3512: Physics of Materials II
Bruce Wessels and Peter GirouardDepartment of Materials Science and EngineeringNorthwestern University
1 Catalog Description (3511,2)
Quantum mechanics; applications to materials and engineering. Band structures and cohesive energy; thermal behavior; electrical conduction; semiconductors; amorphous semiconductors; magnetic behavior of materials; liquid crystals. Lectures, laboratory, problem solving. Prerequisites: GEN ENG 205 4 or equivalent; PHYSICS 135 2,3.
2 Course Outcomes
3 3512: Solid State Physics
At the conclusion of 3512 students will be able to:
 Given basic information about a semiconductor including bandgap and doping level, calculate the magnitudes of currents that result from the application of electric fields and optical excitation, distinguishing between drift and diffusion transport mechanisms.
 Explain how dopant gradients, dopant homojunctions, semiconductorsemiconductor hetero junctions, and semiconductormetal junctions perturb the carrier concentrations in adjacent materials or regions, identify the charge transport processes at the interfaces, and describe how the application of an electric field affects the band profiles and carrier concentrations.
 Represent the microscopic response of dielectrics to electric fields with simple physical models and use the models to predict the macroscopic polarization and the resulting frequency dependence of the real and imaginary components of the permittivity.
 Given the permittivity, calculate the index of refraction, and describe how macroscopic phenomena of propagation, absorption, reflection and transmission of plane waves are affected by the real and imaginary components of the index of refraction.
 Identify the microscopic interactions that lead to magnetic order in materials, describe the classes of magnetism that result from these interactions, and describe the temperature and field dependence of the macroscopic magnetization of bulk crystalline diamagnets, paramagnets, and ferromagnets.
 Specify a material and microstructure that will produce desired magnetic properties illustrated in hysteresis loops including coercivity, remnant magnetization, and saturation magnetization.
 Describe the output characteristics of pn and Schottky junctions in the dark and under illumination and describe their utility in transistors, light emitting diodes, and solar cells.
 For technologies such as cell phones and hybrid electric vehicles, identify key electronic materials and devices used in the technologies, specify basic performance metrics, and relate these metrics to fundamental materials properties.
4 Principles of Semiconductor Devices
Recall that the conductivity of semiconductors is given by
where $n$ is the number of electrons, $e$ is the electronic charge, $\mu $ is the mobility in ${cm}^{2}/V/s$. The following trends for conductivity versus temperature are noted for metals, semiconductors, and insulators:
 Metals: $n$ and $p$ are constant with temperature. Mobility is related to temperature as $\mu \propto {T}^{a}$, where $T$ is temperature and $a$ is a constant. Conductivity decreases with increasing temperature. The resistivity $\rho $ $\left(\rho =1/\sigma \right)$ can be written as a sum of contributing factors using Matthiessen's rule as$$\rho ={\rho}_{0}+\rho \left(T\right)$$where ${\rho}_{0}$ is a constant and $\rho \left(T\right)$ is the temperature dependent resistivity. A typical carrier concentration for a metal is $1{0}^{23}{cm}^{3}$. Metals do not have a gap between the conduction and valence bands.
 Semiconductors: $n$ and $p$ are not constant with temperature but are thermally activated. Mobility is related to temperature as $\mu \propto {T}^{a}$. The typical range of carrier concentrations for semiconductors is $1{0}^{14}1{0}^{20}{cm}^{3}$. The range of bandgaps for semiconductors is typically $0.13.0\phantom{\rule{6px}{0ex}}eV$.
 Insulators: $n$ and $p$ are much lower than they are in metals and semiconductors. A typical carrier concentration for an insulator is $<1{0}^{7}{cm}^{3}$. The conductivity is generally a function of temperature. The bandgap for insulators is $>3.0\phantom{\rule{6px}{0ex}}eV$.
4.1 Law of Mass Action
The equilibrium concentration of electrons and holes can be determined by treating them as chemical species. At equilibrium,
$$\begin{array}{cc}\left[np\right]\leftrightharpoons \left[n\right]+\left[p\right]& (4.2)\end{array}$$
The rate constant is given by
$$\begin{array}{cc}\frac{\left[n\right]\left[p\right]}{\left[np\right]}=K\left(T\right)& (4.3)\end{array}$$
where
$$\begin{array}{cc}K\left(T\right)=\left[F\right(T){]}^{2}exp[\Delta E/kT]& (4.4)\end{array}$$Consider the intrinsic case, that is, when the semiconductor is not doped with chemical impurities. For this case,
$$\begin{array}{cc}\left[n\right]=\left[p\right]=\left[{n}_{i}\right]=Aexp\left[\Delta E/2kT\right]& (4.5)\end{array}$$
where ${n}_{i}$ is the intrinsic carrier concentration. For an intrinsic semiconductor, the conductivity is
$$\begin{array}{cc}\sigma =2{n}_{i}e\mu \cong {\sigma}_{0}exp\left[\Delta E/2kT\right]& (4.6)\end{array}$$
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 IIIAVIA is shown in Fig.
4.2.
On the left side (group IIIA) are metals. The right hand side (group VIA) contains nonmetals. In between these two groups are elemental semiconductors and elements that form compound semiconductors. At the bottom of the table (In, Sn, and Sb) are more metallic elements. Group IVA consists of the covalent semiconductors Si, Ge, and gray Sn.
Electronic bands are made from an assembly of atoms with individual quantum states. The spectroscopic notation for quantum states is given by
$$\begin{array}{cc}1{s}^{2}2{s}^{2}2{p}^{6}3{s}^{2}3{p}^{6}3{d}^{10}4{s}^{2}\mathrm{...}& (4.7)\end{array}$$
4.3 Reciprocal Lattice and the Brillouin Zone
 Reciprocal space: kspace, momentum space
$$\overrightarrow{p}=\hslash \overrightarrow{k}$$
 $\overrightarrow{k}$ is the reciprocal lattice vector with units $1/cm$
Reciprocal 2D Square Lattice
4.4 Nearly Free Electron Model
 Fermi wavevector: ${k}_{F}=(3{\pi}^{2}n{)}^{1/3}$
 Fermi energy: ${E}_{F}=\frac{{\hslash}^{2}{k}_{F}^{2}}{2m}$
 Conductivity is proportional to the Fermi surface area, ${S}_{F}\mathrm{.}$ $\sigma \propto {S}_{F}$
4.5 Two Level Model
 Parabolic bands: $E={\hslash}^{2}{k}^{2}/2{m}^{*}$.
 Origin? Solution to the Schrodinger Equation, $H\Psi =E\Psi $ for free electrons
5 Semiconductor Band Diagrams
Indirect Bandgap
 Direct bandgap semiconductors: IIIV's
 Indirect bandgap semiconductors: Si, Ge
5.1 P and NType Semiconductors
5.2 PN Junctions
5.3 Boltzmann Statistics: Review
For ${E}_{v}$ as the reference energy (${E}_{V}=0$),
For fully ionized acceptors, $p\cong {N}_{A}$. For fully ionized donors, $n\cong {N}_{D}$.
${N}_{s}=Densityofstates$
$$\begin{array}{cc}Electrons:\phantom{\rule{1em}{0ex}}{E}_{F}& ={E}_{c}kTln\left({N}_{S}/n\right)\\ Holes:\phantom{\rule{1em}{0ex}}{E}_{F}& ={E}_{V}+kTln\left({N}_{S}/p\right)\end{array}$$
Note: these equations are true for nongenerate semiconductors.
5.4 PN Junction Equilibrium
Poisson's equation  describes potential distribution $\phi \left(x\right)$:
$$\frac{{d}^{2}\phi \left(x\right)}{d{x}^{2}}=\frac{\rho \left(x\right)}{\epsilon}$$
$$\begin{array}{cc}\phi \left(x\right)& \equiv Potential\\ \epsilon & \equiv DielectricConstantofMaterial\\ \rho \left(x\right)& \equiv ChargeDensity\end{array}$$
Assume that the donors and acceptors are fully ionized
$$\begin{array}{cc}ptype:\phantom{\rule{1em}{0ex}}{N}_{A}& \to {N}_{A}^{}+{p}^{+}\\ ntype:\phantom{\rule{1em}{0ex}}{N}_{D}& \to {N}_{D}^{+}+{e}^{}\end{array}$$
Note: the space charge region is positive on the nside and negative on the pside.
5.5 Charge Profile of the pn Junction
 Total charge: $\rho \left(x\right)=e\left[p\left(x\right)+{N}_{D}\left(x\right)n\left(x\right){N}_{A}\left(x\right)\right]$
 Space charge width: $w={x}_{n}+{x}_{p}$
 Charge balance: ${N}_{A}{x}_{p}={N}_{D}{x}_{n}$
 Note that $x=0$ at the junction interface.
5.6 Calculation of the Electric Field
$$\frac{{d}^{2}\phi}{d{x}^{2}}=\frac{\rho}{\epsilon}\phantom{\rule{1em}{0ex}}\frac{d\phi}{dx}=E$$ $$\Rightarrow \frac{dE}{dx}=\frac{\rho}{\epsilon}$$
$$E\left(x\right)=\int \frac{\rho}{\epsilon}dx$$
Separate the integral for the n and pregions:
$$E\left(x\right)=\underset{pregion}{\underset{\u23df}{{\int}_{{x}_{p}}^{0}\frac{\rho \left(x\right)}{\epsilon}dx}}+\underset{nregion}{\underset{\u23df}{{\int}_{0}^{{x}_{n}}\frac{\rho \left(x\right)}{\epsilon}dx}}$$
Important assumptions made:
 $\rho \approx e{N}_{D}$ in the $n$ region, and $\rho \approx e{N}_{A}$ in the $p$ region.
 $\rho $ is constant in the separate $n$ and $p$ regions.
Boundary conditions:
 $E\left(x={x}_{p}\right)=E\left(x={x}_{n}\right)=0$
 $E$ is continuous at the junction
Resulting electric field across the junction:
$$E\left(x\right)=\{\begin{array}{cc}\frac{e{N}_{A}}{\epsilon}\left(x+{x}_{p}\right),& {x}_{p}\le x\le 0\\ \frac{e{N}_{D}}{\epsilon}\left(x{x}_{n}\right),& 0\le x\le {x}_{n}\end{array}$$
$$E\left(x\right)=\{\begin{array}{cc}\frac{e{N}_{A}}{\epsilon}\left(x+{x}_{p}\right),& {x}_{p}\le x\le 0\\ \frac{e{N}_{D}}{\epsilon}\left(x{x}_{n}\right),& 0\le x\le {x}_{n}\end{array}$$
Electric Field in the Junction
5.7 Calculation of the Electric Potential
Recall
$$\phi =\int Edx$$
Boundary conditions:
 $\phi $ is continuous at the junction boundary
 $\phi =0$ at $x={x}_{p}$ (chosen arbitrarily)
Resulting potential across the junction:
$$\phi \left(x\right)=\{\begin{array}{cc}\frac{e{N}_{A}}{\epsilon}\left(\frac{1}{2}{x}^{2}+x\cdot {x}_{p}+\frac{1}{2}{x}_{p}^{2}\right),& {x}_{p}\le x\le 0\\ \frac{e{N}_{D}}{\epsilon}\left(\frac{{N}_{A}}{2{N}_{D}}{x}_{p}^{2}+x\cdot {x}_{n}\frac{1}{2}{x}^{2}\right),& 0\le x\le {x}_{n}\end{array}$$
Electric Potential in the Junction
5.8 Junction Capacitance
$$w={x}_{p}+{x}_{n}={\left[\frac{2\epsilon {\phi}_{0}}{e\left({N}_{A}+{N}_{D}\right)}\right]}^{1/2}\left[{\left(\frac{{N}_{A}}{{N}_{D}}\right)}^{1/2}+{\left(\frac{{N}_{D}}{{N}_{A}}\right)}^{1/2}\right]$$
$${\phi}_{0}\equiv Builtinpotential$$
 Note that, from charge balance, ${N}_{A}{x}_{p}={N}_{D}{x}_{n}$
 Junction capacitance:
$$C=\frac{\epsilon}{w}\phantom{\rule{1em}{0ex}}{F/cm}^{2}$$
 Onesided abrupt junction: heavily doped $p$ or $n$
 Consider the case of a ${p}^{+}$$n$ junction. For ${N}_{A}\gg {N}_{D}$,
$$w={\left(\frac{2\epsilon {\phi}_{0}}{e{N}_{D}}\right)}^{1/2}$$
$$C={\left(\frac{e\epsilon {N}_{D}}{2{\phi}_{0}}\right)}^{1/2}$$
5.9 Rectification
 Larger ${E}_{g}$, larger ${\phi}_{0}$
 ${I}_{D}={I}_{0D}exp\left(e{\phi}_{0}/kT\right)$

${E}_{g}$

${n}_{i}$ (cm${}^{3}\times 1{0}^{14}$)

Ge

0.66

0.24

Si

1.08

0.00015

GaAs

1.4



GaP

2.25



GaN

3.4



 Under a forward bias $+{\phi}_{1}$, the built in potential is reduced by ${\phi}_{1}$ and there is a higher probability of electrons going over the barrier.
 Barrier height under forward bias: $e\left({\phi}_{0}\phi \right)$
 Current increases exponentially with forward bias voltage.
 Under a reverse bias $+{\phi}_{1}$, the built in potential is increased by ${\phi}_{1}$ and there is a lower probability of electrons going over the barrier.
5.10 Junction Capacitance
Definition of capacitance:
$$C\equiv \frac{\partial Q}{\partial \phi}$$
For a onesided abrupt junction, recall that $Q=e{N}_{D}{x}_{n}$, where
$${x}_{n}={\left[\frac{2\epsilon \left({\phi}_{0}+{\phi}_{1}\right){N}_{A}}{e{N}_{D}\left({N}_{D}+{N}_{A}\right)}\right]}^{1/2}$$
The total charge Q for a one sided junction is then
$$Q={\left[2e\epsilon \left({\phi}_{0}+{\phi}_{1}\right)\frac{{N}_{A}{N}_{D}}{{N}_{A}+{N}_{D}}\right]}^{1/2}$$
Calculate the capacitance as a function of bias voltage ${\phi}_{1}$:
$$C=\frac{\partial Q}{\partial {\phi}_{1}}={\left[\frac{e\epsilon}{2\left({\phi}_{0}+{\phi}_{1}\right)}\frac{{N}_{A}{N}_{D}}{{N}_{A}+{N}_{D}}\right]}^{1/2}$$
A plot of $1/{C}^{2}$ versus ${\phi}_{1}$ is called a Mott Schottky plot and yields ${N}_{A}$ or ${N}_{D}$.
6 Transistors: Semiconductor Amplifiers
 Three terminal device
 Bipolar Junction Transistor (BJT): consists of two pn junctions backtoback
 The three seminconductor regions are the emitter, base, and collector
 pnp and npn devices
6.1 pnp Device
 EmitterBase junction is forward biased, and the CollectorBase junction is reverse biased
 Little recombination in the Base region
 Holes are injected into the Base from the Emitter and collected at the Collector.
Device Diagram
Circuit Diagram
6.2 pnp Device
Device Diagram
6.3 Amplifier Gain
Voltage across the junctions:
$${i}_{c}{\phi}_{r}\gg {i}_{e}{\phi}_{f}\phantom{\rule{1em}{0ex}}\frac{{\phi}_{r}}{{\phi}_{f}}\approx 10\phantom{\rule{6px}{0ex}}V/V$$
Nodal analysis at the base for pnp and npn devices:
$${i}_{c}\approx {i}_{e}\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}{i}_{c}=\alpha {i}_{e}$$
From nodal analysis at the base, ${i}_{c}={i}_{e}{i}_{b}$.
$$\Rightarrow {i}_{c}=\frac{\alpha {i}_{b}}{1\alpha}={h}_{fe}{i}_{b}\phantom{\rule{1em}{0ex}}with\phantom{\rule{1em}{0ex}}{h}_{fe}=\frac{\alpha}{1\alpha}$$
${h}_{fe}$ is the current gain parameter. $\alpha \sim 0.9$.
6.4 Circuit Configurations
 Circuit configurations include the commonbase, commoncollector, and commonemitter.
 The term following “common” indicates which terminal is common to the input and output.
CommonEmitter Diagram
CommonBase IV Characteristic
6.5 MOSFETs
 MOSFET: etal xide emiconductor ield ffect ransistor
 An applied electric field induces a channel between the source and drain through which current conducts.
6.6 OxideSemiconductor Interface
Changing the gate bias causes the following:
 Band bending at the oxidesemiconductor interface.
 Accumulation of charge at the interface
 An increase in channel conductance.
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 semiconductor intercepts the valence (conduction) band due to band bending.
 The channel in a MOSFET is an inversion layer.
6.8 Channel PinchOff
 A threshold voltage ${V}_{T}$ is required for channel inversion.
 For a positive gate voltage ${V}_{G}$, the potential across the oxidesemiconductor junction is ${V}_{G}{V}_{T}\mathrm{.}$
 For a positive voltage between the drain and source (${V}_{DS}$), the potential near the drain is ${V}_{G}{V}_{T}{V}_{DS}$.
 For ${V}_{DS}\ge {V}_{G}{V}_{T}$, the channel is depleted near the drain, resulting in “pinchoff.”
 In saturation mode (${V}_{DS}\ge {V}_{G}{V}_{T}$), ${I}_{D}\propto {V}_{GS}^{2}$
 A small change in ${V}_{DS}$ causes a large change in ${I}_{D}\mathrm{.}$
$$\begin{array}{cc}G& =Gate\\ S& =Source\\ D& =Drain\end{array}$$
6.9 Tunnel Diodes
 Consists of a heavily doped ${p}^{+}$${n}^{+}$ structure
 Tunneling current: $I\propto {N}_{eff}{e}^{x/a}$
 Electrons tunnel from filled to empty states across the junction
 Electrons can go over the barrier or through the barrier (tunneling)
6.10 Gunn Effect
 Excitation of electrons under a high field to a higher energy conduction band with larger effective mass
$$\sigma =ne\mu \phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}\mu =\frac{e\tau}{{m}^{*}},\phantom{\rule{1em}{0ex}}{m}^{*}={\left(\frac{1}{{\hslash}^{2}}\underset{bandcurvature}{\underset{\u23df}{\frac{{\partial}^{2}E}{\partial {k}^{2}}}}\right)}^{1}$$
6.11 Gunn Diodes
 Negative resistance due to increased effective mass
$$J=\sigma E=\frac{n{e}^{2}\tau}{{m}^{*}}E$$
 Transfer of electrons from one valley to another (ransferred lectron evice, )
7 Heterojunctions and Quantum Wells
 Composed of semiconductor superlattices (alternating layers of different semiconductors)
 Multilayers grown by Molecular Beam Epitaxy (MBE) and MetalOrganic Chemical Vapor Deposition (MOCVD)
 Used in the construction of lasers, detectors, and modulators
 Utilizes band offset, can be used to confine carriers
7.1 Quantum Wells
 From the particle in a box solution, minibands are formed within the well
$$\begin{array}{ccc}H\Psi & =& E\Psi ;\phantom{\rule{1em}{0ex}}\Psi =Aexp\left({k}_{n}x\right)\end{array}$$
$${k}_{n}=\frac{n\pi}{{L}_{x}}$$
$${E}_{x}=\frac{{\hslash}^{2}{k}^{2}}{2{m}^{*}}=\frac{{\hslash}^{2}}{2{m}^{*}}\left(\frac{{\pi}^{2}{n}_{x}^{2}}{{L}_{x}^{2}}\right)$$
 Smaller ${L}_{x}$, larger energy of miniband
 ${L}_{x}$ can be tuned to engineer the band gap
7.2 Heterostructures and Heterojunctions
 Devices that use heterojunctions and heterostructures: lasers, modulationdoped field effect transistors (MODFETs)
 Types of heterostructures: quantum wells (2D), wires (1D), and dots (0D)
7.3 Layered Structures: Quantum Wells
 “Cladding” layer can confine carriers due to difference in bandgap and light due to difference in refractive index.
 Examples: GaAs/GaAlAs, InP/InGaAsP, GaN/InGaN
 Semiconductor 1 is “lattice matched” (same lattice constant as substrate)
 Semiconductor 2 is the strained layer (different lattice constant)
7.4 Transitions between Minibands
 Allowed initial and final states are governed by “selection rules.”
 Will have conduction in minibands
 Energy of states:
$$E=\frac{{\hslash}^{2}{k}^{2}}{2{m}^{*}}=\frac{{\hslash}^{2}}{2{m}^{*}}{\left(\frac{n\pi}{{L}_{x}}\right)}^{2}$$
7.5 Quantum Dots
 Electrons are confined in all dimensions to “dots.”
 Quantum Dots: clusters of atoms 310 nm in diameter.
 QDs can be created by colloidal chemical synthesis or island growth epitaxy.
 Modeled after the hydrogen atom with radius $R$. Quantized energies:
$${E}_{i}=\frac{{\hslash}^{2}}{2{m}_{i}}{\left(\frac{{\alpha}_{i}}{R}\right)}^{2}$$
$${\alpha}_{i}\equiv quantumnumber$$
7.6 Quantum Dot Example: Biosensor
 Solution synthesized colloidal quantum dots (CQD)
 Fluorescence energy depends on surface and size of dot
 Binding of molecules to the surface of the QD quenches the photoluminescent intensity.
7.7 Quantum Cascade Devices
 Consist of multiple quantum wells
 Band bending occurs with bias
8 Optoelectronic Devices: Photodetectors and Solar Cells
 Light incident on a pn junction generates electronhole pairs which are separated by the built in potential.
 Separated carriers contribute to the photocurrent.
8.1 IV Characteristics
 Diode equation: $I={I}_{0}\left({e}^{qV/kT}1\right)$
 Short circuit current: ${I}_{sc}=Aq\left({L}_{e}+{L}_{h}\right)G$
$$\begin{array}{cc}{L}_{e}& \equiv Electrondiffusionlength\\ {L}_{h}& \equiv Holediffusionlength\\ G& \equiv Carriergenerationrate\left(dependsonlightintensity\right)\end{array}$$
 Current under illumination: $I={I}_{0}\left({e}^{qV/kT}1\right){I}_{sc}$
 Open circuit voltage: $V\left(I=0\right)={V}_{oc}=\frac{kT}{q}ln\left(\frac{{I}_{sc}}{{I}_{0}}+1\right)$
8.2 Power Generation
 Quadrant IV is the power quadrant.
 Theoretical maximum power: ${P}_{max,\phantom{\rule{6px}{0ex}}theory}={V}_{oc}{I}_{sc}$
 Maximum obtainable power: ${P}_{max}=ff\left({I}_{sc}{V}_{oc}\right)$
$$FillFactor:ff\equiv \frac{{I}_{m}{V}_{m}}{{I}_{sc}{V}_{oc}}$$
8.2.1 Solar Cell Efficiency
 Efficiency is governed by the fill factor and bandgap
 Want larger fill factors and absorption that matches the solar spectrum
8.2.2 Other Types of Solar Cells
 Multijunction solar cells: GaAlAs/GaAs. More efficient absorption.
 Thin films: amorphous silicon (aSi), CdTe. Inexpensive.
 Metalsemiconductor solar cells
8.2.3 MetalSemiconductor Solar Cells
 Energy barrier ${\phi}_{B}$ at metalsemiconductor junction
 Current density, J:
$$J={J}_{00}exp\left(\frac{{\phi}_{B}}{kT}\right)\left[exp\left(\frac{qV}{kT}\right)1\right]$$
$$J\equiv \frac{current}{area}=\frac{I}{A}$$
8.2.4 Schottky Barrier and Photo Effects
 Barrier height is obtained from $log\left(J\right)$ versus $1/T$ plot
 Electrons can tunnel across the barrier
 Photoresponse is characterized by two regions: metal photoemission, and bandtoband excitation.
 $Photoresponse\propto \sqrt{{\phi}_{B}}$
8.2.5 Dependence of ${\phi}_{B}$ on Work Function
$\phi \equiv $ energy required to take an electron from the Fermi Level to the Vacuum Level
$${\phi}_{B}={\phi}_{m}{\phi}_{s}$$
8.2.6 Pinned Surfaces
 Surface states can influence the amount of band bending at a metal/semiconductor interface
 For “pinned” surfaces, the barrier height is nearly independent of the metal work function.
 Total amount of band bending is equal to the band gap energy:
$${\phi}_{B\left(n\right)}+{\phi}_{B\left(p\right)}={E}_{g}$$
 Example, InP (n): ${\phi}_{B\left(n\right)}\sim 0.5\phantom{\rule{6px}{0ex}}\phantom{\rule{10px}{0ex}}eV\phantom{\rule{1em}{0ex}}{\phi}_{B\left(p\right)}\sim 0.8\phantom{\rule{6px}{0ex}}eV\phantom{\rule{1em}{0ex}}{E}_{g}\sim 1.27\phantom{\rule{6px}{0ex}}eV$
 For making an Ohmic contact to an ntype semiconductor, ${\phi}_{B}={\phi}_{m}{\phi}_{s}<0$
8.2.7 Light Emitting Diodes (LEDs)
 Forward biasing a pn junction results in electronhole recombination at the junction
 Recombination results in light emission. The wavelength depends on ${E}_{g}$
$$h\nu ={E}_{g}$$
8.2.8 Light Emitting Diodes (LEDs)
Material

${E}_{g}$ at 300K (eV)

Comments

GaP

2.25


GaAsP

1.72.25

Alloy

GaAs

1.43


InGaAsP

0.81.4


ZnSe

2.58

Blue

GaN

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$${E}_{g}\approx x{E}_{g1}+\left(1x\right){E}_{g2}$$
8.2.10 LED Efficiency
 LED efficiency is the ratio of the radiative recombination probability (${W}_{R}$) to the total recombination probability (${W}_{total}$)
$$\eta =\frac{{W}_{R}}{{W}_{total}}=\frac{{W}_{R}}{{W}_{R}+{W}_{NR}}$$
 ${W}_{NR}$: the probability of nonradiative recombination
 Nonradiative processes limit the efficiency
 Recombination probability is related to the lifetime ($\tau $)
$$W\propto \frac{1}{\tau}{\phantom{\rule{6px}{0ex}}\phantom{\rule{10px}{0ex}}[s}^{1}]$$
$$\frac{1}{{\tau}_{total}}=\frac{1}{{\tau}_{R}}+\frac{1}{{\tau}_{NR}}$$
 Efficiency in terms of probability ($W$) and lifetime ($\tau $)
$$\eta =\frac{{W}_{R}}{{W}_{total}}=\frac{{W}_{R}}{{W}_{R}+{W}_{NR}}$$
$$\eta =\frac{1/{\tau}_{R}}{1/{\tau}_{total}}=\frac{1}{1+{\tau}_{R}/{\tau}_{NR}}$$
 For ${\tau}_{R}\gg {\tau}_{NR}$, ${\tau}_{NR}$ determines the radiation efficiency $\therefore $ want as few nonradiative defects (dislocations, impurities,etc.) as possible
$$\eta =\frac{{\tau}_{NR}}{{\tau}_{R}}$$
 GaN example: $1{0}^{7}$ dislocations per cm${}^{2}$
9 Lasers
 LASER: ight mplification through timulated mission of adiation
 Laser light has narrow divergence and consists of photons of the same frequency and phase
 Photons of a certain frequency and phase stimulate emission of photons with the same frequency and phase
9.1 Emission Rate and Laser Intensity
 Calculate emission rate and laser intensity by accounting for both emission and absorption processes using Boltzmann Statistics
 Calculate the transition rate between states and Einstein $A$ and $B$ coefficients
 Consider 2 and 3 level systems
9.2 Two Level System
 At equilibrium, the population of levels is determined by the Boltzmann distribution
$${N}_{2}\propto N{e}^{{E}_{2}/kT}\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}\frac{{N}_{2}}{N}=\frac{{e}^{{E}_{2}/kT}}{Z}$$
$$Z\equiv sumoverstates$$
 Ratio of populations in a two level system:
$$\frac{{N}_{2}}{{N}_{1}}={e}^{\left({E}_{2}{E}_{1}\right)/kT}$$
At equilibrium, fewer electrons are in level 2
Triggers of Stimulated Emission
 Internal or external radiation can cause the (stimulated) transition to occur
 Internal sources are described by “black body” radiation.
9.3 Planck Distribution Law
 Number of photons inside the cavity is determined by Planck's Distribution Law
 Spectral density of photons of frequency $\nu $ in linewidth $d\nu $
$$\rho \left(\nu \right)d\nu =\underset{Densityofstates}{\underset{\u23df}{\frac{8\pi {n}^{3}h{\nu}^{3}}{{c}^{3}}}}\underset{Planckdistribution}{\underset{\u23df}{\frac{1}{exp\left(h\nu /kT\right)1}}}d\nu $$
9.4 Transition Rates
 Transition rate ${R}_{ij}$: rate at which electrons transition from state $i$ to state $j$.
 Absorption (“up”) rate ${R}_{12}$:
$${R}_{12}=N{}_{1}B{}_{12}\rho \left(\nu {}_{21}\right)d\nu $$
$${N}_{1}\equiv Numberofphotonsinlowerstate$$
$$\rho \left({\nu}_{21}\right)\equiv numberofphotonsavailablehavingtransitionenergy\phantom{\rule{10px}{0ex}}h\nu $$
 Emission (“down”) rate ${R}_{21}$:
$${R}_{21}=N{}_{2}[\underset{Spont\mathrm{.}}{\underset{\u23df}{{A}_{21}}}+\underset{Stim\mathrm{.}}{\underset{\u23df}{{B}_{21}\rho \left({\nu}_{21}\right)]}}d\nu $$
 ${A}_{21}$ and ${B}_{21}$ are the Einstein $A$ and $B$ coefficients for spontaneous and stimulated transitions, respectively, between states 2 and 1.
9.5 Two Level System
 At equilibrium (steady state):
$${R}_{12}={R}_{21}$$
$${N}_{1}{B}_{12}\rho \left({\nu}_{21}\right)d\nu =N{}_{2}\left[{A}_{21}+B{}_{21}\rho \left(\nu {}_{21}\right)\right]d\nu $$
 Substitute in $\frac{{N}_{2}}{{N}_{1}}={e}^{\left({E}_{2}{E}_{1}\right)/kT}$, simplifying and solving for $\rho \left({\nu}_{21}\right):$
$$\rho \left({\nu}_{21}\right)d\nu =\frac{{A}_{21}d\nu}{{B}_{12}exp\left[\left(h{\nu}_{21}\right)/kT\right]{B}_{21}}$$
 Note: same form as Planck Distribution Law with ${B}_{12}={B}_{21}$
 In a two level system, the stimulated emission and absorption rates are equal, i.e., ${B}_{12}={B}_{21}$
 Comparing to the Planck Distribution Law,
$${A}_{21}={B}_{21}\frac{8\pi {n}^{3}h{\nu}^{3}}{{c}^{3}}$$
 The coefficient of spontaneous emission ${A}_{21}$ is related to the stimulated emission rate
Two Level System  Summary
 Number of photons inside the cavity is determined by Planck's Distribution Law
 The stimulated emission and absorption rates are equal
 The coefficient of spontaneous emission ${A}_{21}$ is related to the stimulated emission rate
 A population inversion is required for lasing
9.6 Three Level System
 A three level system consists of a ground state (E${}_{1}$), excited state (E${}_{2}$), and intermediate state (E${}_{3}$).
 Electrons are pumped from E${}_{1}$ to E${}_{3}$ by a pump source with energy $\ge h{\nu}_{31}$; ${N}_{2}$ population is unaffected
 More electrons are in the upper state E${}_{3}$ than the lower (metastable) state E${}_{2}$ $\Rightarrow $ a population inversion is achieved
Carrier Population
 The carrier population is given by the Boltzmann distribution
$$\begin{array}{ccc}N\left(E\right)& =& {N}_{0}exp\left(E/kT\right)\\ logN& =& E/kT+log{N}_{0}\end{array}$$
 Lower energy states have higher carrier populations at equilibrium
Spontaneous Emission
 A${}_{31}$: spontaneous emission rate; measure of the spontaneous depopulation
 Assuming first order kinetics,
$$\frac{d{N}_{3}}{dt}={A}_{31}{N}_{3}$$
$${A}_{31}=\frac{1}{{t}_{spont}}$$
 Solution for initial population ${N}_{3,i}$: ${N}_{3}\left(t\right)={N}_{3,i}exp\left({A}_{31}t\right)$
Populations under High Pumping
 Under high pumping at steady state, the populations in levels 1 and 3 are$${N}_{1}^{*}={N}_{3}^{*}=\frac{{N}_{1}+{N}_{3}}{2}$$
 Equal probabilities of emission and absorption: material is “transparent”
 From energy balance,
$$\underset{Stim\mathrm{.}absorption}{\underset{\u23df}{{B}_{23}{N}_{2}\rho \left(\nu \right)}}=\underset{Spont\mathrm{.}emission}{\underset{\u23df}{{A}_{32}{N}_{3}}}+\underset{Stim\mathrm{.}emission}{\underset{\u23df}{{B}_{32}{N}_{3}\rho \left(\nu \right)}}$$
Laser Gain
 The net number of photons gained per second per unit volume is
$$N{W}_{12}=\left({N}_{2}{N}_{1}\right){W}_{12}$$
$$\begin{array}{cc}{W}_{12}& \equiv Transitionratefromstate1to2\end{array}$$
 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
 Stimulatedemission photons travel in the same direction ($z$) as the incident photons and further stimulate emission
 $\Rightarrow $ photon flux and intensity increase in $z$ direction
 Increase in photon intensity through a volume of thickness $dz$:
$$dI=\left({N}_{2}{N}_{1}\right){W}_{12}h{\nu}_{12}dz$$
$$\therefore \frac{dI}{dz}=\left({N}_{2}{N}_{1}\right){W}_{12}h{\nu}_{12}$$
 Transition probability is defined as
$${W}_{12}=\frac{I}{h\nu}\frac{{c}^{2}g\left(\nu \right)}{8\pi {n}^{2}{\nu}^{2}{t}_{spont}}$$
 $g\left(\nu \right)$ is the lineshape factor
 Laser gain factor $\gamma \left(\nu \right)$ is defined by the equation
$$\frac{dI}{dz}=\gamma \left(\nu \right)I$$
$$\therefore \gamma \left(\nu \right)=\left({N}_{2}{N}_{1}\right)\frac{{c}^{2}g\left(\nu \right)}{8\pi {n}^{2}{\nu}^{2}{t}_{spont}}$$
 The change in laser intensity with distance $z$ is
$$\frac{dI}{dz}=\gamma \left(\nu \right)I$$
 Solution for $I\left(z\right)$:
$$I\left(z\right)=I\left(0\right)exp\left[\gamma \left(\nu \right)z\right]$$
 Recall laser gain factor from previous slide:
$$\gamma \left(\nu \right)=\left({N}_{2}{N}_{1}\right)\frac{{c}^{2}g\left(\nu \right)}{8\pi {n}^{2}{\nu}^{2}{t}_{spont}}$$
 For amplification, $I\left(z\right)/I\left(0\right)>1$ is required $\therefore $ ${N}_{2}>{N}_{1}$. A population version is required for amplification
 Laser threshold: operation at which $I\left(z\right)/I\left(0\right)=1$
9.7 Lasing Modes
 Lasing frequencies are chosen by confining the stimulatedemission photons inside a cavity called a Fabry Perot resonator
 Standing waves of certain discrete frequencies are established inside the cavity
 Laser intensity after a round trip in the cavity:
$$I={I}_{0}{R}_{1}{R}_{2}exp\left[\left(\gamma \alpha \right)2l\right]$$
$$\begin{array}{ccc}\alpha & \equiv & absorptionandscatteringlosses\\ \gamma & \equiv & lasergainfactor\end{array}$$
 Criteria for net gain: gain $\ge $ sum of losses
$${R}_{1}{R}_{2}exp\left[\left(\gamma \alpha \right)2l\right]\ge 1$$
 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 (Al${}_{2}$O${}_{3}$)
 3 level system  first demonstrated laser
 Optically pumped by a “discharge lamp”
9.8.2 Others Examples
Semiconductor:
 Emission from carrier recombination in a forward biased pn junction
 Heterostructures form both pn junction and waveguiding region (cavity)
Others:
 Ceramic: Nd^{3+} :YAG, 4level system
 Glass fiber laser Er${}^{3+}$ :SiO${}_{2}$ Erbium Doped Fiber Amplifier (EDFA), basis of modern optical communication
9.9 Threshold Current Density
 Number of electrons injected into lasing region per unit volume: ${N}_{e}$
 Recombination rate:
$$\frac{{N}_{e}lwd}{{t}_{rec}}=\frac{{I}_{i}\eta}{e}$$
$$\begin{array}{ccc}\eta & \equiv & Efficency\\ {I}_{i}& \equiv & Injectioncurrent\\ e& \equiv & Fundamentalcharge\end{array}$$
 Recall the gain coefficient, $\gamma \left(\nu \right)$
$$\gamma \left(\nu \right)=\left({N}_{3}{N}_{2}\right)\frac{{c}^{2}g\left(\nu \right)}{8\pi {n}^{2}{\nu}^{2}{t}_{spont}}$$
$${t}_{rec}\cong {t}_{spont}$$
$${N}_{e}={N}_{3}{N}_{2}$$
 Substituting in ${N}_{3}{N}_{2}={N}_{e}=\frac{{I}_{i}\eta}{e}\frac{{t}_{rec}}{lwd}$ and ${t}_{spont}\cong {t}_{rec}$,
$$\gamma \left(\nu \right)=\frac{{c}^{2}g\left(\nu \right)\eta}{8\pi {n}^{2}{\nu}^{2}elwd}{I}_{i}\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\left(c{m}^{1}\right)$$
Relates gain coefficient to ${I}_{i}$
 The linewidth factor is $g\left(\nu \right)\cong 1/\Delta \nu $
 Let the mirror reflectivity be ${R}_{1}{R}_{2}=R$. Recall the lasing threshold condition
$$Rexp\left(\gamma \alpha \right)l\text{'}=1$$
$$\left(\gamma \alpha \right)l\text{'}+lnR=0$$
 Define threthold current density: ${j}_{th}={I}_{i,th}/wl$
 Substituting in for $\gamma $ and solving for ${j}_{th}$:
$${j}_{th}=\underset{{\beta}^{1}}{\underset{\u23df}{\frac{8\pi {n}^{2}{\nu}^{2}ed}{{c}^{2}\eta g\left(\nu \right)}}}\left(\alpha \frac{1}{l\text{'}}lnR\right)\phantom{\rule{1em}{0ex}}\left[{A/cm}^{2}\right]$$
 Lasing threshold current density:
$${j}_{th}=\frac{\alpha}{\beta}\frac{1}{\beta l\text{'}}lnR\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\left(R<1\right)$$
Lasing Threshold Condition:$${j}_{th}=\frac{\alpha}{\beta}\frac{1}{\beta l\text{'}}lnR\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\left(R<1\right)$$
9.10 Comparison of Emission Types
Emission

Intensity

Linewidth

Directional

Coherent

Spontaneous

Weak

Broad

No

No

Stimulated

Intense

Narrow

Not

No

(SuperRadiant)

Necessarily

Stimulated

Intense

Very

Yes

Yes

Lasing

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 cavity will lase. This condition is given by
$$Integernumberofwavelengths=Opticalpathlength$$
$$m\lambda =2nl$$
$$m\frac{\lambda}{2n}=l\phantom{\rule{1em}{0ex}}m=1,2,\mathrm{3...}$$
 $n$ is the refractive index of the cavity and $l$ is the cavity length
9.11.1 Longitudinal Laser Modes
 Modes whose condition for lasing depends on the length of the cavity are called longitudinal modes
 Wavelength spacing between modes:
$$\Delta \lambda =\frac{{\lambda}^{2}}{2l\left(n\lambda \frac{dn}{d\lambda}\right)}$$
 $dn/d\lambda $ is the dispersion of the refractive index
9.11.2 Transverse Laser Modes
 If the other facets are flat, then transverse modes are supported in the cavity.
 The effect of transverse modes is visible in the angular dependence of the laser output intensity
9.12 Semiconductor Lasers
 Light is confined to regions of high refractive index
 Semiconductor heterostructure is designed to confine light in the junction
9.13 Photonic Bandgap Materials
 Light interacts with periodic structures with periodicity on the order of the wavelength
 Photonic crystals: structures with spatial periodic variation in the refractive index in 1, 2, or 3 dimensions. The periodicity is on the order of optical wavelengths.
 Photonic crystals can be engineered to slow down the propagation of optical pluses (“slow light”)
 The speed (${v}_{g})$ of a pulse is described by the group index (${n}_{g}$)
$${v}_{g}=c/{n}_{g}\phantom{\rule{1em}{0ex}}as{n}_{g}\to \infty ,\phantom{\rule{1em}{0ex}}{v}_{g}\to 0$$
10 Band Diagrams
 Energy band levels are represented in real space
 Energy levels of interest: Conduction band edge, valence band edge, and Fermi Level
 Energies of interest: work function and electron affinity
$\phi \equiv $Work Function. Energy required to move an electron from the fermi level to the vacuum level.
$\chi \equiv $Electron Affinity. Energy required to move an electron from the conduction band edge to the vacuum level.
${E}_{c}\equiv $Conduction band energy.
${E}_{v}\equiv $Valence band energy.
${E}_{F}\equiv $Fermi level.
10.1 Band Diagrams
10.2 Heterojunctions and the Anderson Model
 Heterojunction: metallurgical junction between two different types of semiconductors
 Anderson Model: junction between different semiconductors results in discontinuities in the energy bands.
 Example: nGaAs (${E}_{g}=1.45$ eV) and pGe (${E}_{g}=0.7$ eV)
10.3 Band Bending at pn Junctions
Band diagrams for isolated p and ntype semiconductors are shown in Fig.
10.1below.
When brought into contact without application of an external bias field, thermal equilibrium requires that the Fermi levels equilibrate, i.e.
$$\begin{array}{cc}{E}_{F,n}={E}_{F,p}& (10.1)\end{array}$$
Substituting in known values,
$$\begin{array}{ccc}q\left({V}_{D,n}+{V}_{D,p}\right)& =\left(4.130.1+0.7\right)\left(4.07+0.1\right)& (10.4)\\ =& 0.56\phantom{\rule{6px}{0ex}}eV<{E}_{g,Ge}& \end{array}$$
where
$$\begin{array}{cc}\frac{{V}_{D,n}}{{V}_{D,p}}=\frac{{N}_{A}{\epsilon}_{Ge}}{{N}_{D}{\epsilon}_{GaAs}}& (10.5)\end{array}$$
Let the Fermi level be zero on the energy scale.
$${\delta}_{GaAs}+q{V}_{D,n}=\left({E}_{g,Ge}\delta {}_{Ge}\right)q{V}_{D,p}+\Delta {E}_{c}$$
$$\Delta {E}_{c}={\delta}_{GaAs}+q{V}_{D,n}\left({E}_{g,Ge}\delta {}_{Ge}\right)+q{V}_{D,p}$$
$$\Delta {E}_{c}={\chi}_{Ge}{\chi}_{GaAs}=+0.6$$
The positive value for $\Delta {E}_{c}$ indicates a spike in the conduction band.
10.3.2 Calculate the Valence Band Discontinuity
The valence band discontinuity $\Delta {E}_{v}$ is calculated in a similar manner. In this case, a negative value for $\Delta {E}_{v}$ indicates a spike in the valence band. $\Delta {E}_{v}$ is calculated as follows:
$$\Delta {E}_{v}=\Delta {E}_{g}\Delta \chi $$
$$\Delta {E}_{v}=\left({E}_{g,GaAs}{E}_{g,Ge}\right)\left({\chi}_{Ge}{\chi}_{GaAs}\right)$$
Substituting in known values,
$$\Delta {E}_{v}=\left(1.450.7\right)\left(4.134.07\right)$$
$$\Delta {E}_{v}=+0.69$$
Since $\Delta {E}_{v}$ is positive, there is no spike in the valence band.
$$q\left({V}_{D,n}+{V}_{D,p}\right)=\underset{p}{\underset{\u23df}{\left({\chi}_{GaAs}+{E}_{g,GaAs}{\delta}_{GaAs}\right)}}\underset{n}{\underset{\u23df}{\left({\chi}_{Ge}+{\delta}_{Ge}\right)}}$$
$$=\left(4.07+1.450.1\right)\left(4.13+0.1\right)$$
$$=1.19<{E}_{g,GaAs}$$
For calculating the valence band discontinuity,
$$\Delta {E}_{v}=\left({E}_{g,Ge}{E}_{g,GaAs}\right)\left({\chi}_{GaAs}{\chi}_{Ge}\right)$$
$$\begin{array}{cc}\Delta {E}_{v}=0.69\Rightarrow \phantom{\rule{6px}{0ex}}spike& (10.6)\end{array}$$
11 Dielectric Materials
 Dielectrics are insulating materials which include glasses, ceramics, polymers, and ferroelectrics
 Dielectric constant ($\epsilon $) describes how well a material stores charge
 $\epsilon $ is related to capacitance ($C$)
Parallel Plate Capacitance
$$\begin{array}{ccc}C& =& \frac{\epsilon A}{d}\\ d& \equiv & distancebetweenplates\\ A& \equiv & platearea\end{array}$$
 From Maxwell's equations,
$$D=\epsilon E\phantom{\rule{1em}{0ex}}\left[{C/m}^{2}\right]$$
$$\begin{array}{ccc}D& \equiv & Electricfluxdensity\\ \epsilon & \equiv & Dielectricconstant\\ E& \equiv & Electricfield\end{array}$$
 Note: $\epsilon ={\epsilon}_{r}{\epsilon}_{0}$
$$\begin{array}{ccc}{\epsilon}_{r}& \equiv & Relativedielectricconstant\left(unitless\right)\\ {\epsilon}_{0}& \equiv & Permittivityoffreespace\left[F/m\right]\end{array}$$
 Total charge stored in a capacitor in vacuum:
$$Q={\epsilon}_{0}\frac{V}{d}\phantom{\rule{1em}{0ex}}\left[C\right]$$with
$$\begin{array}{ccc}V& \equiv & Voltage\left[V\right]\\ d& \equiv & Distance\left[m\right]\end{array}$$
$$Q\text{'}={\epsilon}_{0}{\epsilon}_{r}\frac{V}{d}\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}C=Q\text{'}/V$$
 An increase in charge is stored in the capacitor with insertion of the dielectric
 $D$ (electric flux density) is equal to the surface charge
 $P$ (polarization density) is the additional surface charge
$$\begin{array}{ccc}P& \equiv & D{\epsilon}_{0}E\\ P& \equiv & \left({\epsilon}_{r}{\epsilon}_{0}{\epsilon}_{0}\right)E={\epsilon}_{0}\underset{\chi}{\underset{\u23df}{\left({\epsilon}_{r}1\right)}}E\\ P& \equiv & {\epsilon}_{0}\chi E\end{array}$$
 $\chi =\left({\epsilon}_{r}1\right)$ is the dielectric susceptibility
11.2 Relation of $\epsilon $ to the Microscopic Structure
 Knowledge of the microscopic structure is needed to explain dielectric phenomena. Example: ferroelectric ${\epsilon}_{r}\left(T\right)$
 In the microscopic approach, we consider 1) Atomic behavior and 2) Deformation of the atomic charge cloud (orbitals) when a field is applied.
11.2.1 Relation of Macroscopic to Microscopic
 Induced dipole moment due to an applied field: $\mu =q\delta $
 For ${N}_{m}$ molecules per unit volume, the polarization density is $P={N}_{m}q\delta $ or
$$P={N}_{m}\mu $$
 For small fields, $\mu =\alpha E\text{'}$ and $P={N}_{m}\alpha E\text{'}$
$$\begin{array}{ccc}E\text{'}& \equiv & Localelectricfield\\ \alpha & \equiv & Polarizabilityofspecies\left(atoms,molecules\right)\end{array}$$
11.3 Contributions to the Polarizability
 Electronic: Deformation of the electronic cloud (orbitals) due to optical fields.
 Molecular: Bonds between atoms can stretch, bend, and rotate
 Orientational: Polymers, liquids, and gases can be reoriented in an applied electric field. Example: poled polymers, useful for electrooptic devices.
11.4 Polarization in Solids
 In an applied field, some molecules will be aligned, others will not be completely aligned. The orientation energy is:
$$E=\mu Ecos\theta $$
$$\theta \equiv Anglebetween\mu andE$$
 The number of dipoles with a certain $E$ will depend on the Boltzmann factor $exp\left(E/kT\right)$. Increasing temperature will favor more random alignment.
11.5 Calculation of the Average Dipole Moment
 Average Dipole Moment ($\u27e8\mu \u27e9)$
$$\u27e8\mu \u27e9=\frac{weightedaverageenergyforeachorientation}{totalenergyforallorientations}$$
 Consider the 3dimensional case. In a sphere with radius $r$, the total number of molecules with energy $E$ in a volume element $2\pi {r}^{2}sin\theta d\theta $ is
$$\#ofdipoleswithenergyE=2\pi Asin\theta d\theta exp\left(E/kT\right)$$
$$A=Constantincludingradius$$
 Recall the orientation energy is given by $E=\mu Ecos\theta $. Substituting in for $E$,
$$\#ofdipoleswithenergyE=2\pi Asin\theta d\theta exp\left(\frac{\mu Ecos\theta}{kT}\right)$$
 The total energy for all orientations is then
$${\int}_{0}^{\pi}Aexp\left(\frac{\mu Ecos\theta}{kT}\right)2\pi sin\theta d\theta $$
 The weighted average for each orientation is
$${\int}_{0}^{\pi}Aexp\left(\frac{\mu Ecos\theta}{kT}\right)\left(\mu cos\theta \right)2\pi sin\theta d\theta $$
 The average dipole moment is then
$$\u27e8\mu \u27e9=\frac{{\int}_{0}^{\pi}Aexp\left(\frac{\mu Ecos\theta}{kT}\right)\left(\mu cos\theta \right)2\pi sin\theta d\theta}{{\int}_{0}^{\pi}Aexp\left(\frac{\mu Ecos\theta}{kT}\right)2\pi sin\theta d\theta}$$
$$\frac{\u27e8\mu \u27e9}{\mu}=L\left(a\right)=cotha\frac{1}{a}$$
$$witha=\frac{\mu E}{kT}$$
 For small $a$, $cotha\approx \frac{1}{a}+\frac{a}{3}+\cdots $
$$\therefore \frac{\u27e8\mu \u27e9}{\mu}=\frac{a}{3}\phantom{\rule{1em}{0ex}}forsmalla$$
$$\u27e8\mu \u27e9=\frac{{\mu}^{2}E}{3kT}\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}\frac{\u27e8\mu \u27e9}{\mu}=\frac{\mu E}{3kT}$$
11.6 Polarizability of Solids
 The polarizability of the solid is
$$P={N}_{m}\u27e8\mu \u27e9$$
 $P$ is proportional to the field $E$ and inversely proportional to the temperature $T$
$$P={N}_{m}\frac{{\mu}^{2}E}{3kT}$$
 The definition for the electric susceptibility is $\chi =\frac{\partial P}{\partial E}\Rightarrow \chi \propto \frac{1}{T}$
11.7 Dielectric Constant for a Solid
 Under an applied field ${E}_{0}$, induced dipoles in the solid create an opposing field ${E}_{1}$. The field inside the material is $E={E}_{0}+{E}_{1}$.
 For a crystal, the polarizability will depend on the interaction of dipoles in the lattice
$$P={\sum}_{j}{N}_{j}{P}_{j}={\sum}_{j}{N}_{j}{\alpha}_{j}{E}_{loc,j}$$
 ${E}_{loc,j}$ is the local field of atom $j$ due to interactions with other dipoles. For cubic materials,
$${E}_{loc}=E+\frac{P}{3{\epsilon}_{0}}$$
 The polarizability is then
$$P=\left({\sum}_{j}{N}_{j}{\alpha}_{j}\right)\left(E+\frac{P}{3{\epsilon}_{0}}\right)$$
11.8 Claussius Mossotti Relation
$$\chi =\left({\sum}_{j}{N}_{j}{\alpha}_{j}\right)\frac{E+\frac{P}{3{\epsilon}_{0}}}{{\epsilon}_{0}E}$$
 Recall that $\chi ={\epsilon}_{r}1$ and $P=\chi {\epsilon}_{0}E={\epsilon}_{0}E\left({\epsilon}_{r}1\right)$
$${\epsilon}_{r}1=\left({\sum}_{j}{N}_{j}{\alpha}_{j}\right)\frac{E+\frac{1}{3{\epsilon}_{0}}{\epsilon}_{0}E\left({\epsilon}_{r}1\right)}{{\epsilon}_{0}E}$$
Simplifying and rearranging gives the Claussius Mossotti Relation:
$$\frac{{\epsilon}_{r}1}{{\epsilon}_{r}+2}=\frac{1}{3{\epsilon}_{0}}{\sum}_{j}{N}_{j}{\alpha}_{j}\phantom{\rule{1em}{0ex}}(SIUnits)$$
 The Claussius Mossotti relation relates the dielectric constant to the atomic polarizability
11.9 Frequency Dependence of the Polarizability
 Dipolar, ionic, and electronic contributions to the polarizability
$$\alpha \left(\omega \right)=\sum {\alpha}_{i}\left(\omega \right)$$
 Classical approach: treat electrons as harmonic oscillators
 Apply Hooke's Law
$$e{E}_{loc}=Cx=m{\omega}_{0}^{2}x$$
$$C=m{\omega}_{0}^{2}\equiv Forceconstant$$
$${E}_{loc}=\frac{m{\omega}_{0}^{2}x}{e}$$
 Note: ${\omega}_{0}$ is the natural or resonance frequency
 The electronic contribution to the polarizability is $P={N}_{m}{\alpha}_{el}{E}_{loc}$. Recall that $P={N}_{m}q\delta =Nex$
$${\alpha}_{el}=\frac{P}{N{E}_{loc}}=\frac{Nex}{N{E}_{loc}}=\frac{{e}^{2}}{m{\omega}_{0}^{2}}$$
 For an applied field with frequency $\omega $, the equation of motion for the electron response is
$$m\frac{{d}^{2}x}{d{t}^{2}}+m{\omega}_{0}^{2}x=e{E}_{loc}sin\omega t$$
 The solution is $x={x}_{0}sin\omega t$. Substituting the solution into the equation of motion,
$$m\left({\omega}^{2}+{\omega}_{0}^{2}\right){x}_{0}=e{E}_{loc}$$
$${x}_{0}=\frac{e{E}_{loc}}{m\left({\omega}_{0}^{2}{\omega}^{2}\right)}$$
 The electronic dipole moment is
$${\mu}_{elec}=e{x}_{0}=\frac{{e}^{2}{E}_{loc}}{m\left({\omega}_{0}^{2}{\omega}^{2}\right)}$$
 The electronic polarizability is
$${\alpha}_{elec}=\frac{{\mu}_{elec}}{{E}_{loc}}=\frac{{e}^{2}}{m\left({\omega}_{0}^{2}{\omega}^{2}\right)}$$
 ${\alpha}_{elec}$ becomes large at resonances where ${\omega}_{0}^{2}{\omega}^{2}\to 0$
 Calculate the atomic polarizability:
$${\mu}_{0}=e{x}_{0}=\frac{{e}^{2}{E}_{loc}}{m\left({\omega}_{0}^{2}{\omega}^{2}\right)}$$
$${\alpha}_{elec}=\frac{{\mu}_{0}}{{E}_{loc}}=\frac{{e}^{2}}{m\left({\omega}_{0}^{2}{\omega}^{2}\right)}$$
11.10 Quantum Theory of Polarizability
$${\alpha}_{elec}=\frac{{e}^{2}}{m}{\sum}_{j}\frac{{f}_{ij}}{{\omega}_{0}^{2}{\omega}^{2}}$$
$${f}_{ij}\equiv Oscillatorstrengthforatransitionfromstateitoj$$
$${P}_{elec}=N{\alpha}_{elec}E$$
11.11 Advanced Dielectrics: Ferroelectrics
 Ferroelectrics: materials that are noncentrosymmetric and have a resulting net polarization.
 The polarization is a function of the applied field and displays hysteresis.
$$\begin{array}{ccc}{P}_{S}& \equiv & Spontaneouspolarization\\ {E}_{C}& \equiv & Coercivefield\end{array}$$
12 Phase Transitions
 The transition from a ferroelectric (net polarization) to pyroelectric (no net polarization) structure occurs at the Curie Temperature (${T}_{C}$)
 Dielectric constant:
$$\epsilon =\frac{\xi}{T{T}_{C}}\phantom{\rule{1em}{0ex}}\xi \equiv Curieconstant$$
12.1 Lattice Instabilities
 Recall the ClausiusMossotti relation:
$$\frac{{\epsilon}_{r}1}{{\epsilon}_{r}+2}=\frac{1}{3{\epsilon}_{0}}N\alpha $$
 Solving for ${\epsilon}_{r}$,
$${\epsilon}_{r}1=\frac{N\alpha}{3{\epsilon}_{0}}\left({\epsilon}_{r}+2\right)=\frac{N\alpha {\epsilon}_{r}}{3{\epsilon}_{0}}+\frac{2}{3}\frac{N\alpha}{\epsilon}$$
$${\epsilon}_{r}\left(1\frac{N\alpha}{3{\epsilon}_{0}}\right)=\frac{2N\alpha}{3{\epsilon}_{0}}+1$$
$${\epsilon}_{r}=\frac{2N\alpha /{\epsilon}_{0}+3}{3N\alpha /{\epsilon}_{0}}$$
 Singularity (polarization catastrophe) at the condition $N\alpha /{\epsilon}_{0}=3$
12.2 Curie Weiss Law
 What happens close to the singularity? Consider a small deviation $2s$
$$\frac{N\alpha}{{\epsilon}_{0}}\cong 32s$$
 Rewriting the equation for ${\epsilon}_{r}$,
$${\epsilon}_{r}=\frac{2N\alpha /{\epsilon}_{0}+3}{3N\alpha /{\epsilon}_{0}2s}\propto \frac{1}{s}\phantom{\rule{1em}{0ex}}nearthesingularity$$
 Near the critical temperature, suppose $s\propto T$ such that $s\cong \left(T{T}_{C}\right)/\xi $
$$\therefore {\epsilon}_{r}\cong \frac{\xi}{\left(T{T}_{C}\right)}\phantom{\rule{1em}{0ex}}or\frac{1}{{\epsilon}_{r}}\cong \frac{T{T}_{C}}{\xi}$$
12.3 Ferroelectric Phase Transitions: Landau Theory
 Consider the Helmholtz free energy $\stackrel{\u02c6}{F}$ for a ferroelectric material as a function of polarization ($P$), temperature ($T$), and applied field ($E$).
 Taking a Taylor expansion in the order parameter $s$ about $s=0$,
$$\stackrel{\u02c6}{F}\left(P,T,E\right)=E\cdot P+{s}_{0}+\frac{1}{2}{s}_{2}{P}^{2}+\frac{1}{4}{s}_{4}{P}^{4}+\frac{1}{6}{s}_{6}{P}^{6}+\cdots $$
 Note that there are no odd powers if the unpolarized crystal has a center of inversion.
 At thermal equilibrium, the minimum in $\stackrel{\u02c6}{F}$ with respect to $P$ is
$$\frac{\partial \stackrel{\u02c6}{F}}{\partial P}=0=E+{s}_{2}P+{s}_{4}{P}^{3}+{s}_{6}{P}^{5}$$
 Thermal equilibrium condition:
$$\frac{\partial \stackrel{\u02c6}{F}}{\partial P}=0=E+{s}_{2}P+{s}_{4}{P}^{3}+{s}_{6}{P}^{5}$$
 For a ferroelectric state transition, the coefficient of the first order $P$ term must pass through zero at some temperature ${T}_{0}$ for no applied field.
$$\therefore assume{s}_{2}=\gamma \left(T{T}_{0}\right),\phantom{\rule{1em}{0ex}}with\gamma apositiveconstant$$
 ${s}_{2}$ can be positive or negative:
 ${s}_{2}<0$: lattice is “soft” close to the instability
 ${s}_{2}>0$: unpolarized lattice is unstable
 Rewriting the equilibrium condition,
$$\frac{\partial \stackrel{\u02c6}{F}}{\partial P}=E+\gamma \left(T{T}_{0}\right)P+{s}_{4}{P}^{3}=0$$
 Second order transition: no volume change, smooth transition.
 For $E=0$ and $P={P}_{S}$ at equilibrium,
$$\gamma \left(T{T}_{0}\right){P}_{S}+{s}_{4}{P}_{S}^{3}=0$$
$${P}_{S}\left[\gamma \left(T{T}_{0}\right)+{s}_{4}{P}_{S}^{2}\right]=0$$
 Solutions: ${P}_{S}=0$, or
 For $T\ge {T}_{0}:$ ${P}_{S}^{2}=\left(\gamma /{s}_{4}\right)\left({T}_{0}T\right)\Rightarrow {P}_{S}$ is imaginary; the only real root is ${P}_{S}=0$.
 For $T<{T}_{0}:$
$$\left{P}_{S}\right={\left(\gamma /{s}_{4}\right)}^{1/2}{\left({T}_{0}T\right)}^{1/2}$$
 Second order transition for $T<{T}_{0}:$
$${P}_{S}={\left(\gamma /{s}_{4}\right)}^{1/2}{\left({T}_{0}T\right)}^{1/2}$$
 The phase transition is second order since the polarization goes continuously to zero
12.3.1 First versus Second Order Transitions
 The change in ${P}_{s}$ is discontinuous for first order transitions and continuous for second order transitions
 First order transition results from a structure change. Example: cubic (pyroelectric) $\to $ tetragonal (ferroelectric)
 Recall the Helmholtz free energy:
$$\stackrel{\u02c6}{F}=E\cdot P+{s}_{0}+\frac{1}{2}{s}_{2}{P}^{2}+\frac{1}{4}{s}_{4}{P}^{4}+\frac{1}{6}{s}_{6}{P}^{6}+\cdots $$
 For first order transitions, ${s}_{4}$ is negative and ${s}_{6}$ cannot be neglected
 Equilibrium condition
$$\frac{\partial \stackrel{\u02c6}{F}}{\partial P}=\gamma \left(T{T}_{0}\right){P}_{S}\left{s}_{4}\right{P}_{S}^{3}+{s}_{6}{P}_{S}^{5}=0$$
 Solutions: ${P}_{S}=0$ or
$$\gamma \left(T{T}_{0}\right)\left{s}_{4}\right{P}_{S}^{2}+{s}_{6}{P}_{S}^{4}=0$$
12.3.2 Ferroelectric Example: $BaTi{O}_{3}$
 Spontaneous polarization: ${P}_{s}={N}_{m}q\delta =26\phantom{\rule{0ex}{0ex}}\mu C/c{m}^{2}\approx 3\times 1{0}^{1}C\cdot {m}^{2}$
 Dipole moment of a unit cell: $\mu ={P}_{s}\cdot volume$. For a lattice constant $a\approx 0.4\phantom{\rule{6px}{0ex}}nm$
$$\mu =\left(3\times 1{0}^{1}C\cdot {m}^{2}\right)\times 64\times 1{0}^{30}{m}^{3}\approx 2\times 1{0}^{29}C\cdot m$$
Material

${T}_{C}$ (K)

$BaTi{O}_{3}$

408

$BaTi{O}_{3}$

1480

12.4 Other Instabilities
 Pyroelectrics: heat causes distortion and electrical signal, “pyroelectric detectors”
12.5 Piezoelectrics
 Piezoelectric materials: an applied electric field causes a mechanical deformation
 Piezoelectric materials lack inversion symmetry
$$\begin{array}{ccc}Stress:\phantom{\rule{1em}{0ex}}T& =& CSeE\\ Displacement:\phantom{\rule{1em}{0ex}}D& =& \epsilon E+eS\end{array}$$
$$\begin{array}{ccc}C& \equiv & Elasticconstant\\ e& \equiv & Piezoelectriccoefficient\\ E& \equiv & Electricfield\\ S& \equiv & Strain\end{array}$$
 Note: when $E=0$, $D\ne 0$ for nonzero $e$ and $S$.
 Hooke's Law for $E=0$.
 Applications: electromechanical transducers, microphones, micromotors, microelectromechanical structures (MEMS), nanoelectromechanical structures (NEMS).
13 Diamagnetism and Paramagnetism
 Current and magnetism related through Maxwell's Equations
 Magnetic susceptibility:
$$\chi =\frac{{\mu}_{0}M}{B}$$
$$\begin{array}{ccc}M& \equiv & Magnetization\\ B& \equiv & Magneticfieldintensity\\ {\mu}_{0}& \equiv & Permeabilityoffreespace\end{array}$$
13.1 Diamagnetism
 In a magnetic field, induced current field opposite to applied current
 Electron precesses around field axis with frequency
$$\omega =\frac{eB}{2m}\left[{sec}^{1}\right]$$
 Magnetic field will induce net electric current around the nucleus.
 Consider multielectron atom. The “electron current” for $Z$ electrons is
$$\begin{array}{ccc}I& =& \left(charge\right)\times \left(revolutionspersecond\right)=Ze\frac{\omega}{2\pi}\\ I& =& Ze\left(\frac{1}{2\pi}\frac{eB}{2m}\right)=\frac{Z{e}^{2}B}{4\pi m}\end{array}$$
$$\mu \equiv \left(current\right)\times \left(areaofloop\right)=I\times A$$
 For a loop area $A=\pi {\rho}^{2}$,
$$\mu =\frac{Z{e}^{2}B}{4m}\u27e8{\rho}^{2}\u27e9$$
 $\u27e8{\rho}^{2}\u27e9$ is the mean square of the perpendicular distance from the field axis through charge
$$\u27e8{\rho}^{2}\u27e9=\u27e8{x}^{2}\u27e9+\u27e8{y}^{2}\u27e9$$
 The atomic radius is $\u27e8{r}^{2}\u27e9=\u27e8{x}^{2}\u27e9+\u27e8{y}^{2}\u27e9+\u27e8{z}^{2}\u27e9$
$$\therefore \u27e8{\rho}^{2}\u27e9=\frac{2}{3}\u27e8{r}^{2}\u27e9$$
 Diamagnetic susceptibility of electrons: $M=N\mu $
$$LangevinResult:\chi =\frac{N{\mu}_{0}\mu}{B}=\frac{{\mu}_{0}NZ{e}^{2}}{6m}\u27e8{r}^{2}\u27e9$$
$$\left(notenegativesign\right)$$
 Note: need to calculate $\u27e8{r}^{2}\u27e9$ from quantum mechanics
13.2 Paramagnetism
 Positive $\chi $ occurs for
 atoms, molecules, and lattice defects with an odd number of electrons
 free atoms with partially filled inner shells
 metals
 Magnetic moment $\mu $:
$$\mu \equiv \gamma \hslash J=g{\mu}_{B}J$$
$$\begin{array}{ccc}\gamma & \equiv & gyromagneticratio\\ \hslash J& \equiv & angularmomentum\\ g& \equiv & gfactor(\sim 2.00forelectronspin)\\ {\mu}_{B}& \equiv & Bohrmagneton\left(atomicunit\right)\\ J& \equiv & angularquantumnumberforelectrons\end{array}$$
 From quantum mechanics, only certain alignments of $\mu $ with $B$ are allowed
 Consider single electron case. Energy $U$ of an electron in a magnetic field:
$$U=\overrightarrow{\mu}\cdot \overrightarrow{B}={m}_{J}g{\mu}_{B}B$$
 For single electron atom with no orbital momentum, ${m}_{J}={m}_{S}=\pm 1/2$
 For a two spin orientation case, ${m}_{J}=\pm \frac{1}{2}$, $g=2$, and $U=\pm {\mu}_{B}B$
 Probability of occupying state $i$:
$${P}_{i}=\frac{exp\left({E}_{i}/\tau \right)}{Z}$$
$$\begin{array}{ccc}Z& \equiv & sumoverallstates\left(normalization\right)\\ \tau & \equiv & kT\end{array}$$
$$\begin{array}{ccc}\frac{{N}_{1}}{N}& =& \frac{exp\left(\mu B/\tau \right)}{exp\left(\mu B/\tau \right)+exp\left(\mu B/\tau \right)}\\ \frac{{N}_{2}}{N}& =& \frac{exp\left(\mu B/\tau \right)}{exp\left(\mu B/\tau \right)+exp\left(\mu B/\tau \right)}\end{array}$$
 Magnetization $M\equiv \left(netspindensity\parallel B\right)\times \left(magneticmomentperelectron\right)$
$$M=\left({N}_{1}{N}_{2}\right)\mu =N\mu \frac{{e}^{x}{e}^{x}}{{e}^{x}+{e}^{x}}=N\mu tanhx$$
$$x\equiv \mu B/kT$$
 For $x\ll 1$: high $T$, small field, no interaction.
$$tanhx\approx x\Rightarrow M\cong N\mu \left(\mu B/kT\right)$$
 For multiple electron atoms, the azimuthal quantum number ${m}_{J}$ takes values of $J$, $J1$, $\dots \phantom{\rule{6px}{0ex}},J$
 For an atom with quantum number $J$, there are $2J+1$ energy levels
 CurieBrillouin Law:
$$M=NgJ{\mu}_{B}{B}_{J}\left(x\right)$$
$$x\equiv gJ{\mu}_{B}B/kT$$
 Brillouin function (obtained from $\u27e8\mu \u27e9$ over all quantum configurations)
$${B}_{J}\left(x\right)\frac{2J+1}{2J}coth\left[\frac{\left(2J+1\right)x}{2J}\right]\frac{1}{2J}coth\frac{x}{2J}$$
$$cothx\approx \frac{1}{x}+\frac{x}{3}$$
13.2.1 Calculation of Susceptibility
For small fields and high temperatures,$$\chi =\frac{M}{B}\cong \frac{NJ\left(J+1\right){g}^{2}{\mu}_{B}^{2}}{3kT}=\frac{N{p}^{2}{\mu}_{B}^{2}}{3kT}=C/T$$$$\begin{array}{ccc}{p}^{2}& =& {g}^{2}J\left(J+1\right)\Rightarrow p=g{\left[J\left(J+1\right)\right]}^{1/2}\\ C& \equiv & Curieconstant\end{array}$$
13.2.2 Calculation of Total Angular Momentum $J$
 Atoms with filled shells have no magnetic moment. Example:
$$1{s}^{2}2{s}^{2}2{p}^{6}3{s}^{2}3{p}^{6}3{d}^{10}$$
 The spins arrange themselves so as to give the maximum possible $S$ consistent with Hund's rule
 Pauli Principle: no two electrons in the same system can have the same quantum numbers $n$, $l$, ${m}_{l}$, and ${m}_{s}$
13.2.3 Spectroscopic Notation
Example Orbital Configurations
13.2.4 Paramagnetic Susceptibility
Curie Law for paramagnetic susceptibility: $\chi =C/T$
13.2.5 Calculation of $J$
 Rules for calculating $J$:
Value of $J$

Condition

$LS$

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

$$L=\sum {m}_{l}\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}S=\sum {m}_{s}$$
 $L$ is the orbital angular momentum
13.2.6 Spin Orbit Interactions
 For multielectron atoms,
 Orbital angular momentum: $L={\sum}_{i}{L}_{i}$
 Total spin angular momentum: $S={\sum}_{i}{S}_{i}$
 Total angular momentum: $\hslash J$, where $J=L+S$
 Treat $J$, $L$, and $S$ as vectors. $L$ and $S$ precess around $J$; $J$ remains constant (“conserved”)
 Note: for iron metal group, $J=S$; only the spin contributes. Orbital momentum is “quenched.”
13.2.7 Effective Magnetic Number
 Effective magnetic number:
$$P=g{\left[J\left(J+1\right)\right]}^{1/2}$$
$$J\sim S\phantom{\rule{1em}{0ex}}Fortransitionmetals,"orbitalangularmomentumquenched"$$
$$P=2{\left[S\left(S+1\right)\right]}^{1/2}$$
$$P=2{\left[\frac{1}{2}\cdot \left(\frac{3}{2}\right)\right]}^{1/2}=1.73$$
$$P=2{\left[2\left(3\right)\right]}^{1/2}=4.90$$
13.2.8 Paramagnetic Properties of Metals
 For paramagnetic solids with $N$ magnetic ions
$$M=\frac{N{\mu}^{2}B}{k{T}_{F}}$$
 For metals, not all spins are accessible
$$M\cong \frac{{N}_{eff}{\mu}^{2}B}{kT}\Rightarrow {N}_{eff}\cong \frac{NkT}{k{T}_{F}}$$
$$M\cong \frac{N{\mu}^{2}B}{k{T}_{F}}\phantom{\rule{1em}{0ex}}Temperatureindependent$$
13.2.9 Band Model
 Total magnetization: $M=\left({N}_{+}{N}_{}\right)\mu $
 Total number of electrons with “up” spins (${N}_{+}$)
$$\begin{array}{ccc}{N}_{+}& =& \frac{1}{2}{\int}_{{\mu}_{B}}^{{E}_{F}}dE\underset{Fermifunction}{\underset{\u23df}{f\left(E\right)}}\underset{Densityofstates}{\underset{\u23df}{D\left(E+\mu B\right)}}\\ & & \end{array}$$
 Taking a Taylor expansion of the density of states for ${N}_{+},$
$$\begin{array}{ccc}{N}_{+}& \approx & \frac{1}{2}{\int}_{0}^{{E}_{F}}dEf\left(E\right)D\left(E\right)+\frac{1}{2}{\int}_{0}^{{E}_{F}}dEf\left(E\right)\frac{dD\left(E\right)}{dE}\mu B\\ & \approx & \frac{1}{2}{\int}_{0}^{{E}_{F}}dEf\left(E\right)D\left(E\right)+\frac{1}{2}{\int}_{0}^{{E}_{F}}f\left(E\right)dD\left(E\right)\mu B\\ & \approx & \frac{1}{2}{\int}_{0}^{{E}_{F}}dEf\left(E\right)D\left(E\right)+\frac{1}{2}D\left({E}_{F}\right)\mu B\end{array}$$
 Repeat the procedure for ${N}_{}$
 Taylor series expansion of the density of states for ${N}_{}$
$$\begin{array}{ccc}{N}_{}& =& \frac{1}{2}{\int}_{{\mu}_{B}}^{{E}_{F}}dEf\left(E\right)D\left(E{\mu}_{b}\right)\\ & \approx & \frac{1}{2}{\int}_{0}^{{E}_{F}}d\left(E\right)f\left(E\right)D\left(E\right)\frac{1}{2}D\left({E}_{F}\right)\mu B\end{array}$$
$$M=\left({N}_{+}{N}_{}\right)\mu =D\left({E}_{F}\right){\mu}^{2}B$$
$$D\left({E}_{F}\right)=\frac{3N}{2{E}_{F}}=\frac{3N}{2k{T}_{F}}$$
$$M=\frac{3N{\mu}^{2}B}{2k{T}_{F}}\phantom{\rule{1em}{0ex}}\chi =\frac{\partial M}{\partial B}=\frac{3N{\mu}^{2}}{2k{T}_{F}}\phantom{\rule{1em}{0ex}}IndependentofT!$$
13.2.10 Multivalent Effects
 High density of states near Fermi level.
 dband contribution to $\chi $
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” ${B}_{E}$
$$H\Psi =E\Psi \phantom{\rule{1em}{0ex}}with\phantom{\rule{1em}{0ex}}H=K+{V}_{ex}$$
$${V}_{ex}\equiv ExchangePotential$$
 Mean field approximation: ${B}_{E}=\lambda M$ where $\lambda \equiv Weissconstant$
14.1 Ferromagnetic Phase Transition
 Above the transition temperature, ferromagnetism gives way to paramagnetism
$$T>{T}_{C}\phantom{\rule{1em}{0ex}}Paramagnetism$$
$$T<{T}_{C}\phantom{\rule{1em}{0ex}}Ferromagnetism$$
 Magnetization: $M={\chi}_{p}\left({B}_{A}+{B}_{E}\right)$
$$\begin{array}{ccc}{\chi}_{P}& \equiv & Paramagneticsusceptibility\\ {B}_{A}& \equiv & Appliedfield\end{array}$$
14.2 Molecular Field
 For paramagnetism, we had ${\chi}_{p}=C/T$
 “Molecular” or “exchange” field for ferromagnetism: ${B}_{E}=\lambda M$
$$M={\chi}_{P}\left({B}_{A}+\lambda M\right)\Rightarrow M=\frac{C}{T}\left({B}_{A}+\lambda M\right)$$
$$M\left(1\frac{C\lambda}{T}\right)=\frac{C{B}_{A}}{T}$$
$$M\left(TC\lambda \right)=C{B}_{A}$$
$$\chi =\frac{M}{{B}_{A}}=\frac{C}{\left(TC\lambda \right)}$$
 Taking $C\lambda ={T}_{C}$,
$$\chi =\frac{C}{T{T}_{C}}\phantom{\rule{1em}{0ex}}forT{T}_{C}$$
$$Weissconstant:\lambda ={T}_{C}/C$$
$$\chi \propto \frac{1}{(T{T}_{C}{)}^{1.33}}\phantom{\rule{1em}{0ex}}Scalingtheory$$
 Calculation of field parameter:
$$\lambda =\frac{{T}_{C}}{C}={T}_{C}\underset{CurieConstant}{\underset{\u23df}{\frac{3k}{N{g}^{2}S\left(S+1\right){\mu}_{b}^{2}}}}$$
 For iron, ${T}_{C}\cong 1000K$, $g=2$, $S=1$, $(3{d}^{6}4{s}^{2}\to 3{d}^{8}$), $\lambda =5000$. For a saturation magnetization of ${M}_{S}=1700$ Gauss,
$${B}_{E}=\lambda {M}_{S}$$
$${B}_{E}=5000\times 1700\approx 1{0}^{7}Gauss=1{0}^{3}Tesla$$
 Highest superconducting magnet: 40 Tesla
Material

${T}_{C}$ (K)

Fe (bcc)

1043

Co

1388

Ni

627

Note: fcc iron is not magnetic
14.2.1 Prediction of ${T}_{C}$
 Heisenberg model. For atoms $i$ and $j$ with spins ${S}_{i}$ and ${S}_{j}$, the ordering energy $U$ is
$$U=2{J}_{ex}\overrightarrow{{S}_{i}}\cdot \overrightarrow{{S}_{j}}$$
$${J}_{ex}\equiv Exchangeenergyintegral$$
$${J}_{ex}=\frac{3k{T}_{C}}{2ZS\left(S+1\right)}\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}{T}_{C}=\frac{2ZS\left(S+1\right){J}_{ex}}{3k}$$
$$Z\equiv Numberofneighbors$$
14.2.2 Temperature Dependence of $M\left(T\right)$ for Ferromagnetism
 Consider spin $1/2$ system. Just as for paramagnetism, $M=N\u27e8\mu \u27e9$
$$M=N\mu tanh\left(\frac{\mu B}{kT}\right)$$
$$WhereB={B}_{A}+{B}_{E}$$
 For a ferromagnet, $B\approx {B}_{E}=\lambda M,where{B}_{E}\equiv Molecularfield$.
$$\therefore M=N\mu tanh\left(\frac{\mu \lambda M}{kT}\right)=N\mu tanh\left(x\right)\phantom{\rule{1em}{0ex}}with\phantom{\rule{1em}{0ex}}x\equiv \frac{\mu \lambda M}{kT}$$
14.2.3 FerromagneticParamegnetic Transition
 Define reduced magnetization $m$ and reduced temperature $t$
$$m\equiv \frac{M}{N\mu}\phantom{\rule{1em}{0ex}}t\equiv \frac{kT}{N{\mu}^{2}\lambda}$$
$$\therefore \underset{LHS}{\underset{\u23df}{m}}=\underset{RHS}{\underset{\u23df}{tanh\left(m/t\right)}}$$
 Solve the transcendental equation graphically by plotting both the left hand side (LHS) and right hand side (RHS) versus $m/t$
14.2.4 FerromagneticParamagnetic Transition
 From the Landau model, as $T$ increases, $M\left(T\right)$ decreases
 Second order phase transition: continuous change in $M\left(T\right)/M\left(0\right)$ versus $T$ to ${T}_{C}$
14.2.5 FerromagneticParamagnetic Transition
$$tanh\left(m/t\right)=tanh\xi \cong 12{e}^{2\xi}$$
 Define change in magnetization with temperature, $\Delta M\equiv M\left(0\right)M\left(T\right)$. For large $\xi $ (low $T$),
$$tanh\xi \cong 12{e}^{2\xi}\approx 1$$
$$\therefore \Delta M=2N\mu {e}^{2\xi}=2N\mu {e}^{2\mu \lambda M/kT}$$
$$\Delta M=2N\mu {e}^{2{\mu}^{2}\lambda N/kT}$$
$$\Delta M=2N\mu exp\left(2\lambda N{\mu}^{2}/kT\right)=2N\mu exp\left(2{T}_{C}/T\right)$$
14.2.6 FerromagneticParamagnetic Transition
Change in magnetization with respect to $T=0$ value versus temperature
$$\Delta M=2N\mu exp\left(2\lambda N{\mu}^{2}/kT\right)=2N\mu exp\left(2{T}_{C}/T\right)$$
14.2.7 Ferromagnetism of Alloys
14.2.8 Transition Metals
$d$ Orbital Configurations of Transition Metals
14.2.9 Ni Alloys
 Atomic moment of nickel transition metal alloys changes with composition
 For nickel alloyed with copper, magnetization disappears at 60% Cu
14.2.10 Band Model
 For nickel, $3d$ bands and $4s$ bands are partially occupied
 Net spin for Ni is $0.6{\mu}_{B}$. Only the $d$ bands contribute in this case
 Magneton numbers
 Fe: 2.2${\mu}_{B}$
 Co: 1.7${\mu}_{B}$
 For iron,
 Can copper be ferromagnetic?
superconducting properties?
$$YB{a}_{2}CuO$$
14.2.11 Spin Waves
 Spin waves: low energy excitations
 Lower energy excited state is obtained by having spins arranged at an angle $\theta $ with respect to neighboring spins (“spin waves” or “magnons”)
 Precession angle depends on angle of neighbor
14.2.12 Magnon Dispersion
 Magnons are particles and have an $\omega $ versus $q$ dispersion relation
$$\omega =\frac{4{J}_{ex}{S}^{2}}{\hslash}{sin}^{2}\frac{aq}{2}$$
$$\begin{array}{ccc}{J}_{ex}& \equiv & Exchangeenergyterm\\ S& \equiv & Totalspinangularmomentum\\ a& \equiv & Distancebetweenatomswithmagneticmoments\\ q& \equiv & Wavevector\end{array}$$
14.3 Ferrimagnetic Order in Magnetic Oxides
 Iron oxide ($F{e}_{3}{O}_{4}$) consists of two sublattices:
$$F{e}_{3}{O}_{4}\to FeO\cdot F{e}_{2}{O}_{3}$$
Sublattice

Fe Oxid. State

e${}^{}$ Config.

$S$

$gJ$ (${\mu}_{B})$

(A) FeO

2+

3d${}^{6}$

2

4

(B)$F{e}_{3}{O}_{4}$

3+

3d${}^{5}$

$\frac{5}{2}$

5

 Total expected magnetic moment per formula unit: $2\times 5+4=14\phantom{\rule{6px}{0ex}}{\mu}_{B}$
 Measured magnetic moment per formula unit: $4.1\phantom{\rule{6px}{0ex}}{\mu}_{B}$ (due to ferrimagnetic ordering)
14.3.1 Magnetic Oxides
 Calculation of saturation magnetization for NiOFe${}_{2}$O${}_{3}$ (nickel ferrite)
 Unit cell: inverse spinel with 8 Ni${}^{2+}$ ions (B sites) and 16 Fe${}^{3+}$ ions (A sites)
Ion

$g\sqrt{J\left(J+1\right)}\phantom{\rule{0ex}{0ex}}\left({\mu}_{B}\right)$

Ni${}^{2+}$

2

Fe${}^{3+}$

5

 The Fe${}^{3+}$ ions have antiparallel magnetic moments
$$\therefore net2{\mu}_{B}{performulaunitduetoNi}^{2+}$$
 For 8 Ni${}^{2+}$ ions per formula unit each with $2{\mu}_{B}$ magnetic moment,
$${M}_{S}=\frac{8\times 2{\mu}_{B}}{cellvolume}=\frac{\left(16\right)\left(9.27\times 1{0}^{24}\right)}{(8.37\times 1{0}^{10}{)}^{3}}=2.5\times 1{0}^{5}A\cdot {m}^{1}$$
 Calculation of saturation magnetization for Fe${}_{3}$O${}_{4}$ (magnetite)
 Unit cell: inverse spinel with 8 Fe${}^{3+}$ ions on tetrahedral sites, 8 Fe${}^{3+}$ ions on octahedral sites, and 8 Fe${}^{2+}$ ions on octahedral sites.
Ion

Site

$g\sqrt{J\left(J+1\right)}\phantom{\rule{0ex}{0ex}}\left({\mu}_{B}\right)$

Fe${}^{3+}$

Tetrahedral

5

Fe${}^{3+}$

Octahedral

5

Fe${}^{2+}$

Octahedral

4

 Fe${}^{3+}$ ions on tetrahedral sites have magnetic moments aligned antiparallel to those of Fe${}^{3+}$on the octahedral sites
$$\therefore net2{\mu}_{B}{performulaunitduetoFe}^{2+}$$
 For 8 Fe${}^{2+}$ ions per formula unit each with $4{\mu}_{B}$ magnetic moment,
$${M}_{S}=\frac{8\times 4{\mu}_{B}}{cellvolume}=\frac{\left(32\right)\left(9.27\times 1{0}^{24}\right)}{(8.40\times 1{0}^{10}{)}^{3}}=5.0\times 1{0}^{5}A\cdot {m}^{1}$$
14.3.2 Magnetization and Hysteresis
 In ferromagnetic materials, the magnetization is a nonlinear function of the applied magnetic field.
 Important terms describing the hysteretic behavior:
 ${H}_{C}$: coercive field. The applied magnetic field for which the flux density $B$ disappears
 ${B}_{S}$: saturation flux density. The maximum flux density measured when all of the atomic moments are aligned with the applied magnetic field.
 ${B}_{R}$: remnant flux density. The magnetization remaining in the material after reaching the saturation flux density and removing the field.
 Ferromagnetic domain: a region in a ferromagnetic material over which all magnetic moments are aligned
 Domains exist in in the demagnetized state in order to minimize the large magnetostatic energy associated with single domains
14.4 Domains and Walls
 Domain wall: region between adjacent domains over which the direction of the local magnetic moment changes
 It takes energy to form and to move domain walls
14.4.1 Energy of Bloch Domain Walls
 Energy $U$ (Heisenberg Model):
$$U=2{J}_{ex}{\overrightarrow{S}}_{i}\cdot {\overrightarrow{S}}_{j}$$
 Consider the spin vectors by a classical model, i.e. ${\overrightarrow{S}}_{i}\cdot {\overrightarrow{S}}_{j}=S{}^{2}cos\phi $. Take a Taylor series expansion of $cos\phi $ about $\phi =0$
$$U=2{J}_{ex}S{}^{2}cos\phi \approx 2{J}_{ex}S{}^{2}\left(1\frac{1}{2}{\phi}^{2}\right)$$
 The angledependent exchange energy between two adjacent spins at an angle $\phi $ is then
$${w}_{ex}={J}_{ex}S{}^{2}{\phi}^{2}$$
 For a rotation of the magnetic moment by $\pi $ radians in $N$ steps, ${w}_{ex}={J}_{ex}\leftS{}^{2}\right(\pi /N{)}^{2}$. For a Bloch wall of $N$ spins, the total energy is then
$$N{w}_{ex}={J}_{ex}S{}^{2}\frac{{\pi}^{2}}{N}$$
 Anisotropy energy limits the Bloch wall size since the spins are arranged away from the direction of easy magnetization. (The more that the spins point the “wrong way,” the higher the energy).
14.5 Anisotropy of Magnetization
 Alignment of magnetic moments is more favorable for certain crystalline directions, 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 field with respect to the crystalline direction
 “Soft” direction: crystalline direction with low coercivity (${H}_{C}$)
 “Hard” direction: crystalline direction with high coercivity (${H}_{C}$)
14.5.1 Anisotropy of Magnetization
 Rotation of domains changes the overlap of the electron clouds of neighboring atoms, resulting in different exchange energy ${J}_{ex}$
 Anisotropy energy ${U}_{k}$
$${U}_{k}={k}_{1}^{\text{'}}{sin}^{2}\theta +{k}_{2}^{\text{'}}{sin}^{4}\theta $$
 Easy axis depends on values of ${k}_{1}^{\text{'}}$ and ${k}_{2}^{\text{'}}$
14.6 Ferrimagnetic Ordering  Exchange Terms
 Exchange term for neighboring Fe atoms is negative for atoms on different sublattices and positive for atoms on the same sublattice.
 Different sign of exchange term and different number of Fe atoms per unit volume on each sublattice account for ferrimagnetic ordering in Fe${}_{3}$O${}_{4}$
 Exchange terms for $A$ and $B$ sublattices:
$${J}_{AA}>0\phantom{\rule{1em}{0ex}}{J}_{AB}<0\phantom{\rule{1em}{0ex}}{J}_{BB}>0$$
14.6.1 Exchange Terms and Susceptibility
$$\begin{array}{ccc}{T}_{C}& \equiv & Curietemperature\\ {T}_{N}& \equiv & Neeltemperature\end{array}$$
14.6.2 Structure Dependence of ${J}_{ex}$
15 Electrooptic and Nonlinear Optical Materials
 Nonlinear optical materials have a nonlinear relation between the polarization density ($P$) and electric field ($E$)
 Expand electric susceptibility in higher order terms
$$P={\epsilon}_{0}\left[{\chi}^{\left(1\right)}E+{\chi}^{\left(2\right)}{E}^{2}+{\chi}^{\left(3\right)}{E}^{3}\right]$$
$$\begin{array}{ccc}{\chi}^{\left(1\right)}& \equiv & Linearsusceptibility\\ {\chi}^{\left(2\right)}& \equiv & Quadraticsusceptibility\\ {\chi}^{\left(3\right)}& \equiv & Cubicsusceptibility\end{array}$$
 ${\chi}^{\left(2\right)}$ is nonzero in noncentrosymmetric materials. Examples of ${\chi}^{\left(2\right)}$ effects: linear (Pockels) electrooptic effect, second harmonic generation (frequency doubling)
 The lowest order nonlinearity for cubic materials is ${\chi}^{\left(3\right)}$. Examples of ${\chi}^{\left(3\right)}$ effects: quadratic (Kerr) electrooptic effect, third harmonic generation (frequency tripling)
15.1 Frequency Doubling
 In frequency doubling, two photons with energy $h{\nu}_{1}$ combine to form one photon with energy $h{\nu}_{2}$. The total energy is conserved.
 Sending a beam with photon energy $h{\nu}_{1}$ into a nonlinear optical (NLO) material results in two output beams with energies $h{\nu}_{1}$ and $h{\nu}_{2}$.
 Examples:
 Nd:YAG laser with 1.10$\mu $m (infrared) radiation $\underset{NLO}{\overset{}{}}$ 0.55$\mu $m (green)
 Blue laser by frequency doubling GaAs lasers
15.2 Nonlinearity in Refractive Index
 Velocity of light in matter: $v=c/n$
$$\begin{array}{ccc}c& \equiv & {Speedoflightinfreespace(3\times 10}^{8}m/s)\\ n& \equiv & Refractiveindex\end{array}$$
 Refractive index ($n$) in nonlinear material:
$${n}^{2}=1+{\chi}^{\left(1\right)}+{\chi}^{\left(2\right)}+{\chi}^{\left(3\right)}+\cdots $$
 Dependence of refractive index on light intensity:
$$n={n}_{0}+{n}_{2}I$$
$$\begin{array}{ccc}{n}_{0}& \equiv & Linearand\chi ={\epsilon}_{r}1\\ {n}_{2}& \equiv & Nonlinearindex\\ I& \equiv & Intensity\end{array}$$
 Dielectric constant: ${\epsilon}_{r}\cong {n}^{2}$
15.3 ElectroOptic Modulators
 The optical properties of noncentrosymmetric crystals are modified by applying an external electric field.
$$\Delta \left(\frac{1}{{\epsilon}_{r}}\right)=\Delta \left(\frac{1}{{n}^{2}}\right)=rE$$
 Expanding $n=n\left(E\right)$ in a Taylor series expansion about $E=0$, the following expression is obtained
$$\Delta n=\frac{1}{2}{n}^{3}rE$$
$$r\equiv Electroopticcoefficient\left[pm/V\right]$$
 Example material: LiNbO${}_{3}$, $r=31$ pm/V, $n=$ 2.29. Fields of $1{0}^{6}1{0}^{7}$ V/m are easily obtained in microphotonic devices (110 V across $\sim 1\phantom{\rule{10px}{0ex}}\mu $m)
 Other materials: electrooptic polymers ($r=2050$ pm/V), BaTiO${}_{3}$ ($r=300$ pm/V)
15.4 Optical Memory Devices
 Photorefractivity: change in refractive index by light. Light empties trap states, creating a space charge region that locally changes the refractive index.
 Volume holography:
 Two plane waves (object and reference beams) incident upon a photosensitive material. A standing wave is formed which is preserved in the material as an interference pattern
 Material contains intensity and phase information and the object can be reconstructed
 The difference between hologram and a photograph is that phase information is preserved in a hologram
15.5 Volume Holography
 Pattern in material creates a dielectric grating or spatial modulation in the refractive index and dielectric constant
$$\begin{array}{ccc}{\epsilon}_{r}& =& {\epsilon}_{r0}+{\epsilon}_{r1}cos\left(2kysin\theta \right)\\ k& \equiv & Magnitudeoflightwavevector\end{array}$$
 Period of dielectric grating:
$$\Lambda =\frac{2\pi}{2ksin\theta}=\frac{\lambda}{2nsin\theta}\phantom{\rule{1em}{0ex}}with\phantom{\rule{1em}{0ex}}k=2\pi n/\lambda $$
15.6 Photorefractive Crystals
 Electrooptic and photoconductive  organic and inorganic materials
 Mechanism for photorefraction:
 Generation of electrons and holes
 Transport charges through the material
 Trap the charges
 Local field of trapped charges changes the refractive index $n$
 Spatially varying charge distribution results in spatially varying field. Field changes $n$ through electrooptic coefficient $r$; larger $r$, larger photorefractive effect.
$$\Delta n=\frac{1}{2}{n}^{3}rE$$
15.7 Photorefractive Crystals
Steps in creation of periodic dielectric constant
15.8 Phase Conjugation  All Optical Switching
 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 and the photorefractive effect
 Beam 2 is diffracted by the grating, forming beam 3
15.9 AcoustoOptic Modulators
 Photoelastic effects: strain causes a change in the refractive index
$$\Delta \left(\frac{1}{{\epsilon}_{r}}\right)=pS\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}\Delta \frac{1}{{n}^{2}}=pS$$
$$\begin{array}{ccc}S& \equiv & Strain\\ p& \equiv & Photoelasticconstant\\ n& \equiv & Refractiveindex\end{array}$$
 Acoustooptic effect is used in Bragg cells to create a grating with period $\Lambda $ by launching an acoustic wave. Light is diffracted from the grating.
$$\Lambda =\frac{\lambda}{2nsin\theta}$$
15.10 Integrated Optics
 Integrated optics: passive and active optical elements incorporated onto a wafer of material for local manipulation of light.
 Passive elements: elements that do not require power, such as waveguides, couplers, and filters
 Active elements: elements that require power, such as switches, modulators, directional couplers, amplifiers, lasers
 Photonics: analogue of electronics. Photonics encompasses the control of photons.
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 lower losses
15.12 Example: Phase Shifter
 An electrooptic 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 the electrooptic effect
$$\Delta n=\frac{1}{2}{n}^{3}rE=\frac{1}{2}{n}^{3}r\frac{V}{d}$$
 Phase delay of optical wave due to the electrooptic effect:
$$\Delta \phi =\frac{2\pi}{\lambda}\Delta nL$$
15.13 Example: LiNbO${}_{3}$ Phase Shifter
 Electric field of $1{0}^{6}$ V/m (10 V across 10$\phantom{\rule{6px}{0ex}}\mu $m) yields $\Delta n=1.86\times 1{0}^{4}$ for LiNbO${}_{3}$
 For a phase shift $\Delta \phi =\pi $ with operation at $\lambda =1.5\phantom{\rule{6px}{0ex}}\mu $m,
$$\Delta \phi =\pi =\frac{2\pi}{\lambda}L\Delta n$$
$$L=0.4\phantom{\rule{6px}{0ex}}cm$$
 Long interaction lengths are needed; materials with higher electrooptic coefficients are needed for more compact devices
 The voltage required for a $\pi $ phase shift is the halfwave voltage ${V}_{\pi}$
$${V}_{\pi}=\frac{\lambda d}{{n}^{3}rL}$$
16 Superconductivity
 Superconductivity: a phenomena in which a material undergoes a transition to zero resistivity with zero magnetic induction (Meissner effect) below a critical temperature ${T}_{C}$.
 At temperatures below ${T}_{C}$ application of a critical magnetic field ${H}_{C}$ will cause a transition to the normal state
16.1 BCS Theory of Superconductivity
 BCS Theory: quantum mechanical theory of superconductivity formulated by Bardeen, Cooper, and Schreiffer in 1957 (Nobel Prize in 1972).
 Electronphonon coupling leads to the formation of electron pairs (Cooper pairs)
 BCS Theory accounts for observed properties of superconductivity, including the energy gap, Meissner effect, critical temperature, and quantization of magnetic flux through a superconducting ring
16.2 Density of States and Energy Gap
 The energy gap in superconductors is caused by electronelectron interactions rather than electronlattice interactions for dielectrics
 The energy gap is maximum at 0 K and decreases to zero at the transition temperature.
16.3 Heat Capacity of Superconductors
 The entropy of superconductors decreases upon cooling below ${T}_{C}$, resulting in electronic ordering.
 In the superconducting state, for $T\ll {T}_{C}$,
$${H}_{C}={H}_{0}\left[11.07(T/{T}_{C}{)}^{2}\right]$$
16.4 Tunneling in Superconductor Junctions
16.5 SemiconductorInsulatorSemiconductor Junction
 Tunneling across a thin oxide junction between two superconductors with energy gaps $2{\Delta}_{1}$ and $2{\Delta}_{2}$
 Nobel Prize in physics, Ivan Giaever, 1973
16.6 Type I and Type II Superconductors
 Type I: superconductor is diamagnetic in an applied magnetic field up to the critical field ${H}_{c}$
 Type II: superconductor exhibits two magnetic behaviors. Up to a lower critical field ${H}_{C1}$ the superconductor is diamagnetic, and between ${H}_{C1}$ and an upper critical field ${H}_{C2}$, the flux density is not equal to zero (“incomplete” Meissner effect).
 Type II superconductors tend to be alloys or transition metals with high room temperature resistivity
 Type II superconductors are used in high field magnets, magnetic resonance imaging (MRI) applications
16.7 High ${T}_{C}$ Superconductors
 Highest values for ${T}_{C}$ have been reported in cuprates
 2 layer cuprate compounds support 2D conduction
16.8 Cuprate Superconductors
 Theory: pairing is involved. Nature of pairing? Spin Waves?
 $2\Delta \left(0\right)\approx 6k{T}_{C}$
 Future applications: current carrying wires, Maglev trains
17 3512 Problems
 An abrupt Si pn junction has${N}_{a}=1{0}^{18}c{m}^{3}$ on one side and${N}_{d}=1{0}^{15}c{m}^{3}$on the other.
 Calculate the Fermi level position at 300K on both sides.
 Draw an equilibrium band diagram for the junction.
 Determine the contact potential${\Phi}_{o}$ for this junction.
 A silicon${p}^{+}n$ junction$1{0}^{2}c{m}^{2}$ in area had${N}_{d}=1{0}^{15}c{m}^{3}$ doping on the nside. Calculate the junction capacitance with a reverse bias of 10V.
 For metallic aluminum, calculate:
 The valence electron density.
 The radius of the Fermi sphere${k}_{F}\mathrm{.}$
 Fermi energy in eV.
 From the Schrodinger equation for a quantum well, show that the wave vector is equal to$n\pi /L$ where L is the well width.
 Calculate the energy of light emitted from a 10 nm wide AlGaAs/GaAs quantum well structure that is photoexcited with 2.5 eV laser light.
 What is the luminescent energy for a CdSe quantum dot with a 2 nm radius.
 For a MOSFET device briefly describe how the three types of device work: a) enhancement mode b) depletion mode c) inversion mode.
 Calculate the capacitance of an MOS capacitor with a 10 nm thick$Hf{O}_{2}$ dielectric oxide. What is the ratio of capacitances for${C}_{Hf{O}_{2}}/{C}_{Si{O}_{2}}$. The relatie dielectric constant for$Hf{O}_{2}$is 25.
 Problem 9.9 in Solymar and Walsh
 Problem 9.14 in Solymar and Walsh
 Problem 9.16 in Solymar and Walsh
 Problem 12.10 in Solymar and Walsh
 Consider a quantum cascade laser (QCL) made from GaAs and GaAlAs. What well thickness is needed for laser emission at 3 microns?
 Derive the expression for the average value of the dipole moment. Show that it is given by:
$<\mu >=\mu [$coth$a\frac{1}{a}]$
 The saturation polarization${P}_{s}$ of$PbTi{O}_{3}$, a ferroelectric, is$0.8$ coulombs/${m}^{2}$. The lattice constant is 4.1$\dot{A}$. Calculate the dipole moment of unit cell.
 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.
 Consider the frequency dependence of the atomic polarizability. The polarizability and its frequency dependence can be modeled as a damped harmonic oscillator. Derive the expression for$\alpha $ in this case.
The expression is given by:
$m\frac{dx}{d{t}^{2}}+b\frac{dx}{dt}+{\omega}_{0}^{2}x=e{\u03f5}_{loc}sin\omega t$
Plot$\alpha $ vs.$\omega $ for this case.
 Problem 4.6 in Solymar and Walsh.
 Problem 4.7 in Solymar and Walsh.
 Problem 4.8 in Solymar and Walsh.
 Problem 4.9 in Solymar and Walsh.
 Calculate the magnetic susceptibility of metallic copper. How does it compare to the measured value of 1.0?
 Calculate the effective magneton number p for$M{n}^{2+},C{o}^{2+}$. Show work.
 Consider Mn doped GaP. There are$1{0}^{20}M{n}^{2+}$ ions.
 What is the electron configuration$M{n}^{2+}$ in spectroscopic notation.
 Calculate its magnetic moment at saturation in Bohr magnetons.
 Calculate its magnetic susceptibility.
 For metallic Co, which has a Curie temperature of 1388 K, calculate the Weiss constant$\lambda \mathrm{.}$ Calculate the exchange constant in meV.
18 3512 Laboratories
18.1 Laboratory 1: Measurement of Charge Carrier Transport Parameters Using the Hall Effect
18.1.1 Objective
The purpose of this lab is to measure the electronic transport properties of semiconductors and semiconducting thin films using the ECOPIA Hall apparatus.
18.1.2 Outcomes
Upon completion of the laboratory, the student will be able to:
 Use a Hall effect apparatus to measure the mobility and carrier concentration in a semiconductor.
 Derive the equations that enable the extraction of fundamental materials parameters using the Hall effect.
 Describe the dependence of mobility on carrier concentration and temperature, and explain the origins of differences in mobilities between different semiconductors.
18.1.3 References
(1) M. Ali Omar, Elementary Solid State Physics;
(2) Solymar & Walsh, Electrical Properties of Materials;
(3) MSE 3511 Lecture Notes; and
(4) the NIST web page: http://www.nist.gov/pml/div683/hall.cfm
18.1.4 PreLab Questions
 What is the Lorentz Force?
 What is the Hall effect and when was it discovered?
 Write the equation describing the force, FM, on a particle of charge q and with velocity v in a uniform magnetic field, B.
 Does the velocity of a charged particle (with nonzero initial velocity) in a uniform magnetic field change as a result of that field? If so, how? Does its speed change?
 What is the righthandrule?
 For a particle with negative charge, q, in the situation below, in what direction will the particle be deflected?

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

 By convention, current is defined as the flow of what sign of charge carrier?
 What is the electric field, E, in the situation below? What is the electrostatic force, ${F}_{E}$ on the particle if it has a charge q?

 Why are both a resistivity measurement and a Hall measurement needed in order to extract fundamental material parameters?
18.1.5 Experimental Details
The samples to be characterized include:
 “bulk” Si, GaAs, InAs (i.e. substrates ~ 400 microns thick)
 “thin films” of InAs and doped GaAs, grown on semiinsulating GaAs substrates
 Indium tin oxide (ITO) on glass.
You may have the opportunity to make additional samples. For contacts on ntype GaAs, use InSn solder; for contacts on ptype GaAs, use InZn solder. Most samples are mounted on minicircuit boards for easy insertion into the apparatus.
18.1.6 Instructions/Methods
See Instructor
18.1.7 Link to Google Form for Data Entry
https://docs.google.com/forms/d/1lrAokPI1vIJa8pmLx4FVCqHh80PKgGr8oDD6eVzx_w/viewform
18.1.8 Lab Report Template
 Balance the forces (magnetic and electric) acting on a charged particle in a Hall apparatus to derive the equation that describes the Hall Coefficient in terms of the applied current and magnetic field and measured Hall voltage. Show your work. See hints at the end.
 Apply data from a sample measurement to test the equation derived in (1). Show your work.
 How is the carrier concentration related to the Hall Coefficient? What is the difference between bulk and sheet concentration?
 Calculate the carrier concentration (bulk and sheet) using the sample data and the equations derived above. Show your work.
 Does the mobility exhibit any dependence on the carrier concentration? Discuss briefly; include observations from lab for the same material & type (e.g. nGaAs).
 What is the origin of the difference in mobilities between the ntype and ptype samples, assuming that the doping levels are similar?
 How do carrier mobilities compare for different materials? Use the pooled data to compare mobility as a function of material (as well as carrier type). Explain your observations.
 When the magnetic field, current, and sample thickness are known, the carrier concentration and type may be determined. Conversely, if the current, sample thickness and carrier concentration are known, the magnetic field may be determined. A device that measures these parameters, known as a Hall Probe, provides a way to measure magnetic fields. Write an expression that relates these parameters. Which is the more sensitive (higher ratio of mV/Tesla) Hall probe – the bulk InAs or bulk GaAs sample? Show your work. (Note – you should have recorded average Hall voltage for a given current. How do these compare?)
18.1.9 Hints for derivation
 For the example below, what VH would you need to apply to make the particle continue on in a straight line throughout the sample? (B is uniform throughout the specimen. Hint: Balance the magnetic and electric forces on the particle.)
 The current density, $J$, can be expressed in two ways: J=i/(A), and J=nqv, where i is the total current passing through a crosssectional area A, n is the concentration of carriers per unit volume in the material passing the current, q is the charge on each carrier, and v is the drift velocity of the carriers. If you knew i, A, VH, d, B, and q from the situation illustrated above, what expression tells you n?
 The “Hall coefficient” of a material, ${R}_{H}$, is defined as the Hall electric field, ${E}_{H}$, per current density, per magnetic field, ${R}_{H}$=${E}_{H}$/(J*B) Using the equations given and derived thus far, express the Hall coefficient in terms of just q and n.
18.2 Laboratory 2: Diodes
18.2.1 Objective
The purpose of this lab is to explore the IV characteristics of semiconductor diodes (including light emitting diodes (LEDs) and solar cells), and the spectral response of LEDs and lasers.
18.2.2 Outcomes
Upon completion of the laboratory, the student will be able to:
 Measure diode IV characteristics and relate them to band diagrams.
 Fit IV data to the diode equation, extract relevant parameters, and relate these to materials constants.
 Determine the open circuit voltage and shortcircuit current of the solar cell.
 Describe the dependence of emission wavelength on bandgap and describe origins of spectral broadening.
 Describe how lasers differ from LEDs in design and performance.
18.2.3 Prelab Questions
 What expression describes the IV characteristics of a diode?
 Sketch the IV characteristics of a diode and label the sections of the curve corresponding to zero bias (1), forward bias (beyond the builtin voltage) (2), and reverse bias(3), and then sketch the corresponding band diagrams for 1, 2, and 3.
 Sketch the IV characteristics of a Zener diode and the corresponding banddiagram for reverse bias.
 Sketch the IV characteristics of a pn junction with and without illumination. Label ${V}_{OC}$ and ${I}_{SC}$.
 Take pictures (use your phone) of lights around campus; try to get pictures of LEDs. Which do you think are LEDs (why?)?
 Why are fluorescent lights “white?” How white are they?
Experimental Details
The devices to be characterized include:
 Si diode
 Zener diode
 LEDs (different colors)
 Si solar cell
18.2.4 References
MSE 3512 Lecture Notes, Omar Chapter 7, Solymar & Walsh Chapter 9, 12, 13.
Instructions/Methods
Use multimeters and the Tektronix curve tracer for IV measurements. Use the Ocean Optics spectrometer to obtain spectral responses.
Station 1  CURVE TRACER
pn junction diode
Attach diode to “diode” slot on Curve Tracer. Measure both forward and reverse bias characteristics.
 (In lab) Measure and record the IV (currentvoltage) characteristics of the silicon diode over the current range 2μA to 50 mA in the forward bias condition. Pay particular attention to the region from 200 to 800 mV.
 (In lab) Measure the IV characteristics of the diode in the reverse bias condition.
 (Postlab) The ideal diode equation is: $I={I}_{sat}\left[exp\left(qV/kT\right)1\right]$. Note that for V>>kT/q, $I={I}_{sat}exp\left(qV/kT\right)$; however recombination of carriers in the space charge region leads to a departure from ideality by a factor m, where$I={I}_{0}\left[exp\left(qV/mkT\right)1\right]$ .
 Plot the data using $ln\left(I\right)$ vs. $V$ plot to determine $m$.
 Use the value of the applied voltage corresponding to ~ 1mA to solve for ${I}_{0}$ (which is too small to measure in our case.)
 (In lab) Repeat the forward bias measurement using “Store.” Now cool the diode and repeat. Sketch the two curves. (Postlab) Explain what you observe.
 Test red and green LEDs. Record the color and the turnon voltage.
 (Postlab) Compare the IV characteristics for the silicon diode and LEDs. Why would you choose Si over Ge for a rectifier? (Omar 7.21)
 (InLab) Test the Zener diode. Sketch. (Postlab) Compare to figure 9.28 (S&W).
Station 2: Solar Cells
 Use the potentiometer, an ammeter and voltmeter to determine the IV characteristics of a solar cell at different light levels. Determine the values of ${V}_{oc}$ (the open circuit voltage) and ${I}_{sc}$ (the shortcircuit current) under ambient light, then use the potentiometer to determine additional points on the IV curve. Record the results.
 Repeat the IV measurements for a higher light level. Record the light intensity measured with the photodiode meter. Area~1 cm2 : ____ Measure the area of the solar cell:______
 (PostLab) Estimate the fill – factor and the conversion efficiency of the solar cell.
Station 3: Spectrometer and Power Supply
Light Emitting Diodes
 Measure the spectral response (intensity vs. wavelength) of the light emitting diodes using the spectrometer.
 Cool the LEDs and observe the response. Qualitatively, what do the results suggest about the change in Eg vs. temperature? About emission efficiency vs. temperature? How do these results compare to the IV response of the cooled silicon diode?
 Record the peakwavelength and the fullwidthhalfmaximum (FWHM) for each diode.
 Calculate the bandgap that would correspond to the peak wavelength.
LED Color

Peak wavelength

${E}_{g}$(in eV, from peak wavelength)

Intensity

FW (left)

FW (right)

FWHM




































Laser














Table 1: Spectrometer and Power Supply
Semiconductor Lasers
 Attach the laser to the power meter and setup the power meter to measure intensity from the device. Slowly increase the voltage and record the IV characteristics vs. power output for the laser.
 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 IVpower data?
 Plot the light output (intensity) as a function of current. What is the threshold current? Label the regions of spontaneous and stimulated emission.
 Calculate the efficiency of the laser.
 Determine the bandgap of the laser material.
 Explain the change in the width of the spectral emission.
Station 4: Stereomicroscope
 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 emission when the device is on. Sketch what the device structure might look like.
 Sketch the structure of the laser observed under the stereomicroscope. Sketch what the device structure might look like.
18.3 Laboratory 3: Transistors
18.3.1 Objective
The purpose of this lab is explore the input/output characteristics of transistors and understand how they are used in common technologies.
18.3.2 Outcomes
Upon completion of the laboratory, the student will be able to:
 Measure the output characteristics of a few important transistors using a “curve tracer.”
 Qualitatively relate the characteristics to the pn junctions in the devices.
 Describe transistor function and performance in terms of appropriate gains.
 Identify applications of these devices in common technologies.
18.3.3 Prelab questions
Bipolar Transistor
 Sketch and label the banddiagrams for npn and pnp transitors.
 Sketch IV characteristics for this device as a function of base current.
 In what technologies are these devices used?
MOSFETS
 Sketch and label a MOSFET structure.
 Sketch IV characteristic for this device as a function of gate voltage. In your sketch of the ${I}_{SourceDrain}\u2013{V}_{SourceDrain}$ characteristics vs. gate voltage, ${V}_{gate}$, label the linear and saturated regions.
 What is threshold voltage? What structural and materials properties determine the threshold voltage?
 In what technologies are these devices used?
“Current Events”
 What materials developments have changed transistor technology in the past decade? (Hint: Search Intel highk; Intel transistors)
 What new types of devices are on the horizon? Hint: search IBM nanotubes graphene
18.3.4 Experimental Details
The devices to be characterized include the following
 pnp and npn bipolar junction transistors
 phototransistor
 metal oxide semiconductor field effect transistor (MOSFET).
References
MSE 3512 Lecture Notes, Omar Chapter 7; Solymar and Walsh Chapter 9
Instructions/Methods
Use the Tektronix curve tracer for IV measurements.
 Bipolar Transistors (npn and pnp)
 Attach an npn transistor to the Tshaped Transistor slot on the Curve Tracer (making sure that E, B, & C all connect as indicated on the instrument). Set the menu parameters to generate a family of IV curves. Keep the collectoremitter voltage ${V}_{CE}$ below 30V to avoid damaging the device.
 Sketch how E,B,C are configured in the device.
 Measure ${I}_{collector}$ (output current) for ${V}_{CE}>{V}_{saturation}$ as a function of base (input) current, ${I}_{base}$.
 PostLab: determine the transistor gains, $\alpha ={I}_{collector}/{I}_{emitter}$ and $\beta ={I}_{collector}/{I}_{base}$.
 Repeat for a pnp transistor.
 Phototransistor
 Measure the IV characteristics of the phototransistor under varying illumination intensity (using the microscope light source). Compare qualitatively to what you observed for the pnp and npn transistors.
 Postlab: discuss the mechanism by which the light influences the device current, and compare with the bipolar transistor operation.
 MOSFETS
 Attach MOSFET Device to the linear FET slot on the Curve Tracer, making sure that S,G,D are connected appropriately. Observe both the linear and saturated regions for ${I}_{SourceDrain}vs\mathrm{.}{V}_{SourceDrain}$.
 Vary the gate voltage step size and offset to estimate the threshold value of the gate voltage (the voltage at which the device turns on).
 Measure the ${V}_{SourceDrain}$ and ${I}_{SourceDrain}$ to extract ${R}_{SourceDrain}$ in the linear region as a function of ${V}_{Gate}$. You should take at least 4 measurements.
 Postlab, using the data from B, plot the conductance of the channel, ${G}_{SourceDrain}=1/{R}_{SourceDrain}$, vs. ${V}_{Gate}$ . You should observe a linear relationship following the equation ,$${G}_{SD}=\left(\frac{{\mu}_{n}W}{L}\right)\left(\frac{{C}_{OX}}{A}\right)\left({V}_{G}{V}_{T}\right)$$
where
${\mu}_{n}$ is the electron mobility and W, L, and A are the width, length and area of the gate respectively (A = W x L).
${C}_{OX}$ is the capacitance of the oxide, which can be measured using an impedance analyzer as a function of gate voltage. A plot is shown in Figure
18.5.
 Using ${\mu}_{n}$=1500 cm2/Vsec, C = $4.8x1{0}^{10}F$ (from the attached plot), plot of conductance vs. ${V}_{Gate}$, calculate L, the length of the gate.
 Determine ${V}_{threshold}$ from the above values. Compare with your observations from the curve tracer (part a).
18.4 Laboratory 4: Dielectric Materials
18.4.1 Objective
The objectives of this lab are to measure capacitance and understand the dependence on geometry and the dielectric constant, which may vary with temperature and frequency.
18.4.2 Outcomes
Upon completion of the laboratory, the student will be able to:
 Use an impedance analyzer to measure capacitance.
 Given the capacitance of a parallel plate capacitor, calculate the dielectric constant.
 For a known material, explain the microscopic origins of the temperature dependence of the dielectric constant.
 Understand how the dielectric constant and the index of refraction are related. Use the reflectance spectroscopy to fit the thickness, index of refraction and extinction coefficient of several thin films. Explore how the index of refraction is affected by composition and how this informs design of heterostructure devices.
18.4.3 Prelab questions
 What distinctions can you make between capacitance and the dielectric constant? What are the units of each?
 What is the relationship between the dielectric constant and the ‘relative’ dielectric constant?
 What is the dielectric constant (or permittivity) of ‘free space’ (or vacuum)?
 What is the relative dielectric constant of air? Water? Glass? $Si{O}_{2}$? Explain their relative magnitudes.
 For the capacitor shown in Figure 18.6, what is the expression that relates the capacitance and relative dielectric constant?
 What relationship determines the amount of charge stored on the plates of a capacitor when a specific voltage is applied to it?
 With the above answer in mind how could you measure the capacitance of a structure?
 Name two materials for which the capacitance (charge per unit voltage) is fixed. Name a material type or structure for which the capacitance is dependent on V.
 What is the relationship that gives the amount of energy stored on a capacitor?
 Why are batteries the primary storage medium for electric vehicles, rather than capacitors?
 Intel and their competitors are interested in both “lowk” and “highk” dielectrics.
 What defines the boundary between “low” and “high?”
 What drives the need for “highk” dielectrics? What other properties besides the dielectric constant are important characteristics of these materials?
 What drives the need for “lowk” dielectrics?
 Find one or two additional examples of technologies/devices that incorporate capacitors, and explain the function of the capacitor in that context.
18.4.4 Experimental Details
The dielectric constant of a number of different materials will be probed through measurements of parallel plate capacitance. Capacitors of solid inorganic dielectrics (in thin slabs) is accomplished through the evaporation of metal contacts on either side of the material. The total capacitance of a pn junction diode is measured through the device contacts. The capacitance of liquids is determined by filling a rectangular container between two electrodes, and neglecting the contributions of the container.
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 attached table.
18.4.7 Lab Report Template
Part I  Dielectric constants; Measuring capacitance with Impedance Analyzer
 Using the default frequency of 100kHz and zero bias voltage, measure capacitance and contact dimensions and then calculate the relative dielectric constant of the following materials:
Material

Area

Thickness

Capacitance

${\u03f5}_{r}\left(lab\right)$

${\u03f5}_{r}\left(lit\right)$

Glass slide (sample 1)






Lithium niobate ($LiNb{O}_{3})$






Glass sheet (sample 2)






Plexiglass






Table 2: Dielectric Constants
 Using the default frequency of 100 kHz, measure capacitance vs. voltage for several LEDs
LED=

LED=

LED=

Bias

Capacitance

Bias

Capacitance

Bias

Capacitance

















































Table 3: Capacitance
Determine the builtin voltage of the corresponding pn junction, ${\varphi}_{0}$, for each diode, and discuss the difference. See equation 7.64, pg. 362, Omar. Note that the equation may be rewritten as follows:
$$\frac{1}{{C}^{2}}=2\frac{\left({\varphi}_{0}{V}_{0}\right)\left({N}_{a}{N}_{d}\right)}{\u03f5e{N}_{a}{N}_{d}}$$
Plot $1/{C}^{2}$vs. ${V}_{0}$, and extrapolate the linear region (reverse bias) to the xintercept to determine the value of interest. Note that the slope depends on the doping concentration.
 Calculate the relative dielectric constant, based on capacitance measured in lab for:
 air (4.0inch x 3.5 inch plates, separated by _________ (measure this))
 water at room temperature (4.0 inch x 3.5 inch plate partially submerged; measured dimension: Height:________ x width___________x separation____________________
 ethanol
 Fill in the table below, for water at the different temperatures measured in lab, using the dimensions measured above. Determine the corresponding values of the relative dielectric constant,${\epsilon}_{r}$. Plot $\left({\u03f5}_{r}\u20131\right)/\left({\epsilon}_{r}+2\right)$ 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)

${\u03f5}_{r}$





























Comparison from CRC handbook
Temp (C)

${\u03f5}_{r}$

0

87.74

20

80.14

40

73.15

60

66.8

80

61

100

55.65

 Measure ice as a function of frequency & compare to Omar figure 8.10. Discuss.
Dimension: Height:____ x width_____x separation___
Freq

C (pF)

${\u03f5}_{r}$

Freq

C (pF)

${\u03f5}_{r}$





































 Practical applications: inspect the temperature/humidity meter. Look for the devices used to measure each. Measure the standalone capacitor that resembles that in the meter as a function of exposure.
Part II  Reflectance Spectroscopy of Thin Films
 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. What does this indicate about how you would “build” a waveguide?
18.5 Laboratory 5: Magnetic Properties
Objective:
The objectives of this lab are to measure the response of a material to an applied magnetic field and understand the atomic origins of macroscopic magnetic behavior.
Outcomes:
Upon completion of the laboratory, the student will be able to:
 Use a magnetometer to measure magnetic response with applied field.
 Given the saturation magnetization, solve for the number of Bohr magnetons.
 For a known material, explain the microscopic origins of magnetism.
 Distinguish qualitatively between “hard” and “soft” ferromagnetic materials.
Experimental Details
The purpose of this experiment is to investigate the response of different materials to an applied magnetic field, H. Plotting the measured response, B, vs. the applied field, H, indicates the type of magnetization possible, i.e. whether the material is diamagnetic, paramagnetic or ferromagnetic.
References
Solymar and Walsh Chapter 11, Omar Chapter 9
1. Magnetization of metal wires
Qualitatively observe the hysteresis behavior of the following metals. Indicate whether they are ferromagnetic or not, and why. (Hint: Look at which electronic shells are filled/unfilled in each case.)
Metal wires:

Hysteresis? Y/N

3d

4s

Cu








Fe








Co








Ni








W








Steel








Stainless Steel








Quantify the observations on the metal wires (Fe, Co, Ni, and W 1 mm diameter and Cu 0.5 mm diameter – the cross sectional area must be entered in the Bprobe dimensions). Record hysteresis curves in the “intrinsic” mode (without background subtraction). Also record “initial” curves for the first three metals. Fill in the following table.

${B}_{sat}$ observed

${B}_{sat}$ literature

# Bohr magnetons, ${n}_{B}*$

Fe


2.158 T


Co


1.87 T


Ni


0.616 T


Question 1(ac):
Solve for the number of Bohr magnetons per atom in Fe, Ni and Co. (Use experimental ${B}_{sat}$ if the sample saturates; otherwise use literature value.)
Note:
${\mu}_{0}=4\pi \times 1{0}^{7}Tm/A$
${B}_{S}={\mu}_{0}{M}_{S};{M}_{s}$in A/m or Oe
${M}_{S}=n{M}_{A};$ n = # atoms/volume
${M}_{A}={n}_{B}{\mu}_{B};{\mu}_{B}=\frac{e\hslash}{2me}=9.27\times 1{0}^{24}A{m}^{2}=$Bohr magneton
Question 2: Hard magnet
The sintered CuNiFe pellets provide an example of a “hard” magnet. Record (plot) a hysteresis curve and compare to the “soft” ferromagnetic materials. Comment on the observed behavior.
2. Faraday Rotation
Setup:
 Red laser pointer (operated under 3.5V!!! and 40 mA!!!)
 Glass rod (inside the solenoid coil): Diam = 5mm, length = 10 cm, SF59.
 Solenoid coil: L = 150 mm, turns/layer = 140, layers = 10, DC Resist = 2.6 Ohm
 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 zero magnetic field. (Measure and record the intensity every two degrees or so.) Apply a magnetic field and remeasure the intensity vs. rotation. Plot on the same graph to compare. Find the difference in angle (average) between the on/off magnetic field states. Calculate the Verdet constant, $V$, where $\theta \left(radians\right)=VB\ell $, $B$ = magnetic field strength (millitesla), and $\ell $= length glass rod = 10 cm.
3. Demonstration
Heating a magnetized iron wire is one way to observe a second order phase transition involving the spin degree of freedom (upon cooling, one observes the onset of spontaneous magnetization). What is this transition called? Estimate the value of the transition temperature for this system and compare to literature values.