351-2 Lecture Slides
351-2: Physics of Materials II
Bruce Wessels and Peter Girouard Department of Materials Science and Engineering Northwestern University October 1, 2019
Contents
1 Catalog Description (351-1,2)6
2 Course Outcomes6
3 351-2: Solid State Physics6
4 Semiconductor Devices7 4.1 Law of Mass Action...... 7 4.2 Chemistry and Bonding...... 8 4.3 Reciprocal Lattice...... 11 4.4 Nearly Free Electron Model...... 11 4.5 Two Level Model...... 11
5 Band Diagrams 12 5.1 P- and N-Type Semiconductors...... 13 5.2 P-N Junctions...... 13 5.3 Boltzmann Statistics: Review...... 13 5.4 P-N Junction Equilibrium...... 14 5.5 Charge Profile of the p-n Junction...... 14 5.6 Calculation of the Electric Field...... 15 5.7 Calculation of the Electric Potential...... 16 5.8 Junction Capacitance...... 17 5.9 Rectification...... 18 5.10 Junction Capacitance...... 19
6 Transistors 19 6.1 pnp Device...... 20 6.2 pnp Device...... 21 6.3 Amplifier Gain...... 21
1 CONTENTS CONTENTS
6.4 Circuit Configurations...... 22 6.5 MOSFETs...... 23 6.6 Oxide-Semiconductor Interface...... 23 6.7 Depletion and Inversion...... 24 6.8 Channel Pinch-Off...... 24 6.9 Tunnel Diodes...... 25 6.10 Gunn Effect...... 26 6.11 Gunn Diodes...... 26
7 Heterojunctions 26 7.1 Quantum Wells...... 27 7.2 Heterostructures and Heterojunctions...... 28 7.3 Layered Structures: Quantum Wells...... 28 7.4 Transitions between Minibands...... 29 7.5 Quantum Dots...... 29 7.6 Quantum Dot Example: Biosensor...... 30 7.7 Quantum Cascade Devices...... 30
8 Optoelectronics 31 8.1 I-V Characteristics...... 32 8.2 Power Generation...... 32 8.2.1 Solar Cell Efficiency...... 33 8.2.2 Other Types of Solar Cells...... 33 8.2.3 Metal-Semiconductor Solar Cells...... 34 8.2.4 Schottky Barrier and Photo Effects...... 34 8.2.5 Dependence of φB on Work Function...... 35 8.2.6 Pinned Surfaces...... 36 8.2.7 Light Emitting Diodes (LEDs)...... 37 8.2.8 Light Emitting Diodes (LEDs)...... 37 8.2.9 Solid Solution Alloys...... 37 8.2.10 LED Efficiency...... 38
9 Lasers 38 9.1 Emission Rate and Laser Intensity...... 39 9.2 Two Level System...... 39 9.3 Planck Distribution Law...... 40 9.4 Transition Rates...... 40 9.5 Two Level System...... 41 9.6 Three Level System...... 41 9.7 Lasing Modes...... 45 9.8 Laser Examples...... 45 9.8.1 Ruby...... 45 9.8.2 Others Examples...... 46 9.9 Threshold Current Density...... 46 9.10 Comparison of Emission Types...... 48
2 CONTENTS CONTENTS
9.11 Cavities and Modes...... 48 9.11.1 Longitudinal Laser Modes...... 49 9.11.2 Transverse Laser Modes...... 49 9.12 Semiconductor Lasers...... 49 9.13 Photonic Bandgap Materials...... 50
10 Band Diagrams 50 10.1 Band Diagrams...... 51 10.2 Heterojunctions and the Anderson Model...... 51 10.3 Band Bending at p-n Junctions...... 51 10.3.1 Calculate the Conduction Band Discontinuity...... 52 10.3.2 Calculate the Valence Band Discontinuity...... 53 10.3.3 Example: p-GaAs/n-Ge...... 54
11 Dielectric Materials 54 11.1 Macroscopic Dielectric Theory...... 55 11.2 Microscopic Structure...... 56 11.2.1 Relation of Macroscopic to Microscopic...... 57 11.3 Polarizability...... 57 11.4 Polarization in Solids...... 57 11.5 Dipole Moment...... 58 11.6 Polarizability of Solids...... 59 11.7 Dielectric Constant...... 59 11.8 Claussius Mossotti Relation...... 60 11.9 Frequency Dependence...... 60 11.10Quantum Theory of Polarizability...... 62 11.11Ferroelectrics...... 62
12 Phase Transitions 63 12.1 Lattice Instabilities...... 63 12.2 Curie Weiss Law...... 63 12.3 Ferroelectrics...... 64 12.3.1 First versus Second Order Transitions...... 66 12.3.2 Ferroelectric Example: BaTiO3 ...... 66 12.4 Other Instabilities...... 67 12.5 Piezoelectrics...... 67
13 Diamagnetism and Paramagnetism 68 13.1 Diamagnetism...... 69 13.2 Paramagnetism...... 70 13.2.1 Calculation of Susceptibility...... 72 13.2.2 Calculation of Total Angular Momentum J ...... 73 13.2.3 Spectroscopic Notation...... 73 13.2.4 Paramagnetic Susceptibility...... 73 13.2.5 Calculation of J ...... 73 13.2.6 Spin Orbit Interactions...... 74
3 CONTENTS CONTENTS
13.2.7 Effective Magnetic Number...... 74 13.2.8 Paramagnetic Properties of Metals...... 75 13.2.9 Band Model...... 75 13.2.10 Multivalent Effects...... 76
14 Ferromagnetism 77 14.1 Ferromagnetic Phase Transition...... 77 14.2 Molecular Field...... 78 14.2.1 Prediction of TC ...... 79 14.2.2 Temperature Dependence of M(T ) for Ferromagnetism. 80 14.2.3 Ferromagnetic-Paramegnetic Transition...... 80 14.2.4 Ferromagnetic-Paramagnetic Transition...... 80 14.2.5 Ferromagnetic-Paramagnetic Transition...... 81 14.2.6 Ferromagnetic-Paramagnetic Transition...... 81 14.2.7 Ferromagnetism of Alloys...... 82 14.2.8 Transition Metals...... 82 14.2.9 Ni Alloys...... 82 14.2.10 Band Model...... 83 14.2.11 Spin Waves...... 83 14.2.12 Magnon Dispersion...... 84 14.3 Ferrimagnetic Order...... 84 14.3.1 Magnetic Oxides...... 85 14.3.2 Magnetization and Hysteresis...... 86 14.4 Domains and Walls...... 86 14.4.1 Energy of Bloch Domain Walls...... 87 14.5 Anisotropy of Magnetization...... 87 14.5.1 Anisotropy of Magnetization...... 88 14.6 Ferrimagnetic Ordering...... 88 14.6.1 Exchange Terms and Susceptibility...... 89 14.6.2 Structure Dependence of Jex ...... 89
15 Optical Materials 89 15.1 Frequency Doubling...... 90 15.2 Nonlinearity in Refractive Index...... 90 15.3 Electro-Optic Modulators...... 91 15.4 Optical Memory Devices...... 91 15.5 Volume Holography...... 92 15.6 Photorefractive Crystals...... 92 15.7 Photorefractive Crystals...... 93 15.8 All-Optical Switching...... 93 15.9 Acousto-Optic Modulators...... 94 15.10Integrated Optics...... 94 15.11Dielectric Waveguides...... 95 15.12Phase Shifter...... 95 15.13LiNbO Phase Shifter...... 96
4 CONTENTS CONTENTS
16 Superconductivity 96 16.1 BCS Theory...... 96 16.2 Density of States and Energy Gap...... 97 16.3 Heat Capacity...... 97 16.4 Superconductor Junctions...... 98 16.5 Junctions...... 98 16.6 Type I and Type II Superconductors...... 98 16.7 High TC Superconductors...... 99 16.8 Cuprate Superconductors...... 99
17 351-2 Problems 100
18 351-2 Laboratories 101 18.1 Laboratory 1: Measurement of Charge Carrier Transport Param- eters Using the Hall Effect...... 101 18.1.1 Objective...... 101 18.1.2 Outcomes...... 102 18.1.3 References...... 102 18.1.4 Pre-Lab Questions...... 102 18.1.5 Experimental Details...... 103 18.1.6 Instructions/Methods...... 104 18.1.7 Link to Google Form for Data Entry...... 104 18.1.8 Lab Report Template...... 104 18.1.9 Hints for derivation...... 105 18.2 Laboratory 2: Diodes...... 106 18.2.1 Objective...... 106 18.2.2 Outcomes...... 106 18.2.3 Pre-lab Questions...... 106 18.2.4 References...... 107 18.3 Laboratory 3: Transistors...... 110 18.3.1 Objective...... 110 18.3.2 Outcomes...... 110 18.3.3 Pre-lab questions...... 110 18.3.4 Experimental Details...... 111 18.4 Laboratory 4: Dielectric Materials...... 113 18.4.1 Objective...... 113 18.4.2 Outcomes...... 113 18.4.3 Pre-lab questions...... 113 18.4.4 Experimental Details...... 114 18.4.5 References...... 115 18.4.6 Instructions/Methods...... 115 18.4.7 Lab Report Template...... 115 18.5 Laboratory 5: Magnetic Properties...... 118
5 3 351-2: SOLID STATE PHYSICS
1 Catalog Description (351-1,2)
Quantum mechanics; applications to materials and engineering. Band struc- tures and cohesive energy; thermal behavior; electrical conduction; semicon- ductors; amorphous semiconductors; magnetic behavior of materials; liquid crystals. Lectures, laboratory, problem solving. Prerequisites: GEN ENG 205 4 or equivalent; PHYSICS 135 2,3.
2 Course Outcomes 3 351-2: Solid State Physics
At the conclusion of 351-2 students will be able to: 1. Given basic information about a semiconductor including bandgap and doping level, calculate the magnitudes of currents that result from the application of electric fields and optical excitation, distinguishing be- tween drift and diffusion transport mechanisms. 2. Explain how dopant gradients, dopant homojunctions, semiconductor- semiconductor hetero junctions, and semiconductor-metal junctions per- turb the carrier concentrations in adjacent materials or regions, identify the charge transport processes at the interfaces, and describe how the application of an electric field affects the band profiles and carrier con- centrations. 3. 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. 4. 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 com- ponents of the index of refraction. 5. Identify the microscopic interactions that lead to magnetic order in ma- terials, describe the classes of magnetism that result from these interac- tions, and describe the temperature and field dependence of the macro- scopic magnetization of bulk crystalline diamagnets, paramagnets, and ferromagnets. 6. Specify a material and microstructure that will produce desired magnetic properties illustrated in hysteresis loops including coercivity, remnant magnetization, and saturation magnetization.
6 4 SEMICONDUCTOR DEVICES
7. Describe the output characteristics of p-n and Schottky junctions in the dark and under illumination and describe their utility in transistors, light emitting diodes, and solar cells. 8. For technologies such as cell phones and hybrid electric vehicles, iden- tify key electronic materials and devices used in the technologies, specify basic performance metrics, and relate these metrics to fundamental ma- terials properties.
4 Principles of Semiconductor Devices
Recall that the conductivity of semiconductors is given by
σ = neµ + peµ (4.1) where n is the number of electrons, e is the electronic charge, µ is the mobility in cm2/V/s. The following trends for conductivity versus temperature are noted for metals, semiconductors, and insulators:
1. Metals: n and p are constant with temperature. Mobility is related to temperature as µ ∝ T −a, where T is temperature and a is a con- stant. Conductivity decreases with increasing temperature. The resis- tivity ρ (ρ = 1/σ) can be written as a sum of contributing factors using Matthiessen’s rule as ρ = ρ0 + ρ(T )
where ρ0 is a constant and ρ(T ) is the temperature dependent resistivity. A typical carrier concentration for a metal is 1023cm−3. Metals do not have a gap between the conduction and valence bands. 2. Semiconductors: n and p are not constant with temperature but are ther- mally activated. Mobility is related to temperature as µ ∝ T −a. The typi- cal range of carrier concentrations for semiconductors is 1014 −1020cm−3. The range of bandgaps for semiconductors is typically 0.1 − 3.0 eV.
3. Insulators: n and p are much lower than they are in metals and semicon- ductors. A typical carrier concentration for an insulator is < 107cm−3. The conductivity is generally a function of temperature. The bandgap for insulators is > 3.0 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,
[np] [n] + [p] (4.2)
7 4.2 Chemistry and Bonding 4 SEMICONDUCTOR DEVICES
The rate constant is given by [n][p] = K(T ) (4.3) [np] where
K(T ) = [F (T )]2 exp[−∆E/kT ] (4.4) Consider the intrinsic case, that is, when the semiconductor is not doped with chemical impurities. For this case,
[n] = [p] = [ni] = A exp[−∆E/2kT ] (4.5) where ni is the intrinsic carrier concentration. For an intrinsic semiconductor, the conductivity is
∼ σ = 2nieµ = σ0 exp[−∆E/2kT ] (4.6) A log σ versus 1/T plot gives a straight line as shown in Fig. 4.1a.
IOS IOS
1/T K 1/T (a) (b)
Figure 4.1: (a) Log versus 1/T plot of conductivity and resistivity in semicon- ductors. (b) Log of conductivity versus 1/T plot for a semiconductor spanning the regimes of extrinsic and intrinsic conduction.
At lower temperatures where ∆Eg > kT , the conductivity is dominated by carriers contributed from ionized impurities. For temperatures where ∆Eg < kT , thermal energy is sufficient to excite electrons from the valence to conduc- tion band. In this regime, the intrinsic carriers dominate the conduction. The two regimes of conduction are illustrated in Fig. 4.1b.
4.2 Chemistry and Bonding From the periodic table, we know the valence, atomic weight, and atomic size. A portion of the periodic table with groups IIIA-VIA is shown in Fig. 4.2. On the left side (group IIIA) are metals. The right hand side (group VIA) con- tains nonmetals. In between these two groups are elemental 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.
8 4.2 Chemistry and Bonding 4 SEMICONDUCTOR DEVICES
IIIA IVA VA VIA
B CN O
Al Si P more Ga Ge As non- metallic In Sn Sb
more metallic Figure 4.2: Portion of the periodic table showing elements in typical elemental and compound semiconductors.
Electronic bands are made from an assembly of atoms with individual quan- tum states. The spectroscopic notation for quantum states is given by
1s22s22p63s23p63d104s2... (4.7) By the Pauli exclusion principle, no two electrons can share the same quantum state. This results in a formation of electronic bands when atoms are brought together in a periodic lattice. The atomic bonding energy as a function of dis- tance between atoms is shown in Fig. 4.3. In this figure, deq is the equilibrium
E
d
Figure 4.3: Potential energy versus distance between atoms. The equilibrium distance, deq corresponds to the minimum in the energy curve. The different energies corresponding to deq form discrete levels in an electronic band. distance between atoms with the lowest potential energy and is equal to the lattice constant. The energies correponding to deq form the allowed energies in electronic bands. A diagram showing the energy levels in electronic bands is shown in Fig. 4.4. Core electrons do not contribute to conduction. Conduction of charge carri- ers occurs in only in partially filled bands. A diagram showing the filling of energy states in the valence band is given in Fig. 4.5. Note that two electrons occupy each energy state having different values of the spin quantum number, that is, “spin up” or “spin down.” Recall that semiconductors have band gap energies of 0.1−3.0 eV that separate the conduction from the valence band. An electron can be promoted from the valence to conduction band if it is supplied with energy greater than or equal to the band gap energy. The source of this energy may be thermal, optical, or
9 4.2 Chemistry and Bonding 4 SEMICONDUCTOR DEVICES
E 2p etc anti-bonding states in a forbidden periodic energy lattice band single atom 2s valence elctrons bonding
core 1s electrons
Figure 4.4: Energy diagram showing the electronic bands formed in a crys- talline solid.
bond filled
spin up spin down valence partially filled metallic (a) (b)
Figure 4.5: (a) Diagram of energy states in a filled valence band. Up arrows indicate the “spin up” state and down arrows indicate the “spin down” state.. (b) Filling of states in a partially filled band. Since the band is partially filled, the electrons can contribute to conduction. electrical in nature. A simple two-level band model for a semiconductor show- ing the conduction and valence band levels is shown in Fig. 4.6a Also shown (Fig. 4.6b) is a diagram illustrating the process of promoting an electron from the valence to conduction band. When an electron is excited to the conduc- tion band, it leaves behind a “hole” in the valence band, which represents an unfilled energy state. As discussed earlier with regards to the conductivity in semiconductors, both electrons and holes contribute to conduction (see Eqn. 4.1).
e (n)
h (p) (a) (b)
Figure 4.6: (a) Energy band diagram for a simple two-level system. (b) Dia- gram illustrating the generation of an electron and hole pair.
10 4.3 Reciprocal Lattice 4 SEMICONDUCTOR DEVICES
4.3 Reciprocal Lattice and the Brillouin Zone • Reciprocal space: k-space, momentum space
~p = ~~k
• ~k is the reciprocal lattice vector with units 1/cm
Reciprocal 2D Square Lattice
Irreducible Brillouin Zone
4.4 Nearly Free Electron Model
2 1/3 • Fermi wavevector: kF = (3π n)
2 2 ~ kF • Fermi energy: E = F 2m
• Conductivity is proportional to the Fermi surface area, SF . σ ∝ SF
Constant energy surface for free electrons
Fermi surface of second BZ
First Brillouin Zone
4.5 Two Level Model
11 5 BAND DIAGRAMS
E
K
• Parabolic bands: E = ~2k2/2m∗. m∗ ≡ effective mass £ • Origin? Solution to the Schrodinger¢ Equation, HΨ = ¡EΨ for free elec- trons
5 Semiconductor Band Diagrams
Direct Bandgap
E
K
Indirect Bandgap
12 5.1 P- and N-Type Semiconductors 5 BAND DIAGRAMS
E
K
• Direct bandgap semiconductors: III-V’s • Indirect bandgap semiconductors: Si, Ge
5.1 P- and N-Type Semiconductors
E Intrinsic
X
n-type
p-type
5.2 P-N Junctions
p-n Junction n-type n-type
p-type
transition region W (space change region) p-type
5.3 Boltzmann Statistics: Review
For Ev as the reference energy (EV = 0), Ec − EF n = N exp − (5.1) s kT
13 5.4 P-N Junction Equilibrium 5 BAND DIAGRAMS
EF − EV −EF p = N exp − = N exp (5.2) s kT S kT ∼ ∼ For fully ionized acceptors, p = NA. For fully ionized donors, n = ND.
Ns = Density of states § ¤ The Fermi Level for electrons¦ and holes are calculated¥ from Eq.5.1 and 5.2, respectively:
Electrons: EF = Ec − kT ln(NS/n)
Holes: EF = EV + kT ln(NS/p)
Note: these equations are true for non-generate semiconductors.
5.4 P-N Junction Equilibrium Poisson’s equation - describes potential distribution φ(x):
d2φ(x) ρ(x) = − dx2
φ(x) ≡ Potential ≡ Dielectric Constant of Material ρ(x) ≡ Charge Density
Assume that the donors and acceptors are fully ionized
− + p-type: NA → NA + p + − n-type: ND → ND + e
Note: the space charge region is positive on the n-side and negative on the p-side.
5.5 Charge Profile of the p-n Junction
+ neutral -
W
14 5.6 Calculation of the Electric Field 5 BAND DIAGRAMS
• Total charge: ρ(x) = e [p(x) + ND(x) − n(x) − NA(x)]
• Space charge width: w = xn + xp
• Charge balance: NAxp = NDxn • Note that x = 0 at the junction interface.
5.6 Calculation of the Electric Field d2φ ρ dφ = − = −E dx2 dx dE ρ ⇒ = dx Z ρ E(x) = dx Separate the integral for the n- and p-regions:
Z 0 ρ(x) Z xn ρ(x) E(x) = dx + dx −xp 0 | {z } | {z } p-region n-region Important assumptions made:
1. ρ ≈ eND in the n region, and ρ ≈ −eNA in the p region. 2. ρ is constant in the separate n and p regions. Boundary conditions:
1. E(x = −xp) = E(x = xn) = 0 2. E is continuous at the junction
Resulting electric field across the junction: ( − eNA (x + x ), −x ≤ x ≤ 0 E(x) = p p eND (x − xn), 0 ≤ x ≤ xn ( − eNA (x + x ), −x ≤ x ≤ 0 E(x) = p p eND (x − xn), 0 ≤ x ≤ xn
Electric Field in the Junction
15 5.7 Calculation of the Electric Potential 5 BAND DIAGRAMS
5.7 Calculation of the Electric Potential Recall Z φ = − Edx
Boundary conditions: 1. φ is continuous at the junction boundary
2. φ = 0 at x = −xp (chosen arbitrarily) Resulting potential across the junction:
( eNA 1 2 1 2 x + x · xp + xp , −xp ≤ x ≤ 0 φ(x) = 2 2 eND NA 2 1 2 x + x · xn − x , 0 ≤ x ≤ xn 2ND p 2
Electric Potential in the Junction
16 5.8 Junction Capacitance 5 BAND DIAGRAMS
5.8 Junction Capacitance • Space charge width: 1/2 " 1/2 1/2# 2φ0 NA ND w = xp + xn = + e(NA + ND) ND NA
φ0 ≡ Built-in potential
• Note that, from charge balance, NAxp = NDxn • Junction capacitance: C = F/cm2 w • One-sided abrupt junction: heavily doped p or n + • Consider the case of a p -n junction. For NA ND,
2φ 1/2 w = 0 eND
eN 1/2 C = D 2φ0
17 5.9 Rectification 5 BAND DIAGRAMS
5.9 Rectification Band Diagram Current-Voltage Characteristic
I
V
Reverse Forward Bias Bias
transition region
• Larger Eg, larger φ0
• ID = I0D exp(−eφ0/kT )
−3 14 Eg ni (cm × 10 ) Ge 0.66 0.24 Si 1.08 0.00015 GaAs 1.4 - GaP 2.25 - GaN 3.4 -
• Under a forward bias +φ1, the built in potential is reduced by φ1 and there is a higher probability of electrons going over the barrier.
• Barrier height under forward bias: e(φ0 − φ) • Current increases exponentially with forward bias voltage.
p n Forward Bias
p n Reverse Bias
• Under a reverse bias +φ1, the built in potential is increased by φ1 and there is a lower probability of electrons going over the barrier.
18 5.10 Junction Capacitance 6 TRANSISTORS
5.10 Junction Capacitance Definition of capacitance: ∂Q C ≡ ∂φ
For a one-sided abrupt junction, recall that Q = eNDxn, where
1/2 2(φ0 + φ1)NA xn = eND(ND + NA) The total charge Q for a one sided junction is then
1/2 NAND Q = 2e(φ0 + φ1) NA + ND
Calculate the capacitance as a function of bias voltage φ1:
∂Q e N N 1/2 C = = A D ∂φ1 2(φ0 + φ1) NA + ND
2 A plot of 1/C versus φ1 is called a Mott Schottky plot and yields NA or ND.
U
6 Transistors: Semiconductor Amplifiers
• Three terminal device
19 6.1 pnp Device 6 TRANSISTORS
• Bipolar Junction Transistor (BJT): consists of two p-n junctions back-to- back • The three seminconductor regions are the emitter, base, and collector
• pnp and npn devices
emitter base collector
p n p +
input output
6.1 pnp Device • Emitter-Base junction is forward biased, and the Collector-Base 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
base (n) + +
emitter collector (p) (p)
forward reversed biased junction biased junction
Circuit Diagram
20 6.2 pnp Device 6 TRANSISTORS
6.2 pnp Device Device Diagram
base (n) + +
emitter collector (p) (p)
forward reversed biased junction biased junction
Band Diagram
emitter collector
base
6.3 Amplifier Gain Voltage across the junctions:
φr icφr ieφf ≈ 10 V/V φf Nodal analysis at the base for pnp and npn devices:
21 6.4 Circuit Configurations 6 TRANSISTORS
npn e
b
c
ic ≈ ie and ic = αie
From nodal analysis at the base, ic = ie − ib.
αib α ⇒ i = = h i with h = c 1 − α fe b fe 1 − α hfe is the current gain parameter. α ∼ 0.9.
6.4 Circuit Configurations • Circuit configurations include the common-base, common-collector, and common-emitter.
• The term following “common” indicates which terminal is common to the input and output.
Common-Emitter Diagram
n + p + n
Common-Base I-V Characteristic
22 6.5 MOSFETs 6 TRANSISTORS
2mA 3mA = 4mA
6.5 MOSFETs • MOSFET: Metal Oxide Semiconductor Field Effect Transistor • An applied electric field induces a channel between the source and drain through which current conducts.
gate source drain
oxide
n conduction channel
body
6.6 Oxide-Semiconductor Interface Changing the gate bias causes the following: 1. Band bending at the oxide-semiconductor interface.
2. Accumulation of charge at the interface 3. An increase in channel conductance.
23 6.7 Depletion and Inversion 6 TRANSISTORS
filled states (conducting electrons)
oxide semiconductor
6.7 Depletion and Inversion • Band bending in the opposite sense leads to depletion of charge carriers. • Inversion occurs when the Fermi level of an n (p) type semiconductor intercepts the valence (conduction) band due to band bending. • The channel in a MOSFET is an inversion layer.
Depletion Inversion
filled states (conducting holes)
6.8 Channel Pinch-Off
• A threshold voltage VT is required for channel inversion.
• For a positive gate voltage VG, the potential across the oxide- semiconductor junction is VG − VT .
• For a positive voltage between the drain and source (VDS), the potential near the drain is VG − VT − VDS.
• For VDS ≥ VG − VT , the channel is depleted near the drain, resulting in “pinch-off.” source gate drain
pinched-off channel channel p
24 6.9 Tunnel Diodes 6 TRANSISTORS
2 • In saturation mode (VDS ≥ VG − VT ), ID ∝ VGS
• A small change in VDS causes a large change in ID.
Increasing Gate Bias
G = Gate S = Source D = Drain
6.9 Tunnel Diodes • Consists of a heavily doped p+-n+ structure −x/a • Tunneling current: I ∝ Neff e • Electrons tunnel from filled to empty states across the junction
insulating region (barrier)
• Electrons can go over the barrier or through the barrier (tunneling)
Rectifying Tunneling Thin Junction negative resistance region I I I
V V V
25 6.10 Gunn Effect 7 HETEROJUNCTIONS
6.10 Gunn Effect • Excitation of electrons under a high field to a higher energy conduction band with larger effective mass
−1 2 eτ ∗ 1 ∂ E σ = neµ and µ = , m = m∗ ~2 ∂k2 | {z } band curvature
GaAs E band diagram
heavy electron light electron
K
6.11 Gunn Diodes • Negative resistance due to increased effective mass
ne2τ J = σE = E m∗ • Transfer of electrons from one valley to another (Transferred Electron Device, TED)
I
negative positive A resistance resistance B
0 V
7 Heterojunctions and Quantum Wells
• Composed of semiconductor super-lattices (alternating layers of differ- ent semiconductors) • Multilayers grown by Molecular Beam Epitaxy (MBE) and Metal- Organic Chemical Vapor Deposition (MOCVD)
26 7.1 Quantum Wells 7 HETEROJUNCTIONS
• Used in the construction of lasers, detectors, and modulators • Utilizes band offset, can be used to confine carriers
GaAlAs GaAs
7.1 Quantum Wells • From the particle in a box solution, minibands are formed within the well
GaAlAs GaAs GaAlAs
miniband
HΨ = EΨ; Ψ = A exp(−knx)
nπ kn = Lx
2 2 2 2 2 ~ k ~ π nx Ex = ∗ = ∗ 2 2m 2m Lx
I II
27 7.2 Heterostructures and Heterojunctions 7 HETEROJUNCTIONS
• Smaller Lx, larger energy of miniband
• Lx can be tuned to engineer the band gap
7.2 Heterostructures and Heterojunctions • Devices that use heterojunctions and heterostructures: lasers, modulation-doped field effect transistors (MODFETs) • Types of heterostructures: quantum wells (2D), wires (1D), and dots (0D)
Well Wire Dot
2D 1D 0D (confinement in (confinement in (confinement in 1 direction) 2 directions) 3 directions)
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)
semiconductor 2 semiconductor 1 Z semiconductor 2
28 7.4 Transitions between Minibands 7 HETEROJUNCTIONS
7.4 Transitions between Minibands • Allowed initial and final states are governed by “selection rules.” • Will have conduction in minibands
• Energy of states:
2 2 2 2 ~ k ~ nπ E = ∗ = ∗ 2m 2m Lx
GaAlAs GaAs GaAlAs
n=2
n=1 miniband
n=1
n=2
7.5 Quantum Dots Quantum Dots
Substrate
29 7.6 Quantum Dot Example: Biosensor 7 HETEROJUNCTIONS
• Electrons are confined in all dimensions to “dots.” • Quantum Dots: clusters of atoms 3-10 nm in diameter. • QDs can be created by colloidal chemical synthesis or island growth epi- taxy. • Modeled after the hydrogen atom with radius R. Quantized energies:
2 2 ~ αi Ei = 2mi R
αi ≡ quantum number
7.6 Quantum Dot Example: Biosensor
bioactive material CdTe
ZnSe
• Solution synthesized colloidal quantum dots (CQD) • Fluorescence energy depends on surface and size of dot • Binding of molecules to the surface of the QD quenches the photolumi- nescent intensity.
7.7 Quantum Cascade Devices • Consist of multiple quantum wells
• Band bending occurs with bias
30 8 OPTOELECTRONICS
8 Optoelectronic Devices: Photodetectors and So- lar Cells
• Light incident on a p-n junction generates electron-hole pairs which are separated by the built in potential. • Separated carriers contribute to the photocurrent.
+
p n
I
V
with light
e
h
31 8.1 I-V Characteristics 8 OPTOELECTRONICS
8.1 I-V Characteristics qV/kT • Diode equation: I = I0 e − 1
• Short circuit current: Isc = Aq(Le + Lh)G
Le ≡ Electron diffusion length Lh ≡ Hole diffusion length G ≡ Carrier generation rate (depends on light intensity)
qV/kT • Current under illumination: I = I0 e − 1 − Isc kT Isc • Open circuit voltage: V (I = 0) = Voc = ln + 1 q I0
8.2 Power Generation • Quadrant IV is the power quadrant.
• Theoretical maximum power: Pmax, theory = VocIsc
• Maximum obtainable power: Pmax = ff(IscVoc)
ImVm Fill Factor: ff ≡ § IscVoc ¤ ¦ I ¥
V
32 8.2 Power Generation 8 OPTOELECTRONICS
8.2.1 Solar Cell Efficiency • Efficiency is governed by the fill factor and bandgap • Want larger fill factors and absorption that matches the solar spectrum
Intensity
23
19 GaAs
Si a-Si
1.1 1.30 1.40 1.70 Band Gap (eV)
8.2.2 Other Types of Solar Cells 1. Multi-junction solar cells: GaAlAs/GaAs. More efficient absorption.
2. Thin films: amorphous silicon (a-Si), CdTe. Inexpensive. 3. Metal-semiconductor solar cells
Heterojunction Cell
GaAlAs
GaAs
33 8.2 Power Generation 8 OPTOELECTRONICS
8.2.3 Metal-Semiconductor Solar Cells
• Energy barrier φB at metal-semiconductor junction • Current density, J: −φ qV J = J exp B exp − 1 00 kT kT
current I J ≡ = area A
Slope =
Intercept =
E
metal
8.2.4 Schottky Barrier and Photo Effects • Barrier height is obtained from log(J) versus 1/T plot • Electrons can tunnel across the barrier
slope = log J
1/T
34 8.2 Power Generation 8 OPTOELECTRONICS
E
(Barrier height) tunnel
metal
semiconductor
• Photo-response is characterized by two regions: metal photo-emission, and band-to-band excitation. √ • Photo-response ∝ φB
Photoemission band-to-band from metal Response
over barrier
Photon Energy (eV) ) 1/2 ( μ A 1/2 Response
Photon Energy
8.2.5 Dependence of φB on Work Function
φ ≡ energy required to take an electron from the Fermi Level ¨ to the Vacuum Level
© φB = φm − φs
35 8.2 Power Generation 8 OPTOELECTRONICS
Au In Mg ZnS unpinned
Al As Pt Au
GaAs nearly metal independent
Electronegativity Difference
metal semiconductor
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:
φB(n) + φB(p) = Eg
• Example, InP (n): φB(n) ∼ 0.5 eV φB(p) ∼ 0.8 eV Eg ∼ 1.27 eV
• For making an Ohmic contact to an n-type semiconductor, φB = φm − φs < 0
metal semiconductor
electronic surface states
36 8.2 Power Generation 8 OPTOELECTRONICS
8.2.7 Light Emitting Diodes (LEDs) • Forward biasing a p-n junction results in electron-hole recombination at the junction
• Recombination results in light emission. The wavelength depends on Eg
hν = Eg
p-n Junction under Forward Bias
8.2.8 Light Emitting Diodes (LEDs)
Material Eg at 300K (eV) Comments GaP 2.25 GaAsP 1.7-2.25 Alloy GaAs 1.43 InGaAsP 0.8-1.4 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
Eg ≈ xEg1 + (1 − x)Eg2
Eg (eV) 2.25
Indirect
1.43 Direct
GaAs X GaP
37 9 LASERS
8.2.10 LED Efficiency • LED efficiency is the ratio of the radiative recombination probability (WR) to the total recombination probability (Wtotal)
W W η = R = R Wtotal WR + WNR
• WNR: the probability of non-radiative recombination • Non-radiative processes limit the efficiency • Recombination probability is related to the lifetime (τ) 1 W ∝ [s−1] τ
1 1 1 = + τtotal τR τNR
• Efficiency in terms of probability (W ) and lifetime (τ) W W η = R = R Wtotal WR + WNR 1/τ 1 η = R = 1/τtotal 1 + τR/τNR
• For τR τNR, τNR determines the radiation efficiency ∴ want as few non-radiative defects (dislocations, impurities,etc.) as possible τ η = NR τR
• GaN example: 107 dislocations per cm2
9 Lasers
• LASER: Light Amplification through Stimulated Emission of Radiation • Laser light has narrow divergence and consists of photons of the same frequency and phase • Photons of a certain frequency and phase stimulate emission of photons with the same frequency and phase
same wavelength same phase "coherent"
Intensity Output Input Intensity
38 9.1 Emission Rate and Laser Intensity 9 LASERS
9.1 Emission Rate and Laser Intensity • Calculate emission rate and laser intensity by accounting for both emis- sion and absorption processes using Boltzmann Statistics • Calculate the transition rate between states and Einstein A and B coeffi- cients
• Consider 2 and 3 level systems
Two Level System Three Level System
9.2 Two Level System • At equilibrium, the population of levels is determined by the Boltzmann distribution
−E2/kT N2 e N ∝ Ne−E2/kT or = 2 N Z
Z ≡ sum over states
• Ratio of populations in a two level system:
N2 = e−(E2−E1)/kT N1
At equilibrium, fewer electrons are in level 2
Triggers of Stimulated Emission
• Internal or external radiation can cause the (stimulated) transition to oc- cur
• Internal sources are described by “black body” radiation.
int. ext.
stimulated emission
39 9.3 Planck Distribution Law 9 LASERS
9.3 Planck Distribution Law • Number of photons inside the cavity is determined by Planck’s Distribu- tion Law • Spectral density of photons of frequency ν in linewidth dν
8πn3hν3 1 ρ(ν)dν = dν c3 exp(hν/kT ) − 1 | {z } | {z } Density of states Planck distribution
9.4 Transition Rates
• Transition rate Rij: rate at which electrons transition from state i to state j.
• Absorption (“up”) rate R12:
R12 = N 1B12ρ(ν21)dν
N1 ≡ Number of photons in lower state
ρ(ν21) ≡ number of photons available having transition energy hν
• Emission (“down”) rate R21:
R21 = N 2[ A21 + B21ρ(ν21)] dν |{z} | {z } Spont. Stim.
• A21 and B21 are the Einstein A and B coefficients for spontaneous and stimulated transitions, respectively, between states 2 and 1.
40 9.5 Two Level System 9 LASERS
9.5 Two Level System • At equilibrium (steady state):
R12 = R21
N1B12ρ(ν21)dν = N 2 [A21 + B21ρ(ν21)] dν
N2 −(E2−E1)/kT • Substitute in = e , simplifying and solving for ρ(ν21): N1
A21dν ρ(ν21)dν = B12 exp [(hν21)/kT ] − B21
• Note: same form as Planck Distribution Law with B12 = B21 • In a two level system, the stimulated emission and absorption rates are equal, i.e., B12 = B21 • Comparing to the Planck Distribution Law, 8πn3hν3 A = B 21 21 c3
• The coefficient of spontaneous emission A21 is related to the stimulated emission rate
Two Level System - Summary • Number of photons inside the cavity is determined by Planck’s Distribu- tion Law • The stimulated emission and absorption rates are equal
• The coefficient of spontaneous emission A21 is related to the stimulated emission rate • A population inversion is required for lasing
9.6 Three Level System
• A three level system consists of a ground state (E1), excited state (E2), and intermediate state (E3).
• Electrons are pumped from E1 to E3 by a pump source with energy ≥ hν31; N2 population is unaffected
• More electrons are in the upper state E3 than the lower (metastable) state E2 ⇒ a population inversion is achieved
41 9.6 Three Level System 9 LASERS
Carrier Population • The carrier population is given by the Boltzmann distribution
N(E) = N0 exp(−E/kT )
log N = −E/kT + log N0 • Lower energy states have higher carrier populations at equilibrium
thermal non-equilibrium equilibrium (excited) Energy
log N
Spontaneous Emission
•A 31: spontaneous emission rate; measure of the spontaneous depopula- tion • Assuming first order kinetics, dN 3 = −A N dt 31 3 1 A31 = tspont
• Solution for initial population N3,i: N3(t) = N3,i exp (−A31t)
3 slope = 1/tspont log N
t
finite lifetime
42 9.6 Three Level System 9 LASERS
Populations under High Pumping
• Under high pumping at steady state, the populations in levels 1 and 3 are N + N N ∗ = N ∗ = 1 3 1 3 2 • Equal probabilities of emission and absorption: material is “transparent” • From energy balance,
B23N2ρ(ν) = A32N3 + B32N3ρ(ν) | {z } | {z } | {z } Stim. absorption Spont. emission Stim. emission
Laser Gain
• The net number of photons gained per second per unit volume is
NW12 = (N2 − N1)W12
W12 ≡ Transition rate from state 1 to 2
• For N > 0: population inversion, material can act as an amplifier • For N = 0: the material is transparent • For N < 0: light is attenuated
For amplification, a population inversion must exist.
Derivation of Laser Gain Factor
• Stimulated-emission photons travel in the same direction (z) as the inci- dent photons and further stimulate emission • ⇒ photon flux and intensity increase in z direction
mirror
lasing medium
• Increase in photon intensity through a volume of thickness dz:
43 9.6 Three Level System 9 LASERS
dI = (N2 − N1)W12hν12dz
dI = (N − N )W hν ∴ dz 2 1 12 12 • Transition probability is defined as
I c2g(ν) W12 = 2 2 hν 8πn ν tspont
• g(ν) is the lineshape factor • Laser gain factor γ(ν) is defined by the equation dI = γ(ν)I dz
c2g(ν) ∴ γ(ν) = (N2 − N1) 2 2 8πn ν tspont
• The change in laser intensity with distance z is dI = γ(ν)I dz • Solution for I(z):
I(z) = I(0) exp [γ(ν)z]
• Recall laser gain factor from previous slide:
c2g(ν) γ(ν) = (N2 − N1) 2 2 8πn ν tspont
• For amplification, I(z)/I(0) > 1 is required ∴ N2 > N1. A population version is required for amplification
• Laser threshold: operation at which I(z)/I(0) = 1
44 9.7 Lasing Modes 9 LASERS
9.7 Lasing Modes • Lasing frequencies are chosen by confining the stimulated-emission pho- tons inside a cavity called a Fabry Perot resonator • Standing waves of certain discrete frequencies are established inside the cavity
lasing medium
mirror partially transmissive mirror
• Laser intensity after a round trip in the cavity:
I = I0R1R2 exp [(γ − α)2l]
α ≡ absorption and scattering losses γ ≡ laser gain factor
• Criteria for net gain: gain ≥ sum of losses
R1R2 exp [(γ − α)2l] ≥ 1
• Modes that are in phase after a round trip are stabilized
9.8 Laser Examples 9.8.1 Ruby
• Transition metal ion (Cr) doped alumina (Al2O3) • 3 level system - first demonstrated laser • Optically pumped by a “discharge lamp”
pump levels 3
2
laser transition
1
45 9.9 Threshold Current Density 9 LASERS
9.8.2 Others Examples Semiconductor: • Emission from carrier recombination in a forward biased p-n junction • Heterostructures form both p-n junction and waveguiding region (cav- ity)
Others: • Ceramic: Nd^{3+} :YAG, 4-level system
3+ • Glass fiber laser Er :SiO2 Erbium Doped Fiber Amplifier (EDFA), basis of modern optical communication
p n
V
active region thickness
9.9 Threshold Current Density
• Number of electrons injected into lasing region per unit volume: Ne • Recombination rate: N lwd I η e = i trec e
η ≡ Efficency
Ii ≡ Injection current e ≡ Fundamental charge
• Recall the gain coefficient, γ(ν)
c2g(ν) γ(ν) = (N3 − N2) 2 2 8πn ν tspont ∼ trec = tspont
Ne = N3 − N2
Iiη trec ∼ • Substituting in N3 − N2 = Ne = e lwd and tspont = trec,
46 9.9 Threshold Current Density 9 LASERS
c2g(ν)η γ(ν) = I (cm−1) 8πn2ν2elwd i
Relates gain coefficient to Ii
• The linewidth factor is g(ν) =∼ 1/∆ν
• Let the mirror reflectivity be R1R2 = R. Recall the lasing threshold con- dition
R exp(γ − α)l0 = 1
(γ − α)l0 + ln R = 0
• Define threthold current density: jth = Ii,th/wl
• Substituting in for γ and solving for jth:
8πn2ν2ed 1 j = α − ln R [A/cm2] th c2ηg(ν) l0 | {z } β−1
• Lasing threshold current density:
α 1 j = − ln R (R < 1) th β βl0
Lasing Threshold Condition: α 1 j = − ln R (R < 1) th β βl0
0 100 200
47 9.10 Comparison of Emission Types 9 LASERS
Lasing Threshold Lasing Spectrum
spontaneous stimulated Intensity Intensity LED Laser
Energy (eV)
9.10 Comparison of Emission Types Emission Intensity Linewidth Directional Coherent Spontaneous Weak Broad No No Stimulated Not Intense Narrow No (Super-Radiant) Necessarily Stimulated Very Intense 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
Integer number of wavelengths = Optical path length
mλ = 2nl
λ m = l m = 1, 2, 3... 2n • n is the refractive index of the cavity and l is the cavity length
these will standing waves are be amplified formed
48 9.12 Semiconductor Lasers 9 LASERS
9.11.1 Longitudinal Laser Modes • Modes whose condition for lasing depends on the length of the cavity are called longitudinal modes • Wavelength spacing between modes:
λ2 ∆λ = dn 2l n − λ dλ
• dn/dλ is the dispersion of the refractive index
modes modulate intensity I
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
Transverse Modes
m = 1 m = 2
Output Intensity
m=0
m=1 m=2 Emitted Intensity
0 20 40 degrees
9.12 Semiconductor Lasers • Light is confined to regions of high refractive index • Semiconductor heterostructure is designed to confine light in the junc- tion
49 9.13 Photonic Bandgap Materials 10 BAND DIAGRAMS
GaAs Laser Example
GaAs
Ga0.94Al0.06As
Ga0.98Al0.02As
Ga0.94Al0.06As GaAs Substrate
index
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 refrac- tive 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 (vg) of a pulse is described by the group index (ng)
vg = c/ng as ng → ∞, vg → 0
10 Band Diagrams
• Energy band levels are represented in real space • Energy levels of interest: Conduction band edge, valence band edge, and Fermi Level • Energies of interest: work function and electron affinity
φ ≡Work Function. Energy required to move an electron from the fermi level to the vacuum level. χ ≡Electron Affinity. Energy required to move an electron from the conduction band edge to the vacuum level. Ec ≡Conduction band energy. Ev ≡Valence band energy. EF ≡Fermi level.
50 10.1 Band Diagrams 10 BAND DIAGRAMS
10.1 Band Diagrams
Semiconductor Band Diagram Metal Band Diagram
Vacuum Level Vacuum Level
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: n-GaAs (Eg = 1.45 eV) and p-Ge (Eg = 0.7 eV)
n-GaAs p-Ge
10.3 Band Bending at p-n Junctions Band diagrams for isolated p- and n-type semiconductors are shown in Fig. 10.1below. When brought into contact without application of an external bias field, ther- mal equilibrium requires that the Fermi levels equilibrate, i.e.
EF,n = EF,p (10.1) This results in a bending of the conduction and valence bands as shown in Fig. 10.2. With the Fermi levels equilibrated, the following expression is true considering the path A-B in Fig. 10.2:
EF,n −EF,p = 0 = (χGaAs +δGaAs +qVD,n +qVD,p)−(χGe +Eg,Ge −δGe) (10.2)
51 10.3 Band Bending at p-n Junctions 10 BAND DIAGRAMS
Vacuum Level
n-GaAs p-Ge
= 4.07 = 4.13 eV
0.7 eV 1.45 eV
Figure 10.1: Band diagrams of isolated n-GaAs and p-Ge.
Rearranging Eqn. 10.2,
q(VD,n + VD,p) = (χGe − δGe + Eg,Ge) − (χGaAs + δGaAs) (10.3) | {z } | {z } p n
Substituting in known values,
q(VD,n + VD,p) = (4.13 − 0.1 + 0.7) − (4.07 + 0.1) (10.4)
=0.56 eV < Eg,Ge where
VD,n NAGe = (10.5) VD,p NDGaAs
10.3.1 Calculate the Conduction Band Discontinuity Let the Fermi level be zero on the energy scale.
52 10.3 Band Bending at p-n Junctions 10 BAND DIAGRAMS
n-GaAs p-Ge
Vacuum Level
A B
Figure 10.2: Band diagram of a n-GaAs/p-Ge heterojunction.
δGaAs + qVD,n = (Eg,Ge − δGe) − qVD,p + ∆Ec
∆Ec = δGaAs + qVD,n − (Eg,Ge − δGe) + qVD,p On substituting in Eqn. 10.3, we get
∆Ec = χGe − χGaAs = +0.6
The positive value for ∆Ec indicates a spike in the conduction band.
10.3.2 Calculate the Valence Band Discontinuity
The valence band discontinuity ∆Ev is calculated in a similar manner. In this case, a negative value for ∆Ev indicates a spike in the valence band. ∆Ev is calculated as follows:
∆Ev = ∆Eg − ∆χ
∆Ev = (Eg,GaAs − Eg,Ge) − (χGe − χGaAs) Substituting in known values,
∆Ev = (1.45 − 0.7) − (4.13 − 4.07)
∆Ev = +0.69
Since ∆Ev is positive, there is no spike in the valence band.
53 11 DIELECTRIC MATERIALS
Vacuum Level p-GaAs Vacuum Level n-Ge
= 4.07 eV = 4.13 eV
= 1.45 eV = 0.7
(a) (b)
Figure 10.3: Isolated band diagrams of (a) p-GaAs and (b) n-Ge.
10.3.3 Example: p-GaAs/n-Ge The isolated band diagrams for p-GaAs and n-Ge are shown in Figs. 10.3a and 10.3b. Here, we calculate the conduction and valence band discontinuities using the approaches outlined in Sections 10.3.1 and 10.3.2. For the conduction band discontinuity,
q(VD,n + VD,p) = (χGaAs + Eg,GaAs − δGaAs) − (χGe + δGe) | {z } | {z } p n
= (4.07 + 1.45 − 0.1) − (4.13 + 0.1)
= 1.19 < Eg,GaAs For calculating the valence band discontinuity,
∆Ev = (Eg,Ge − Eg,GaAs) − (χGaAs − χGe)
∆Ev = −0.69 ⇒ spike (10.6) The resulting band diagram is shown in Fig. 10.4.
11 Dielectric Materials
• Dielectrics are insulating materials which include glasses, ceramics, polymers, and ferroelectrics • Dielectric constant () describes how well a material stores charge • is related to capacitance (C) Parallel Plate Capacitance A C = d d ≡ distance between plates A ≡ plate area
54 11.1 Macroscopic Dielectric Theory 11 DIELECTRIC MATERIALS
p-GaAs n-Ge
Vacuum Level
Figure 10.4: Band diagram of a p-GaAs/n-Ge heterojunction as predicted us- ing the approach in Section10.3.3. Note that there is a spike in the valence band which impedes the conduction of holes.
Dielectric + + + + + V
11.1 Macroscopic Dielectric Theory • From Maxwell’s equations, D = E [C/m2]
D ≡ Electric flux density ≡ Dielectric constant E ≡ Electric field
• Note: = r0
r ≡ Relative dielectric constant (unitless) 0 ≡ Permittivity of free space [F/m]
• Total charge stored in a capacitor in vacuum: V Q = [C] 0 d with
55 11.2 Microscopic Structure 11 DIELECTRIC MATERIALS
V ≡ Voltage [V] d ≡ Distance [m]
• Putting in a dielectric, V Q0 = and C = Q0/V 0 r d • An increase in charge is stored in the capacitor with insertion of the di- electric • D (electric flux density) is equal to the surface charge • P (polarization density) is the additional surface charge
P ≡ D − 0E
P ≡ (r0 − 0)E = 0 (r − 1) E | {z } χ
P ≡ 0χE
• χ = (r − 1) is the dielectric susceptibility
11.2 Relation of to the Microscopic Structure • Knowledge of the microscopic structure is needed to explain dielectric phenomena. Example: ferroelectric r(T )
T
• In the microscopic approach, we consider 1) Atomic behavior and 2) De- formation of the atomic charge cloud (orbitals) when a field is applied.
No Applied Field With Applied Field
56 11.3 Polarizability 11 DIELECTRIC MATERIALS
11.2.1 Relation of Macroscopic to Microscopic • Induced dipole moment due to an applied field: µ = qδ
• For Nm molecules per unit volume, the polarization density is P = Nmqδ or
P = Nmµ
0 0 • For small fields, µ = αE and P = NmαE
E0 ≡ Local electric field α ≡ Polarizability of species (atoms, molecules)
Single Dipole Polarized Dielectric
11.3 Contributions to the Polarizability 1. Electronic: Deformation of the electronic cloud (orbitals) due to optical fields. 2. Molecular: Bonds between atoms can stretch, bend, and rotate
3. Orientational: Polymers, liquids, and gases can be reoriented in an ap- plied electric field. Example: poled polymers, useful for electro-optic 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 = −µE cos θ
θ ≡ Angle between µ and E
• The number of dipoles with a certain E will depend on the Boltzmann factor exp(−E/kT ). Increasing temperature will favor more random alignment.
57 11.5 Dipole Moment 11 DIELECTRIC MATERIALS
Dipoles in a Solid
+
- can align with field
11.5 Calculation of the Average Dipole Moment • Average Dipole Moment (hµi) weighted average energy for each orientation hµi = total energy for all orientations • Consider the 3-dimensional case. In a sphere with radius r, the total number of molecules with energy E in a volume element 2πr2 sin θdθ is
# of dipoles with energy E = 2πA sin θdθ exp(−E/kT )
A = Constant including radius
• Recall the orientation energy is given by E = −µE cos θ. Substituting in for E, µE cos θ # of dipoles with energy E = 2πA sin θdθ exp kT
• The total energy for all orientations is then Z π µE cos θ A exp 2π sin θdθ 0 kT • The weighted average for each orientation is Z π µE cos θ A exp (µ cos θ) 2π sin θdθ 0 kT • The average dipole moment is then
R π µE cos θ 0 A exp kT (µ cos θ) 2π sin θdθ hµi = R π µE cos θ 0 A exp kT 2π sin θdθ
• Integrating, hµi 1 = L(a) = coth a − µ a µE with a = kT
58 11.6 Polarizability of Solids 11 DIELECTRIC MATERIALS
1 a • For small a, coth a ≈ a + 3 + ··· hµi a = for small a ∴ µ 3
µ2E hµi µE hµi = or = 3kT µ 3kT
11.6 Polarizability of Solids • The polarizability of the solid is
P = Nmhµi
• P is proportional to the field E and inversely proportional to the temper- ature T µ2E P = N m 3kT
∂P 1 • The definition for the electric susceptibility is χ = ∂E ⇒ χ ∝ T
11.7 Dielectric Constant for a Solid
• Under an applied field E0, induced dipoles in the solid create an oppos-
ing field E1. The field inside the material is E = E0 + E1 . + _ + _ + + _ + + _ _ + + __ _ + _ _ + _ + _
• For a crystal, the polarizability will depend on the interaction of dipoles in the lattice X X P = NjPj = NjαjEloc,j j j
59 11.8 Claussius Mossotti Relation 11 DIELECTRIC MATERIALS
• Eloc,j is the local field of atom j due to interactions with other dipoles. For cubic materials, P Eloc = E + 30 • The polarizability is then X P P = N α E + j j 3 j 0
11.8 Claussius Mossotti Relation • From the relation for the macroscopic electric susceptibility (see slide 11.1), χ = P/0E
P X E + 3 χ = N α 0 j j E j 0
• Recall that χ = r − 1 and P = χ0E = 0E(r − 1)
1 X E + 3 0E(r − 1) − 1 = N α 0 r j j E j 0 Simplifying and rearranging gives the Claussius Mossotti Relation:
r − 1 1 X = N α (SI Units) + 2 3 j j r 0 j
• 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 X α(ω) = αi(ω)
60 11.9 Frequency Dependence 11 DIELECTRIC MATERIALS
• Classical approach: treat electrons as harmonic oscillators • Apply Hooke’s Law
2 eEloc = Cx = mω0x
2 C = mω0 ≡ Force constant
mω2x E = 0 loc e
• Note: ω0 is the natural or resonance frequency
• The electronic contribution to the polarizability is P = NmαelEloc. Recall that P = Nmqδ = Nex P Nex e2 αel = = = 2 NEloc NEloc mω0 • For an applied field with frequency ω, the equation of motion for the electron response is
d2x m + mω2x = −eE sin ωt dt2 0 loc
• The solution is x = x0 sin ωt. Substituting the solution into the equation of motion,
2 2 m(−ω + ω0)x0 = −eEloc
• Solving for x0
−eEloc x0 = 2 2 m(ω0 − ω )
• The electronic dipole moment is
2 e Eloc µelec = −ex0 = 2 2 m(ω0 − ω ) • The electronic polarizability is
2 µelec e αelec = = 2 2 Eloc m(ω0 − ω )
2 2 • αelec becomes large at resonances where ω0 − ω → 0 • Calculate the atomic polarizability:
61 11.10 Quantum Theory of Polarizability 11 DIELECTRIC MATERIALS
2 e Eloc µ0 = −ex0 = 2 2 m(ω0 − ω ) 2 µ0 e αelec = = 2 2 Eloc m(ω0 − ω )
11.10 Quantum Theory of Polarizability 2 e X fij α = elec m ω2 − ω2 j 0
fij ≡ Oscillator strength for a transition from state i to j
Pelec = NαelecE
11.11 Advanced Dielectrics: Ferroelectrics • Ferroelectrics: materials that are non-centrosymmetric and have a result- ing net polarization. • The polarization is a function of the applied field and displays hysteresis.
+ ion moves, forming a net dipole
PS ≡ Spontaneous polarization
EC ≡ Coercive field
62 12 PHASE TRANSITIONS
12 Phase Transitions
• The transition from a ferroelectric (net polarization) to pyroelectric (no net polarization) structure occurs at the Curie Temperature (TC ) • Dielectric constant: ξ = ξ ≡ Curie constant T − TC
Ferroelectric Pyroelectric Phase Phase Phase Transition Heat Capacity Dielectric Constant
Temperature Temperature
12.1 Lattice Instabilities • Recall the Clausius-Mossotti relation: − 1 1 r = Nα r + 2 30
• Solving for r,
Nα Nαr 2 Nα r − 1 = (r + 2) = + 30 30 3 Nα 2Nα r 1 − = + 1 30 30
2Nα/0 + 3 r = 3 − Nα/0
• Singularity (polarization catastrophe) at the condition Nα/0 = 3
12.2 Curie Weiss Law • What happens close to the singularity? Consider a small deviation 2s Nα =∼ 3 − 2s 0
• Rewriting the equation for r,
63 12.3 Ferroelectrics 12 PHASE TRANSITIONS
2Nα/0 + 3 1 r = ∝ near the singularity 3 − Nα/0 − 2s s ∼ • Near the critical temperature, suppose s ∝ T such that s = (T − TC )/ξ
∼ ξ 1 ∼ T − TC ∴ r = or = (T − TC ) r ξ
12.3 Ferroelectric Phase Transitions: Landau Theory
• Consider the Helmholtz free energy Fˆ 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, 1 1 1 Fˆ(P,T, E) = −E · P + s + s P 2 + s P 4 + s P 6 + ··· 0 2 2 4 4 6 6 • Note that there are no odd powers if the unpolarized crystal has a center of inversion. • At thermal equilibrium, the minimum in Fˆ with respect to P is
∂Fˆ = 0 = −E + s P + s P 3 + s P 5 ∂P 2 4 6 • Thermal equilibrium condition:
∂Fˆ = 0 = −E + s P + s P 3 + s P 5 ∂P 2 4 6 • For a ferroelectric state transition, the coefficient of the first order P term must pass through zero at some temperature T0 for no applied field.
∴ assume s2 = γ(T − T0), with γ a positive constant
• s2 can be positive or negative:
• s2 < 0: lattice is “soft” close to the instability
• s2 > 0: unpolarized lattice is unstable
64 12.3 Ferroelectrics 12 PHASE TRANSITIONS
two minima
• Rewriting the equilibrium condition,
∂Fˆ = −E + γ(T − T )P + s P 3 = 0 ∂P 0 4 • Second order transition: no volume change, smooth transition.
• For E = 0 and P = PS at equilibrium,
3 γ(T − T0)PS + s4PS = 0
2 PS γ(T − T0) + s4PS = 0
• Solutions: PS = 0, or
2 • For T ≥ T0 : PS = (γ/s4)(T0 − T ) ⇒ PS is imaginary; the only real root is PS = 0.
• For T < T0 :
1/2 1/2 |PS| = (γ/s4) (T0 − T )
• Second order transition for T < T0 :
1/2 1/2 PS| = (γ/s4) (T0 − T )
• The phase transition is second order since the polarization goes continuously to zero
dipoles aligned
order parameter dipoles randomly aligned
65 12.3 Ferroelectrics 12 PHASE TRANSITIONS
12.3.1 First versus Second Order Transitions
• The change in Ps is discontinuous for first order transitions and continuous for second order transitions • First order transition results from a structure change. Example: cubic (pyroelectric) → tetragonal (ferroelectric)
First Order Second Order
• Recall the Helmholtz free energy: 1 1 1 Fˆ = −E · P + s + s P 2 + s P 4 + s P 6 + ··· 0 2 2 4 4 6 6
• For first order transitions, s4 is negative and s6 cannot be neglected • Equilibrium condition
∂Fˆ = γ(T − T )P − |s |P 3 + s P 5 = 0 ∂P 0 S 4 S 6 S
• Solutions: PS = 0 or
2 4 γ(T − T0) − |s4|PS + s6PS = 0
First Order Second Order
12.3.2 Ferroelectric Example: BaTiO3 2 −1 −2 • Spontaneous polarization: Ps = Nmqδ = 26 µC/cm ≈ 3 × 10 C · m
• Dipole moment of a unit cell: µ = Ps · volume. For a lattice constant a ≈ 0.4 nm
66 12.4 Other Instabilities 12 PHASE TRANSITIONS
µ = (3 × 10−1C · m−2) × 64 × 10−30m3 ≈ 2 × 10−29C · m
Material TC (K)
BaTiO3 408 BaTiO3 1480
12.4 Other Instabilities
ferroelectric + + + + + +
anti-ferroelectric + + + + + +
• Pyroelectrics: heat causes distortion and electrical signal, “pyroelectric detectors”
12.5 Piezoelectrics • Piezoelectric materials: an applied electric field causes a mechanical de- formation • Piezoelectric materials lack inversion symmetry
67 13 DIAMAGNETISM AND PARAMAGNETISM
no inversion symmetry
+ + +
+ + + + + + + + + + + + +
• Stress and displacement:
Stress: T = CS − eE Displacement: D = E + eS
C ≡ Elastic constant e ≡ Piezoelectric coefficient E ≡ Electric field S ≡ Strain
• Note: when E = 0, D 6= 0 for non-zero e and S.
• Hooke’s Law for E = 0. • Applications: electro-mechanical transducers, microphones, micro- motors, micro-electro-mechanical structures (MEMS), nano-electro- mechanical structures (NEMS).
13 Diamagnetism and Paramagnetism
• Current and magnetism related through Maxwell’s Equations
• Magnetic susceptibility:
µ M χ = 0 B
M ≡ Magnetization B ≡ Magnetic field intensity
µ0 ≡ Permeability of free space
68 13.1 Diamagnetism 13 DIAMAGNETISM AND PARAMAGNETISM
+ Paramagnetism
Pauli paramagnetism 0 Temperature _ Diamagnetism
13.1 Diamagnetism • In a magnetic field, induced current field opposite to applied current • Electron precesses around field axis with frequency eB ω = [sec−1] 2m • Magnetic field will induce net electric current around the nucleus.
B electron
• Consider multi-electron atom. The “electron current” for Z electrons is
ω I = (charge) × (revolutions per second) = −Ze 2π 1 eB Ze2B I = −Ze = − 2π 2m 4πm
• Magnetic moment:
µ ≡ (current) × (area of loop) = I × A
• For a loop area A = πρ2,
Ze2B µ = − hρ2i 4m • hρ2i is the mean square of the perpendicular distance from the field axis through charge
hρ2i = hx2i + hy2i
69 13.2 Paramagnetism 13 DIAMAGNETISM AND PARAMAGNETISM
• The atomic radius is hr2i = hx2i + hy2i + hz2i 2 hρ2i = hr2i ∴ 3 • Diamagnetic susceptibility of electrons: M = Nµ
2 Nµ0µ µ0NZe Langevin Result: χ = = − hr2i B 6m
(note negative sign)
• Note: need to calculate hr2i from quantum mechanics
13.2 Paramagnetism • Positive χ occurs for • atoms, molecules, and lattice defects with an odd number of electrons • free atoms with partially filled inner shells
• metals • Magnetic moment µ:
µ ≡ γ~J = −gµBJ
' $ γ ≡ gyromagnetic ratio ~J ≡ angular momentum g ≡ g factor (∼2.00 for electron spin)
µB ≡ Bohr magneton (atomic unit) J ≡ angular quantum number for electrons
• From& quantum mechanics, only certain alignments of µ with%B are al- lowed • Consider single electron case. Energy U of an electron in a magnetic field:
U = −~µ · B~ = mJ gµBB
• For single electron atom with no orbital momentum, mJ = mS = ±1/2
70 13.2 Paramagnetism 13 DIAMAGNETISM AND PARAMAGNETISM
only certain alignments allowed lower energy state
upper energy
Population state
1 • For a two spin orientation case, mJ =± 2 , g = 2, and U = ±µBB • Probability of occupying state i:
exp(−E /τ) P = i i Z
Z ≡ sum over all states (normalization) τ ≡ kT
• For N total atoms,
N exp(µB/τ) 1 = N exp(µB/τ) + exp(−µB/τ) N exp(−µB/τ) 2 = N exp(µB/τ) + exp(−µB/τ)
2
1
• Magnetization M ≡ (net spin density kB) × (magnetic moment per electron)
ex − e−x M = (N − N )µ = Nµ = Nµ tanh x 1 2 ex + e−x
x ≡ µB/kT
• For x 1: high T , small field, no interaction.
tanh x ≈ x ⇒ M =∼ Nµ(µB/kT )
71 13.2 Paramagnetism 13 DIAMAGNETISM AND PARAMAGNETISM
two level system
all spins aligned
few spins aligned
• For multiple electron atoms, the azimuthal quantum number mJ takes values of J, J − 1, ..., −J • For an atom with quantum number J, there are 2J + 1 energy levels • Curie-Brillouin Law:
M = NgJµBBJ (x)
x ≡ gJµBB/kT
• Brillouin function (obtained from hµi over all quantum configurations) 2J + 1 (2J + 1)x 1 x B (x) coth − coth J 2J 2J 2J 2J • For x 1 1 x coth x ≈ + x 3
13.2.1 Calculation of Susceptibility For small fields and high temperatures, M NJ(J + 1)g2µ2 Np2µ2 χ = =∼ B = B = C/T B 3kT 3kT
p2 = g2J(J + 1) ⇒ p = g [J(J + 1)]1/2 C ≡ Curie constant
s = 7/2 7 s = 5/2
s = 3/2
B/T
72 13.2 Paramagnetism 13 DIAMAGNETISM AND PARAMAGNETISM
13.2.2 Calculation of Total Angular Momentum J • Atoms with filled shells have no magnetic moment. Example:
1s22s22p63s23p63d10
• The spins arrange themselves so as to give the maximum possible S con- sistent with Hund’s rule
• Pauli Principle: no two electrons in the same system can have the same quantum numbers n, l, ml, and ms
13.2.3 Spectroscopic Notation Example Orbital Configurations
1s 2s 2p B
1s 2s 2p 3s 3p 3d5 4s1 Cr
1s 2s 2p 3s 3p 3d5 4s2
Mn
1s 2s 2p 3s 3p 3d6 4s2 Fe
1s 2s 2p 3s 3p 3d1 Ti3+
13.2.4 Paramagnetic Susceptibility Curie Law for paramagnetic susceptibility: χ = C/T
Curie Law
free spin Will depend on valence electron and J
13.2.5 Calculation of J • Rules for calculating J:
73 13.2 Paramagnetism 13 DIAMAGNETISM AND PARAMAGNETISM
Value of J Condition |L − S| When the shell is less than half full |L + S| When the shell is more than half full S When the shell is half full
• Values for L and S: X X L = ml and S = ms
• L is the orbital angular momentum
13.2.6 Spin Orbit Interactions • For multi-electron atoms, P • Orbital angular momentum: L = i Li P • Total spin angular momentum: S = i Si • Total angular momentum: ~J, where J = L + S • Treat J, L, and S as vectors. L and S precess around J; J remains con- stant (“conserved”) • Note: for iron metal group, J = S; only the spin contributes. Orbital momentum is “quenched.”
vector diagram J
S L
13.2.7 Effective Magnetic Number • Effective magnetic number:
P = g [J(J + 1)]1/2
• Example: Ti3+
J ∼ S For transition metals, "orbital angular momentum quenched"
3d1 S=1/2
74 13.2 Paramagnetism 13 DIAMAGNETISM AND PARAMAGNETISM
P = 2 [S(S + 1)]1/2
1 31/2 P = 2 · = 1.73 2 2
• Example: Fe2+
3d6 S=2
P = 2 [2 (3)]1/2 = 4.90
13.2.8 Paramagnetic Properties of Metals • For paramagnetic solids with N magnetic ions
Nµ2B M = kTF • For metals, not all spins are accessible
2 ∼ Neff µ B ∼ NkT M = ⇒ Neff = kT kTF
Nµ2B M =∼ Temperature independent kTF
13.2.9 Band Model
• Total magnetization: M = (N+ − N−)µ
• Total number of electrons with “up” spins (N+)
1 Z EF N = dE f(E) D(E + µB) + 2 −µB | {z } | {z } Fermi function Density of states
75 13.2 Paramagnetism 13 DIAMAGNETISM AND PARAMAGNETISM
Density of States B=0 B>0
Spins Spins parallel anti-parallel to B to B
• Taking a Taylor expansion of the density of states for N+,
1 Z EF 1 Z EF dD(E) N+ ≈ dEf(E)D(E) + dEf(E) µB 2 0 2 0 dE 1 Z EF 1 Z EF ≈ dEf(E)D(E) + f(E)dD(E)µB 2 0 2 0 1 Z EF 1 ≈ dEf(E)D(E) + D(EF )µB 2 0 2
• Repeat the procedure for N−
• Taylor series expansion of the density of states for N−
1 Z EF N− = dEf(E)D(E − µb) 2 µB 1 Z EF 1 ≈ d(E)f(E)D(E) − D(EF )µB 2 0 2
• Total magnetization:
2 M = (N+ − N−)µ = D(EF )µ B
• For a metal, 3N 3N D(EF ) = = 2EF 2kTF
3Nµ2B ∂M 3Nµ2 M = χ = = Independent of T! 2kTF ∂B 2kTF
13.2.10 Multivalent Effects • High density of states near Fermi level. • d-band contribution to χ
76 14 FERROMAGNETISM
Pd (Experimental)
V
Mo
K z=1 Rb z=1
14 Ferromagnetism
• Magnetic ordering in solids is due to interaction of spins • Below a transition temperature, an internal field tends to line up spins. This is called the “exchange field” BE
HΨ = EΨ with H=K+Vex
Vex ≡ Exchange Potential
• Mean field approximation: BE = λM where λ ≡ Weiss constant
Alignment of Spins
Ferromagnet Simple Antiferromagnet Ferrimagnet
14.1 Ferromagnetic Phase Transition • Above the transition temperature, ferromagnetism gives way to param- agnetism
T > TC Paramagnetism
T < TC Ferromagnetism
• Magnetization: M = χp(BA + BE)
χP ≡ Paramagnetic susceptibility
BA ≡ Applied field
77 14.2 Molecular Field 14 FERROMAGNETISM
Paramagnetic
Ferro- magnetic
14.2 Molecular Field
• For paramagnetism, we had χp = C/T
• “Molecular” or “exchange” field for ferromagnetism: BE = λM C M = χ (B + λM) ⇒ M = (B + λM) P A T A • Solving for M,
Cλ CB M 1 − = A T T
M(T − Cλ) = CBA
M C χ = = BA (T − Cλ)
• Taking Cλ = TC ,
C χ = for T > TC T − TC
Weiss constant: λ = TC /C
• Ferromagnetic behavior:
1 χ ∝ 1.33 Scaling theory (T − TC )
• Calculation of field parameter:
TC 3k λ = = TC 2 2 C Ng S(S + 1)µb | {z } Curie Constant ∼ 6 2 8 • For iron, TC = 1000K, g = 2, S = 1, (3d 4s → 3d ), λ = 5000. For a saturation magnetization of MS = 1700 Gauss,
78 14.2 Molecular Field 14 FERROMAGNETISM
BE = λMS
7 3 BE = 5000 × 1700 ≈ 10 Gauss = 10 Tesla
• Highest superconducting magnet: 40 Tesla
Ferromagnetic (Ordered)
Paramagnetic (Disordered)
1040 K
Material TC (K) Fe (bcc) 1043 Co 1388 Ni 627
Note: fcc iron is not magnetic
14.2.1 Prediction of TC
• Heisenberg model. For atoms i and j with spins Si and Sj, the ordering energy U is
U = −2JexS~i · S~j
Jex ≡ Exchange energy integral
• From mean field theory,
3kTC 2ZS(S + 1)Jex J = or T = ex 2ZS(S + 1) C 3k
Z ≡ Number of neighbors
79 14.2 Molecular Field 14 FERROMAGNETISM
14.2.2 Temperature Dependence of M(T ) for Ferromagnetism • Consider spin 1/2 system. Just as for paramagnetism, M = Nhµi
µB M = Nµ tanh kT
Where B = BA + BE
• For a ferromagnet, B ≈ BE = λM, where BE ≡ Molecular field.
µλM µλM M = Nµ tanh = Nµ tanh(x) with x ≡ ∴ kT kT
At low temperatures, Saturation all magnetic spins Magnetization line up
14.2.3 Ferromagnetic-Paramegnetic Transition • Define reduced magnetization m and reduced temperature t M kT m ≡ t ≡ Nµ Nµ2λ
∴ m = tanh(m/t) |{z} | {z } LHS RHS
• Solve the transcendental equation graphically by plotting both the left hand side (LHS) and right hand side (RHS) versus m/t
14.2.4 Ferromagnetic-Paramagnetic Transition • From the Landau model, as T increases, M(T ) decreases • Second order phase transition: continuous change in M(T )/M(0) versus T to TC
80 14.2 Molecular Field 14 FERROMAGNETISM
Ordered
Disordered
14.2.5 Ferromagnetic-Paramagnetic Transition • At low temperatures,
tanh(m/t) = tanh ξ =∼ 1 − 2e−2ξ
• Define change in magnetization with temperature, ∆M ≡ M(0)−M(T ). For large ξ (low T ),
tanh ξ =∼ 1 − 2e−2ξ ≈ 1
−2ξ −2µλM/kT ∴ ∆M = 2Nµe = 2Nµe
• Noting that M = Nµ,
2 ∆M = 2Nµe−2µ λN/kT
2 ∆M = 2Nµ exp(−2λNµ /kT ) = 2Nµ exp(−2TC /T )
14.2.6 Ferromagnetic-Paramagnetic Transition Change in magnetization with respect to T = 0 value versus temperature
2 ∆M = 2Nµ exp(−2λNµ /kT ) = 2Nµ exp(−2TC /T )
For small
81 14.2 Molecular Field 14 FERROMAGNETISM
14.2.7 Ferromagnetism of Alloys
3
2
1 0.6
Cr Mn Fe Co Ni 7 8 9 3d Electron Concentration
14.2.8 Transition Metals d Orbital Configurations of Transition Metals
3d64s2 Fe 8 3d 2 unpaired spin
3d74s2 Co 3d9 1 unpaired
3d84s2 Ni 3d9.44s0.6 0.6 unpaired
Cu 3d104s1
Others: Mn 3d54s2 Pd 4d10
14.2.9 Ni Alloys • Atomic moment of nickel transition metal alloys changes with composi- tion • For nickel alloyed with copper, magnetization disappears at 60% Cu
Mn
Pd 0.6 magentism Cu disappears Zn at 60% Cu 0.2 0.4 0.6 Alloy Composition of Second Component
82 14.2 Molecular Field 14 FERROMAGNETISM
14.2.10 Band Model • For nickel, 3d bands and 4s bands are partially occupied
3d10 3d9.44s0.6 5 up Paired spins 4.4 down (Pauli paramagnetism)
• Net spin for Ni is 0.6µB. Only the d bands contribute in this case
Filled States d band
s band
• Magneton numbers
• Fe: 2.2µB
• Co: 1.7µB • For iron,
3d8 3d7.84s0.2
• Can copper be ferromagnetic?
3d104s1 3d94s2 (yes)
superconducting properties?
YBa2CuO
14.2.11 Spin Waves • Spin waves: low energy excitations • Lower energy excited state is obtained by having spins arranged at an angle θ with respect to neighboring spins (“spin waves” or “magnons”)
• Precession angle depends on angle of neighbor
83 14.3 Ferrimagnetic Order 14 FERROMAGNETISM
B
Ground State Excited State, Spin Wave Higher Energy "Magnons"
Spin Wave - Top View
14.2.12 Magnon Dispersion • Magnons are particles and have an ω versus q dispersion relation 4J S2 aq ω = ex sin2 ~ 2
Jex ≡ Exchange energy term S ≡ Total spin angular momentum a ≡ Distance between atoms with magnetic moments q ≡ Wavevector
14.3 Ferrimagnetic Order in Magnetic Oxides
• Iron oxide (Fe3O4) consists of two sublattices:
Fe3O4 → FeO · Fe2O3
Sublattice A (1 Fe2+ ion)
Sublattice B (2 Fe3+ ions)
Net Spin: S=-2
− Sublattice Fe Oxid. State e Config. S gJ (µB) (A) FeO 2+ 3d6 2 4 5 5 (B)Fe3O4 3+ 3d 2 5
84 14.3 Ferrimagnetic Order 14 FERROMAGNETISM
• Total expected magnetic moment per formula unit: 2 × 5 + 4 = 14 µB
• Measured magnetic moment per formula unit: 4.1 µB (due to ferrimag- netic ordering)
14.3.1 Magnetic Oxides
• Calculation of saturation magnetization for NiOFe2O3 (nickel ferrite) • Unit cell: inverse spinel with 8 Ni2+ ions (B sites) and 16 Fe3+ ions (A sites) p Ion g J(J + 1) (µB) Ni2+ 2 Fe3+ 5
• The Fe3+ ions have anti-parallel magnetic moments
2+ ∴ net 2µB per formula unit due to Ni
2+ • For 8 Ni ions per formula unit each with 2µB magnetic moment,
−24 8 × 2µB (16)(9.27 × 10 ) 5 −1 MS = = = 2.5 × 10 A · m cell volume (8.37 × 10−10)3
• Calculation of saturation magnetization for Fe3O4 (magnetite) • Unit cell: inverse spinel with 8 Fe3+ ions on tetrahedral sites, 8 Fe3+ ions on octahedral sites, and 8 Fe2+ ions on octahedral sites. p Ion Site g J(J + 1) (µB) Fe3+ Tetrahedral 5 Fe3+ Octahedral 5 Fe2+ Octahedral 4
• Fe3+ ions on tetrahedral sites have magnetic moments aligned anti- parallel to those of Fe3+on the octahedral sites
2+ ∴ net 2µB per formula unit due to Fe
2+ • For 8 Fe ions per formula unit each with 4µB magnetic moment,
−24 8 × 4µB (32)(9.27 × 10 ) 5 −1 MS = = = 5.0 × 10 A · m cell volume (8.40 × 10−10)3
85 14.4 Domains and Walls 14 FERROMAGNETISM
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:
• HC : coercive field. The applied magnetic field for which the flux density B disappears
• BS: saturation flux density. The maximum flux density measured when all of the atomic moments are aligned with the applied magnetic field.
• BR: 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
Magnetized Demagnetized
N NS N SNS
S SN S N S N Closure field lines Domain Wall 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
86 14.5 Anisotropy of Magnetization 14 FERROMAGNETISM
Bloch Wall
Wall Region
14.4.1 Energy of Bloch Domain Walls • Energy U (Heisenberg Model):
U = −2JexS~i · S~j
2 • Consider the spin vectors by a classical model, i.e. S~i · S~j = |S| cos φ. Take a Taylor series expansion of cos φ about φ = 0
1 U = −2J |S|2 cos φ ≈ −2J |S|2 1 − φ2 ex ex 2
• The angle-dependent exchange energy between two adjacent spins at an angle φ is then
2 2 wex = Jex|S| φ
• For a rotation of the magnetic moment by π radians in N steps, wex = 2 2 Jex|S| (π/N) . For a Bloch wall of N spins, the total energy is then
π2 Nw = J |S|2 ex ex 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 (HC )
87 14.6 Ferrimagnetic Ordering 14 FERROMAGNETISM
• “Hard” direction: crystalline direction with high coercivity (HC )
[100] Parallel to C axis
[111]
Basal plane
Co
14.5.1 Anisotropy of Magnetization • Rotation of domains changes the overlap of the electron clouds of neigh- boring atoms, resulting in different exchange energy Jex
• Anisotropy energy Uk
0 2 0 4 Uk = k1 sin θ + k2 sin θ
0 0 • Easy axis depends on values of k1 and k2
c-axis M Electron Applied Clouds Field
Basal Plane
14.6 Ferrimagnetic Ordering - Exchange Terms • Exchange term for neighboring Fe atoms is negative for atoms on differ- ent sublattices and positive for atoms on the same sublattice. • Different sign of exchange term and different number of Fe atoms per unit volume on each sublattice account for ferrimagnetic ordering in Fe3O4 • Exchange terms for A and B sublattices:
JAA > 0 JAB < 0 JBB > 0
Sublattice A
Sublattice B
88 15 OPTICAL MATERIALS
14.6.1 Exchange Terms and Susceptibility
Paramagnetism Ferromagnetism Antiferromagnetism
TC ≡ Curie temperature
TN ≡ Neel temperature
14.6.2 Structure Dependence of Jex Cr (bcc): Antiferromagnetic Fe (bcc): Ferromagnetic
+ Co Fe Ferromagnetism Ni Gd
Fe (FCC) Mn Antiferromagnetism _ Distance between atoms
15 Electro-optic and Nonlinear Optical Materials
• Nonlinear optical materials have a nonlinear relation between the polar- ization density (P ) and electric field (E) • Expand electric susceptibility in higher order terms
89 15.1 Frequency Doubling 15 OPTICAL MATERIALS
h (1) (2) 2 (3) 3i P = 0 χ E + χ E + χ E
χ(1) ≡ Linear susceptibility χ(2) ≡ Quadratic susceptibility χ(3) ≡ Cubic susceptibility
• χ(2) is nonzero in non-centrosymmetric materials. Examples of χ(2) ef- fects: linear (Pockels) electro-optic effect, second harmonic generation (frequency doubling) • The lowest order nonlinearity for cubic materials is χ(3). Examples of χ(3) effects: quadratic (Kerr) electro-optic effect, third harmonic genera- tion (frequency tripling)
15.1 Frequency Doubling
• In frequency doubling, two photons with energy hν1 combine to form one photon with energy hν2. The total energy is conserved.
• Sending a beam with photon energy hν1 into a nonlinear optical (NLO) material results in two output beams with energies hν1 and hν2. • Examples:
NLO • Nd:YAG laser with 1.10µm (infrared) radiation −−−→ 0.55µm (green) • Blue laser by frequency doubling GaAs lasers
Nonlinear Optical (NLO) Material
15.2 Nonlinearity in Refractive Index • Velocity of light in matter: v = c/n
c ≡ Speed of light in free space (3×108 m/s) n ≡ Refractive index
• Refractive index (n) in nonlinear material:
90 15.3 Electro-Optic Modulators 15 OPTICAL MATERIALS
n2 = 1 + χ(1) + χ(2) + χ(3) + ···
• Dependence of refractive index on light intensity:
n = n0 + n2I
n0 ≡ Linear and χ = r − 1
n2 ≡ Nonlinear index I ≡ Intensity
∼ 2 • Dielectric constant: r = n
15.3 Electro-Optic Modulators • The optical properties of non-centrosymmetric crystals are modified by applying an external electric field. 1 1 ∆ = ∆ 2 = rE r n • Expanding n = n(E) in a Taylor series expansion about E = 0, the fol- lowing expression is obtained 1 ∆n = − n3rE 2
r ≡ Electro-optic coefficient [pm/V]
6 • Example material: LiNbO3, r = 31 pm/V, n = 2.29. Fields of 10 − 107 V/m are easily obtained in micro-photonic devices (1-10 V across ∼ 1 µm)
• Other materials: electro-optic polymers (r = 20 − 50 pm/V), BaTiO3 (r = 300 pm/V)
15.4 Optical Memory Devices • Photorefractivity: change in refractive index by light. Light empties 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 photo- sensitive material. A standing wave is formed which is preserved in the material as an interference pattern
91 15.5 Volume Holography 15 OPTICAL MATERIALS
• Material contains intensity and phase information and the object can be reconstructed • The difference between hologram and a photograph is that phase infor- mation is preserved in a hologram
15.5 Volume Holography • Pattern in material creates a dielectric grating or spatial modulation in the refractive index and dielectric constant
r = r0 + r1 cos(2ky sin θ) k ≡ Magnitude of light wavevector
• Period of dielectric grating: 2π λ Λ = = with k = 2πn/λ 2k sin θ 2n sin θ
Record Hologram Reconstruct Hologram
Object Reference Reference
Object
15.6 Photorefractive Crystals • Electro-optic and photoconductive - organic and inorganic materials
• Mechanism for photorefraction:
1. Generation of electrons and holes
2. Transport charges through the material 3. Trap the charges 4. Local field of trapped charges changes the refractive index n
• Spatially varying charge distribution results in spatially varying field. Field changes n through electro-optic coefficient r; larger r, larger pho- torefractive effect. 1 ∆n = − n3rE 2
92 15.7 Photorefractive Crystals 15 OPTICAL MATERIALS
light no light
+ Incident donor crystal V Beams Defect states in the band gap
15.7 Photorefractive Crystals Steps in creation of periodic dielectric constant
Light Net Charge Intensity Density
Donor Electric Concentration Field
Electron Dielectric Density Constant
15.8 Phase Conjugation - All Optical Switching beam 3 beam 2
beam 4
beam 1
• 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
Beam 4 (Probe) Beam 3 Beam 2
phase-conjugate beam
Beam 1
93 15.9 Acousto-Optic Modulators 15 OPTICAL MATERIALS
15.9 Acousto-Optic Modulators • Photoelastic effects: strain causes a change in the refractive index
1 1 ∆ = pS and ∆ 2 = pS r n
S ≡ Strain p ≡ Photoelastic constant n ≡ Refractive index
• Acousto-optic effect is used in Bragg cells to create a grating with period Λ by launching an acoustic wave. Light is diffracted from the grating. λ Λ = 2n sin θ
Acoustic wave off: Acoustic wave on: Beam transmitted Beam diffracted
Acousto-Optic Transmitted Material Beam
Input Input Beam Beam Diffracted Beam
Acoustic Wave ("Sound")
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 waveg- uides, couplers, and filters
• Active elements: elements that require power, such as switches, modu- lators, directional couplers, amplifiers, lasers • Photonics: analogue of electronics. Photonics encompasses the control of photons.
94 15.11 Dielectric Waveguides 15 OPTICAL MATERIALS
15.11 Dielectric Waveguides • In integrated optics, dielectrics are used to guide light in waveguides. Waveguides act as “optical wires” • Light is confined to the region of highest refractive index • Why dielectric materials instead of metals? Ease of fabrication and lower losses
Types of Dielectric Waveguides
Slab Ridge Fiber
Darker shading = higher index
15.12 Example: Phase Shifter • An electro-optic dielectric waveguide confines and guides light • An electric field is applied to electrodes surrounding the waveguide. The refractive index of the dielectric waveguide is then modified by the electro-optic effect 1 1 V ∆n = − n3rE = − n3r 2 2 d • Phase delay of optical wave due to the electro-optic effect: 2π ∆φ = ∆nL λ
-V +
Au Electrode
LiNbO3 Substrate Ti diffused waveguide
95 15.13 LiNbO Phase Shifter 16 SUPERCONDUCTIVITY
15.13 Example: LiNbO3 Phase Shifter • Electric field of 106 V/m (10 V across 10 µm) yields ∆n = 1.86 × 10−4 for LiNbO3 • For a phase shift ∆φ = π with operation at λ = 1.5 µm, 2π ∆φ = π = L∆n λ
L = 0.4 cm • Long interaction lengths are needed; materials with higher electro-optic coefficients are needed for more compact devices
• The voltage required for a π phase shift is the half-wave voltage Vπ λd V = π n3rL 16 Superconductivity
• Superconductivity: a phenomena in which a material undergoes a tran- sition to zero resistivity with zero magnetic induction (Meissner effect) below a critical temperature TC .
• At temperatures below TC application of a critical magnetic field HC will cause a transition to the normal state Resistivity Critical Field
Metal 800
600 Pb 400
200 Superconductor Tl 2 4 6 8
16.1 BCS Theory of Superconductivity • BCS Theory: quantum mechanical theory of superconductivity formu- lated by Bardeen, Cooper, and Schreiffer in 1957 (Nobel Prize in 1972). • Electron-phonon coupling leads to the formation of electron pairs (Cooper pairs) • BCS Theory accounts for observed properties of superconductivity, in- cluding the energy gap, Meissner effect, critical temperature, and quan- tization of magnetic flux through a superconducting ring
96 16.2 Density of States and Energy Gap 16 SUPERCONDUCTIVITY
16.2 Density of States and Energy Gap • The energy gap in superconductors is caused by electron-electron inter- actions rather than electron-lattice interactions for dielectrics • The energy gap is maximum at 0 K and decreases to zero at the transition temperature.
Density of States Temperature Dependence of
1.0
1.0
16.3 Heat Capacity of Superconductors
• The entropy of superconductors decreases upon cooling below TC , re- sulting in electronic ordering.
• In the superconducting state, for T TC ,
2 HC = H0 1 − 1.07(T/TC )
Heat Capacity of Gallium
1.5
1.0 Superconducting
Normal 0.5
0 0.5 1.0 1.5
97 16.4 Superconductor Junctions 16 SUPERCONDUCTIVITY
16.4 Tunneling in Superconductor Junctions
Metal-Insulator-Metal
V = 0 V > 0 I-V Characteristic I + - Metal Metal
V Insulator Superconductor-Insulator-Metal
V = 0 V > 0 I-V Characteristic
I
Metal
Superconductor V Insulator
16.5 Semiconductor-Insulator-Semiconductor Junction • Tunneling across a thin oxide junction between two superconductors with energy gaps 2∆1 and 2∆2 • Nobel Prize in physics, Ivan Giaever, 1973
Superconductor 1 Superconductor 2 I-V Characteristic
I
Negative resistance
Insulator V
16.6 Type I and Type II Superconductors • Type I: superconductor is diamagnetic in an applied magnetic field up to the critical field Hc • Type II: superconductor exhibits two magnetic behaviors. Up to a lower critical field HC1 the superconductor is diamagnetic, and between HC1 and an upper critical field HC2, the flux density is not equal to zero (“in- complete” Meissner effect). • Type II superconductors tend to be alloys or transition metals with high room temperature resistivity
98 16.7 High TC Superconductors 16 SUPERCONDUCTIVITY
• Type II superconductors are used in high field magnets, magnetic reso- nance imaging (MRI) applications
Type 1 Type II -M -M
H H
16.7 High TC Superconductors
• Highest values for TC have been reported in cuprates • 2 layer cuprate compounds support 2D conduction
YBaCuO BiSrCaCoO
16.8 Cuprate Superconductors • Theory: pairing is involved. Nature of pairing? Spin Waves?
• 2∆(0) ≈ 6kTC • Future applications: current carrying wires, Maglev trains
Phase Diagram for Cuprates
Semiconductor Paramagnetic metal Strange metal Temperature insulator Antiferromagnetic Superconductor
0.1 0.2 Hole Doping Level
99 17 351-2 PROBLEMS
17 351-2 Problems
18 −3 1. An abrupt Si p-n junction hasNa = 10 cm on one side 15 −3 andNd = 10 cm on the other.
(a) Calculate the Fermi level position at 300K on both sides.
(b) Draw an equilibrium band diagram for the junction.
(c) Determine the contact potentialΦo for this junction.
+ −2 2 15 −3 2. A siliconp − n junction10 cm in area hadNd = 10 cm doping on the n-side. Calculate the junction capacitance with a reverse bias of 10V. 3. For metallic aluminum, calculate:
(a) The valence electron density.
(b) The radius of the Fermi spherekF. (c) Fermi energy in eV.
4. From the Schrodinger equation for a quantum well, show that the wave vector is equal tonπ/L where L is the well width. 5. 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. 6. What is the luminescent energy for a CdSe quantum dot with a 2 nm radius. 7. For a MOSFET device briefly describe how the three types of device work: a) enhancement mode b) depletion mode c) inversion mode.
8. Calculate the capacitance of an MOS capacitor with a 10 nm thickHfO2
dielectric oxide. What is the ratio of capacitances forCHfO2 /CSiO2 . The relatie dielectric constant forHfO2is 25. 9. Problem 9.9 in Solymar and Walsh 10. Problem 9.14 in Solymar and Walsh 11. Problem 9.16 in Solymar and Walsh 12. Problem 12.10 in Solymar and Walsh 13. Consider a quantum cascade laser (QCL) made from GaAs and GaAlAs. What well thickness is needed for laser emission at 3 microns? 14. Derive the expression for the average value of the dipole moment. Show that it is given by: 1 < µ >= µ[cotha − a ]
100 18 351-2 LABORATORIES
15. The saturation polarizationPs ofPbTiO3, a ferroelectric, is0.8 coulombs/m2. The lattice constant is 4.1A˙. Calculate the dipole moment of unit cell. 16. Calculate the polarization P of one liter of argon gas at 273 K and 1 atm. The diameter of an argon atom is 0.3 nm. 17. Consider the frequency dependence of the atomic polarizability. The po- larizability and its frequency dependence can be modeled as a damped harmonic oscillator. Derive the expression forα in this case. The expression is given by: dx dx 2 m dt2 + b dt + ω0x = −eεlocsinωt Plotα vs.ω for this case.
18. Problem 4.6 in Solymar and Walsh. 19. Problem 4.7 in Solymar and Walsh. 20. Problem 4.8 in Solymar and Walsh.
21. Problem 4.9 in Solymar and Walsh. 22. Calculate the magnetic susceptibility of metallic copper. How does it compare to the measured value of -1.0? 23. Calculate the effective magneton number p forMn2+, Co2+. Show work.
24. Consider Mn doped GaP. There are1020Mn2+ ions. 25. What is the electron configurationMn2+ in spectroscopic notation.
(a) Calculate its magnetic moment at saturation in Bohr magnetons. (b) Calculate its magnetic susceptibility.
26. For metallic Co, which has a Curie temperature of 1388 K, calculate the Weiss constantλ. Calculate the exchange constant in meV.
18 351-2 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 appa- ratus.
101 18.1 Laboratory 1: Measurement of Charge Carrier Transport Parameters Using the Hall Effect 18 351-2 LABORATORIES
Figure 18.1: Part 7; B is uniform throughout the area of interest
18.1.2 Outcomes Upon completion of the laboratory, the student will be able to: 1. Use a Hall effect apparatus to measure the mobility and carrier concen- tration in a semiconductor. 2. Derive the equations that enable the extraction of fundamental materials parameters using the Hall effect. 3. Describe the dependence of mobility on carrier concentration and tem- perature, and explain the origins of differences in mobilities between dif- ferent semiconductors.
18.1.3 References (1) M. Ali Omar, Elementary Solid State Physics; (2) Solymar & Walsh, Electrical Properties of Materials; (3) MSE 351-1 Lecture Notes; and (4) the NIST web page: http://www.nist.gov/pml/div683/hall.cfm
18.1.4 Pre-Lab Questions 1. What is the Lorentz Force?
2. What is the Hall effect and when was it discovered? 3. Write the equation describing the force, FM, on a particle of charge q and with velocity v in a uniform magnetic field, B. 4. Does the velocity of a charged particle (with non-zero initial velocity) in a uniform magnetic field change as a result of that field? If so, how? Does its speed change? 5. What is the right-hand-rule? 6. For a particle with negative charge, q, in the situation below, in what direction will the particle be deflected?
(a)
102 18.1 Laboratory 1: Measurement of Charge Carrier Transport Parameters Using the Hall Effect 18 351-2 LABORATORIES
P = + P = - M = M = N = N =
Figure 18.2: Part 6
Figure 18.3: Part 9
7. For the example below, what will be the sign of the charge built up 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.)
(a)
8. By convention, current is defined as the flow of what sign of charge car- rier?
9. What is the electric field, E, in the situation below? What is the electro- static force, FE on the particle if it has a charge q?
(a)
10. Why are both a resistivity measurement and a Hall measurement needed in order to extract fundamental material parameters?
18.1.5 Experimental Details The samples to be characterized include: 1. “bulk” Si, GaAs, InAs (i.e. substrates ~ 400 microns thick) 2. “thin films” of InAs and doped GaAs, grown on semi-insulating GaAs substrates 3. Indium tin oxide (ITO) on glass.
103 18.1 Laboratory 1: Measurement of Charge Carrier Transport Parameters Using the Hall Effect 18 351-2 LABORATORIES
You may have the opportunity to make additional samples. For contacts on n-type GaAs, use In-Sn solder; for contacts on p-type GaAs, use In-Zn solder. Most samples are mounted on mini-circuit boards for easy insertion into 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/1lrAokPI1vIJ- a8pmLx4FVCqHh80PKgGr8oDD6eVzx_w/viewform
18.1.8 Lab Report Template 1. 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. 2. Apply data from a sample measurement to test the equation derived in (1). Show your work. 3. How is the carrier concentration related to the Hall Coefficient? What is the difference between bulk and sheet concentration? 4. Calculate the carrier concentration (bulk and sheet) using the sample data and the equations derived above. Show your work. 5. Does the mobility exhibit any dependence on the carrier concentration? Discuss briefly; include observations from lab for the same material & type (e.g. n-GaAs). 6. What is the origin of the difference in mobilities between the n-type and p-type samples, assuming that the doping levels are similar? 7. 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. 8. 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 mag- netic field may be determined. A device that measures these parameters, known as a Hall Probe, provides a way to measure magnetic fields. Write
104 18.1 Laboratory 1: Measurement of Charge Carrier Transport Parameters Using the Hall Effect 18 351-2 LABORATORIES
Figure 18.4: Example
an expression that relates these parameters. Which is the more sensitive (higher ratio of mV/Tesla) Hall probe – the bulk InAs or bulk GaAs sam- ple? Show your work. (Note – you should have recorded average Hall voltage for a given current. How do these compare?)
18.1.9 Hints for derivation 1. For the example below, what VH would you need to apply to make the particle continue on in a straight line throughout the sample? (B is uni- form throughout the specimen. Hint: Balance the magnetic and electric forces on the particle.)
2. 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 cross-sectional area A, n is the concentration of carriers per unit volume in the material pass- ing the current, q is the charge on each carrier, and v is the drift velocity of the carriers. If you knew i, A, VH, d, B, and q from the situation illus- trated above, what expression tells you n?
3. The “Hall coefficient” of a material, RH , is defined as the Hall electric field, EH , per current density, per magnetic field, RH =EH /(J*B) Using the equations given and derived thus far, express the Hall coefficient in terms of just q and n.
105 18.2 Laboratory 2: Diodes 18 351-2 LABORATORIES
18.2 Laboratory 2: Diodes 18.2.1 Objective The purpose of this lab is to explore the I-V characteristics of semiconductor diodes (including light emitting diodes (LEDs) and solar cells), and the spec- tral response of LEDs and lasers.
18.2.2 Outcomes Upon completion of the laboratory, the student will be able to: 1. Measure diode I-V characteristics and relate them to band diagrams.
2. Fit I-V data to the diode equation, extract relevant parameters, and relate these to materials constants. 3. Determine the open circuit voltage and short-circuit current of the solar cell.
4. Describe the dependence of emission wavelength on bandgap and de- scribe origins of spectral broadening. 5. Describe how lasers differ from LEDs in design and performance.
18.2.3 Pre-lab Questions 1. What expression describes the I-V characteristics of a diode?
2. Sketch the I-V characteristics of a diode and label the sections of the curve corresponding to zero bias (1), forward bias (beyond the built-in volt- age) (2), and reverse bias(3), and then sketch the corresponding band diagrams for 1, 2, and 3. 3. Sketch the I-V characteristics of a Zener diode and the corresponding band-diagram for reverse bias. 4. Sketch the I-V characteristics of a p-n junction with and without illumi- nation. Label VOC and ISC . 5. Take pictures (use your phone) of lights around campus; try to get pic- tures of LEDs. Which do you think are LEDs (why?)? 6. Why are fluorescent lights “white?” How white are they?
106 18.2 Laboratory 2: Diodes 18 351-2 LABORATORIES
Experimental Details The devices to be characterized include: 1. Si diode 2. Zener diode
3. LEDs (different colors) 4. Si solar cell
18.2.4 References MSE 351-2 Lecture Notes, Omar Chapter 7, Solymar & Walsh Chapter 9, 12, 13.
Instructions/Methods Use multimeters and the Tektronix curve tracer for I-V measurements. Use the Ocean Optics spectrometer to obtain spectral responses.
Station 1 - CURVE TRACER p-n junction diode
Attach diode to “diode” slot on Curve Tracer. Measure both forward and reverse bias characteristics.
1. (In lab) Measure and record the I-V (current-voltage) 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. 2. (In lab) Measure the I-V characteristics of the diode in the reverse bias condition.
3. (Post-lab) The ideal diode equation is: I = Isat[exp(qV/kT ) − 1]. Note that for V>>kT/q, I = Isat exp(qV/kT ); however recombination of car- riers in the space charge region leads to a departure from ideality by a factor m, whereI = I0[exp(qV/mkT ) − 1] .
(a) Plot the data using ln(I) vs. V plot to determine m. (b) Use the value of the applied voltage corresponding to ~ 1mA to solve for I0 (which is too small to measure in our case.)
4. (In lab) Repeat the forward bias measurement using “Store.” Now cool the diode and repeat. Sketch the two curves. (Post-lab) Explain what you observe.
107 18.2 Laboratory 2: Diodes 18 351-2 LABORATORIES
5. Test red and green LEDs. Record the color and the turn-on voltage. 6. (Post-lab) Compare the I-V characteristics for the silicon diode and LEDs. Why would you choose Si over Ge for a rectifier? (Omar 7.21) 7. (In-Lab) Test the Zener diode. Sketch. (Post-lab) Compare to figure 9.28 (S&W).
Station 2: Solar Cells 1. Use the potentiometer, an ammeter and voltmeter to determine the I- V characteristics of a solar cell at different light levels. Determine the values of Voc (the open circuit voltage) and Isc (the short-circuit current) under ambient light, then use the potentiometer to determine additional points on the I-V curve. Record the results. 2. Repeat the I-V measurements for a higher light level. Record the light intensity measured with the photodiode meter. Area~1 cm2 : ____ Mea- sure the area of the solar cell:______3. (Post-Lab) Estimate the fill – factor and the conversion efficiency of the solar cell.
Station 3: Spectrometer and Power Supply Light Emitting Diodes 1. Measure the spectral response (intensity vs. wavelength) of the light emitting diodes using the spectrometer. 2. Cool the LEDs and observe the response. Qualitatively, what do the re- sults suggest about the change in Eg vs. temperature? About emission efficiency vs. temperature? How do these results compare to the I-V response of the cooled silicon diode? 3. Record the peak-wavelength and the full-width-half-maximum (FWHM) for each diode. 4. Calculate the bandgap that would correspond to the peak wavelength.
Semiconductor Lasers 1. Attach the laser to the power meter and setup the power meter to mea- sure intensity from the device. Slowly increase the voltage and record the I-V characteristics vs. power output for the laser. 2. Measure the spectral response of the laser: i) below threshold, and ii) above threshold. Note the FWHM of the peaks. How do they differ? How do they correspond to the I-V-power data?
108 18.2 Laboratory 2: Diodes 18 351-2 LABORATORIES
LED Color Peak Eg(in eV, Intensity FW FW FWHM wave- from (left) (right) length peak wave- length)
Laser
Table 18.1: Spectrometer and Power Supply
3. Plot the light output (intensity) as a function of current. What is the threshold current? Label the regions of spontaneous and stimulated emission. 4. Calculate the efficiency of the laser. 5. Determine the bandgap of the laser material.
6. Explain the change in the width of the spectral emission.
Station 4: Stereomicroscope 1. Sketch the structure of the LED observed under the stereomicroscope. Note the color of the chip when the device is off and the color of emission when the device is on. Sketch what the device structure might look like.
2. Sketch the structure of the laser observed under the stereomicroscope. Sketch what the device structure might look like.
109 18.3 Laboratory 3: Transistors 18 351-2 LABORATORIES
18.3 Laboratory 3: Transistors 18.3.1 Objective The purpose of this lab is explore the input/output characteristics of transis- tors and understand how they are used in common technologies.
18.3.2 Outcomes Upon completion of the laboratory, the student will be able to: 1. Measure the output characteristics of a few important transistors using a “curve tracer.” 2. Qualitatively relate the characteristics to the p-n junctions in the devices. 3. Describe transistor function and performance in terms of appropriate gains. 4. Identify applications of these devices in common technologies.
18.3.3 Pre-lab questions Bipolar Transistor 1. Sketch and label the band-diagrams for npn and pnp transitors. 2. Sketch I-V characteristics for this device as a function of base current. 3. In what technologies are these devices used?
MOSFETS 4. Sketch and label a MOSFET structure.
1. Sketch I-V characteristic for this device as a function of gate voltage. In your sketch of the ISource−Drain–VSource−Drain characteristics vs. gate voltage, Vgate, label the linear and saturated regions. 2. What is threshold voltage? What structural and materials properties de- termine the threshold voltage? 3. In what technologies are these devices used?
“Current Events” 1. What materials developments have changed transistor technology in the past decade? (Hint: Search Intel high-k; Intel transistors) 2. What new types of devices are on the horizon? Hint: search IBM nan- otubes graphene
110 18.3 Laboratory 3: Transistors 18 351-2 LABORATORIES
18.3.4 Experimental Details The devices to be characterized include the following 1. pnp and npn bi-polar junction transistors 2. phototransistor
3. metal oxide semiconductor field effect transistor (MOSFET).
References MSE 351-2 Lecture Notes, Omar Chapter 7; Solymar and Walsh Chapter 9
Instructions/Methods Use the Tektronix curve tracer for I-V measurements. 1. Bipolar Transistors (npn and pnp)
(a) Attach an npn transistor to the T-shaped Transistor slot on 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 I-V curves. Keep the collector-emitter voltage VCE below 30V to avoid damaging the device. (b) Sketch how E,B,C are configured in the device.
(c) Measure Icollector (output current) for VCE > Vsaturation as a func- tion of base (input) current, Ibase.
(d) Post-Lab: determine the transistor gains, α = −Icollector/Iemitter and β = Icollector/Ibase. (e) Repeat for a pnp transistor.
2. Phototransistor
(a) Measure the I-V 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. (b) Post-lab: discuss the mechanism by which the light influences the device current, and compare with the bipolar transistor operation.
3. MOSFETS
(a) Attach MOSFET Device to the linear FET slot on the Curve Tracer, making sure that S,G,D are connected appropriately. Observe both the linear and saturated regions for ISource−Drainvs.VSource−Drain. (b) 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).
111 18.3 Laboratory 3: Transistors 18 351-2 LABORATORIES
Figure 18.5: n-Channel MOFSET.
(c) Measure the VSource−Drain and ISource−Drain to extract RSource−Drain in the linear region as a function of VGate. You should take at least 4 measurements. (d) Post-lab, using the data from B, plot the conductance of the channel, GSource−Drain = 1/RSource−Drain, vs. VGate . You should observe a linear relationship following the equation ,
µ W C G = n OX (V − V ) SD L A G T
where µn is the electron mobility and W, L, and A are the width, length and area of the gate respectively (A = W x L). COX is the ca- pacitance of the oxide, which can be measured using an impedance analyzer as a function of gate voltage. A plot is shown in Figure 18.5. −10 (e) Using µn=1500 cm2/V-sec, C = 4.8x10 F (from the attached plot), plot of conductance vs. VGate, calculate L, the length of the gate.
(f) Determine Vthreshold from the above values. Compare with your observations from the curve tracer (part a).
112 18.4 Laboratory 4: Dielectric Materials 18 351-2 LABORATORIES
18.4 Laboratory 4: Dielectric Materials 18.4.1 Objective The objectives of this lab are to measure capacitance and understand the de- pendence on geometry and the dielectric constant, which may vary with tem- perature and frequency.
18.4.2 Outcomes Upon completion of the laboratory, the student will be able to: 1. Use an impedance analyzer to measure capacitance.
2. Given the capacitance of a parallel plate capacitor, calculate the dielectric constant. 3. For a known material, explain the microscopic origins of the temperature dependence of the dielectric constant.
4. 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 Pre-lab questions 1. What distinctions can you make between capacitance and the dielectric constant? What are the units of each? 2. What is the relationship between the dielectric constant and the ‘relative’ dielectric constant?
3. What is the dielectric constant (or permittivity) of ‘free space’ (or vac- uum)?
4. What is the relative dielectric constant of air? Water? Glass? SiO2? Ex- plain their relative magnitudes. 5. For the capacitor shown in Figure 18.6, what is the expression that relates the capacitance and relative dielectric constant? 6. What relationship determines the amount of charge stored on the plates of a capacitor when a specific voltage is applied to it? 7. With the above answer in mind how could you measure the capacitance of a structure?
113 18.4 Laboratory 4: Dielectric Materials 18 351-2 LABORATORIES
Figure 18.6: Capacitor.
8. Name two materials for which the capacitance (charge per unit voltage) is fixed. Name a material type or structure for which the capacitance is dependent on V. 9. What is the relationship that gives the amount of energy stored on a ca- pacitor? 10. Why are batteries the primary storage medium for electric vehicles, rather than capacitors? 11. Intel and their competitors are interested in both “low-k” and “high-k” dielectrics.
(a) What defines the boundary between “low” and “high?” (b) What drives the need for “high-k” dielectrics? What other proper- ties besides the dielectric constant are important characteristics of these materials? (c) What drives the need for “low-k” dielectrics?
12. Find one or two additional examples of technologies/devices that in- corporate capacitors, and explain the function of the capacitor in 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 in- organic dielectrics (in thin slabs) is accomplished through the evaporation of metal contacts on either side of the material. The total capacitance of a p-n junction diode is measured through the device contacts. The capacitance of liq- uids is determined by filling a rectangular container between two electrodes, and neglecting the contributions of the container.
114 18.4 Laboratory 4: Dielectric Materials 18 351-2 LABORATORIES
Material Area ThicknessCapacitanceεr(lab) εr(lit) Glass slide (sam- ple 1) Lithium niobate (LiNbO3) Glass sheet (sam- ple 2) Plexiglass
Table 18.2: Dielectric Constants
18.4.5 References Solymar and Walsh Chapter 10, Omar Chapter 8, Kittel Chapter 16
18.4.6 Instructions/Methods Use the HP Impedance analyzer to measure the materials provided in lab. Note the sample dimensions and the measured capacitance values on the at- tached table.
18.4.7 Lab Report Template Part I - Dielectric constants; Measuring capacitance with Impedance Ana- lyzer
1. Using the default frequency of 100kHz and zero bias voltage, measure capacitance and contact dimensions and then calculate the relative di- electric constant of the following materials: 2. Using the default frequency of 100 kHz, measure capacitance vs. voltage for several LEDs
Determine the built-in voltage of the corresponding p-n junction, ϕ0, for each diode, and discuss the difference. See equation 7.64, pg. 362, Omar. Note that the equation may be re-written as follows:
1 (ϕ0 − V0)(Na − Nd) 2 = 2 C εeNaNd
115 18.4 Laboratory 4: Dielectric Materials 18 351-2 LABORATORIES
LED= LED= LED= Bias CapacitanceBias CapacitanceBias Capacitance
Table 18.3: Capacitance
2 Plot 1/C vs. V0, and extrapolate the linear region (reverse bias) to the x- intercept to determine the value of interest. Note that the slope depends on the doping concentration.
3. Calculate the relative dielectric constant, based on capacitance measured in lab for:
(a) air (4.0inch x 3.5 inch plates, separated by ______(measure this)) (b) water at room temperature (4.0 inch x 3.5 inch plate par- tially submerged; measured dimension: Height:______x width______x separation______(c) ethanol
4. Fill in the table below, for water at the different temperatures measured in lab, using the dimensions measured above. Determine the corre- sponding values of the relative dielectric constant,εr. Plot (εr–1)/(εr +2) vs 1/T (Kelvin) for both sets of data below, and compare to figure 8.13 (p. 388, Omar).
Class data: Area = ______; separation = ______
Temp (C) Freq C (pF) εr
116 18.4 Laboratory 4: Dielectric Materials 18 351-2 LABORATORIES
Comparison from CRC handbook
Temp (C) εr 0 87.74 20 80.14 40 73.15 60 66.8 80 61 100 55.65
5. Measure ice as a function of frequency & compare to Omar figure 8.10. Discuss. Dimension: Height:____ x width_____x separation___
Freq C εr Freq C εr (pF) (pF)
6. Practical applications: inspect the temperature/humidity meter. Look for the devices used to measure each. Measure the stand-alone capacitor that resembles that in the meter as a function of exposure.
Part II - Reflectance Spectroscopy of Thin Films
1. Measure the reflectance spectra to determine n, k and thickness of: • ~ micron thick ~Al(0.3)Ga(0.7)As on GaAs substrate • ~ 1 micron thick ~ Al(0.6)Ga(0.4)As on GaAs substrate • SiO2 on Si substrate • Si3N4/Si substrate
Compare the index of refraction of the two different AlGaAs samples. What does this indicate about how you would “build” a waveguide?
117 18.5 Laboratory 5: Magnetic Properties 18 351-2 LABORATORIES
18.5 Laboratory 5: Magnetic Properties Objective: The objectives of this lab are to measure the response of a 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: 1. Use a magnetometer to measure magnetic response with applied field. 2. Given the saturation magnetization, solve for the number of Bohr mag- netons.
3. For a known material, explain the microscopic origins of magnetism. 4. Distinguish qualitatively between “hard” and “soft” ferromagnetic ma- terials.
Experimental Details The purpose of this experiment is to investigate 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 magne- tization 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 be- havior 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? 3d 4s Y/N Cu Fe Co Ni W Steel Stainless Steel
Quantify the observations on the metal wires (Fe, Co, Ni, and W 1 mm diam- eter and Cu 0.5 mm diameter – the cross sectional area must be entered in the
118 18.5 Laboratory 5: Magnetic Properties 18 351-2 LABORATORIES
B-probe dimensions). Record hysteresis curves in the “intrinsic” mode (with- out background subtraction). Also record “initial” curves for the first three metals. Fill in the following table.
Bsat observed Bsat literature # Bohr magnetons, nB∗ Fe 2.158 T Co 1.87 T Ni 0.616 T
Question 1(a-c): Solve for the number of Bohr magnetons per atom in Fe, Ni and Co. (Use experimental Bsat if the sample saturates; otherwise use litera- ture value.)
−7 Note: µ0 = 4π × 10 T − m/A BS = µ0MS; Msin A/m or Oe MS = nMA; n = # atoms/volume e~ −24 2 MA = nBµB; µB = 2me = 9.27 × 10 A − m =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
Set-up:
• Red laser pointer (operated under 3.5V!!! and 40 mA!!!)
• Glass rod (inside the solenoid coil): Diam = 5mm, length = 10 cm, SF-59. • Solenoid coil: L = 150 mm, turns/layer = 140, layers = 10, DC Resist = 2.6 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 re-measure 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 θ(radians) = V B`, B = magnetic field strength (millitesla), and `= length glass rod = 10 cm.
119 18.5 Laboratory 5: Magnetic Properties 18 351-2 LABORATORIES
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 sys- tem and compare to literature values.
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