ELECTRONICS II VLSI DESIGNFALL 2013

LECTURE 1INSTRUCTOR: L.M. HEAD, PhDELECTRICAL & COMPUTER ENGINEERINGROWAN UNIVERSITY

Semiconductors are Crystalline Materials 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

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 body-centered face centered

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/sili2.html

Planes and Directions

a

bc

x

y

z

(214)

http://jas2.eng.buffalo.edu/applets/education/solid/unitCell/home.html

Blackbody Radiation

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 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. 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

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:

set 1 (Lyman)

set 2 (Balmer)

set 3 (Paschen)

c = speed of light, R = Rydberg constant = 109,678 cm-1

......4,3,2,1

1

122

n

ncR

......5,4,3,1

2

122

n

ncR

......6,5,4,1

3

122

n

ncR

Photon Energies in the Hydrogen Spectrum

n = 5

n = 4

n = 3

n = 2

n = 1

Lyman

Balmer

Paschen

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 )

np

The Bohr Model

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.

Kn

q

Kn

qnv

mr

nv

mq

Knr

r

n

mrKr

q

r

nvm

nmvrp

r

mv

Kr

q

n

n

nnn

n

2

22

2

2

22

2

22

2

2

2

2222

2

2

2

1

+q

-q

r

o4K

Calculating Bohr Energies

Determine both kinetic and potential energies.

Calculate total energy.

22

21

22

4

1n2n

222

4

n

222

4

n

2

222

42

n

1

n

1

K2

mqEE

nK2

mq.E.P.E.KE

nK

mq

Kr

q.E.P

nK2

mqmv

2

1.E.K

Electron Orbits and Transitions in the Bohr Model of the Hydrogen Atom

n = 5

4

3

2

1 Lyman

Balmer

Paschen

Quantum Mechanics 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:

xpx

Probabilities

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:

normalization condition

1dx)x(P

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:

for a normalized pdf

dx)x(P)x(f

dx)x(P

dx)x(P)x(f

)x(f

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, , are continuous, finite, and single valued.

2. Classical quantities such as energy and momentum must be related to abstract quantum mechanical operators:

zyx ///

E

)x(p

)x(f

x

tj

xj

)x(f

x

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:

and the average value <Q> of any variable Q is calculated from by using the operators defined in postulate 2

1dxdydz

dxdydzQQ op

Schrödinger’s Wave Equation

Kinetic Energy + Potential Energy = Total Energy

2

2

2

2

2

22

22

2

22

2

2

),(),()(

),(

2

2

1

zyx

tjV

m

t

tx

jtxxV

x

tx

m

EVpm

Apply Separation of Variables

0)x()x(VEm2

dx

)x(d

0)t(jE

dt

)t(d

t

)t()x(

j)t()x()x(V)t(

x

)x(

m2

22

2

2

22

Potential Well Problem

V=0

V= V=

x=0 x=L

Lx0,0)x(Em2

dx

)x(d

0)L(,0)0(

22

2

Possible solutions are

mE2

k

)kxcos(),kxsin(

)kxsin(AFrom BC 1choose:

From BC 2 choose: 3,2,1n,L

nk

Boundary conditions:

2

222

n

n

mL2

nE

L

nmE2

Note that the energy is quantized.The integer n is called the quantum number.

We still need to determine the constant A.

For this we use the normalizationcondition.

xL

n

L

LA

LAdxx

L

nAdx

n

L

sin2

2

12

sin 2

2

0

2

Tunneling

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.