351-2: Physics of Materials II

Bruce Wessels and Peter GirouardDepartment of Materials Science and EngineeringNorthwestern University

Table of Contents

1 Catalog Description (351-1,2)

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

2 Course Outcomes

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 between drift and diffusion transport mechanisms.
  2. Explain how dopant gradients, dopant homojunctions, semiconductor-semiconductor hetero junctions, and semiconductor-metal junctions perturb the carrier concentrations in adjacent materials or regions, identify the charge transport processes at the interfaces, and describe how the application of an electric field affects the band profiles and carrier concentrations.
  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 components of the index of refraction.
  5. Identify the microscopic interactions that lead to magnetic order in materials, describe the classes of magnetism that result from these interactions, and describe the temperature and field dependence of the macroscopic magnetization of bulk crystalline diamagnets, paramagnets, and ferromagnets.
  6. Specify a material and microstructure that will produce desired magnetic properties illustrated in hysteresis loops including coercivity, remnant magnetization, and saturation magnetization.
  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, identify key electronic materials and devices used in the technologies, specify basic performance metrics, and relate these metrics to fundamental materials properties.

4 Principles of Semiconductor Devices

Recall that the conductivity of semiconductors is given by
σ =ne μ +pe μ (4.1)
where n is the number of electrons, e is the electronic charge, μ is the mobility in cm 2 / 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 constant. Conductivity decreases with increasing temperature. The resistivity ρ ( ρ =1/ σ ) can be written as a sum of contributing factors using Matthiessen's rule as ρ = ρ 0 + ρ ( T ) where ρ 0 is a constant and ρ ( T ) is the temperature dependent resistivity. A typical carrier concentration for a metal is 1 0 23 cm -3 . Metals do not have a gap between the conduction and valence bands.
  2. Semiconductors: n and p are not constant with temperature but are thermally activated. Mobility is related to temperature as μ T -a . The typical range of carrier concentrations for semiconductors is 1 0 14 -1 0 20 cm -3 . The range of bandgaps for semiconductors is typically 0.1-3.0 eV .
  3. Insulators: n and p are much lower than they are in metals and semiconductors. A typical carrier concentration for an insulator is <1 0 7 cm -3 . The conductivity is generally a function of temperature. The bandgap for insulators is >3.0 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)
The rate constant is given by
[ n ] [ p ] [ np ] =K( T ) (4.3)
where
K( T ) =[F( T ) ] 2 exp [ - Δ E/kT ] (4.4) Consider the intrinsic case, that is, when the semiconductor is not doped with chemical impurities. For this case,
[ n ] =[ p ] =[ n i ] =A exp [ - Δ E/2kT ] (4.5)
where n i is the intrinsic carrier concentration. For an intrinsic semiconductor, the conductivity is
σ =2 n i e μ σ 0 exp [ - Δ E/2kT ] (4.6)
A log σ versus 1/T plot gives a straight line as shown in Fig. a.
image: 4_home_ken_Mydocs_MSEcore_351-2_figures_figure7.svg
Sub-Figure a:
image: 5_home_ken_Mydocs_MSEcore_351-2_figures_figure8.svg
Figure 4.1: (a) Log versus 1/T plot of conductivity and resistivity in semiconductors. (b) Log of conductivity versus 1/T plot for a semiconductor spanning the regimes of extrinsic and intrinsic conduction.
At lower temperatures where Δ E g >kT , the conductivity is dominated by carriers contributed from ionized impurities. For temperatures where Δ E g <kT , thermal energy is sufficient to excite electrons from the valence to conduction band. In this regime, the intrinsic carriers dominate the conduction. The two regimes of conduction are illustrated in Fig. b.

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.
image: 6_home_ken_Mydocs_MSEcore_351-2_figures_figure10.svg
Figure 4.2: Portion of the periodic table showing elements in typical elemental and compound semiconductors.
On the left side (group IIIA) are metals. The right hand side (group VIA) contains nonmetals. In between these two groups are elemental semiconductors and elements that form compound semiconductors. At the bottom of the table (In, Sn, and Sb) are more metallic elements. Group IVA consists of the covalent semiconductors Si, Ge, and gray Sn.
Electronic bands are made from an assembly of atoms with individual quantum states. The spectroscopic notation for quantum states is given by
1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 ... (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 distance between atoms is shown in Fig. 4.3. In this figure, d eq is the equilibrium distance between atoms with the lowest potential energy and is equal to the lattice constant. The energies correponding to d eq form the allowed energies in electronic bands. A diagram showing the energy levels in electronic bands is shown in Fig. 4.4.
image: 7_home_ken_Mydocs_MSEcore_351-2_figures_figure12.svg
Figure 4.3: Potential energy versus distance between atoms. The equilibrium distance, d eq corresponds to the minimum in the energy curve. The different energies corresponding to d eq form discrete levels in an electronic band.
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Figure 4.4: Energy diagram showing the electronic bands formed in a crystalline solid.
Core electrons do not contribute to conduction. Conduction of charge carriers 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.”
image: 9_home_ken_Mydocs_MSEcore_351-2_figures_figure13.svg Sub-Figure a:
image: 10_home_ken_Mydocs_MSEcore_351-2_figures_figure14.svg
Sub-Figure 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.
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 electrical in nature. A simple two-level band model for a semiconductor showing the conduction and valence band levels is shown in Fig. a Also shown (Fig. b) is a diagram illustrating the process of promoting an electron from the valence to conduction band. When an electron is excited to the conduction 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).
image: 11_home_ken_Mydocs_MSEcore_351-2_figures_figure16edit1.svg
Sub-Figure a:
image: 12_home_ken_Mydocs_MSEcore_351-2_figures_figure15.svg
Figure 4.6: (a) Energy band diagram for a simple two-level system. (b) Diagram illustrating the generation of an electron and hole pair.

4.3 Reciprocal Lattice and the Brillouin Zone

p = k
Reciprocal 2D Square Lattice
image: 13_home_ken_Mydocs_MSEcore_351-2_figures_figure17.svg

4.4 Nearly Free Electron Model

image: 14_home_ken_Mydocs_MSEcore_351-2_figures_figure18edit1.svg

4.5 Two Level Model


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Semiconductor Band Diagrams

Direct Bandgap
image: 17_home_ken_Mydocs_MSEcore_351-2_figures_figure20.svg

Indirect Bandgap
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5.1 P- and N-Type Semiconductors

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5.2 P-N Junctions

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5.3 Boltzmann Statistics: Review

For E v as the reference energy ( E V =0 ),
n= N s exp [ - E c - E F kT ] (5.1)
p= N s exp [ - E F - E V kT ] = N S exp ( - E F kT ) (5.2)
For fully ionized acceptors, p N A . For fully ionized donors, n N D .
N s = Density of states
The Fermi Level for electrons and holes are calculated from Eq.5.1 and 5.2, respectively:
Electrons: E F = E c -kT ln ( N S /n ) Holes: E F = E V +kT ln ( N S /p )
Note: these equations are true for non-generate semiconductors.

5.4 P-N Junction Equilibrium

Poisson's equation - describes potential distribution φ ( x ) :
d 2 φ ( x ) d x 2 =- ρ ( x ) ε
φ ( x ) Potential ε Dielectric Constant of Material ρ ( x ) Charge Density
Assume that the donors and acceptors are fully ionized
p-type: N A N A - + p + n-type: N D N D + + 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

image: 21_home_ken_Mydocs_MSEcore_351-2_figures_figure25edit1.svg

5.6 Calculation of the Electric Field

d 2 φ d x 2 =- ρ ε d φ dx =- E d E dx = ρ ε
E ( x ) = ρ ε x
Separate the integral for the n- and p-regions:
E ( x ) = - x p 0 ρ ( x ) ε x p-region + 0 x n ρ ( x ) ε x n-region
Important assumptions made:
  1. ρ e N D in the n region, and ρ -e N A in the p region.
  2. ρ is constant in the separate n and p regions.
Boundary conditions:
  1. E ( x=- x p ) = E ( x= x n ) =0
  2. E is continuous at the junction
Resulting electric field across the junction:
E ( x ) ={ - e N A ε ( x+ x p ) , - x p x0 e N D ε ( x- x n ) , 0x x n
E ( x ) ={ - e N A ε ( x+ x p ) , - x p x0 e N D ε ( x- x n ) , 0x x n
Electric Field in the Junction
image: 22_home_ken_Mydocs_MSEcore_351-2_figures_figure26.svg

5.7 Calculation of the Electric Potential

Recall
φ =- E x
Boundary conditions:
  1. φ is continuous at the junction boundary
  2. φ =0 at x=- x p (chosen arbitrarily)
Resulting potential across the junction:
φ ( x ) ={ e N A ε ( 1 2 x 2 +x x p + 1 2 x p 2 ) , - x p x0 e N D ε ( N A 2 N D x p 2 +x x n - 1 2 x 2 ) , 0x x n
Electric Potential in the Junction
image: 23_home_ken_Mydocs_MSEcore_351-2_figures_figure27edit1.svg

5.8 Junction Capacitance

w= x p + x n = [ 2 ε φ 0 e( N A + N D ) ] 1/2 [ ( N A N D ) 1/2 + ( N D N A ) 1/2 ]
φ 0 Built-in potential
C= ε w F / cm 2
w= ( 2 ε φ 0 e N D ) 1/2
C= ( e ε N D 2 φ 0 ) 1/2

5.9 Rectification

image: 24_home_ken_Mydocs_MSEcore_351-2_figures_figure28edit2.svg
E g
n i (cm -3 × 1 0 14 )
Ge
0.66
0.24
Si
1.08
0.00015
GaAs
1.4
-
GaP
2.25
-
GaN
3.4
-

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5.10 Junction Capacitance

Definition of capacitance:
C Q φ
For a one-sided abrupt junction, recall that Q=e N D x n , where
x n = [ 2 ε ( φ 0 + φ 1 ) N A e N D ( N D + N A ) ] 1/2
The total charge Q for a one sided junction is then
Q= [ 2e ε ( φ 0 + φ 1 ) N A N D N A + N D ] 1/2
Calculate the capacitance as a function of bias voltage φ 1 :
C= Q φ 1 = [ e ε 2( φ 0 + φ 1 ) N A N D N A + N D ] 1/2
A plot of 1/ C 2 versus φ 1 is called a Mott Schottky plot and yields N A or N D .
image: 28_home_ken_Mydocs_MSEcore_351-2_figures_figure33.svg

6 Transistors: Semiconductor Amplifiers

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6.1 pnp Device


Device Diagram
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Circuit Diagram
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6.2 pnp Device


Device Diagram
image: 30_home_ken_Mydocs_MSEcore_351-2_figures_figure36.svg

Band Diagram
image: 32_home_ken_Mydocs_MSEcore_351-2_figures_figure37.svg

6.3 Amplifier Gain

Voltage across the junctions:
i c φ r i e φ f φ r φ f 10 V/V
Nodal analysis at the base for pnp and npn devices:
image: 33_home_ken_Mydocs_MSEcore_351-2_figures_figure37b.svg
i c i e and i c = α i e
From nodal analysis at the base, i c = i e - i b .
i c = α i b 1- α = h fe i b with h fe = α 1- α
h fe is the current gain parameter. α 0.9 .

6.4 Circuit Configurations

Common-Emitter Diagram
image: 34_home_ken_Mydocs_MSEcore_351-2_figures_figure39.svg

Common-Base I-V Characteristic
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6.5 MOSFETs

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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.
image: 37_home_ken_Mydocs_MSEcore_351-2_figures_figure42.svg

6.7 Depletion and Inversion

image: 38_home_ken_Mydocs_MSEcore_351-2_figures_figure43edit1.svg

6.8 Channel Pinch-Off

image: 39_home_ken_Mydocs_MSEcore_351-2_figures_figure44.svg
image: 40_home_ken_Mydocs_MSEcore_351-2_figures_figure45edit1.svg
G = Gate S = Source D = Drain

6.9 Tunnel Diodes


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6.10 Gunn Effect

σ =ne μ and μ = e τ m * , m * = ( 1 2 2 E k 2 band curvature ) -1
image: 44_home_ken_Mydocs_MSEcore_351-2_figures_figure49.svg

6.11 Gunn Diodes

J= σ E = n e 2 τ m * E
image: 45_home_ken_Mydocs_MSEcore_351-2_figures_figure50.svg

7 Heterojunctions and Quantum Wells

image: 46_home_ken_Mydocs_MSEcore_351-2_figures_figure51.svg

7.1 Quantum Wells


image: 47_home_ken_Mydocs_MSEcore_351-2_figures_figure53edit1.svg
H Ψ = E Ψ ; Ψ =A exp ( - k n x )
k n = n π L x
E x = 2 k 2 2 m * = 2 2 m * ( π 2 n x 2 L x 2 ) image: 48_home_ken_Mydocs_MSEcore_351-2_figures_figure52.svg

image: 49_home_ken_Mydocs_MSEcore_351-2_figures_figure54.svg

7.2 Heterostructures and Heterojunctions

image: 50_home_ken_Mydocs_MSEcore_351-2_figures_figure54a.svg

7.3 Layered Structures: Quantum Wells

image: 51_home_ken_Mydocs_MSEcore_351-2_figures_figure55.svg
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7.4 Transitions between Minibands

E= 2 k 2 2 m * = 2 2 m * ( n π L x ) 2
image: 53_home_ken_Mydocs_MSEcore_351-2_figures_figure57edit1.svg

7.5 Quantum Dots

image: 54_home_ken_Mydocs_MSEcore_351-2_figures_figure58_edit1.svg
E i = 2 2 m i ( α i R ) 2
α i quantum number

7.6 Quantum Dot Example: Biosensor

image: 55_home_ken_Mydocs_MSEcore_351-2_figures_figure59.svg

7.7 Quantum Cascade Devices

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8 Optoelectronic Devices: Photodetectors and Solar Cells


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8.1 I-V Characteristics

L e Electron diffusion length L h Hole diffusion length G Carrier generation rate ( depends on light intensity )

8.2 Power Generation

Fill Factor:  ff I m V m I sc V oc

image: 60_home_ken_Mydocs_MSEcore_351-2_figures_figure64edit1.svg

image: 61_home_ken_Mydocs_MSEcore_351-2_figures_figure66.svg

8.2.1 Solar Cell Efficiency


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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
image: 64_home_ken_Mydocs_MSEcore_351-2_figures_figure69edit1.svg

8.2.3 Metal-Semiconductor Solar Cells

J= J 00 exp ( - φ B kT ) [ exp ( qV kT ) -1 ]
J current area = I A

image: 65_home_ken_Mydocs_MSEcore_351-2_figures_figure70edit1.svg

image: 66_home_ken_Mydocs_MSEcore_351-2_figures_figure71.svg

8.2.4 Schottky Barrier and Photo Effects


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8.2.5 Dependence of φ B on Work Function

φ energy required to take an electron from the Fermi Level to the Vacuum Level
φ B = φ m - φ s

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image: 72_home_ken_Mydocs_MSEcore_351-2_figures_figure77.svg

8.2.6 Pinned Surfaces

φ B( n ) + φ B( p ) = E g
image: 73_home_ken_Mydocs_MSEcore_351-2_figures_figure78.svg

8.2.7 Light Emitting Diodes (LEDs)

h ν = E g
image: 74_home_ken_Mydocs_MSEcore_351-2_figures_figure82edit1.svg

8.2.8 Light Emitting Diodes (LEDs)

Material
E g 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

image: 75_home_ken_Mydocs_MSEcore_351-2_figures_figure83.svg

8.2.10 LED Efficiency

η = W R W total = W R W R + W NR
W 1 τ [s -1 ]
1 τ total = 1 τ R + 1 τ NR
η = W R W total = W R W R + W NR
η = 1/ τ R 1/ τ total = 1 1+ τ R / τ NR
η = τ NR τ R

9 Lasers

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9.1 Emission Rate and Laser Intensity

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9.2 Two Level System

N 2 N e - E 2 /kT or N 2 N = e - E 2 /kT Z
Z sum over states
N 2 N 1 = e -( E 2 - E 1 ) /kT
At equilibrium, fewer electrons are in level 2
Triggers of Stimulated Emission
image: 78_home_ken_Mydocs_MSEcore_351-2_figures_figure89edit1.svg

9.3 Planck Distribution Law

ρ ( ν ) d ν = 8 π n 3 h ν 3 c 3 Density of states 1 exp ( h ν /kT ) -1 Planck distribution d ν
image: 79_home_ken_Mydocs_MSEcore_351-2_figures_figure90edit1.svg

9.4 Transition Rates

R 12 =N 1 B 12 ρ ( ν 21 ) d ν
N 1 Number of photons in lower state
ρ ( ν 21 ) number of photons available having transition energy h ν
R 21 =N 2 [ A 21 Spont. + B 21 ρ ( ν 21 ) ] Stim. d ν

9.5 Two Level System

R 12 = R 21
N 1 B 12 ρ ( ν 21 ) d ν =N 2 [ A 21 +B 21 ρ ( ν 21 ) ] d ν
ρ ( ν 21 ) d ν = A 21 d ν B 12 exp [ ( h ν 21 ) /kT ] - B 21
A 21 = B 21 8 π n 3 h ν 3 c 3
Two Level System - Summary

9.6 Three Level System

image: 80_home_ken_Mydocs_MSEcore_351-2_figures_figure92edit1.svg
Carrier Population
N( E ) = N 0 exp ( -E/kT ) log N = -E/kT+ log N 0
image: 81_home_ken_Mydocs_MSEcore_351-2_figures_figure93edit1.svg
Spontaneous Emission
d N 3 dt =- A 31 N 3
A 31 = 1 t spont

image: 82_home_ken_Mydocs_MSEcore_351-2_figures_figure95edit1.svg

image: 83_home_ken_Mydocs_MSEcore_351-2_figures_figure96edit1.svg
Populations under High Pumping
B 23 N 2 ρ ( ν ) Stim. absorption = A 32 N 3 Spont. emission + B 32 N 3 ρ ( ν ) Stim. emission
Laser Gain
N W 12 =( N 2 - N 1 ) W 12
W 12 Transition rate from state 1 to 2
For amplification, a population inversion must exist.
Derivation of Laser Gain Factor
image: 84_home_ken_Mydocs_MSEcore_351-2_figures_figure98edit1.svg
dI=( N 2 - N 1 ) W 12 h ν 12 dz
dI dz =( N 2 - N 1 ) W 12 h ν 12
W 12 = I h ν c 2 g( ν ) 8 π n 2 ν 2 t spont
dI dz = γ ( ν ) I
γ ( ν ) =( N 2 - N 1 ) c 2 g( ν ) 8 π n 2 ν 2 t spont
dI dz = γ ( ν ) I
I( z ) =I( 0 ) exp [ γ ( ν ) z ]
γ ( ν ) =( N 2 - N 1 ) c 2 g( ν ) 8 π n 2 ν 2 t spont

9.7 Lasing Modes

image: 85_home_ken_Mydocs_MSEcore_351-2_figures_figure99edit1.svg
I= I 0 R 1 R 2 exp [ ( γ - α ) 2l ]
α absorption and scattering losses γ laser gain factor
R 1 R 2 exp [ ( γ - α ) 2l ] 1

9.8 Laser Examples

9.8.1 Ruby

image: 86_home_ken_Mydocs_MSEcore_351-2_figures_figure101edit1.svg

9.8.2 Others Examples

Semiconductor:
Others:
image: 87_home_ken_Mydocs_MSEcore_351-2_figures_figure102edit.svg

9.9 Threshold Current Density

N e lwd t rec = I i η e
η Efficency I i Injection current e Fundamental charge
γ ( ν ) =( N 3 - N 2 ) c 2 g( ν ) 8 π n 2 ν 2 t spont
t rec t spont
N e = N 3 - N 2
γ ( ν ) = c 2 g( ν ) η 8 π n 2 ν 2 elwd I i ( c m -1 )
Relates gain coefficient to I i
R exp ( γ - α ) l'=1
( γ - α ) l'+ ln R=0
j th = 8 π n 2 ν 2 ed c 2 η g( ν ) β -1 ( α - 1 l' ln R ) [ A/cm 2 ]
j th = α β - 1 β l' ln R( R<1 )

Lasing Threshold Condition: j th = α β - 1 β l' ln R( R<1 )

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9.10 Comparison of Emission Types

Emission
Intensity
Linewidth
Directional
Coherent
Spontaneous
Weak
Broad
No
No
Stimulated
Intense
Narrow
Not
No
(Super-Radiant)
Necessarily
Stimulated
Intense
Very
Yes
Yes
Lasing
Narrow

9.11 Cavities and Modes

Integer number of wavelengths =  Optical path length
m λ =2nl
m λ 2n =lm=1,2,3...
image: 90_home_ken_Mydocs_MSEcore_351-2_figures_figure108.svg

9.11.1 Longitudinal Laser Modes

Δ λ = λ 2 2l( n- λ dn d λ )
image: 91_home_ken_Mydocs_MSEcore_351-2_figures_figure109.svg

9.11.2 Transverse Laser Modes

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9.12 Semiconductor Lasers

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9.13 Photonic Bandgap Materials

v g =c/ n g as  n g , v g 0
image: 94_home_ken_Mydocs_MSEcore_351-2_figures_figure116.svg

10 Band Diagrams

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.
E c Conduction band energy.
E v Valence band energy.
E F Fermi level.

10.1 Band Diagrams

image: 95_home_ken_Mydocs_MSEcore_351-2_figures_figure81edit1.svg

10.2 Heterojunctions and the Anderson Model

image: 96_home_ken_Mydocs_MSEcore_351-2_figures_figure79.svg

10.3 Band Bending at p-n Junctions

Band diagrams for isolated p- and n-type semiconductors are shown in Fig. 10.1below.
image: 97_home_ken_Mydocs_MSEcore_351-2_figures_figure1edit1.svg
Figure 10.1: Band diagrams of isolated n-GaAs and p-Ge.
When brought into contact without application of an external bias field, thermal equilibrium requires that the Fermi levels equilibrate, i.e.
E F,n = E F,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:
image: 98_home_ken_Mydocs_MSEcore_351-2_figures_figure3edit1.svg
Figure 10.2: Band diagram of a n-GaAs/p-Ge heterojunction.
E F,n - E F,p =0=( χ GaAs + δ GaAs +q V D,n +q V D,p ) -( χ Ge + E g,Ge - δ Ge ) (10.2)
Rearranging Eqn. 10.2, q( V D,n + V D,p ) = ( χ Ge - δ Ge + E g,Ge ) p - ( χ GaAs + δ GaAs ) n (10.3)
Substituting in known values,
q( V D,n + V D,p ) =( 4.13-0.1+0.7 ) -( 4.07+0.1 ) (10.4) = 0.56 eV < E g,Ge
where
V D,n V D,p = N A ε Ge N D ε GaAs (10.5)

10.3.1 Calculate the Conduction Band Discontinuity

Let the Fermi level be zero on the energy scale.
δ GaAs +q V D,n =( E g,Ge - δ Ge ) -q V D,p + Δ E c
Δ E c = δ GaAs +q V D,n -( E g,Ge - δ Ge ) +q V D,p
On substituting in Eqn. 10.3, we get
Δ E c = χ Ge - χ GaAs =+0.6
The positive value for Δ E c indicates a spike in the conduction band.

10.3.2 Calculate the Valence Band Discontinuity

The valence band discontinuity Δ E v is calculated in a similar manner. In this case, a negative value for Δ E v indicates a spike in the valence band. Δ E v is calculated as follows:
Δ E v = Δ E g - Δ χ
Δ E v =( E g,GaAs - E g,Ge ) -( χ Ge - χ GaAs )
Substituting in known values,
Δ E v =( 1.45-0.7 ) -( 4.13-4.07 )
Δ E v =+0.69
Since Δ E v is positive, there is no spike in the valence band.

10.3.3 Example: p-GaAs/n-Ge

The isolated band diagrams for p-GaAs and n-Ge are shown in Figs. a and b. 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,
image: 99_home_ken_Mydocs_MSEcore_351-2_figures_figure4edit1.svg
Sub-Figure a:
image: 100_home_ken_Mydocs_MSEcore_351-2_figures_figure5edit1.svg
Figure 10.3: Isolated band diagrams of (a) p-GaAs and (b) n-Ge.
q( V D,n + V D,p ) = ( χ GaAs + E g,GaAs - δ GaAs ) p - ( χ Ge + δ Ge ) n
=( 4.07+1.45-0.1 ) -( 4.13+0.1 )
=1.19< E g,GaAs
For calculating the valence band discontinuity,
Δ E v =( E g,Ge - E g,GaAs ) -( χ GaAs - χ Ge )
Δ E v =-0.69 spike (10.6)
The resulting band diagram is shown in Fig. 10.4.
image: 101_home_ken_Mydocs_MSEcore_351-2_figures_figure6edit1.svg
Figure 10.4: Band diagram of a p-GaAs/n-Ge heterojunction as predicted using the approach in Section10.3.3. Note that there is a spike in the valence band which impedes the conduction of holes.

Dielectric Materials


Parallel Plate Capacitance
C = ε A d d distance between plates A plate area

image: 102_home_ken_Mydocs_MSEcore_351-2_figures_figure117edit1.svg

11.1 Macroscopic Dielectric Theory

D= ε E [ C/m 2 ]
D Electric flux density ε Dielectric constant E Electric field
ε r Relative dielectric constant ( unitless ) ε 0 Permittivity of free space [ F/m ]
Q= ε 0 V d [ C ] with
V Voltage [ V ] d Distance [ m ]
Q'= ε 0 ε r V d and C=Q'/V
P D- ε 0 E P ( ε r ε 0 - ε 0 ) E = ε 0 ( ε r -1 ) χ E P ε 0 χ E

11.2 Relation of ε to the Microscopic Structure

image: 103_home_ken_Mydocs_MSEcore_351-2_figures_figure118edit1.svg
image: 104_home_ken_Mydocs_MSEcore_351-2_figures_figure119edit1.svg

11.2.1 Relation of Macroscopic to Microscopic

P= N m μ
E ' Local electric field α Polarizability of species ( atoms, molecules )
image: 105_home_ken_Mydocs_MSEcore_351-2_figures_figure120edit1.svg

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 applied electric field. Example: poled polymers, useful for electro-optic devices.

11.4 Polarization in Solids

E=- μ E cos θ
θ Angle between  μ  and  E
image: 106_home_ken_Mydocs_MSEcore_351-2_figures_figure121edit1.svg

11.5 Calculation of the Average Dipole Moment

μ = weighted average energy for each orientation total energy for all orientations
of dipoles with energy  E=2 π A sin θ d θ exp ( -E/kT )
A= Constant including radius
of dipoles with energy  E=2 π A sin θ d θ exp ( μ E cos θ kT )
0 π A exp ( μ E cos θ kT ) 2 π sin θ θ
0 π A exp ( μ E cos θ kT ) ( μ cos θ ) 2 π sin θ θ
μ = 0 π A exp ( μ E cos θ kT ) ( μ cos θ ) 2 π sin θ θ 0 π A exp ( μ E cos θ kT ) 2 π sin θ θ
μ μ =L( a ) = coth a- 1 a
with  a= μ E kT
μ μ = a 3 for small a
μ = μ 2 E 3kT or  μ μ = μ E 3kT

11.6 Polarizability of Solids

P= N m μ
P= N m μ 2 E 3kT
image: 107_home_ken_Mydocs_MSEcore_351-2_figures_figure123.svg

11.7 Dielectric Constant for a Solid

image: 108_home_ken_Mydocs_MSEcore_351-2_figures_figure124.svg
P= j N j P j = j N j α j E loc,j
E loc = E + P 3 ε 0
P=( j N j α j ) ( E + P 3 ε 0 )

11.8 Claussius Mossotti Relation

χ =( j N j α j ) E + P 3 ε 0 ε 0 E
ε r -1=( j N j α j ) E + 1 3 ε 0 ε 0 E ( ε r -1 ) ε 0 E
Simplifying and rearranging gives the Claussius Mossotti Relation:
ε r -1 ε r +2 = 1 3 ε 0 j N j α j ( SI Units)

11.9 Frequency Dependence of the Polarizability

α ( ω ) = α i ( ω )
image: 109_home_ken_Mydocs_MSEcore_351-2_figures_figure125edit1.svg
e E loc =Cx=m ω 0 2 x
C=m ω 0 2 Force constant 
E loc = m ω 0 2 x e
α el = P N E loc = Nex N E loc = e 2 m ω 0 2
m d 2 x d t 2 +m ω 0 2 x=-e E loc sin ω t
m( - ω 2 + ω 0 2 ) x 0 =-e E loc
x 0 = -e E loc m( ω 0 2 - ω 2 )
μ elec =-e x 0 = e 2 E loc m( ω 0 2 - ω 2 )
α elec = μ elec E loc = e 2 m( ω 0 2 - ω 2 )
μ 0 =-e x 0 = e 2 E loc m( ω 0 2 - ω 2 )
α elec = μ 0 E loc = e 2 m( ω 0 2 - ω 2 )
image: 110_home_ken_Mydocs_MSEcore_351-2_figures_figure127edit1.svg

11.10 Quantum Theory of Polarizability

α elec = e 2 m j f ij ω 0 2 - ω 2
f ij Oscillator strength for a transition from state  i  to  j
P elec =N α elec E

11.11 Advanced Dielectrics: Ferroelectrics

image: 111_home_ken_Mydocs_MSEcore_351-2_figures_figure129edit2.svg
P S Spontaneous polarization E C Coercive field

12 Phase Transitions

ε = ξ T- T C ξ Curie constant
image: 112_home_ken_Mydocs_MSEcore_351-2_figures_figure131edit1.svg

12.1 Lattice Instabilities

ε r -1 ε r +2 = 1 3 ε 0 N α
ε r -1= N α 3 ε 0 ( ε r +2 ) = N α ε r 3 ε 0 + 2 3 N α ε
ε r ( 1- N α 3 ε 0 ) = 2N α 3 ε 0 +1
ε r = 2N α / ε 0 +3 3-N α / ε 0

12.2 Curie Weiss Law

N α ε 0 3-2s
ε r = 2N α / ε 0 +3 3-N α / ε 0 -2s 1 s near the singularity
ε r ξ ( T- T C ) or  1 ε r T- T C ξ
image: 113_home_ken_Mydocs_MSEcore_351-2_figures_figure132edit1.svg

12.3 Ferroelectric Phase Transitions: Landau Theory

F ˆ ( P,T, E ) =- E P+ s 0 + 1 2 s 2 P 2 + 1 4 s 4 P 4 + 1 6 s 6 P 6 +
F ˆ P =0=-E+ s 2 P+ s 4 P 3 + s 6 P 5
F ˆ P =0=-E+ s 2 P+ s 4 P 3 + s 6 P 5
assume  s 2 = γ ( T- T 0 ) , with  γ  a positive constant
image: 114_home_ken_Mydocs_MSEcore_351-2_figures_figure137edit1.svg
F ˆ P =- E + γ ( T- T 0 ) P+ s 4 P 3 =0
γ ( T- T 0 ) P S + s 4 P S 3 =0
P S [ γ ( T- T 0 ) + s 4 P S 2 ] =0
| P S |= ( γ / s 4 ) 1/2 ( T 0 -T ) 1/2
P S |= ( γ / s 4 ) 1/2 ( T 0 -T ) 1/2
image: 115_home_ken_Mydocs_MSEcore_351-2_figures_figure138edit1.svg

12.3.1 First versus Second Order Transitions

image: 116_home_ken_Mydocs_MSEcore_351-2_figures_figure139edit1.svg
F ˆ =- E P+ s 0 + 1 2 s 2 P 2 + 1 4 s 4 P 4 + 1 6 s 6 P 6 +
F ˆ P = γ ( T- T 0 ) P S -| s 4 | P S 3 + s 6 P S 5 =0
γ ( T- T 0 ) -| s 4 | P S 2 + s 6 P S 4 =0
image: 117_home_ken_Mydocs_MSEcore_351-2_figures_figure141edit1.svg

12.3.2 Ferroelectric Example: BaTi O 3

μ =( 3 × 1 0 -1 C m -2 ) × 64 × 1 0 -30 m 3 2 × 1 0 -29 Cm

Material
T C (K)
BaTi O 3
408
BaTi O 3
1480
image: 118_home_ken_Mydocs_MSEcore_351-2_figures_figure142edit1.svg

12.4 Other Instabilities

image: 119_home_ken_Mydocs_MSEcore_351-2_figures_figure143edit1.svg

12.5 Piezoelectrics

image: 120_home_ken_Mydocs_MSEcore_351-2_figures_figure144edit1.svg
Stress:  T = CS-e E Displacement:  D = ε E +eS
C Elastic constant  e Piezoelectric coefficient E Electric field S Strain

13 Diamagnetism and Paramagnetism

χ = μ 0 M B
M Magnetization B Magnetic field intensity μ 0 Permeability of free space
image: 121_home_ken_Mydocs_MSEcore_351-2_figures_figure147.svg

13.1 Diamagnetism

ω = eB 2m [ sec -1 ]
image: 122_home_ken_Mydocs_MSEcore_351-2_figures_figure149.svg
I = ( charge ) × ( revolutions per second ) =-Ze ω 2 π I = -Ze( 1 2 π eB 2m ) =- Z e 2 B 4 π m
μ ( current ) × ( area of loop ) =I × A
μ =- Z e 2 B 4m ρ 2
ρ 2 = x 2 + y 2
ρ 2 = 2 3 r 2
Langevin Result:  χ = N μ 0 μ B =- μ 0 NZ e 2 6m r 2
( note negative sign )

13.2 Paramagnetism

μ γ J=-g μ B J
γ gyromagnetic ratio J angular momentum g g factor (2.00  for electron spin) μ B Bohr magneton ( atomic unit ) J angular quantum number for electrons
U=- μ B = m J g μ B B
image: 123_home_ken_Mydocs_MSEcore_351-2_figures_figure153edit3.svg
P i = exp ( - E i / τ ) Z
Z sum over all states ( normalization ) τ kT
N 1 N = exp ( μ B/ τ ) exp ( μ B/ τ ) + exp ( - μ B/ τ ) N 2 N = exp ( - μ B/ τ ) exp ( μ B/ τ ) + exp ( - μ B/ τ )

image: 124_home_ken_Mydocs_MSEcore_351-2_figures_figure155edit1.svg
M=( N 1 - N 2 ) μ =N μ e x - e -x e x + e -x =N μ tanh x
x μ B/kT
tanh xxMN μ ( μ B/kT )
image: 125_home_ken_Mydocs_MSEcore_351-2_figures_figure156edit1.svg
M=NgJ μ B B J ( x )
xgJ μ B B/kT
B J ( x ) 2J+1 2J coth [ ( 2J+1 ) x 2J ] - 1 2J coth x 2J
coth x 1 x + x 3

13.2.1 Calculation of Susceptibility

For small fields and high temperatures, χ = M B NJ( J+1 ) g 2 μ B 2 3kT = N p 2 μ B 2 3kT =C/T p 2 = g 2 J( J+1 ) p=g [ J( J+1 ) ] 1/2 C Curie constant
image: 126_home_ken_Mydocs_MSEcore_351-2_figures_figure158edit1.svg

13.2.2 Calculation of Total Angular Momentum J

1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10

13.2.3 Spectroscopic Notation

Example Orbital Configurations
image: 127_home_ken_Mydocs_MSEcore_351-2_figures_figure159edit1.svg

13.2.4 Paramagnetic Susceptibility

Curie Law for paramagnetic susceptibility: χ =C/T

image: 128_home_ken_Mydocs_MSEcore_351-2_figures_figure160edit1.svg

13.2.5 Calculation of J

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
L= m l and S= m s

13.2.6 Spin Orbit Interactions

image: 129_home_ken_Mydocs_MSEcore_351-2_figures_figure161.svg

13.2.7 Effective Magnetic Number

P=g [ J( J+1 ) ] 1/2
JS For transition metals, "orbital angular momentum quenched"
image: 130_home_ken_Mydocs_MSEcore_351-2_figures_figure161b.svg
P=2 [ S( S+1 ) ] 1/2
P=2 [ 1 2 ( 3 2 ) ] 1/2 =1.73
image: 131_home_ken_Mydocs_MSEcore_351-2_figures_figure161a.svg
P=2 [ 2( 3 ) ] 1/2 =4.90

13.2.8 Paramagnetic Properties of Metals

M= N μ 2 B k T F
M N eff μ 2 B kT N eff NkT k T F
M N μ 2 B k T F Temperature independent
image: 132_home_ken_Mydocs_MSEcore_351-2_figures_figure163edit1.svg

13.2.9 Band Model

N + = 1 2 - μ B E F E f( E ) Fermi function D( E+ μ B ) Density of states
image: 133_home_ken_Mydocs_MSEcore_351-2_figures_figure165edit1.svg
N + 1 2 0 E F E f( E ) D( E ) + 1 2 0 E F E f( E ) dD( E ) dE μ B 1 2 0 E F E f( E ) D( E ) + 1 2 0 E F f( E ) D ( E ) μ B 1 2 0 E F E f( E ) D( E ) + 1 2 D( E F ) μ B
N - = 1 2 μ B E F E f( E ) D( E- μ b ) 1 2 0 E F ( E ) f( E ) D( E ) - 1 2 D( E F ) μ B
M=( N + - N - ) μ =D( E F ) μ 2 B
D( E F ) = 3N 2 E F = 3N 2k T F
M= 3N μ 2 B 2k T F χ = M B = 3N μ 2 2k T F Independent of T!

13.2.10 Multivalent Effects

image: 134_home_ken_Mydocs_MSEcore_351-2_figures_figure168edit1.svg

14 Ferromagnetism

H Ψ =E Ψ withH=K+ V ex
V ex Exchange Potential
image: 135_home_ken_Mydocs_MSEcore_351-2_figures_figure170edit1.svg

14.1 Ferromagnetic Phase Transition

T> T C Paramagnetism
T< T C Ferromagnetism
χ P Paramagnetic susceptibility B A Applied field
image: 136_home_ken_Mydocs_MSEcore_351-2_figures_figure171edit1.svg

14.2 Molecular Field

M= χ P ( B A + λ M ) M= C T ( B A + λ M )
M( 1- C λ T ) = C B A T
M( T-C λ ) =C B A
χ = M B A = C ( T-C λ )
χ = C T- T C for  T> T C
Weiss constant:  λ = T C /C
χ 1 (T- T C ) 1.33 Scaling theory
λ = T C C = T C 3k N g 2 S( S+1 ) μ b 2 Curie Constant
B E = λ M S
B E =5000 × 17001 0 7  Gauss =1 0 3  Tesla

image: 137_home_ken_Mydocs_MSEcore_351-2_figures_figure173edit1.svg

Material
T C (K)
Fe (bcc)
1043
Co
1388
Ni
627
Note: fcc iron is not magnetic

14.2.1 Prediction of T C

U=-2 J ex S i S j
J ex Exchange energy integral
J ex = 3k T C 2ZS( S+1 ) or T C = 2ZS( S+1 ) J ex 3k
Z Number of neighbors
image: 138_home_ken_Mydocs_MSEcore_351-2_figures_figure174edit1.svg

14.2.2 Temperature Dependence of M( T ) for Ferromagnetism

M=N μ tanh ( μ B kT )
Where  B= B A + B E
M=N μ tanh ( μ λ M kT ) =N μ tanh ( x ) with x μ λ M kT
image: 139_home_ken_Mydocs_MSEcore_351-2_figures_figure175edit1.svg

14.2.3 Ferromagnetic-Paramegnetic Transition

m M N μ t kT N μ 2 λ
m LHS = tanh ( m/t ) RHS
image: 140_home_ken_Mydocs_MSEcore_351-2_figures_figure176edit1.svg

14.2.4 Ferromagnetic-Paramagnetic Transition

image: 141_home_ken_Mydocs_MSEcore_351-2_figures_figure177edit1.svg

14.2.5 Ferromagnetic-Paramagnetic Transition

tanh ( m/t ) = tanh ξ 1-2 e -2 ξ
tanh ξ 1-2 e -2 ξ 1
Δ M=2N μ e -2 ξ =2N μ e -2 μ λ M/kT
Δ M=2N μ e -2 μ 2 λ N/kT
Δ M=2N μ exp ( -2 λ N μ 2 /kT ) =2N μ exp ( -2 T C /T )

14.2.6 Ferromagnetic-Paramagnetic Transition

Change in magnetization with respect to T=0 value versus temperature
Δ M=2N μ exp ( -2 λ N μ 2 /kT ) =2N μ exp ( -2 T C /T )
image: 142_home_ken_Mydocs_MSEcore_351-2_figures_figure178edit1.svg

14.2.7 Ferromagnetism of Alloys

image: 143_home_ken_Mydocs_MSEcore_351-2_figures_figure179edit1.svg

14.2.8 Transition Metals

d Orbital Configurations of Transition Metals
image: 144_home_ken_Mydocs_MSEcore_351-2_figures_figure180edit1.svg

14.2.9 Ni Alloys

image: 145_home_ken_Mydocs_MSEcore_351-2_figures_figure181edit1.svg

14.2.10 Band Model

image: 146_home_ken_Mydocs_MSEcore_351-2_figures_figure182a.svg
image: 147_home_ken_Mydocs_MSEcore_351-2_figures_figure182edit1.svg
image: 148_home_ken_Mydocs_MSEcore_351-2_figures_figure182b.svg
image: 149_home_ken_Mydocs_MSEcore_351-2_figures_figure182c.svg
superconducting properties?
YB a 2 CuO

14.2.11 Spin Waves

image: 150_home_ken_Mydocs_MSEcore_351-2_figures_figure183edit1.svg

14.2.12 Magnon Dispersion

ω = 4 J ex S 2 sin 2 aq 2
J ex Exchange energy term S Total spin angular momentum a Distance between atoms with magnetic moments q Wavevector
image: 151_home_ken_Mydocs_MSEcore_351-2_figures_figure185edit1.svg

14.3 Ferrimagnetic Order in Magnetic Oxides

F e 3 O 4 FeOF e 2 O 3
image: 152_home_ken_Mydocs_MSEcore_351-2_figures_figure199edit1.svg
Sublattice
Fe Oxid. State
e - Config.
S
gJ ( μ B )
(A) FeO
2+
3d 6
2
4
(B) F e 3 O 4
3+
3d 5
5 2
5

14.3.1 Magnetic Oxides

Ion
g J( J+1 ) ( μ B )
Ni 2+
2
Fe 3+
5
net  2 μ B  per formula unit due to Ni 2+
M S = 8 × 2 μ B cell volume = ( 16 ) ( 9.27 × 1 0 -24 ) (8.37 × 1 0 -10 ) 3 =2.5 × 1 0 5  A m -1
Ion
Site
g J( J+1 ) ( μ B )
Fe 3+
Tetrahedral
5
Fe 3+
Octahedral
5
Fe 2+
Octahedral
4
net  2 μ B  per formula unit due to Fe 2+
M S = 8 × 4 μ B cell volume = ( 32 ) ( 9.27 × 1 0 -24 ) (8.40 × 1 0 -10 ) 3 =5.0 × 1 0 5  A m -1

14.3.2 Magnetization and Hysteresis

image: 153_home_ken_Mydocs_MSEcore_351-2_figures_figure198edit1.svg
image: 154_home_ken_Mydocs_MSEcore_351-2_figures_figure195edit1.svg

14.4 Domains and Walls

image: 155_home_ken_Mydocs_MSEcore_351-2_figures_figure197edit1.svg

14.4.1 Energy of Bloch Domain Walls

U=-2 J ex S i S j
U=-2 J ex |S | 2 cos φ -2 J ex |S | 2 ( 1- 1 2 φ 2 )
w ex = J ex |S | 2 φ 2
N w ex = J ex |S | 2 π 2 N

14.5 Anisotropy of Magnetization

image: 156_home_ken_Mydocs_MSEcore_351-2_figures_figure188edit1.svg

14.5.1 Anisotropy of Magnetization

U k = k 1 ' sin 2 θ + k 2 ' sin 4 θ
image: 157_home_ken_Mydocs_MSEcore_351-2_figures_figure189a.svg

14.6 Ferrimagnetic Ordering - Exchange Terms

J AA >0 J AB <0 J BB >0
image: 158_home_ken_Mydocs_MSEcore_351-2_figures_figure199a.svg

14.6.1 Exchange Terms and Susceptibility

image: 159_home_ken_Mydocs_MSEcore_351-2_figures_figure200edit1.svg
T C Curie temperature T N Neel temperature

14.6.2 Structure Dependence of J ex

image: 160_home_ken_Mydocs_MSEcore_351-2_figures_figure201edit2.svg
image: 161_home_ken_Mydocs_MSEcore_351-2_figures_figure203edit1.svg

15 Electro-optic and Nonlinear Optical Materials

P= ε 0 [ χ ( 1 ) E + χ ( 2 ) E 2 + χ ( 3 ) E 3 ]
χ ( 1 ) Linear susceptibility χ ( 2 ) Quadratic susceptibility χ ( 3 ) Cubic susceptibility

15.1 Frequency Doubling

image: 162_home_ken_Mydocs_MSEcore_351-2_figures_figure206edit1.svg

15.2 Nonlinearity in Refractive Index

c Speed of light in free space (3 × 10 8  m/s ) n Refractive index
n 2 =1+ χ ( 1 ) + χ ( 2 ) + χ ( 3 ) +
n= n 0 + n 2 I
n 0 Linear and  χ = ε r -1 n 2 Nonlinear index I Intensity

15.3 Electro-Optic Modulators

Δ ( 1 ε r ) = Δ ( 1 n 2 ) =r E
Δ n=- 1 2 n 3 r E
r Electro-optic coefficient [ pm/V ]

15.4 Optical Memory Devices

15.5 Volume Holography

ε r = ε r0 + ε r1 cos ( 2ky sin θ ) k Magnitude of light wavevector
Λ = 2 π 2k sin θ = λ 2n sin θ with k=2 π n/ λ
image: 163_home_ken_Mydocs_MSEcore_351-2_figures_figure208edit1.svg

15.6 Photorefractive Crystals

  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
Δ n=- 1 2 n 3 r E
image: 164_home_ken_Mydocs_MSEcore_351-2_figures_figure209edit1.svg

15.7 Photorefractive Crystals

Steps in creation of periodic dielectric constant
image: 165_home_ken_Mydocs_MSEcore_351-2_figures_figure212edit1.svg

15.8 Phase Conjugation - All Optical Switching

image: 166_home_ken_Mydocs_MSEcore_351-2_figures_figure215edit1.svg
image: 167_home_ken_Mydocs_MSEcore_351-2_figures_figure217edit1.svg

15.9 Acousto-Optic Modulators

Δ ( 1 ε r ) =pS and Δ 1 n 2 =pS
S Strain p Photoelastic constant n Refractive index
Λ = λ 2n sin θ
image: 168_home_ken_Mydocs_MSEcore_351-2_figures_figure218edit1.svg

15.10 Integrated Optics

15.11 Dielectric Waveguides

image: 169_home_ken_Mydocs_MSEcore_351-2_figures_figure219edit1.svg

15.12 Example: Phase Shifter

Δ n=- 1 2 n 3 r E =- 1 2 n 3 r V d
Δ φ = 2 π λ Δ nL
image: 170_home_ken_Mydocs_MSEcore_351-2_figures_figure221edit1.svg

15.13 Example: LiNbO 3 Phase Shifter

Δ φ = π = 2 π λ L Δ n
L=0.4 cm
V π = λ d n 3 rL

16 Superconductivity

image: 171_home_ken_Mydocs_MSEcore_351-2_figures_figure223edit1.svg

16.1 BCS Theory of Superconductivity

16.2 Density of States and Energy Gap

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16.3 Heat Capacity of Superconductors

H C = H 0 [ 1-1.07(T/ T C ) 2 ]
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16.4 Tunneling in Superconductor Junctions

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16.5 Semiconductor-Insulator-Semiconductor Junction

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16.6 Type I and Type II Superconductors

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16.7 High T C Superconductors

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16.8 Cuprate Superconductors

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17 351-2 Problems

  1. An abrupt Si p-n junction has N a =1 0 18 c m -3 on one side and N d =1 0 15 c m -3 on the other.
    1. Calculate the Fermi level position at 300K on both sides.
    1. Draw an equilibrium band diagram for the junction.
    2. Determine the contact potential Φ o for this junction.
  1. A silicon p + -n junction 1 0 -2 c m 2 in area had N d =1 0 15 c m -3 doping on the n-side. Calculate the junction capacitance with a reverse bias of 10V.
  2. For metallic aluminum, calculate:
    1. The valence electron density.
    2. The radius of the Fermi sphere k F .
    3. Fermi energy in eV.
  3. From the Schrodinger equation for a quantum well, show that the wave vector is equal to n π / L where L is the well width.
  4. 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.
  5. What is the luminescent energy for a CdSe quantum dot with a 2 nm radius.
  6. For a MOSFET device briefly describe how the three types of device work: a) enhancement mode b) depletion mode c) inversion mode.
  7. Calculate the capacitance of an MOS capacitor with a 10 nm thick Hf O 2 dielectric oxide. What is the ratio of capacitances for C Hf O 2 / C Si O 2 . The relatie dielectric constant for Hf O 2 is 25.
  8. Problem 9.9 in Solymar and Walsh
  9. Problem 9.14 in Solymar and Walsh
  10. Problem 9.16 in Solymar and Walsh
  11. Problem 12.10 in Solymar and Walsh
  12. Consider a quantum cascade laser (QCL) made from GaAs and GaAlAs. What well thickness is needed for laser emission at 3 microns?
  13. Derive the expression for the average value of the dipole moment. Show that it is given by:
    < μ >= μ [ coth a- 1 a ]
  14. The saturation polarization P s of PbTi O 3 , a ferroelectric, is 0.8 coulombs/ m 2 . The lattice constant is 4.1 A ˙ . Calculate the dipole moment of unit cell.
  15. 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.
  16. Consider the frequency dependence of the atomic polarizability. The polarizability and its frequency dependence can be modeled as a damped harmonic oscillator. Derive the expression for α in this case.
    The expression is given by:
    m dx d t 2 +b dx dt + ω 0 2 x=- e ϵ loc sin ω t
    Plot α vs. ω for this case.
  17. Problem 4.6 in Solymar and Walsh.
  18. Problem 4.7 in Solymar and Walsh.
  19. Problem 4.8 in Solymar and Walsh.
  20. Problem 4.9 in Solymar and Walsh.
  21. Calculate the magnetic susceptibility of metallic copper. How does it compare to the measured value of -1.0?
  22. Calculate the effective magneton number p for M n 2+ ,C o 2+ . Show work.
  23. Consider Mn doped GaP. There are 1 0 20 M n 2+ ions.
  24. What is the electron configuration M n 2+ in spectroscopic notation.
    1. Calculate its magnetic moment at saturation in Bohr magnetons.
    2. Calculate its magnetic susceptibility.
  25. For metallic Co, which has a Curie temperature of 1388 K, calculate the Weiss constant λ . 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 apparatus.

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 concentration 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 temperature, and explain the origins of differences in mobilities between different semiconductors.

18.1.3 References

(1) M. Ali Omar, Elementary Solid State Physics;
(2) Solymar & Walsh, Electrical Properties of Materials;
(3) MSE 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?
    1. image: 179_home_ken_Mydocs_MSEcore_351-2_figures_heffect2.png Figure 18.1: Part 7; B is uniform throughout the area of interest
  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.)
    1. image: 180_home_ken_Mydocs_MSEcore_351-2_figures_hall3.png
      P = +
      P = -
      M =
      M =
      N =
      N =
      Figure 18.2: Part 6
  8. By convention, current is defined as the flow of what sign of charge carrier?
  9. What is the electric field, E, in the situation below? What is the electrostatic force, F E on the particle if it has a charge q?
    1. image: 181_home_ken_Mydocs_MSEcore_351-2_figures_hall4.png Figure 18.3: Part 9
  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.
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 magnetic field may be determined. A device that measures these parameters, known as a Hall Probe, provides a way to measure magnetic fields. Write an expression that relates these parameters. Which is the more sensitive (higher ratio of mV/Tesla) Hall probe – the bulk InAs or bulk GaAs sample? Show your work. (Note – you should have recorded average Hall voltage for a given current. How do these compare?)

18.1.9 Hints for derivation

  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 uniform throughout the specimen. Hint: Balance the magnetic and electric forces on the particle.)
    image: 182_home_ken_Mydocs_MSEcore_351-2_figures_halleffect.png Figure 18.4: Example
  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 passing the current, q is the charge on each carrier, and v is the drift velocity of the carriers. If you knew i, A, VH, d, B, and q from the situation illustrated above, what expression tells you n?
  3. The “Hall coefficient” of a material, R H , is defined as the Hall electric field, E H , per current density, per magnetic field, R H = E H /(J*B) Using the equations given and derived thus far, express the Hall coefficient in terms of just q and n.

18.2 Laboratory 2: Diodes

18.2.1 Objective

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

18.2.2 Outcomes

Upon completion of the laboratory, the student will be able to:
  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 describe 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 voltage) (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 illumination. Label V OC and I SC .
  5. Take pictures (use your phone) of lights around campus; try to get pictures of LEDs. Which do you think are LEDs (why?)?
  6. Why are fluorescent lights “white?” How white are they?

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= I sat [ exp ( qV/kT ) -1 ] . Note that for V>>kT/q, I= I sat exp ( qV/kT ) ; however recombination of carriers in the space charge region leads to a departure from ideality by a factor m, where I= I 0 [ exp ( qV/mkT ) -1 ] .
    1. Plot the data using ln ( I ) vs. V plot to determine m .
    2. Use the value of the applied voltage corresponding to ~ 1mA to solve for I 0 (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.
  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 V oc (the open circuit voltage) and I sc (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 : ____ Measure 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 results 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.
LED Color
Peak wavelength
E g (in eV, from peak wavelength)
Intensity
FW (left)
FW (right)
FWHM
Laser
Table 1: Spectrometer and Power Supply
Semiconductor Lasers
  1. Attach the laser to the power meter and setup the power meter to measure intensity from the device. Slowly increase the voltage and record the 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?
  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.

18.3 Laboratory 3: Transistors

18.3.1 Objective

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

18.3.2 Outcomes

Upon completion of the laboratory, the student will be able to:
  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
  1. 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 I Source-Drain V Source-Drain characteristics vs. gate voltage, V gate , label the linear and saturated regions.
  2. What is threshold voltage? What structural and materials properties determine 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 nanotubes graphene

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)
    1. 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 V CE below 30V to avoid damaging the device.
    2. Sketch how E,B,C are configured in the device.
    3. Measure I collector (output current) for V CE > V saturation as a function of base (input) current, I base .
    4. Post-Lab: determine the transistor gains, α=- I collector / I emitter and β= I collector / I base .
    5. Repeat for a pnp transistor.
  2. Phototransistor
    1. 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.
    2. Post-lab: discuss the mechanism by which the light influences the device current, and compare with the bipolar transistor operation.
  3. MOSFETS
    1. Attach MOSFET Device to the linear FET slot on the Curve Tracer, making sure that S,G,D are connected appropriately. Observe both the linear and saturated regions for I Source-Drain vs. V Source-Drain .
    2. 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).
    3. Measure the V Source-Drain and I Source-Drain to extract R Source-Drain in the linear region as a function of V Gate . You should take at least 4 measurements.
    4. Post-lab, using the data from B, plot the conductance of the channel, G Source-Drain =1/ R Source-Drain , vs. V Gate . You should observe a linear relationship following the equation , G SD =( μ n W L ) ( C OX A ) ( V G - V T )
      where μ n is the electron mobility and W, L, and A are the width, length and area of the gate respectively (A = W x L). C OX is the capacitance of the oxide, which can be measured using an impedance analyzer as a function of gate voltage. A plot is shown in Figure 18.5.
    5. Using μ n =1500 cm2/V-sec, C = 4.8x1 0 -10 F (from the attached plot), plot of conductance vs. V Gate , calculate L, the length of the gate.
    6. Determine V threshold from the above values. Compare with your observations from the curve tracer (part a).
    image: 183_home_ken_Mydocs_MSEcore_351-2_figures_MOFSET.png Figure 18.5: n-Channel MOFSET.

18.4 Laboratory 4: Dielectric Materials

18.4.1 Objective

The objectives of this lab are to measure capacitance and understand the dependence on geometry and the dielectric constant, which may vary with temperature and frequency.

18.4.2 Outcomes

Upon completion of the laboratory, the student will be able to:
  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 vacuum)?
  4. What is the relative dielectric constant of air? Water? Glass? Si O 2 ? Explain their relative magnitudes.
  5. For the capacitor shown in Figure 18.6, what is the expression that relates the capacitance and relative dielectric constant?
    image: 184_home_ken_Mydocs_MSEcore_351-2_figures_dielectric.png
    Figure 18.6: Capacitor.
  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?
  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 capacitor?
  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.
    1. What defines the boundary between “low” and “high?”
    2. What drives the need for “high-k” dielectrics? What other properties besides the dielectric constant are important characteristics of these materials?
    3. What drives the need for “low-k” dielectrics?
  12. Find one or two additional examples of technologies/devices that incorporate capacitors, and explain the function of the capacitor in that context.

18.4.4 Experimental Details

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

18.4.5 References

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

18.4.6 Instructions/Methods

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

18.4.7 Lab Report Template

Part I - Dielectric constants; Measuring capacitance with Impedance Analyzer
  1. Using the default frequency of 100kHz and zero bias voltage, measure capacitance and contact dimensions and then calculate the relative dielectric constant of the following materials:
    Material
    Area
    Thickness
    Capacitance
    ϵ r ( lab )
    ϵ r ( lit )
    Glass slide (sample 1)
    Lithium niobate ( LiNb O 3 )
    Glass sheet (sample 2)
    Plexiglass
    Table 2: Dielectric Constants
  2. Using the default frequency of 100 kHz, measure capacitance vs. voltage for several LEDs
    LED=
    LED=
    LED=
    Bias
    Capacitance
    Bias
    Capacitance
    Bias
    Capacitance
    Table 3: Capacitance
    Determine the built-in voltage of the corresponding p-n junction, ϕ 0 , 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 C 2 =2 ( ϕ 0 - V 0 ) ( N a - N d ) ϵ e N a N d
    Plot 1/ C 2 vs. V 0 , 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:
    1. air (4.0inch x 3.5 inch plates, separated by _________ (measure this))
    2. water at room temperature (4.0 inch x 3.5 inch plate partially submerged; measured dimension: Height:________ x width___________x separation____________________
    3. ethanol
  4. Fill in the table below, for water at the different temperatures measured in lab, using the dimensions measured above. Determine the corresponding values of the relative dielectric constant, ε r . Plot ( ϵ r 1 ) /( ε r +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)
    ϵ r
     
    Comparison from CRC handbook
    Temp (C)
    ϵ r
    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 (pF)
    ϵ r
    Freq
    C (pF)
    ϵ r
  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?

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 magnetons.
  3. For a known material, explain the microscopic origins of magnetism.
  4. Distinguish qualitatively between “hard” and “soft” ferromagnetic materials.
Experimental Details
The purpose of this experiment is to investigate the response of different materials to an applied magnetic field, H. Plotting the measured response, B, vs. the applied field, H, indicates the type of magnetization possible, i.e. whether the material is diamagnetic, paramagnetic or ferromagnetic.
References
Solymar and Walsh Chapter 11, Omar Chapter 9
1. Magnetization of metal wires
Qualitatively observe the hysteresis behavior of the following metals. Indicate whether they are ferromagnetic or not, and why. (Hint: Look at which electronic shells are filled/unfilled in each case.)
 
Metal wires:
Hysteresis? Y/N
3d
4s
Cu
Fe
Co
Ni
W
Steel
Stainless Steel
 
Quantify the observations on the metal wires (Fe, Co, Ni, and W 1 mm diameter and Cu 0.5 mm diameter – the cross sectional area must be entered in the B-probe dimensions). Record hysteresis curves in the “intrinsic” mode (without background subtraction). Also record “initial” curves for the first three metals. Fill in the following table.
B sat observed
B sat literature
# Bohr magnetons, n B *
Fe
2.158 T
Co
1.87 T
Ni
0.616 T
Question 1(a-c):
Solve for the number of Bohr magnetons per atom in Fe, Ni and Co. (Use experimental B sat if the sample saturates; otherwise use literature value.)
Note:
μ 0 =4 π × 1 0 -7 T-m/A
B S = μ 0 M S ; M s in A/m or Oe
M S =n M A ; n = # atoms/volume
M A = n B μ B ; μ B = e 2me =9.27 × 1 0 -24 A- m 2 = Bohr magneton
Question 2: Hard magnet
The sintered CuNiFe pellets provide an example of a “hard” magnet. Record (plot) a hysteresis curve and compare to the “soft” ferromagnetic materials. Comment on the observed behavior.
2. Faraday Rotation
Set-up:
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 ) =VB , B = magnetic field strength (millitesla), and = length glass rod = 10 cm.
3. Demonstration
Heating a magnetized iron wire is one way to observe a second order phase transition involving the spin degree of freedom (upon cooling, one observes the onset of spontaneous magnetization). What is this transition called? Estimate the value of the transition temperature for this system and compare to literature values.