ELECTRONICS II
VLSI DESIGN
FALL 2013
LECTURE 1
INSTRUCTOR: L.M. HEAD, PhD
ELECTRICAL & COMPUTER ENGINEERING
ROWAN UNIVERSITY
Semiconductors are Crystalline Materials
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Three types of solids: crystalline, amorphous, polycrystalline.
The arrangement of atoms in a crystal is called the lattice.
Points within the lattice are indistinguishable if the vector
between the points is: px + qy + sz
 p, q and s are integers
Crystalline
Amorphous
Polycrystalline
Cubic Lattices
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The unit cell is the smallest regularly repeated volume in the
lattice.
The simplest example of the unit cell is the simple cubic
structure.
Others are body-centered cubic and face-centered cubic.
By assuming that the atoms are solid spheres, an estimate of
the material density can be calculated.
Simple cubic
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/sili2.html
body-centered
face centered
Planes and Directions
z
(214)
c
b
y
a
x
http://jas2.eng.buffalo.edu/applets/education/solid/unitCell/home.html
Blackbody Radiation
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When something is heated, it emits light.
Classical physics predicts that the total energy emitted should
increase with frequency, this would result, however, in the
energy density becoming infinite.
Empirically, scientists saw the energy peak at a particular
frequency and then decrease with the peak being a function
of the temperature.
Planck was able to describe these experiments mathematically
by assuming that the emitted energy was quantized in units
of h where  is frequency and h is Planck’s constant
(6.63X10-34 J-s).
Planck did not understand why this worked, he just knew that
it fit the experimental data.
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html#c2
Photoelectric Effect
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Einstein used Planck’s idea to interpret another experiment that
produced “unexplainable” results.
When light is directed at a sample of metal, electrons can be
“dislodged/emitted” from the surface.
However, if the light has a frequency below a certain value (for
a certain metal), no electrons will be emitted, no matter how
intense the light is!
Above this certain frequency (which is different for each metal)
electrons will be emitted. A higher intensity light yields more
electrons but each one has the SAME amount of energy! (i.e.,
each electron absorbs one photon)
The energy associated with the “certain” or characteristic
frequency is call the work function.
Photoelectric Effect – con’t.
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An electron can only absorb a single photon.
If that photon has energy (h from Planck’s work) that is less than
the work function (q ) of the metal, the electron that absorbs
the energy will not be able to break out of the solid structure of
the metal.
If that photon has energy that is greater than the work function of
the metal, the electron will break free and will (perhaps) have
some energy left over (if h - q   0).
The left over energy will be in the form of kinetic energy.
If we measure that leftover energy and plot it versus frequency,
the slope of the straight line fit to the data is h, Planck’s
constant!!
WOW, what a coincidence.
http://www.lon-capa.org/~mmp/kap28/PhotoEffect/photo.htm
Another Piece of
Quantum Evidence
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By the early years of the 20th century, researchers had observed
the spectra of several different atoms such as hydrogen.
The wavelengths of the emitted light (=c/) were not continuous
but were sharp lines.
They turned out to have the following relationships:
 1
1 
, n  2,3,4......
  cR 
2
2
n 
1
 1
1 
, n  3,4,5......
  cR 
2
2
n 
2

set 1 (Lyman)

set 2 (Balmer)

set 3 (Paschen)

c = speed of light, R = Rydberg constant = 109,678 cm-1
 1
1 
, n  4,5,6......
  cR

2
2
n 
3
Photon Energies in the Hydrogen
Spectrum
n=5
n=4
Paschen
n=3
Balmer
n=2
Lyman
n=1
The Bohr Model

Postulates:



Electrons exist in certain stable, circular orbits about
the nucleus. (no radiation from angular acceleration)
The electron may shift to an orbit of higher or lower
energy, thus gaining or losing energy equal to the
difference in the energy levels. (absorption or emission
of a photon of energy h )
The angular momentum of the electron in an orbit is
always an integral multiple of Planck’s constant divided
by 2 (h/(2) is abbreviated  )
p  n
The Bohr Model
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

q2
mv 2


2
r
Kr
p  mvr  n
An electron in a stable orbit with radius r
about the proton of a hydrogen atom.
Equate the electrostatic and centripetal
forces.
Use postulate 3.
Solve for r and v.
-q
r
m 2v 2 
K  4 o
rn2
q2
1 n 2 2


2
Krn mrn rn2
rn 
v
+q
n 2 2
Kn 2  2
mq 2
n
mrn
n q 2
q2
v

2 2
Kn
Kn 
Calculating Bohr Energies
4
1
mq
K .E .  mv 2 
2
2 K 2n2 2
q2
mq 4
P .E .  

Krn
K 2n2 2
E n  K .E .  P .E .  
mq 4
2 K 2n2 2
mq 4  1
1 
E n 2  E n1 

2 2  2
2 K   n1 n22 


Determine both kinetic
and potential
energies.
Calculate total energy.
Electron Orbits and Transitions in the Bohr
Model of the Hydrogen Atom
n=5
4
Paschen
3
2
1
Balmer
Lyman
Quantum Mechanics
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The Bohr Model did not explain all aspects of the
spectrum of even a hydrogen atom.
In the 1920’s physicists developed a new method
for dealing with these discrepancies.
The method, QM, is based on the Heisenberg
Uncertainty Principle:

In any measurement of the position and
momentum of a particle, the uncertainties
in the two measured quantities will be
related by:
xpx   
Probabilities
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Given the uncertainty principle, we cannot determine the
exact position of an electron but must find the probability
that it is at a certain position.
So what we need are probability density functions rather
than exact values of position, velocity, energy.
Given a probability density function, P(x), the probability
of finding an electron between x and x+dx is P(x)dx.
And, since the electron must be somewhere:

 P (

x )dx  1
normalization condition
Expected Value

To find the value of a function of x, we need
only multiply the value of that function in each
increment dx by the probability of finding the
electron in that dx and sum over all x. So the
average value of f(x) is:

 f ( x )P ( x )dx
f(x) 


 P ( x )dx


for a normalized pdf

  f ( x )P ( x )dx

Basic Postulates of Wave Mechanics
1. Each particle in a physical system is described by a
wave function, (x,y,z,t). This function and its space
derivative,  / x   / y   / z , are continuous, finite,
and single valued.
2. Classical quantities such as energy and momentum
must be related to abstract quantum mechanical operators:
x
x
f(x)
f(x)
p( x )
 
j x
E

 
j t
Basic Postulates, con’t.
3. The probability of finding a particle with wave
function  in the volume dx dy dz is * dx dy dz.
The product * is normalized such that:



 dxdydz  1

and the average value <Q> of any variable Q is
calculated from  by using the operators defined in
postulate 2

Q 


 Qopdxdydz

Schrödinger’s Wave Equation

Kinetic Energy + Potential Energy = Total Energy
1 2
p V  E
2m
 2  2  ( x, t )
  ( x, t )

 V ( x )  ( x, t )  
2
2m x
j
t
2 2
 

   V  
2m
j t
2 
2
x
2

2
y
2

2
z 2
Apply Separation of Variables
 2  2 ( x )

 ( t )

 ( t )  V ( x ) ( x ) ( t )    ( x )
2
2 m x
j
t
d ( t ) jE

( t )  0
dt

d 2 ( x )
dx
2

2m

2
E  V ( x ) ( x )  0
Potential Well Problem
Boundary conditions:
( 0 )  0 ,
d 2 ( x )
V=
V=
V=0
x=0
dx 2

( L )  0
2m
2
E( x )  0 ,
Possible solutions are
sin( kx ),
0 xL
cos( kx )
2 mE

  A sin( kx )
k
x=L
From BC 1choose:
From BC 2 choose: k 
n
,
L
n  1 ,2 ,3 
2 mE n

En 
Note that the energy is quantized.
The integer n is called the
quantum number.
n

L
We still need to determine the
constant A.
n  
2 2 2
2 mL2

For this we use the normalization
condition.
2
n 

2 L
 dx  0 A  sin L x  dx  A 2  1

L
2
A
n 
2
L
2
n
sin
x
L
L
Tunneling
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The wave functions are easier to calculate for the potential
well problem with infinite walls since the boundary
conditions force  to be zero at the walls.
If the barrier actually has a finite value we have to address
the probability that the particle/electron will actually pass
through the barrier even though its energy is less than the
barrier height.
This phenomena (tunneling) is a result of Postulate 1, that
the function and its space derivatives must be continuous,
finite and single valued.