Transcript Solid State Electronic Devices

In the name of God
Electronic physics
Solid State Electrical Devices
‫مهدی حریری‬
)‫دانشکدة فنی دانشگاه زنجان(گروه الکترونیک‬
[email protected]
1385-1386‫نیمسال اول‬
SOLID STATE ELECTRONIC
DEVICES
Chapter II


The behavior of solid state devices is
directly related to atomic theory, quantum
mechanics, and electron models.
In this chapter we shall investigate some
of the important properties of electrons,
with special emphasis on two points:
(1) the electronic structure of atoms and
(2) the interaction of atoms and electrons
with excitation, such as the absorption and
emission of light.
1

First, we shall investigate some of the
experimental observations which led to
the modern concept of the atom, and then
we shall give a brief introduction to the
theory of quantum mechanics.
2
INTRODUCTION TO
PHYSICAL MODELS


In the 1920s it became necessary to
develop a new theory to describe
phenomena on the atomic scale.
Physicists discovered that Newtonian
mechanics did not apply when objects
were very small or moved very fast!
3


If things are confined to very small
dimensions (nanometer-scale), then
QUANTUM mechanics is necessary.
If things move very fast (close to the
speed of light), then RELATIVISTIC
mechanics is necessary.
4
The Photoelectric Effect

An important observation by Planck
indicated that radiation from a heated
sample is emitted in discrete units of
energy, called quanta; the energy
units were described by hv, where ν is
the frequency of the radiation, and h is
a quantity now called Planck's constant
(h = 6.63 x 10-34 J-s).
5

Electrons are ejected from the surface of a
metal when exposed to light of frequency
ν in a vacuum.
6


Plot of the maximum
kinetic energy of
ejected electrons vs.
frequency of the
incoming light.
The equation of line is:
Em =hv - qФ
7
Atomic Spectra


Early in the 19th century, Fraunhofer saw dark bands on the solar
spectrum.
In 1885, Balmer observed hydrogen spectrum and saw colored lines :
 Found empirical formula for discrete wavelengths of lines.
 Formula generalized by Rydberg for all one-electron atoms.
8
Atomic Spectra: Modern Physics Lab
Neon Tube
Eyepiece
(to observe lines)
High Voltage
Supply
(to “excite” atoms)
Diffraction
Grating
(to separate light)
9
Emissionline spectra
of Na, H,
Ca, Hg, Ne
(1) Continuous spectrum from an incandescent light bulb.
(2) Absorption-line spectrum (schematic) of sun’s most
prominent
lines: H, Ca, Fe, Na.
10

Some important line in the emission spectrum of hydrogen :

Photon energy hv is then related to wavelength by :
11

The various series in the spectrum were observed to
follow certain empirical forms :
Where R is a constant called the Rydberg constant :
12

Each energy can be obtained by taking sums and differences of
other photon energies in the spectrum :
For example :
E42 =
In the Balmer
series
E 41 - E 21
In the Lyman
series
13
The Bohr Model


Classical model of the electron “orbiting” nucleus is unstable. Why
unstable?
 Electron experiences centripetal acceleration.
 Accelerated electron emits radiation.
 Radiation leads to energy loss.
 Electron eventually “crashes” into nucleus.
In 1913, Bohr proposed quantized model of the H atom to predict the
observed spectrum.
14
To develop the model, Bohr made several postulates :
1-
Electrons exist in certain stable, circular orbits about the nucleus. This
assumption implies that the orbiting electron does not give off radiation as
classical electromagnetic theory would normally require of a charge
experiencing angular acceleration; otherwise, the electron would not be
stable in the orbit but would spiral into the nucleus as it lost energy by
radiation.
2-
The electron may shift to an orbit of higher or lower energy, thereby
gaining or losing energy equal to the difference in the energy levels :
15
3-

The angular momentum Pө of the electron in an orbit is
always an integral multiple of Planck's constant divided by
2π (h / 2π is often abbreviated ħ for convenience) :
If we visualize the electron in a stable orbit of radius r about the proton of
the hydrogen atom, we can equate the electrostatic force between the
charges to the centripetal force:
16
17
18
The energy difference between orbits n 1 and n 2 is given by :
The frequency of light given off by a transition between these two orbits is :
19

The factor in brackets is essentially the Rydberg constant R times the
speed of light c.

Whereas the Bohr model accurately describes the gross features of the
hydrogen spectrum, it does not include many fine points. For example,
experimental evidence indicates some splitting of levels in additon to the
levels predicted by the theory. Also, difficulties arise in extending the
model to atoms more complicated than hydrogen. Attempts were made
to modify the Bohr model for more general cases, but it soon became
obvious that a more comprehensive theory was needed.
Electron orbits and
transitions in the
Bohr model of the
hydrogen atom.
Orbit spacing is not
drawn to scale.
20
Orbital Radii and Energies, cont.



rn=0.0529n2 (nm)
En=-13.6/n2 (eV)
Energy difference between the
levels DE=13.6(1/nf2-1/ni2)
Initial State, ni
DE=10.2 eV
Final state, nf
For example, between n=1 and n=2 (as drawn in the picture)
DE=13.6(1/nf2-1/ni2)=13.6(1/12-1/22)=10.2 eV
21
Bohr’s Correspondence
Principle

Bohr’s Correspondence Principle states
that quantum mechanics is in agreement
with classical physics when the energy
differences between quantized levels are
very small.

Similar to having Newtonian Mechanics be a
special case of relativistic mechanics when
v << c .
22
Successes of the Bohr Theory

Explained several features of the hydrogen spectrum






Accounts for Balmer and other series
Predicts a value for RH that agrees with the experimental value
Gives an expression for the radius of the atom
Predicts energy levels of hydrogen
Gives a model of what the atom looks like and how it behaves
Can be extended to “hydrogen-like” atoms


Those with one electron
Ze2 needs to be substituted for e2 in the Bohr equations
 Z is the atomic number of the element (=number of protons)
23
Pioneers of Quantum Mechanics
Schrodinger
Fermi
Heisenberg
24
Probability and the Uncertainty Principle

In any measurement of the position and momentum of a
particle, the uncertainties in the two measured quantities will
be related by :

Similarly, the uncertainties in an energy measurement will
be related to the uncertainty in the time at which the
measurement was made by :
25
DE Dt Uncertainty Principle Example*
A particular optical fiber transmits light over the range 1300-1600 nm
(corresponding to a frequency range of 2.3x1014 Hz to 1.9x1014 Hz). How
long (approximately) is the shortest pulse that can propagate down this
fiber?
a. 4 ns
b. 2 fs
c. 4 fs
D Dt  1  2 Df Dt  1
d. 2 ns
Dt  1/ 2 Df
 1/(2 0.4 1014 Hz )
 4 1015 s  4 fs
Note: This means the upper limit to data transmission is
~1/(4fs) = 2.5x1014 bits/second = 250 Gb/s
*This problem obviously does not require “quantum mechanics”. However,
due to the Correspondence Principle, the quantum constraints on single
photons also apply at the classical-pulse level.
26

One implication of the uncertainty principle is that we cannot
properly speak of the position of an electron , for example, but
must look for the probability of finding an electron at a certain
position. Thus one of the important results of quantum
mechanics is that a probability density function can be
obtained for a particle in a certain environment, and this
function can be used to find the expectation value of important
quantities such as position, momentum, and energy.

Given a probability density function P(x) for a one-dimensional
problem, the probability of finding the particle in a range from
x to x + dx is P(x) dx. Since the particle will be somewhere,
this definition implies that :
27

To find the average 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 particle in that dx and sum over
all x. Thus the average value of f(x) is :

If the probability density function is not normalized, this
equation should be written :
28
The Schrodinger Wave Equation
29
30
31
32

Now the variables can be separated to obtain the timedependent equation in one dimension

and the time-independent equation,
33
Potential Well Problem

The simplest problem is the potential energy well with
infinite boundaries. Let us assume a particle is trapped in a
potential well with V(x) zero except at the boundaries x = 0
and L, where it is infinitely large :
34
35

From Eqs. (2-30) and (2-31) we can solve for the total
energy E„ for each value of the integer n.
36
The problem of a particle in a potential well :(a) potential
energy diagram; (b) wave functions in the first three
quantum states ;(c) probability density distribution for the
second state.
37
Particle Motion in a Box: Example

Consider the numerical example:
An electron in the infinite square well
potential is initially (at t=0) confined to the left
side of the well, and is described by the
following wavefunction:
2
 
 2
 ( x, t  0)  A
 sin  x   sin 
L
L 
 L

x  

If the well width is L = 0.5 nm, determine the
time to it takes for the particle to “move” to
the right side of the well.
h2
1.505 eV  nm 2
En 

2me n2
n2
n  2L/n
1.505 eV  nm2 1.505 eV  nm 2
E1 

 1.505 eV
4 L2
4(.5nm) 2
|y (x,t=0)|2
U=
0
x
L
|y (x,t0)|2
U=
U=
0
En  E1n 2
U=
x
L
period T = 1/f = 2t0
with f = (E2-E1)/h
T
h
h
4.136 1015 eV  sec
to  


 4.6  1016 sec
2 2  E 2  E1  2  3E1 
2  3 1.5eV 
38
Tunneling
Classical
Wave Function
For Finite Square
Well Potential
Where E<V
Classically, when an object hits a potential that
it doesn’t have enough energy to pass, it will
never go though that potential wall, it always
bounces back.
In English, if you throw a ball at a wall, it will
bounce back at you.
39
Quantum
Wave Function
For Finite Square
Well Potential
Where E<V
In quantum mechanics when a particle hits a
potential that it doesn’t have enough energy
to pass, when inside the square well, the wave
function dies off exponentially.
If the well is short enough, there will be a noticeable
probability of finding the particle on the other side.
40
More graphs of tunneling:
n(r) is the
probability of
finding an electron
V(r) is the potential
An electron tunneling from atom to atom:
41
Quantum mechanical
tunneling :
(a) potential barrier of
height Vo and
thickness W;
(b) probability density
for an electron with
energy E < V0,
indicating a
nonzero value of
the wave function
beyond the barrier.
42
Now looking more in depth at the case of tunneling
from one metal to another. EF represents the Fermi
energy. Creating a voltage drop between the two
metals allows current.
Sample
Tip
43
The Hydrogen Atom

Finding the wave functions for the hydrogen atom requires a solution of
the Schrodinger equation in three dimensions for a coulombic potential
field. Since the problem is spherically symmetric, the spherical
coordinate system is used in the calculation. The term V (x, y, z) in Eq.
(2-24) must be replaced by V (r, ө, Ф) , representing the Coulomb
potential which the electron experiences in the vicinity of the proton. The
Coulomb potential varies only with r in spherical coordinates :
When the separation of variables is made, the
time-independent equation can be written as :
44
Schrödinger Equation in Spherical Coordinates
2  2
2
2 
 2  2  2 y x, y, z   U ( x, y, z )y x, y, z   Ey x, y, z 
3D Cartesian: 
2m  x y
z 
Assuming spherical symmetry, change to spherical coordinate system
z
r sin 
r

Radius :
y  r sin  sin 
Polar :
z  r cos

x
x  r sin  cos 
y
r sin  cos 
r sin  sin 
Ze 2
Potential energy: U (r )  k
r
Azimuthal :
2
r  x2  y2  z 2


1
z

  cos 
2
2
2

x  y  z 
  tan 1  y x 

2

2
Laplacian  
 2 2
2
Operator:
x
y
z
2

1   2  
1
 
 
r

sin





2
2

r

r




r

 r sin 



1
2
r 2 sin 2   2
45
Separation of Variables
Begin with the time-dependant Schrodinger wave equation:
 2  2  x, t 
 x, t 




U
(
x
)

x
,
t

i

 E  x, t 
2
2m
t
x
U(x): Potential Energy
Assume:
 ( x, t ) : Complex wave function
1) 1D, free particle  U(x)=0.
2) a separable wave function,
( x, t ) the
X ( x) T (tSchrodigenr
)
Inserting the trial wave function into
equation
above, and dividing through by Ψ(x,t):
  2 X " ( x)
T ' (t )

i

 2
2m X ( x)
T (t )
X " ( x)   X ( x)  X ( x)  Ae i k x
2m
  ( x, t )  X ( x)T (t )  Ae i [ k x  ( k )t ]
i T ' (t )   T (t )  T (t )  Bei  (t ) t
 2k 2
with  (k ) 
2m
46
Spherical Symmetric Solution of the Schrödinger Equation(1)
 2
2m
 1   2 y
 2 r
 r r  r
1
 
y


sin

 2


 r sin   
1
 2y 

 Uy  Ey
 2 2
2 
 r sin   
Let y r, ,   Rr   
 2  1 d  2 dR  
  2 
1
d 
d 
1
d 2  
 constant
   E  U (r )R 
 sin 
 2 2
 2
 2 r
2 
2m  r dr  dr  
2m  r sin  d 
d  r sin  d 
Since the LHS of the equation is a function of r, the RHS is a function of
Θ and Φ, the only possibility is that both sides equal to a constant Λ.
 2 d  2 dR 
2
r
  E  U (r )r R  
2m dr  dr 
1
d 
d 
1
d 2
2m
  2   '
 sin 

2
2
 sin  d 
d   sin  d

47
Spherical Symmetric Solution of the Schrödinger Equation(2)
2
d

2
  A sin m  B cos m
Consider the  part first:

m
0
2
d
Since      2n ,
m  0,1,2,...
1 d 
d  
m2 
The  part becomes:
 sin 
    ' 2   0
sin  d 
d  
sin  
Change variables:   cos  F     
2




d
dF
m
2
The equation is transformed into
F  0
 1  
    '
2 
d 
d  
1  
If m=0, F= Legendre polynomials. The solution cannot be finite unless
 '  l (l  1) where λ= positive integer or 0.


If m≠0, F= associated Legendre polynomials with m≤ l and  '  l (l  1)
are the only non-singular and physically acceptable solutions.
48
Spherical Symmetric Solution of the Schrödinger Equation(3)
2
2
2




d
R
2
dR
2
m

KZe
l
l

1

The radial part of the equation is



 ER  0

dr 2 r dr  2  r
2mr 2



u
r
Let R r  
r
The radial wave equation is
 2 d 2u   KZe2 l l  1 2 



u  Eu
2
2 
2m dr
2mr 
 r
2
2
m
kZe
U

2
mE
l 1  
 
Let  
r and u r    e w   0 
2
E 
E

d 2w
dw
The equation becomes 



2
l

1


 0  2l  1w  0
2
d
d
The finiteness of w at 0 demands
0  2N  l  1 where l= 0, 1, 2… and N=0, 1, 2…
Define the principal quantum number n  N  l  1  0 2
We find the energy is quantized as
k 2 Z 2e 4 m
En  
2n 2  2
n=1, 2, 3…
49
Quantum Numbers for the H-atom
Principal quantum number
n=1, 2, 3 …
Z 2 E0
Energy En  
where E0=13.6eV
2
n
Orbital quantum number
l=0, 1, 2, 3 …, n-1
Orbital angular momentum L  l l  1
Magnetic quantum number
m=-l, -l+1, …, l-1, l
Angular momentum in Z-direction
Lz  m
• Angular momentum is quantized in magnitude and direction
• Z can be any direction in free space, but will be parallel to the magnetic field.
• Bound state energies are negative.
• Energy depending only on n is a result of spherical symmetry.
50
The Ground State Wave Functions of H-atom
32
1  Z   Zr a0
  e
Ground state n=1, l=m=0 y 100 
  a0 
Normalization condition
y
100
 
2
0
0
dV  

0
y r sin drdd  1
2
2
Radial probability density
pr   4r 2 y
2
2
a0 
 0.0529nm
2
mke
where Bohr radius
2
y
3
2
 z  2  2 Zr a0
 4  r e
 a0 
r a0
1 2
pr 
y r
2 2
Electrons can be anywhere, but most
likely to be at r=a0 for the ground state.
1 2
r a0
51
The First Excited States Wave Functions of H-atom
1st excited state has three
degenerated states
2
32
y
200

Zr   Zr 2 a0
n=2, l=0, m=0 y 200  C200  2   e
a0 

Zr  Zr 2 a0
y

C
e
cos 
210
n=2, l=1, m=0 210
a0
n=2, l=1, m=±1 y 211  C211
y 210
2
Zr  Zr 2 a0
e
sin e i
a0
pr 
y 211
2
r  2a0
y 210 r 2
2
2
y 200 r 2
2
4
r a0
6
Bohr model: r  n 2
a0
Z
52
Meaning of Quantum Numbers
Principle Quantum Number
Principal quantum number n=1, 2, 3 …
Determines energy
En  
Z 2 E0
n2
where E0=13.6eV
S
P
D
F
G
l=0
1
2
3
4
Degenerate states
n=5
n=4
n=3
hf  E3  E2
n=2
hf  E2  E1
n=1
53
Meaning of Quantum Numbers
Orbital Quantum Number
Orbital quantum number
l=0, 1, 2, 3 …, n-1
Determines magnitude of orbital angular momentum
L  l l  1
54
Meaning of Quantum Numbers
Magnetic Quantum Number
Magnetic quantum number m=-l, -l+1, …, l-1, l
Determines angular momentum in Z-direction
Lz  m
or equivalently the direction of L
55
Example
56
Electronic
configurations
for atoms in the
ground state.
57
Noble Gas
Halogen
Group VI
Group V
n
l = 0 (s)
Group IV
Group III
Periodic Table
l = 1 (p)
1
2
l = 2 (d)
3
4
5
6
7
l = 3 (f)
58
Probability Distributions for Hydrogen Atom
59
The end of slide show for “solid State Electronic Devices”.
Directed by :
Nasser Talebi
Electronic Group - Zanjan University
Aban - Azar 1384 (November-December 2005)