Solving for x0

x0 =−eEloc

m(ω20 − ω2)

• The electronic dipole moment is

µelec = −ex0 =e2Eloc

m(ω20 − ω2)

• The electronic polarizability is

αelec =µelecEloc

=e2

m(ω20 − ω2)

• αelec becomes large at resonances where ω20 − ω2 → 0

• Calculate the atomic polarizability:

61

11.10 Quantum Theory of Polarizability 11 DIELECTRIC MATERIALS

µ0 = −ex0 =e2Eloc

m(ω20 − ω2)

αelec =µ0

Eloc=

e2

m(ω20 − ω2)

11.10 Quantum Theory of Polarizability

αelec =e2

m

∑j

fijω20 − ω2

fij ≡ Oscillator strength for a transition from state i to j

Pelec = NαelecE

11.11 Advanced Dielectrics: Ferroelectrics

• Ferroelectrics: materials that are non-centrosymmetric and have a result-ing net polarization.

• The polarization is a function of the applied field and displays hysteresis.

ion moves, forminga net dipole

+

PS ≡ Spontaneous polarizationEC ≡ Coercive field

62

12 PHASE TRANSITIONS

12 Phase Transitions

• The transition from a ferroelectric (net polarization) to pyroelectric (nonet polarization) structure occurs at the Curie Temperature (TC)

• Dielectric constant:

ε =ξ

T − TCξ ≡ Curie constant

Heat

Cap

aci

ty

Phase Transition

Ferroelectric Phase

PyroelectricPhase

Temperature

Die

lect

ric

Const

an

t

Temperature

12.1 Lattice Instabilities

• Recall the Clausius-Mossotti relation:

εr − 1

εr + 2=

1

3ε0Nα

• Solving for εr,

εr − 1 =Nα

3ε0(εr + 2) =

Nαεr3ε0

+2

3

Nα

ε

εr

(1− Nα

3ε0

)=

2Nα

3ε0+ 1

εr =2Nα/ε0 + 3

3−Nα/ε0

• Singularity (polarization catastrophe) at the condition Nα/ε0 = 3

12.2 Curie Weiss Law

• What happens close to the singularity? Consider a small deviation 2s

Nα

ε0∼= 3− 2s

• Rewriting the equation for εr,

63

12.3 Ferroelectrics 12 PHASE TRANSITIONS

εr =2Nα/ε0 + 3

3−Nα/ε0 − 2s∝ 1

snear the singularity

• Near the critical temperature, suppose s ∝ T such that s ∼= (T − TC)/ξ

∴ εr ∼=ξ

(T − TC)or

1

εr∼=T − TC

ξ

12.3 Ferroelectric Phase Transitions: Landau Theory

• Consider the Helmholtz free energy F for a ferroelectric material as afunction of polarization (P ), temperature (T ), and applied field (E).

• Taking a Taylor expansion in the order parameter s about s = 0,

F (P, T, E) = −E · P + s0 +1

2s2P

2 +1

4s4P

4 +1

6s6P

6 + · · ·

• Note that there are no odd powers if the unpolarized crystal has a centerof inversion.

• At thermal equilibrium, the minimum in F with respect to P is

∂F

∂P= 0 = −E + s2P + s4P

3 + s6P5

• Thermal equilibrium condition:

∂F

∂P= 0 = −E + s2P + s4P

3 + s6P5

• For a ferroelectric state transition, the coefficient of the first order P termmust pass through zero at some temperature T0 for no applied field.

∴ assume s2 = γ(T − T0), with γ a positive constant

• s2 can be positive or negative:

• s2 < 0: lattice is “soft” close to the instability

• s2 > 0: unpolarized lattice is unstable

64


two minima

• Rewriting the equilibrium condition,

∂F

∂P= −E + γ(T − T0)P + s4P

3 = 0

• Second order transition: no volume change, smooth transition.

• For E = 0 and P = PS at equilibrium,

γ(T − T0)PS + s4P3S = 0

PS[γ(T − T0) + s4P

2S

]= 0

• Solutions: PS = 0, or

• For T ≥ T0 : P 2S = (γ/s4)(T0 − T ) ⇒ PS is imaginary; the only real root

is PS = 0.

• For T < T0 :

|PS | = (γ/s4)1/2

(T0 − T )1/2

• Second order transition for T < T0 :

PS | = (γ/s4)1/2

(T0 − T )1/2

• The phase transition is second order since the polarization goescontinuously to zero

dipoles aligned

order parameter dipoles

randomly aligned

65


12.3.1 First versus Second Order Transitions

• The change in Ps is discontinuous for first order transitions andcontinuous for second order transitions

• First order transition results from a structure change. Example: cubic(pyroelectric)→ tetragonal (ferroelectric)

First Order Second Order

• Recall the Helmholtz free energy:

F = −E · P + s0 +1

2s2P

2 +1

4s4P

4 +1

6s6P

6 + · · ·

• For first order transitions, s4 is negative and s6 cannot be neglected

• Equilibrium condition

∂F

∂P= γ(T − T0)PS − |s4|P 3

S + s6P5S = 0

• Solutions: PS = 0 or

γ(T − T0)− |s4|P 2S + s6P

4S = 0

First Order Second Order

12.3.2 Ferroelectric Example: BaTiO3

• Spontaneous polarization: Ps = Nmqδ = 26 µC/cm2 ≈ 3× 10−1C ·m−2

• Dipole moment of a unit cell: µ = Ps · volume. For a lattice constanta ≈ 0.4nm

66

12.4 Other Instabilities 12 PHASE TRANSITIONS

µ = (3× 10−1C ·m−2)× 64× 10−30m3 ≈ 2× 10−29C ·m

Material TC (K)BaTiO3 408BaTiO3 1480

12.4 Other Instabilities

+

ferroelectric

anti-ferroelectric +

+ + + +

+ + + +

+ +

• Pyroelectrics: heat causes distortion and electrical signal, “pyroelectricdetectors”

12.5 Piezoelectrics

• Piezoelectric materials: an applied electric field causes a mechanical de-formation

• Piezoelectric materials lack inversion symmetry

67

13 DIAMAGNETISM AND PARAMAGNETISM

+

+

+

++

+

++

+

+

no inversion symmetry

+

+

+

+

+

+

• Stress and displacement:

Stress: T = CS − eEDisplacement: D = εE + eS

C ≡ Elastic constante ≡ Piezoelectric coefficientE ≡ Electric fieldS ≡ Strain

• Note: when E = 0, D 6= 0 for non-zero e and S.

• Hooke’s Law for E = 0.

• Applications: electro-mechanical transducers, microphones, micro-motors, micro-electro-mechanical structures (MEMS), nano-electro-mechanical structures (NEMS).

13 Diamagnetism and Paramagnetism

• Current and magnetism related through Maxwell’s Equations

• Magnetic susceptibility:

χ =µ0M

B

M ≡ MagnetizationB ≡ Magnetic field intensityµ0 ≡ Permeability of free space

68

13.1 Diamagnetism 13 DIAMAGNETISM AND PARAMAGNETISM

Pauli paramagnetism

TemperatureDiamagnetism

Paramagnetism

0_

+

13.1 Diamagnetism

• In a magnetic field, induced current field opposite to applied current

• Electron precesses around field axis with frequency

ω =eB

2m[sec−1]

• Magnetic field will induce net electric current around the nucleus.

B

electron

• Consider multi-electron atom. The “electron current” for Z electrons is

I = (charge)× (revolutions per second) = −Ze ω2π

I = −Ze(

1

2π

eB

2m

)= −Ze

2B

4πm

• Magnetic moment:

µ ≡ (current)× (area of loop) = I ×A

• For a loop area A = πρ2,

µ = −Ze2B

4m〈ρ2〉

• 〈ρ2〉 is the mean square of the perpendicular distance from the field axisthrough charge

〈ρ2〉 = 〈x2〉+ 〈y2〉

69

13.2 Paramagnetism 13 DIAMAGNETISM AND PARAMAGNETISM

• The atomic radius is 〈r2〉 = 〈x2〉+ 〈y2〉+ 〈z2〉

∴ 〈ρ2〉 =2

3〈r2〉

• Diamagnetic susceptibility of electrons: M = Nµ

Langevin Result: χ =Nµ0µ

B= −µ0NZe

2

6m〈r2〉

(note negative sign)

• Note: need to calculate 〈r2〉 from quantum mechanics

13.2 Paramagnetism

• Positive χ occurs for

• atoms, molecules, and lattice defects with an odd number of electrons

• free atoms with partially filled inner shells

• metals

• Magnetic moment µ:

µ ≡ γ~J = −gµBJ'

&

$

%

γ ≡ gyromagnetic ratio~J ≡ angular momentumg ≡ g factor (∼2.00 for electron spin)

µB ≡ Bohr magneton (atomic unit)J ≡ angular quantum number for electrons

• From quantum mechanics, only certain alignments of µ with B are al-lowed

• Consider single electron case. EnergyU of an electron in a magnetic field:

U = −~µ · ~B = mJgµBB

• For single electron atom with no orbital momentum, mJ = mS = ±1/2

70


lower energystate

upper energy state Po

pula

tion

only certainalignmentsallowed

• For a two spin orientation case, mJ=± 12 , g = 2, and U = ±µBB

• Probability of occupying state i:

Pi =exp(−Ei/τ)

Z��

��Z ≡ sum over all states (normalization)

τ ≡ kT

• For N total atoms,

N1

N=

exp(µB/τ)

exp(µB/τ) + exp(−µB/τ)

N2

N=

exp(−µB/τ)

exp(µB/τ) + exp(−µB/τ)

1

2

• Magnetization M ≡ (net spin density ‖B) ×(magnetic moment per electron)

M = (N1 −N2)µ = Nµex − e−x

ex + e−x= Nµ tanhx

x ≡ µB/kT

• For x� 1: high T , small field, no interaction.

tanhx ≈ x⇒M ∼= Nµ(µB/kT )

71


few spins aligned

all spinsaligned

two level system

• For multiple electron atoms, the azimuthal quantum number mJ takesvalues of J , J − 1, . . . ,−J

• For an atom with quantum number J , there are 2J + 1 energy levels

• Curie-Brillouin Law:

M = NgJµBBJ(x)

x ≡ gJµBB/kT

• Brillouin function (obtained from 〈µ〉 over all quantum configurations)

BJ(x)2J + 1

2Jcoth

[(2J + 1)x

2J

]− 1

2Jcoth

x

2J

• For x� 1

cothx ≈ 1

x+x

3

13.2.1 Calculation of Susceptibility

For small fields and high temperatures,

χ =M

B∼=NJ(J + 1)g2µ2

B

3kT=Np2µ2

B

3kT= C/T

p2 = g2J(J + 1)⇒ p = g [J(J + 1)]1/2

C ≡ Curie constant

s = 7/2

s = 5/2

s = 3/2

B/T

7

72


13.2.2 Calculation of Total Angular Momentum J

• Atoms with filled shells have no magnetic moment. Example:

1s22s22p63s23p63d10

• The spins arrange themselves so as to give the maximum possible S con-sistent with Hund’s rule

• Pauli Principle: no two electrons in the same system can have the samequantum numbers n, l, ml, and ms

13.2.3 Spectroscopic Notation

Example Orbital Configurations

1s 2s 2p

B

Ti3+1s 2s 2p 3s 3p 3d1

1s 2s 2p 3s 3p

Cr3d5 4s1

Mn

1s 2s 2p 3s 3p 3d5 4s2

Fe1s 2s 2p 3s 3p 3d6 4s2

13.2.4 Paramagnetic Susceptibility

Curie Law for paramagnetic susceptibility: χ = C/T

Will dependon valenceelectronand J

Curie Law

free spin

13.2.5 Calculation of J

• Rules for calculating J :

73


Value of J Condition

|L− S| When the shell is less than half full

|L+ S| When the shell is more than half full

S When the shell is half full

• Values for L and S:

L =∑

ml and S =∑

ms

• L is the orbital angular momentum

13.2.6 Spin Orbit Interactions

• For multi-electron atoms,

• Orbital angular momentum: L =∑i Li

• Total spin angular momentum: S =∑i Si

• Total angular momentum: ~J , where J = L+ S

• Treat J , L, and S as vectors. L and S precess around J ; J remains con-stant (“conserved”)

• Note: for iron metal group, J = S; only the spin contributes. Orbitalmomentum is “quenched.”

S

J

L

vector diagram

13.2.7 Effective Magnetic Number

• Effective magnetic number:

P = g [J(J + 1)]1/2

• Example: Ti3+

J ∼ S For transition metals, "orbital angular momentum quenched"

3d1 S=1/2

74


P = 2 [S(S + 1)]1/2

P = 2

[1

2·(

3

2

)]1/2= 1.73

• Example: Fe2+

3d6 S=2

P = 2 [2 (3)]1/2

= 4.90

13.2.8 Paramagnetic Properties of Metals

• For paramagnetic solids with N magnetic ions

M =Nµ2B

kTF

• For metals, not all spins are accessible

M ∼=Neffµ

2B

kT⇒ Neff ∼=

NkT

kTF

M ∼=Nµ2B

kTFTemperature independent

13.2.9 Band Model

• Total magnetization: M = (N+ −N−)µ

• Total number of electrons with “up” spins (N+)

N+ =1

2

∫ EF

−µB

dE f(E)︸︷︷︸Fermi function

D(E + µB)︸︷︷︸Density of states

75


Spins parallel

to B

Spins anti-parallel

to B

Density of States B=0 B>0

• Taking a Taylor expansion of the density of states for N+,

N+ ≈ 1

2

∫ EF

0

dEf(E)D(E) +1

2

∫ EF

0

dEf(E)dD(E)

dEµB

≈ 1

2

∫ EF

0

dEf(E)D(E) +1

2

∫ EF

0

f(E)dD(E)µB

≈ 1

2

∫ EF

0

dEf(E)D(E) +1

2D(EF )µB

• Repeat the procedure for N−

• Taylor series expansion of the density of states for N−

N− =1

2

∫ EF

µB

dEf(E)D(E − µb)

≈ 1

2

∫ EF

0

d(E)f(E)D(E)− 1

2D(EF )µB

• Total magnetization:

M = (N+ −N−)µ = D(EF )µ2B

• For a metal,

D(EF ) =3N

2EF=

3N

2kTF

M =3Nµ2B

2kTFχ =

∂M

∂B=

3Nµ2

2kTFIndependent of T!

13.2.10 Multivalent Effects

• High density of states near Fermi level.

• d-band contribution to χ

76

14 FERROMAGNETISM

Pd

V

KRbz=1z=1

Mo

(Experimental)

14 Ferromagnetism

• Magnetic ordering in solids is due to interaction of spins

• Below a transition temperature, an internal field tends to line up spins.This is called the “exchange field” BE

HΨ = EΨ with H=K+Vex

Vex ≡ Exchange Potential

• Mean field approximation: BE = λM where λ ≡Weiss constant

Ferromagnet Simple Antiferromagnet Ferrimagnet

Alignment of Spins

14.1 Ferromagnetic Phase Transition

• Above the transition temperature, ferromagnetism gives way to param-agnetism

T > TC Paramagnetism

T < TC Ferromagnetism

• Magnetization: M = χp(BA +BE)

χP ≡ Paramagnetic susceptibilityBA ≡ Applied field

77

14.2 Molecular Field 14 FERROMAGNETISM

Ferro-magnetic

Paramagnetic

14.2 Molecular Field

• For paramagnetism, we had χp = C/T

• “Molecular” or “exchange” field for ferromagnetism: BE = λM

M = χP (BA + λM)⇒M =C

T(BA + λM)

• Solving for M ,

M

(1− Cλ

T

)=CBAT

M(T − Cλ) = CBA

χ =M

BA=

C

(T − Cλ)

• Taking Cλ = TC ,

χ =C

T − TCfor T > TC

Weiss constant: λ = TC/C

• Ferromagnetic behavior:

χ ∝ 1

(T − TC)1.33Scaling theory

• Calculation of field parameter:

λ =TCC

= TC3k

Ng2S(S + 1)µ2b︸︷︷︸

Curie Constant

• For iron, TC ∼= 1000K, g = 2, S = 1, (3d64s2 → 3d8), λ = 5000. For asaturation magnetization of MS = 1700 Gauss,

78


BE = λMS

BE = 5000× 1700 ≈ 107 Gauss = 103 Tesla

• Highest superconducting magnet: 40 Tesla

Ferromagnetic(Ordered)

Paramagnetic(Disordered)

1040 K

Material TC (K)Fe (bcc) 1043

Co 1388Ni 627

Note: fcc iron is not magnetic

14.2.1 Prediction of TC

• Heisenberg model. For atoms i and j with spins Si and Sj , the orderingenergy U is

U = −2Jex ~Si · ~Sj

Jex ≡ Exchange energy integral

• From mean field theory,

Jex =3kTC

2ZS(S + 1)or TC =

2ZS(S + 1)Jex3k

Z ≡ Number of neighbors

79


14.2.2 Temperature Dependence of M(T ) for Ferromagnetism

• Consider spin 1/2 system. Just as for paramagnetism, M = N〈µ〉

M = Nµ tanh

(µB

kT

)Where B = BA +BE

• For a ferromagnet, B ≈ BE = λM , where BE ≡Molecular field.

∴M = Nµ tanh

(µλM

kT

)= Nµ tanh(x) with x ≡ µλM

kT

SaturationMagnetization

At low temperatures,all magnetic spins line up

14.2.3 Ferromagnetic-Paramegnetic Transition

• Define reduced magnetization m and reduced temperature t

m ≡ M

Nµt ≡ kT

Nµ2λ

∴ m︸︷︷︸LHS

= tanh(m/t)︸︷︷︸RHS

• Solve the transcendental equation graphically by plotting both the lefthand side (LHS) and right hand side (RHS) versus m/t

14.2.4 Ferromagnetic-Paramagnetic Transition

• From the Landau model, as T increases, M(T ) decreases

• Second order phase transition: continuous change in M(T )/M(0) versusT to TC

80


Ordered

Disordered


• At low temperatures,

tanh(m/t) = tanh ξ ∼= 1− 2e−2ξ

• Define change in magnetization with temperature, ∆M ≡M(0)−M(T ).For large ξ (low T ),

tanh ξ ∼= 1− 2e−2ξ ≈ 1

∴ ∆M = 2Nµe−2ξ = 2Nµe−2µλM/kT

• Noting that M = Nµ,

∆M = 2Nµe−2µ2λN/kT

∆M = 2Nµ exp(−2λNµ2/kT ) = 2Nµ exp(−2TC/T )


Change in magnetization with respect to T = 0 value versus temperature

∆M = 2Nµ exp(−2λNµ2/kT ) = 2Nµ exp(−2TC/T )

For small

81


14.2.7 Ferromagnetism of Alloys

Mn Fe Co Ni7 8 9

Cr

1

2

3

0.6

3d Electron Concentration

14.2.8 Transition Metals

d Orbital Configurations of Transition Metals

Fe

Co

Ni

Cu

2 unpaired spin

1 unpaired

0.6 unpaired

Mn

Pd

Others:

3d9

3d74s2

3d64s2

3d8

3d84s2

3d9.44s0.6

3d104s1

3d54s2

4d10

14.2.9 Ni Alloys

• Atomic moment of nickel transition metal alloys changes with composi-tion

• For nickel alloyed with copper, magnetization disappears at 60% Cu

0.6

0.2 0.4

Pd

Cu

Alloy Compositionof Second Component

magentismdisappearsat 60% Cu

Mn

0.6

Zn

82


14.2.10 Band Model

• For nickel, 3d bands and 4s bands are partially occupied

3d10 3d9.44s0.6

5 up4.4 down

Paired spins(Pauli paramagnetism)

• Net spin for Ni is 0.6µB . Only the d bands contribute in this case

d band

s band

Filled States

• Magneton numbers

• Fe: 2.2µB

• Co: 1.7µB

• For iron,

3d8 3d7.84s0.2

• Can copper be ferromagnetic?

3d104s1 3d94s2 (yes)

superconducting properties?

YBa2CuO

14.2.11 Spin Waves

• Spin waves: low energy excitations

• Lower energy excited state is obtained by having spins arranged at anangle θ with respect to neighboring spins (“spin waves” or “magnons”)

• Precession angle depends on angle of neighbor

83

14.3 Ferrimagnetic Order 14 FERROMAGNETISM

Excited State,Higher Energy

Spin Wave"Magnons"

B

Ground State

Spin Wave - Top View

14.2.12 Magnon Dispersion

• Magnons are particles and have an ω versus q dispersion relation

ω =4JexS

2

~sin2 aq

2

Jex ≡ Exchange energy termS ≡ Total spin angular momentuma ≡ Distance between atoms with magnetic momentsq ≡ Wavevector

14.3 Ferrimagnetic Order in Magnetic Oxides

• Iron oxide (Fe3O4) consists of two sublattices:

Fe3O4 → FeO · Fe2O3

Sublattice A(1 Fe2+ ion)

Sublattice B(2 Fe3+ ions)

Net Spin: S=-2

Sublattice Fe Oxid. State e− Config. S gJ (µB)

(A) FeO 2+ 3d6 2 4(B)Fe3O4 3+ 3d5 5

2 5

84

14.3 Ferrimagnetic Order 14 FERROMAGNETISM

• Total expected magnetic moment per formula unit: 2× 5 + 4 = 14µB

• Measured magnetic moment per formula unit: 4.1µB (due to ferrimag-netic ordering)

14.3.1 Magnetic Oxides

• Calculation of saturation magnetization for NiOFe2O3 (nickel ferrite)

• Unit cell: inverse spinel with 8 Ni2+ ions (B sites) and 16 Fe3+ ions (Asites)

Ion g√J(J + 1) (µB)

Ni2+ 2Fe3+ 5

• The Fe3+ ions have anti-parallel magnetic moments

∴ net 2µB per formula unit due to Ni2+

• For 8 Ni2+ ions per formula unit each with 2µB magnetic moment,

MS =8× 2µB

cell volume=

(16)(9.27× 10−24)

(8.37× 10−10)3= 2.5× 105 A ·m−1

• Calculation of saturation magnetization for Fe3O4 (magnetite)

• Unit cell: inverse spinel with 8 Fe3+ ions on tetrahedral sites, 8 Fe3+ ionson octahedral sites, and 8 Fe2+ ions on octahedral sites.

Ion Site g√J(J + 1) (µB)

Fe3+ Tetrahedral 5Fe3+ Octahedral 5Fe2+ Octahedral 4

• Fe3+ ions on tetrahedral sites have magnetic moments aligned anti-parallel to those of Fe3+on the octahedral sites

∴ net 2µB per formula unit due to Fe2+

• For 8 Fe2+ ions per formula unit each with 4µB magnetic moment,

MS =8× 4µB

cell volume=

(32)(9.27× 10−24)

(8.40× 10−10)3= 5.0× 105 A ·m−1

85

14.4 Domains and Walls 14 FERROMAGNETISM

14.3.2 Magnetization and Hysteresis

• In ferromagnetic materials, the magnetization is a nonlinear function ofthe applied magnetic field.

• Important terms describing the hysteretic behavior:

• HC : coercive field. The applied magnetic field for which the flux densityB disappears

• BS : saturation flux density. The maximum flux density measured whenall of the atomic moments are aligned with the applied magnetic field.

• BR: remnant flux density. The magnetization remaining in the materialafter reaching the saturation flux density and removing the field.

• Ferromagnetic domain: a region in a ferromagnetic material over whichall magnetic moments are aligned

• Domains exist in in the demagnetized state in order to minimize the largemagnetostatic energy associated with single domains

N

S

N S

S N

Magnetized

field lines

N S N S

S N S NClosure

Domains

Demagnetized

Domain Wall

14.4 Domains and Walls

• Domain wall: region between adjacent domains over which the directionof the local magnetic moment changes

• It takes energy to form and to move domain walls

86

14.5 Anisotropy of Magnetization 14 FERROMAGNETISM

Bloch Wall

Wall Region

14.4.1 Energy of Bloch Domain Walls

• Energy U (Heisenberg Model):

U = −2Jex~Si · ~Sj

• Consider the spin vectors by a classical model, i.e. ~Si · ~Sj = |S|2 cosφ.Take a Taylor series expansion of cosφ about φ = 0

U = −2Jex|S|2 cosφ ≈ −2Jex|S|2(

1− 1

2φ2)

• The angle-dependent exchange energy between two adjacent spins at anangle φ is then

wex = Jex|S|2φ2

• For a rotation of the magnetic moment by π radians in N steps, wex =Jex|S|2(π/N)2. For a Bloch wall of N spins, the total energy is then

Nwex = Jex|S|2π2

N

• Anisotropy energy limits the Bloch wall size since the spins are arrangedaway from the direction of easy magnetization. (The more that the spinspoint the “wrong way,” the higher the energy).

14.5 Anisotropy of Magnetization

• Alignment of magnetic moments is more favorable for certain crystallinedirections, resulting in anisotropic energy term

• Under an applied field, domains rotate to align with the field

• The hysteretic response is dependent on the direction of the applied fieldwith respect to the crystalline direction

• “Soft” direction: crystalline direction with low coercivity (HC)

87

14.6 Ferrimagnetic Ordering 14 FERROMAGNETISM

• “Hard” direction: crystalline direction with high coercivity (HC)

[111]

[100]

Co

Basal plane

Parallel to C axis

14.5.1 Anisotropy of Magnetization

• Rotation of domains changes the overlap of the electron clouds of neigh-boring atoms, resulting in different exchange energy Jex

• Anisotropy energy Uk

Uk = k′

1 sin2 θ + k′

2 sin4 θ

• Easy axis depends on values of k′

1 and k′

2

ElectronClouds

AppliedField

M

c-axis

Basal Plane

14.6 Ferrimagnetic Ordering - Exchange Terms

• Exchange term for neighboring Fe atoms is negative for atoms on differ-ent sublattices and positive for atoms on the same sublattice.

• Different sign of exchange term and different number of Fe atoms perunit volume on each sublattice account for ferrimagnetic ordering inFe3O4

• Exchange terms for A and B sublattices:

JAA > 0 JAB < 0 JBB > 0

Sublattice B

Sublattice A

88

15 OPTICAL MATERIALS

14.6.1 Exchange Terms and SusceptibilityParamagnetism Antiferromagnetism

Ferromagnetism

��

��

TC ≡ Curie temperatureTN ≡ Neel temperature

14.6.2 Structure Dependence of Jex

Cr (bcc): Antiferromagnetic Fe (bcc): Ferromagnetic

Ferromagnetism

Antiferromagnetism

GdNi

CoFe

Mn

+

_Distance between atoms

Fe (FCC)

15 Electro-optic and Nonlinear Optical Materials

• Nonlinear optical materials have a nonlinear relation between the polar-ization density (P ) and electric field (E)

• Expand electric susceptibility in higher order terms

89

15.1 Frequency Doubling 15 OPTICAL MATERIALS

P = ε0

[χ(1)E + χ(2)E2 + χ(3)E3

]�

�

�

�χ(1) ≡ Linear susceptibility

χ(2) ≡ Quadratic susceptibility

χ(3) ≡ Cubic susceptibility

• χ(2) is nonzero in non-centrosymmetric materials. Examples of χ(2) ef-fects: linear (Pockels) electro-optic effect, second harmonic generation(frequency doubling)

• The lowest order nonlinearity for cubic materials is χ(3). Examples ofχ(3) effects: quadratic (Kerr) electro-optic effect, third harmonic genera-tion (frequency tripling)

15.1 Frequency Doubling

• In frequency doubling, two photons with energy hν1 combine to formone photon with energy hν2. The total energy is conserved.

• Sending a beam with photon energy hν1 into a nonlinear optical (NLO)material results in two output beams with energies hν1 and hν2.

• Examples:

• Nd:YAG laser with 1.10µm (infrared) radiation NLO−−−→ 0.55µm (green)

• Blue laser by frequency doubling GaAs lasers

Nonlinear Optical (NLO) Material

15.2 Nonlinearity in Refractive Index

• Velocity of light in matter: v = c/n

c ≡ Speed of light in free space (3×108 m/s)

n ≡ Refractive index

• Refractive index (n) in nonlinear material:

90

15.3 Electro-Optic Modulators 15 OPTICAL MATERIALS

n2 = 1 + χ(1) + χ(2) + χ(3) + · · ·

• Dependence of refractive index on light intensity:

n = n0 + n2I�

�

�

�n0 ≡ Linear and χ = εr − 1

n2 ≡ Nonlinear indexI ≡ Intensity

• Dielectric constant: εr ∼= n2

15.3 Electro-Optic Modulators

• The optical properties of non-centrosymmetric crystals are modified byapplying an external electric field.

∆

(1

εr

)= ∆

(1

n2

)= rE

• Expanding n = n(E) in a Taylor series expansion about E = 0, the fol-lowing expression is obtained

∆n = −1

2n3rE

r ≡ Electro-optic coefficient [pm/V]

• Example material: LiNbO3, r = 31 pm/V, n = 2.29. Fields of 106 −107 V/m are easily obtained in micro-photonic devices (1-10 V across∼ 1 µm)

• Other materials: electro-optic polymers (r = 20 − 50 pm/V), BaTiO3

(r = 300 pm/V)

15.4 Optical Memory Devices

• Photorefractivity: change in refractive index by light. Light empties trapstates, creating a space charge region that locally changes the refractiveindex.

• Volume holography:

• Two plane waves (object and reference beams) incident upon a photo-sensitive material. A standing wave is formed which is preserved in thematerial as an interference pattern

91

15.5 Volume Holography 15 OPTICAL MATERIALS

• Material contains intensity and phase information and the object can bereconstructed

• The difference between hologram and a photograph is that phase infor-mation is preserved in a hologram

15.5 Volume Holography

• Pattern in material creates a dielectric grating or spatial modulation inthe refractive index and dielectric constant

εr = εr0 + εr1 cos(2ky sin θ)

k ≡ Magnitude of light wavevector

• Period of dielectric grating:

Λ =2π

2k sin θ=

λ

2n sin θwith k = 2πn/λ

Reference

Object

ReferenceObject

Record Hologram Reconstruct Hologram

15.6 Photorefractive Crystals

• Electro-optic and photoconductive - organic and inorganic materials

• Mechanism for photorefraction:

1. Generation of electrons and holes

2. Transport charges through the material

3. Trap the charges

4. Local field of trapped charges changes the refractive index n

• Spatially varying charge distribution results in spatially varying field.Field changes n through electro-optic coefficient r; larger r, larger pho-torefractive effect.

∆n = −1

2n3rE

92

15.7 Photorefractive Crystals 15 OPTICAL MATERIALS

VcrystalIncidentBeams

donorDefect statesin the band gap

light no light

+

15.7 Photorefractive Crystals

Steps in creation of periodic dielectric constant

LightIntensity

ElectronDensity

Donor Concentration

Net ChargeDensity

ElectricField

DielectricConstant

15.8 Phase Conjugation - All Optical Switching

beam 1

beam 4

beam 3beam 2

• Beams 1 and 2 are pump beams

• Beam 4 is a probe beam

• Beam 3 is the phase conjugate beam

• Diffraction grating is created by beams 1 and 4 through interference andthe photorefractive effect

• Beam 2 is diffracted by the grating, forming beam 3

Beam 3 Beam 2

phase-conjugate beam

Beam 4(Probe)

Beam 1

93

15.9 Acousto-Optic Modulators 15 OPTICAL MATERIALS

15.9 Acousto-Optic Modulators

• Photoelastic effects: strain causes a change in the refractive index

∆

(1

εr

)= pS and ∆

1

n2= pS

�

�

�

�S ≡ Strainp ≡ Photoelastic constantn ≡ Refractive index

• Acousto-optic effect is used in Bragg cells to create a grating with periodΛ by launching an acoustic wave. Light is diffracted from the grating.

Λ =λ

2n sin θ

InputBeam Diffracted

Beam

Acoustic Wave("Sound")

InputBeam

Transmitted Beam

Acousto-OpticMaterial

Acoustic wave off: Beam transmitted

Acoustic wave on: Beam diffracted

15.10 Integrated Optics

• Integrated optics: passive and active optical elements incorporated ontoa wafer of material for local manipulation of light.

• Passive elements: elements that do not require power, such as waveg-uides, couplers, and filters

• Active elements: elements that require power, such as switches, modu-lators, directional couplers, amplifiers, lasers

• Photonics: analogue of electronics. Photonics encompasses the controlof photons.

94

15.11 Dielectric Waveguides 15 OPTICAL MATERIALS

15.11 Dielectric Waveguides

• In integrated optics, dielectrics are used to guide light in waveguides.Waveguides act as “optical wires”

• Light is confined to the region of highest refractive index

• Why dielectric materials instead of metals? Ease of fabrication and lowerlosses

Types of Dielectric Waveguides

Slab Ridge Fiber

Darker shading = higher index

15.12 Example: Phase Shifter

• An electro-optic dielectric waveguide confines and guides light

• An electric field is applied to electrodes surrounding the waveguide.The refractive index of the dielectric waveguide is then modified by theelectro-optic effect

∆n = −1

2n3rE = −1

2n3r

V

d

• Phase delay of optical wave due to the electro-optic effect:

∆φ =2π

λ∆nL

Ti diffused waveguide

Au Electrode

V- +

LiNbO3 Substrate

95

15.13 LiNbO Phase Shifter 16 SUPERCONDUCTIVITY

15.13 Example: LiNbO3 Phase Shifter

• Electric field of 106 V/m (10 V across 10µm) yields ∆n = 1.86× 10−4 forLiNbO3

• For a phase shift ∆φ = π with operation at λ = 1.5µm,

∆φ = π =2π

λL∆n

L = 0.4 cm

• Long interaction lengths are needed; materials with higher electro-opticcoefficients are needed for more compact devices

• The voltage required for a π phase shift is the half-wave voltage Vπ

Vπ =λd

n3rL

16 Superconductivity

• Superconductivity: a phenomena in which a material undergoes a tran-sition to zero resistivity with zero magnetic induction (Meissner effect)below a critical temperature TC .

• At temperatures below TC application of a critical magnetic fieldHC willcause a transition to the normal state

Metal

Superconductor

2 4 6

800

Pb

Tl

8

600

400

200

Resistivity Critical Field

16.1 BCS Theory of Superconductivity

• BCS Theory: quantum mechanical theory of superconductivity formu-lated by Bardeen, Cooper, and Schreiffer in 1957 (Nobel Prize in 1972).

• Electron-phonon coupling leads to the formation of electron pairs(Cooper pairs)

• BCS Theory accounts for observed properties of superconductivity, in-cluding the energy gap, Meissner effect, critical temperature, and quan-tization of magnetic flux through a superconducting ring

96

16.2 Density of States and Energy Gap 16 SUPERCONDUCTIVITY

16.2 Density of States and Energy Gap

• The energy gap in superconductors is caused by electron-electron inter-actions rather than electron-lattice interactions for dielectrics

• The energy gap is maximum at 0 K and decreases to zero at the transitiontemperature.

Density of States

1.0

1.0

Temperature Dependence of

16.3 Heat Capacity of Superconductors

• The entropy of superconductors decreases upon cooling below TC , re-sulting in electronic ordering.

• In the superconducting state, for T � TC ,

HC = H0

[1− 1.07(T/TC)2

]

Normal

Heat Capacity of Gallium

Superconducting

1.0 1.50 0.5

0.5

1.0

1.5

97

16.4 Superconductor Junctions 16 SUPERCONDUCTIVITY

16.4 Tunneling in Superconductor Junctions

V = 0

Metal Metal

Insulator

V > 0

-+I

V

I-V Characteristic

Metal-Insulator-Metal

Metal

Superconductor-Insulator-Metal

V > 0V = 0

I

V

I-V Characteristic

InsulatorSuperconductor

16.5 Semiconductor-Insulator-Semiconductor Junction

• Tunneling across a thin oxide junction between two superconductorswith energy gaps 2∆1 and 2∆2

• Nobel Prize in physics, Ivan Giaever, 1973

Insulator

Superconductor 1 Superconductor 2

I

V

Negative resistance

I-V Characteristic

16.6 Type I and Type II Superconductors

• Type I: superconductor is diamagnetic in an applied magnetic field up tothe critical field Hc

• Type II: superconductor exhibits two magnetic behaviors. Up to a lowercritical field HC1 the superconductor is diamagnetic, and between HC1

and an upper critical field HC2, the flux density is not equal to zero (“in-complete” Meissner effect).

• Type II superconductors tend to be alloys or transition metals with highroom temperature resistivity

98

16.7 High TC Superconductors 16 SUPERCONDUCTIVITY

• Type II superconductors are used in high field magnets, magnetic reso-nance imaging (MRI) applications

Type 1

H

-M -M

H

Type II

16.7 High TC Superconductors

• Highest values for TC have been reported in cuprates

• 2 layer cuprate compounds support 2D conduction

BiSrCaCoOYBaCuO

16.8 Cuprate Superconductors

• Theory: pairing is involved. Nature of pairing? Spin Waves?

• 2∆(0) ≈ 6kTC

• Future applications: current carrying wires, Maglev trains

Phase Diagram for Cuprates

0.1 0.2

Paramagneticmetal

Semiconductor

Strangemetal

Antiferromagnetic

insulator

Hole Doping Level

Superconductor

Temperature

99

17 351-2 PROBLEMS

17 351-2 Problems

1. An abrupt Si p-n junction hasNa = 1018cm−3 on one sideandNd = 1015cm−3on the other.

(a) Calculate the Fermi level position at 300K on both sides.

(b) Draw an equilibrium band diagram for the junction.

(c) Determine the contact potentialΦo for this junction.

2. A siliconp+ − n junction10−2cm2 in area hadNd = 1015cm−3 doping onthe n-side. Calculate the junction capacitance with a reverse bias of 10V.

3. For metallic aluminum, calculate:

(a) The valence electron density.

(b) The radius of the Fermi spherekF.

(c) Fermi energy in eV.

4. From the Schrodinger equation for a quantum well, show that the wavevector is equal tonπ/L where L is the well width.

5. Calculate the energy of light emitted from a 10 nm wide AlGaAs/GaAsquantum well structure that is photoexcited with 2.5 eV laser light.

6. What is the luminescent energy for a CdSe quantum dot with a 2 nmradius.

7. For a MOSFET device briefly describe how the three types of devicework: a) enhancement mode b) depletion mode c) inversion mode.

8. Calculate the capacitance of an MOS capacitor with a 10 nm thickHfO2

dielectric oxide. What is the ratio of capacitances forCHfO2/CSiO2

. Therelatie dielectric constant forHfO2is 25.

9. Problem 9.9 in Solymar and Walsh




13. Consider a quantum cascade laser (QCL) made from GaAs and GaAlAs.What well thickness is needed for laser emission at 3 microns?

14. Derive the expression for the average value of the dipole moment. Showthat it is given by:

< µ >= µ[cotha− 1a ]

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18 351-2 LABORATORIES

15. The saturation polarizationPs ofPbTiO3, a ferroelectric, is0.8

coulombs/m2. The lattice constant is 4.1A. Calculate the dipolemoment of unit cell.

16. Calculate the polarization P of one liter of argon gas at 273 K and 1 atm.The diameter of an argon atom is 0.3 nm.

17. Consider the frequency dependence of the atomic polarizability. The po-larizability and its frequency dependence can be modeled as a dampedharmonic oscillator. Derive the expression forα in this case.

The expression is given by:

m dxdt2 + bdxdt + ω2

0x = −eεlocsinωt

Plotα vs.ω for this case.

18. Problem 4.6 in Solymar and Walsh.




22. Calculate the magnetic susceptibility of metallic copper. How does itcompare to the measured value of -1.0?

23. Calculate the effective magneton number p forMn2+,Co2+. Show work.

24. Consider Mn doped GaP. There are1020Mn2+ ions.

25. What is the electron configurationMn2+ in spectroscopic notation.

(a) Calculate its magnetic moment at saturation in Bohr magnetons.

(b) Calculate its magnetic susceptibility.

26. For metallic Co, which has a Curie temperature of 1388 K, calculate theWeiss constantλ. Calculate the exchange constant in meV.

18 351-2 Laboratories

18.1 Laboratory 1: Measurement of Charge Carrier TransportParameters Using the Hall Effect

18.1.1 Objective

The purpose of this lab is to measure the electronic transport properties ofsemiconductors and semiconducting thin films using the ECOPIA Hall appa-ratus.

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18.1 Laboratory 1: Measurement of Charge Carrier Transport ParametersUsing the Hall Effect 18 351-2 LABORATORIES

Figure 18.1: Part 7; B is uniform throughout the area of interest

18.1.2 Outcomes

Upon completion of the laboratory, the student will be able to:

1. Use a Hall effect apparatus to measure the mobility and carrier concen-tration in a semiconductor.

2. Derive the equations that enable the extraction of fundamental materialsparameters using the Hall effect.

3. Describe the dependence of mobility on carrier concentration and tem-perature, and explain the origins of differences in mobilities between dif-ferent semiconductors.

18.1.3 References

(1) M. Ali Omar, Elementary Solid State Physics;(2) Solymar & Walsh, Electrical Properties of Materials;(3) MSE 351-1 Lecture Notes; and(4) the NIST web page: http://www.nist.gov/pml/div683/hall.cfm

18.1.4 Pre-Lab Questions

1. What is the Lorentz Force?

2. What is the Hall effect and when was it discovered?

3. Write the equation describing the force, FM, on a particle of charge q andwith velocity v in a uniform magnetic field, B.

4. Does the velocity of a charged particle (with non-zero initial velocity) ina uniform magnetic field change as a result of that field? If so, how? Doesits speed change?

5. What is the right-hand-rule?

6. For a particle with negative charge, q, in the situation below, in whatdirection will the particle be deflected?

(a)

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P = + P = -M = M =N = N =

Figure 18.2: Part 6

Figure 18.3: Part 9

7. For the example below, what will be the sign of the charge built up onthe surfaces (M) and (N) if the particle P is charged (+)? if it is (-)? (B isinto the page and uniform throughout the specimen.)

(a)

8. By convention, current is defined as the flow of what sign of charge car-rier?

9. What is the electric field, E, in the situation below? What is the electro-static force, FE on the particle if it has a charge q?

(a)

10. Why are both a resistivity measurement and a Hall measurement neededin order to extract fundamental material parameters?

18.1.5 Experimental Details

The samples to be characterized include:

1. “bulk” Si, GaAs, InAs (i.e. substrates ~ 400 microns thick)

2. “thin films” of InAs and doped GaAs, grown on semi-insulating GaAssubstrates

3. Indium tin oxide (ITO) on glass.

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You may have the opportunity to make additional samples. For contacts onn-type GaAs, use In-Sn solder; for contacts on p-type GaAs, use In-Zn solder.Most samples are mounted on mini-circuit boards for easy insertion into theapparatus.

18.1.6 Instructions/Methods

See Instructor

18.1.7 Link to Google Form for Data Entry

https://docs.google.com/forms/d/1lrAokPI1vIJ-a8pmLx4FVCqHh80PKgGr8oDD6eVzx_w/viewform

18.1.8 Lab Report Template

1. Balance the forces (magnetic and electric) acting on a charged particle in aHall apparatus to derive the equation that describes the Hall Coefficientin terms of the applied current and magnetic field and measured Hallvoltage. Show your work. See hints at the end.

2. Apply data from a sample measurement to test the equation derived in(1). Show your work.

3. How is the carrier concentration related to the Hall Coefficient? What isthe difference between bulk and sheet concentration?

4. Calculate the carrier concentration (bulk and sheet) using the sampledata and the equations derived above. Show your work.

5. Does the mobility exhibit any dependence on the carrier concentration?Discuss briefly; include observations from lab for the same material &type (e.g. n-GaAs).

6. What is the origin of the difference in mobilities between the n-type andp-type samples, assuming that the doping levels are similar?

7. How do carrier mobilities compare for different materials? Use thepooled data to compare mobility as a function of material (as well ascarrier type). Explain your observations.

8. When the magnetic field, current, and sample thickness are known, thecarrier concentration and type may be determined. Conversely, if thecurrent, sample thickness and carrier concentration are known, the mag-netic field may be determined. A device that measures these parameters,known as a Hall Probe, provides a way to measure magnetic fields. Write

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Figure 18.4: Example

an expression that relates these parameters. Which is the more sensitive(higher ratio of mV/Tesla) Hall probe – the bulk InAs or bulk GaAs sam-ple? Show your work. (Note – you should have recorded average Hallvoltage for a given current. How do these compare?)

18.1.9 Hints for derivation

1. For the example below, what VH would you need to apply to make theparticle continue on in a straight line throughout the sample? (B is uni-form throughout the specimen. Hint: Balance the magnetic and electricforces on the particle.)

2. The current density, J , can be expressed in two ways: J=i/(A), andJ=nqv, where i is the total current passing through a cross-sectional areaA, n is the concentration of carriers per unit volume in the material pass-ing the current, q is the charge on each carrier, and v is the drift velocityof the carriers. If you knew i, A, VH, d, B, and q from the situation illus-trated above, what expression tells you n?

3. The “Hall coefficient” of a material, RH , is defined as the Hall electricfield, EH , per current density, per magnetic field, RH=EH/(J*B) Usingthe equations given and derived thus far, express the Hall coefficient interms of just q and n.

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18.2 Laboratory 2: Diodes 18 351-2 LABORATORIES

18.2 Laboratory 2: Diodes

18.2.1 Objective

The purpose of this lab is to explore the I-V characteristics of semiconductordiodes (including light emitting diodes (LEDs) and solar cells), and the spec-tral response of LEDs and lasers.

18.2.2 Outcomes


1. Measure diode I-V characteristics and relate them to band diagrams.

2. Fit I-V data to the diode equation, extract relevant parameters, and relatethese to materials constants.

3. Determine the open circuit voltage and short-circuit current of the solarcell.

4. Describe the dependence of emission wavelength on bandgap and de-scribe origins of spectral broadening.

5. Describe how lasers differ from LEDs in design and performance.

18.2.3 Pre-lab Questions

1. What expression describes the I-V characteristics of a diode?

2. Sketch the I-V characteristics of a diode and label the sections of the curvecorresponding to zero bias (1), forward bias (beyond the built-in volt-age) (2), and reverse bias(3), and then sketch the corresponding banddiagrams for 1, 2, and 3.

3. Sketch the I-V characteristics of a Zener diode and the correspondingband-diagram for reverse bias.

4. Sketch the I-V characteristics of a p-n junction with and without illumi-nation. Label VOC and ISC .

5. Take pictures (use your phone) of lights around campus; try to get pic-tures of LEDs. Which do you think are LEDs (why?)?

6. Why are fluorescent lights “white?” How white are they?

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Experimental Details

The devices to be characterized include:

1. Si diode

2. Zener diode

3. LEDs (different colors)

4. Si solar cell

18.2.4 References

MSE 351-2 Lecture Notes, Omar Chapter 7, Solymar & Walsh Chapter 9, 12,13.

Instructions/Methods

Use multimeters and the Tektronix curve tracer for I-V measurements. Use theOcean Optics spectrometer to obtain spectral responses.

Station 1 - CURVE TRACER

p-n junction diode

Attach diode to “diode” slot on Curve Tracer. Measure both forward andreverse bias characteristics.

1. (In lab) Measure and record the I-V (current-voltage) characteristics ofthe silicon diode over the current range 2μA to 50 mA in the forwardbias condition. Pay particular attention to the region from 200 to 800 mV.

2. (In lab) Measure the I-V characteristics of the diode in the reverse biascondition.

3. (Post-lab) The ideal diode equation is: I = Isat[exp(qV/kT ) − 1]. Notethat for V>>kT/q, I = Isat exp(qV/kT ); however recombination of car-riers in the space charge region leads to a departure from ideality by afactor m, whereI = I0[exp(qV/mkT )− 1] .

(a) Plot the data using ln(I) vs. V plot to determine m.

(b) Use the value of the applied voltage corresponding to ~ 1mA tosolve for I0 (which is too small to measure in our case.)

4. (In lab) Repeat the forward bias measurement using “Store.” Now coolthe diode and repeat. Sketch the two curves. (Post-lab) Explain what youobserve.

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5. Test red and green LEDs. Record the color and the turn-on voltage.

6. (Post-lab) Compare the I-V characteristics for the silicon diode and LEDs.Why would you choose Si over Ge for a rectifier? (Omar 7.21)

7. (In-Lab) Test the Zener diode. Sketch. (Post-lab) Compare to figure 9.28(S&W).

Station 2: Solar Cells

1. Use the potentiometer, an ammeter and voltmeter to determine the I-V characteristics of a solar cell at different light levels. Determine thevalues of Voc (the open circuit voltage) and Isc (the short-circuit current)under ambient light, then use the potentiometer to determine additionalpoints on the I-V curve. Record the results.

2. Repeat the I-V measurements for a higher light level. Record the lightintensity measured with the photodiode meter. Area~1 cm2 : ____ Mea-sure the area of the solar cell:______

3. (Post-Lab) Estimate the fill – factor and the conversion efficiency of thesolar cell.

Station 3: Spectrometer and Power Supply

Light Emitting Diodes

1. Measure the spectral response (intensity vs. wavelength) of the lightemitting diodes using the spectrometer.

2. Cool the LEDs and observe the response. Qualitatively, what do the re-sults suggest about the change in Eg vs. temperature? About emissionefficiency vs. temperature? How do these results compare to the I-Vresponse of the cooled silicon diode?

3. Record the peak-wavelength and the full-width-half-maximum (FWHM)for each diode.

4. Calculate the bandgap that would correspond to the peak wavelength.

Semiconductor Lasers

1. Attach the laser to the power meter and setup the power meter to mea-sure intensity from the device. Slowly increase the voltage and recordthe I-V characteristics vs. power output for the laser.

2. Measure the spectral response of the laser: i) below threshold, and ii)above threshold. Note the FWHM of the peaks. How do they differ?How do they correspond to the I-V-power data?

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LED Color Peakwave-length

Eg(in eV,frompeak

wave-length)

Intensity FW(left)

FW(right)

FWHM

Laser

Table 18.1: Spectrometer and Power Supply

3. Plot the light output (intensity) as a function of current. What is thethreshold current? Label the regions of spontaneous and stimulatedemission.

4. Calculate the efficiency of the laser.

5. Determine the bandgap of the laser material.

6. Explain the change in the width of the spectral emission.

Station 4: Stereomicroscope

1. Sketch the structure of the LED observed under the stereomicroscope.Note the color of the chip when the device is off and the color of emissionwhen the device is on. Sketch what the device structure might look like.

2. Sketch the structure of the laser observed under the stereomicroscope.Sketch what the device structure might look like.

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18.3 Laboratory 3: Transistors 18 351-2 LABORATORIES

18.3 Laboratory 3: Transistors

18.3.1 Objective

The purpose of this lab is explore the input/output characteristics of transis-tors and understand how they are used in common technologies.

18.3.2 Outcomes


1. Measure the output characteristics of a few important transistors using a“curve tracer.”

2. Qualitatively relate the characteristics to the p-n junctions in the devices.

3. Describe transistor function and performance in terms of appropriategains.

4. Identify applications of these devices in common technologies.

18.3.3 Pre-lab questions

Bipolar Transistor

1. Sketch and label the band-diagrams for npn and pnp transitors.

2. Sketch I-V characteristics for this device as a function of base current.

3. In what technologies are these devices used?

MOSFETS

4. Sketch and label a MOSFET structure.

1. Sketch I-V characteristic for this device as a function of gate voltage. Inyour sketch of the ISource−Drain–VSource−Drain characteristics vs. gatevoltage, Vgate, label the linear and saturated regions.

2. What is threshold voltage? What structural and materials properties de-termine the threshold voltage?

3. In what technologies are these devices used?

“Current Events”

1. What materials developments have changed transistor technology in thepast decade? (Hint: Search Intel high-k; Intel transistors)

2. What new types of devices are on the horizon? Hint: search IBM nan-otubes graphene

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The devices to be characterized include the following

1. pnp and npn bi-polar junction transistors

2. phototransistor

3. metal oxide semiconductor field effect transistor (MOSFET).

References MSE 351-2 Lecture Notes, Omar Chapter 7; Solymar and WalshChapter 9

Instructions/Methods Use the Tektronix curve tracer for I-V measurements.

1. Bipolar Transistors (npn and pnp)

(a) Attach an npn transistor to the T-shaped Transistor slot on theCurve Tracer (making sure that E, B, & C all connect as indicatedon the instrument). Set the menu parameters to generate a familyof I-V curves. Keep the collector-emitter voltage VCE below 30V toavoid damaging the device.

(b) Sketch how E,B,C are configured in the device.

(c) Measure Icollector (output current) for VCE > Vsaturation as a func-tion of base (input) current, Ibase.

(d) Post-Lab: determine the transistor gains, α = −Icollector/Iemitterand β = Icollector/Ibase.

(e) Repeat for a pnp transistor.

2. Phototransistor

(a) Measure the I-V characteristics of the phototransistor under varyingillumination intensity (using the microscope light source). Comparequalitatively to what you observed for the pnp and npn transistors.

(b) Post-lab: discuss the mechanism by which the light influences thedevice current, and compare with the bipolar transistor operation.

3. MOSFETS

(a) Attach MOSFET Device to the linear FET slot on the Curve Tracer,making sure that S,G,D are connected appropriately. Observe boththe linear and saturated regions for ISource−Drainvs.VSource−Drain.

(b) Vary the gate voltage step size and offset to estimate the thresholdvalue of the gate voltage (the voltage at which the device turns on).

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Figure 18.5: n-Channel MOFSET.

(c) Measure the VSource−Drain and ISource−Drain to extractRSource−Drain in the linear region as a function of VGate. Youshould take at least 4 measurements.

(d) Post-lab, using the data from B, plot the conductance of the channel,GSource−Drain = 1/RSource−Drain, vs. VGate . You should observe alinear relationship following the equation ,

GSD =

(µnW

L

)(COXA

)(VG − VT )

where µn is the electron mobility and W, L, and A are the width,length and area of the gate respectively (A = W x L). COX is the ca-pacitance of the oxide, which can be measured using an impedanceanalyzer as a function of gate voltage. A plot is shown in Figure18.5.

(e) Using µn=1500 cm2/V-sec, C = 4.8x10−10F (from the attached plot),plot of conductance vs. VGate, calculate L, the length of the gate.

(f) Determine Vthreshold from the above values. Compare with yourobservations from the curve tracer (part a).

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18.4 Laboratory 4: Dielectric Materials 18 351-2 LABORATORIES

18.4 Laboratory 4: Dielectric Materials

18.4.1 Objective

The objectives of this lab are to measure capacitance and understand the de-pendence on geometry and the dielectric constant, which may vary with tem-perature and frequency.

18.4.2 Outcomes


1. Use an impedance analyzer to measure capacitance.

2. Given the capacitance of a parallel plate capacitor, calculate the dielectricconstant.

3. For a known material, explain the microscopic origins of the temperaturedependence of the dielectric constant.

4. Understand how the dielectric constant and the index of refraction arerelated. Use the reflectance spectroscopy to fit the thickness, index ofrefraction and extinction coefficient of several thin films. Explore howthe index of refraction is affected by composition and how this informsdesign of heterostructure devices.

18.4.3 Pre-lab questions

1. What distinctions can you make between capacitance and the dielectricconstant? What are the units of each?

2. What is the relationship between the dielectric constant and the ‘relative’dielectric constant?

3. What is the dielectric constant (or permittivity) of ‘free space’ (or vac-uum)?

4. What is the relative dielectric constant of air? Water? Glass? SiO2? Ex-plain their relative magnitudes.

5. For the capacitor shown in Figure 18.6, what is the expression that relatesthe capacitance and relative dielectric constant?

6. What relationship determines the amount of charge stored on the platesof a capacitor when a specific voltage is applied to it?

7. With the above answer in mind how could you measure the capacitanceof a structure?

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Figure 18.6: Capacitor.

8. Name two materials for which the capacitance (charge per unit voltage)is fixed. Name a material type or structure for which the capacitance isdependent on V.

9. What is the relationship that gives the amount of energy stored on a ca-pacitor?

10. Why are batteries the primary storage medium for electric vehicles,rather than capacitors?

11. Intel and their competitors are interested in both “low-k” and “high-k”dielectrics.

(a) What defines the boundary between “low” and “high?”

(b) What drives the need for “high-k” dielectrics? What other proper-ties besides the dielectric constant are important characteristics ofthese materials?

(c) What drives the need for “low-k” dielectrics?

12. Find one or two additional examples of technologies/devices that in-corporate capacitors, and explain the function of the capacitor in thatcontext.


The dielectric constant of a number of different materials will be probedthrough measurements of parallel plate capacitance. Capacitors of solid in-organic dielectrics (in thin slabs) is accomplished through the evaporation ofmetal contacts on either side of the material. The total capacitance of a p-njunction diode is measured through the device contacts. The capacitance of liq-uids is determined by filling a rectangular container between two electrodes,and neglecting the contributions of the container.

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Material Area ThicknessCapacitanceεr(lab) εr(lit)Glassslide(sam-

ple1)

Lithiumniobate(LiNbO3)

Glasssheet(sam-

ple2)

Plexiglass

Table 18.2: Dielectric Constants

18.4.5 References

Solymar and Walsh Chapter 10, Omar Chapter 8, Kittel Chapter 16

18.4.6 Instructions/Methods

Use the HP Impedance analyzer to measure the materials provided in lab.Note the sample dimensions and the measured capacitance values on the at-tached table.

18.4.7 Lab Report Template

Part I - Dielectric constants; Measuring capacitance with Impedance Ana-lyzer

1. Using the default frequency of 100kHz and zero bias voltage, measurecapacitance and contact dimensions and then calculate the relative di-electric constant of the following materials:

2. Using the default frequency of 100 kHz, measure capacitance vs. voltagefor several LEDs

Determine the built-in voltage of the corresponding p-n junction, ϕ0, foreach diode, and discuss the difference. See equation 7.64, pg. 362, Omar.Note that the equation may be re-written as follows:

1

C2= 2

(ϕ0 − V0)(Na −Nd)εeNaNd

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LED= LED= LED=Bias CapacitanceBias CapacitanceBias Capacitance

Table 18.3: Capacitance

Plot 1/C2vs. V0, and extrapolate the linear region (reverse bias) to the x-intercept to determine the value of interest. Note that the slope dependson the doping concentration.

3. Calculate the relative dielectric constant, based on capacitance measuredin lab for:

(a) air (4.0inch x 3.5 inch plates, separated by _________ (measure this))

(b) water at room temperature (4.0 inch x 3.5 inch plate par-tially submerged; measured dimension: Height:________ xwidth___________x separation____________________

(c) ethanol

4. Fill in the table below, for water at the different temperatures measuredin lab, using the dimensions measured above. Determine the corre-sponding values of the relative dielectric constant,εr. Plot (εr–1)/(εr+2)vs 1/T (Kelvin) for both sets of data below, and compare to figure 8.13(p. 388, Omar).

Class data:

Area = __________; separation = __________

Temp (C) Freq C (pF) εr

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Comparison from CRC handbook

Temp (C) εr0 87.74

20 80.1440 73.1560 66.880 61100 55.65

5. Measure ice as a function of frequency & compare to Omar figure 8.10.Discuss.

Dimension: Height:____ x width_____x separation___

Freq C(pF)

εr Freq C(pF)

εr

6. Practical applications: inspect the temperature/humidity meter. Lookfor the devices used to measure each. Measure the stand-alone capacitorthat resembles that in the meter as a function of exposure.

Part II - Reflectance Spectroscopy of Thin Films

1. Measure the reflectance spectra to determine n, k and thickness of:

• ~ micron thick ~Al(0.3)Ga(0.7)As on GaAs substrate

• ~ 1 micron thick ~ Al(0.6)Ga(0.4)As on GaAs substrate

• SiO2 on Si substrate

• Si3N4/Si substrate

Compare the index of refraction of the two different AlGaAs samples. Whatdoes this indicate about how you would “build” a waveguide?

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18.5 Laboratory 5: Magnetic Properties 18 351-2 LABORATORIES

18.5 Laboratory 5: Magnetic Properties

Objective: The objectives of this lab are to measure the response of a materialto an applied magnetic field and understand the atomic origins of macroscopicmagnetic behavior.

Outcomes: Upon completion of the laboratory, the student will be able to:

1. Use a magnetometer to measure magnetic response with applied field.

2. Given the saturation magnetization, solve for the number of Bohr mag-netons.

3. For a known material, explain the microscopic origins of magnetism.

4. Distinguish qualitatively between “hard” and “soft” ferromagnetic ma-terials.

Experimental Details The purpose of this experiment is to investigate theresponse of different materials to an applied magnetic field, H. Plotting themeasured response, B, vs. the applied field, H, indicates the type of magne-tization possible, i.e. whether the material is diamagnetic, paramagnetic orferromagnetic.

References Solymar and Walsh Chapter 11, Omar Chapter 9

1. Magnetization of metal wires Qualitatively observe the hysteresis be-havior of the following metals. Indicate whether they are ferromagnetic ornot, and why. (Hint: Look at which electronic shells are filled/unfilled in eachcase.)

Metal wires:

Hysteresis?Y/N

3d 4s

CuFeCoNiW

SteelStainless Steel

Quantify the observations on the metal wires (Fe, Co, Ni, and W 1 mm diam-eter and Cu 0.5 mm diameter – the cross sectional area must be entered in the

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B-probe dimensions). Record hysteresis curves in the “intrinsic” mode (with-out background subtraction). Also record “initial” curves for the first threemetals. Fill in the following table.

Bsat observed Bsat literature # Bohr magnetons, nB∗Fe 2.158 TCo 1.87 TNi 0.616 T

Question 1(a-c): Solve for the number of Bohr magnetons per atom in Fe, Niand Co. (Use experimental Bsat if the sample saturates; otherwise use litera-ture value.)

Note: µ0 = 4π × 10−7T −m/ABS = µ0MS ;Msin A/m or OeMS = nMA; n = # atoms/volumeMA = nBµB ;µB = e~

2me = 9.27× 10−24A−m2 =Bohr magneton

Question 2: Hard magnet The sintered CuNiFe pellets provide an exampleof a “hard” magnet. Record (plot) a hysteresis curve and compare to the “soft”ferromagnetic materials. Comment on the observed behavior.

2. Faraday Rotation

Set-up:

• Red laser pointer (operated under 3.5V!!! and 40 mA!!!)

• Glass rod (inside the solenoid coil): Diam = 5mm, length = 10 cm, SF-59.

• Solenoid coil: L = 150 mm, turns/layer = 140, layers = 10, DC Resist = 2.6Ohm

• B = 11.1 mT/A x I, where I is in Amperes. Maximum current is 3 Amps!

• Detector: Photodiode in series with 3 resistors.

Measure the intensity of the laser vs. rotation of the polarizer with zeromagnetic field. (Measure and record the intensity every two degrees or so.)Apply a magnetic field and re-measure the intensity vs. rotation. Plot onthe same graph to compare. Find the difference in angle (average) betweenthe on/off magnetic field states. Calculate the Verdet constant, V , whereθ(radians) = V B`,B = magnetic field strength (millitesla), and `= length glassrod = 10 cm.

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3. Demonstration Heating a magnetized iron wire is one way to observea second order phase transition involving the spin degree of freedom (uponcooling, one observes the onset of spontaneous magnetization). What is thistransition called? Estimate the value of the transition temperature for this sys-tem and compare to literature values.

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Documents

351-2: Physics of Materials II - Northwestern Universitymsecore.northwestern.edu/351-2/351-2text.pdf · or equivalent; PHYSICS 135 2,3. 2 Course Outcomes 3 351-2: Solid State Physics