semi-conducting magnetic materials-week 2-jan 16-2012

8/2/2019 Semi-Conducting Magnetic Materials-Week 2-Jan 16-2012

1/93

Semi-conducting & Magnetic Materials

Prof S. B. Sant

Department of Metallurgical & Materials EngineeringIIT Kharagpur

Week 2

MT41016


2/93


Schrdinger Equation:

Time-Independent (Stationary conditions)

&

Time-Dependent (wave equation)


3/93


Schrdinger Equation:Time-Independent (Stationary conditions)

Potential Energy (or potential barrier) depends only on thelocation. Equation of vibration.

Where,

m is the rest mass of an electron (also called mo)

And the total energy of the system is given by,

E = Ekin + V

( ) 02

2

2=+ VE

m

2

2

2

2

2

2

zyx

+

+

=

(3.1)

(3.2)

(3.3)


4/93


Schrdinger Equation:Time-Dependent (wave equation)

(x, y, z, t) = (x, y, z). e it

From equation (2.1), E = v.h =

Differentiating equation (3.4), we have for:

i.e.,

(2.1) & (3.6) gives:

i.e.,

(3.4)

ieit

ti==

(3.5)

t

i

= (3.6)

t

iE

= (3.7)

0

222

2

=

t

mimV

(3.8)


5/93


Special Properties of Vibrational Problems

Boundary conditions yield the constants to solve equations. (e.g. = 0 at x = 0)

Consider a vibrating string: Fixed ends do not undergo vibration.

When boundary conditions are used for vibrating problems we call them boundary oreigenvalue problems.

Not all frequency values are possible and since E = vh, not all values for energy are allowed.

The function belonging to the Eigenvalues, are solutions of the vibration equation & satisfythe boundary conditions are called Eigenfunctions of the differential equation.

We saw that . * is the probability of finding a particle at a certain location. Likewise, theprobability of finding a particle somewhere in space is one or

1*.

2==


6/93


Solutions to the Schrdinger Equation:

Four specific problems

1.0 Free Electrons:

Electrons that propagate freely potential-free space in +ve x-direction No potential barrier (V).

2.0 Bound Electrons: In a potential Well.

Electrons bound to its atomic nucleus.3.0 Finite Potential Barrier: Tunnel effect

Free electrons encounters a potential barrier with potential energy,

V0 > total energy E of the electron.4.0 Electron in a Periodic Field of a Crystal (the Solid State)

Atoms in a crystal are arrange periodically.


7/93


Solutions to the Schrdinger Equation:

Four specific problems

1.0 Free Electrons: Electrons that propagate freely potential-

free space in +ve x-direction No potential barrier (V).From equation (3.1),

Or,

Differential equation for an undamped vibration with spatial

periodicity whose solution is:

Where

(3.1)

( ) xiAex =

(4.1)

(4.2)

( ) 02

2

2=+ VE

m

02 22

2

=+ Emx

2

2

mE= (4.3)


8/93


1.0 Free Electrons: continued

In equation (3.4), we saw that (x, y, z, t) = (x, y, z). e it

Combining (3.4) with equation (4.2) above,

Since we only consider propagation in +ve x-direction,

From equation (4.3),

For free-flying electron, there is no boundary condition all

values of energy are allowed we have energy continuum

(4.5)

(4.6)

tixi eAex .)( =

2

2

2

mE =


9/93



This yields:

(2.3)

(4.3)

k

pmE

====

222

(4.7)

22

2

m

E

=

m

pEk

2

2

= (1.4)

2

2

mE=

=p

(4.6)

22

2k

mE

= (4.8)


10/93



The term 2/was defined to be the wave number, k same as .

From (4.7) we see that

where p is the momentum, and p = m.v, therefore, proportional to the velocity ofelectrons.

Since momentum & velocity are vectors, kis a vector too.

The vectorkwith components kx, k

yand k

zis:

(4.9)

pk

2k=

Since k is inversely proportional to the wavelength, , it is also called the wave

vector and describes the wave properties of electrons.

K and p are mutually proportional, the proportionality factor is 1/.


11/93


Solutions to the Schrdinger Equation:2.0 Bound Electrons: In a potential Well.

Electrons bound to its atomic nucleus.

Figure 4.2: One-dimensional potential well. Walls are infinitely high potentialbarriers.

Electrons move freely between 2 infinitely high potentialbarriers that do not allow the electrons to escape.

= 0 at x 0 and x a

V

xo a

nucleus


12/93


2.0 Bound Electrons: In a potential Well. continuedWe can begin with the one-dimensional case, but, as the electrons arereflected by the walls, we will have propagation in the +ve and vex-direction.

The Potential Energy inside the well is zero, hence,

Due to the 2 propagating directions, the solution to (4.10) is:

Where,

(4.10)02

22

2

=+

E

m

x

xixi BeAe +=

2

2

mE=

(4.12)

(4.11)


13/93


2.0 Bound Electrons: In a potential Well. continued

The constants A and B are determined by the boundary

conditions mentioned earlier, i.e.,

= 0 at x 0 and x aThis is similar to that of a vibrating string where there is no

vibration at the clamped ends.

Thus, from (4.11), using = 0 at x 0; B = - A

Using = 0 at x a, and (4.13), we have

(4.13)

aiaiaiai eeABeAe =+== 0 (4.14)


14/93



Using the Euler equation

We rewrite equation (4.14) as:

Equation (4.16) is only valid ifsin a = 0, i.e., if

a = n, where, n = 0, 1, 2, 3

Substituting from (4.12) into (4.17) gives:

(4.15)

(4.16)

( )

ii

eei

=

2

1

sin

0sin.2 == aAieeA aiai

(4.17)

2

2

222

2

22n

mam

En

== where, n = 0, 1, 2, 3 (4.18)


15/93



Because of the boundary conditions, only certain solutions of theSchrdinger Equation exist where n is an integer.

The energy assumes only those values determined by (4.18) calledEnergy Levels.

Therefore, when electrons are excited or absorbed, they possesdiscrete values called energy quantization and the lowest energyis called zero-point energy when n=1. This is not the at the

bottom of the well, rather, somewhat higher than the bottom.


16/93



The Wave function and the probability * for finding an

electron within the potential well.

According to (4.11), (4.13) and the Euler equation (4.15) we obtainwithin the well:

= 2Ai.sin x

And the complex conjugate of

* = -2Ai.sin x

The product * = 4A2.sin2 x

(4.19)

(4.20)

(4.21)


17/93



i.e., 2r = n or

Therefore only distinct orbits are allowed, viz-a-viz allowed energylevels discussed earlier.

Consider the wave mechanical properties of a Hydrogen atom.Electron with charge, -e is bound to its nucleus.

Potential,V, in which the electron propagates is taken as the

Coulombic potential:

nr

2=

r

eV

o4

2

=


18/93



Equations (4.19) and (4.21) are plotted in figure 4.4

In figure 4.4(a) the standing electron waves are created between thewalls of the potential well. The integer multiples of half a wavelengthare equal to the length, a, of the potential well.

The probability of finding the electron at a certain place within thewell, * is shown in figure 4.4(b).For n=1, * is largest at the middle of the well while

For n=2, * is largest at 1/4a and 3/4a.


19/93


2.0 Bound Electrons: In a potential Well. continuedWe know that the electrons move in distinct orbits around apositively charged nucleus.

As in figure (4.4)a, the electron waves associated with an orbitingelectron have to be standing waves. Why?

If not, after one orbit the electron wave would be out of phase with

itself. If this continues, then the waves would annihilate itself bydestructive interference.

Therefore there is a radius for the orbiting electron that results in a

continuum of the wave pattern such that the circumference has to bean integral multiple, n, of the wavelength .


20/93


2.0 Bound Electrons: In a potential Well. continuedSolving this we obtain Discrete Energy Levels:

( ) )(1

6.131

42 2220

4

eVnn

meE == (4.18a)

o

n = 1

n =

n = 3

n = 2

(ionization energy)

-13.6 eV

Energy is -1/n2 (and not n2)

Crowding of energy

levels at higher energies

E

Energy at lowest level

that needed to remove

an electron from its nucleus

Energy diagrams

Common in

Spectroscopy

Origin - arbitrary

Ionization

Energies

Counted

Negative


21/93


2.0 Bound Electrons: In a potential Well. continuedFor a 3-Dimensional Well:

Smallest allowed energy in a 3-D potential well occurs when

nx = ny = nz = 1

For the next higher energy 3 possibilities for combinations of

n-values, i.e., (nx ,ny ,nz ) is (1,1,2), (1,2,1), (2,1,1)

States that have the same energy, but, different Quantum Numbers,are called Degenerate States

( )222222

2zyxn nnn

maE ++=

where, n = 0, 1, 2, 3

(4.26)


22/93


3.0 Finite Potential Barrier: Tunnel effectAssume that a free electron, propagating in the +ve x-direction,encounter a potential barrier whose potential energy, V0 (height ofthe barrier) is larger than the total energy of the electron, E.

O X

Fig 4.6: Finite potential barrier

V

V0I

II


23/93


3.0 Finite Potential Barrier: Tunnel effect - continuedWe have 2 Schrdinger Equations for the 2 different areas:

(I) Region I: (x < 0) Electron is assumed to be free.

(II) Region II: (x > 0)

(4.27)0

222

2

=+

E

m

x

( ) 02 0222

=+

VEmx

(4.28)


24/93


3.0 Finite Potential Barrier: Tunnel effect - continuedThe solution to these 2 Schrdinger Equations are:

(I) Region I: (x < 0) Electron is assumed to be free.

Solution

where

(II) Region II: (x > 0)

(4.27)0222

2=+

Emx

( ) 02

022

2

=+

VE

m

x

(4.28)

xixi

I BeAe +=

2

2

mE=

(4.29)

(4.31)Solution

xixi

II DeCe +=

where( )

2 0

2

VEm =

(4.30)

(4.32)


25/93


3.0 Finite Potential Barrier: Tunnel effect - continuedNote of caution:

We stipulated that V0 > E

Therefore, (E-V0) is Negative and becomes imaginary.

To prevent this, we define a new parameter:

= i becomes (4.34)

(4.33)

(4.32)( )

202

EVm =

Rearranging (4.33): i=(4.35)

Inserting (4.35) into (4.31):xx

II DeCe

+= (4.36)


26/93


3.0 Finite Potential Barrier: Tunnel effect - continued

Using boundary conditions, we determine one of the constantsC or D.

For x , from we see that:

(4.37)

implies that and therefore could be infinity.

xx

II DeCe

+=

(4.36)

0.. DCII +=

(4.36)

11 1111 *(4.37)

The probability can never be >1 and thusis no solution. Therefore C has to go to zero.

1111 * 11

C 0 (4.38)


27/93



x

II

De

= (4.39)

and (4.36) becomes:

This means that the -function exponentially decreases in Region IIAs shown in fig (4.7)below:

The Electron Wave (x, t) is then given by (dashed curve above):

( )kxtix

eDe

=

.


28/93



Equation (4.39) provides the envelope for the electron wave thatpropagates in the finite potential barrier decreasing amplitude.

If the potential barrier is moderately high and sufficiently narrow,the electron wave can continue on the opposite side of the barrier.

This is called tunneling.

Analogous to: Light wave that penetrates to a certain degree into

a material with exponentially decreasing amplitude.


29/93


3.0 Finite Potential Barrier: Tunnel effectComplete solution of the behaviour of an electron wave thatpenetrates a finite potential barrier need extra boundary conditions:

O X

Fig 4.6 Finite potential barrier

V

V0

I II

(1)The functions and are continuous at x = 0.

Thus, at x = 0.

1 11

111


30/93



xxixi DeBeAe =+

When x= 0, we have: A + B = D (4.40)

Combining (4.29), (4.36) and (4.38):

(2) The slopes of the wave functions in Regions I and II are

continuous at x = 0, i.e.,( ) ( )dxddxd // 111

Therefore,xxixi DeeBieAi

= (4.41)

When x = 0, we have: DBiAi = (4.42)

xx

II DeCe

+=(4.29) (4.38)(4.36) C 0

xixi

I BeAe

+=


31/93



(4.43)

Inserting (4.42) into (4.40):

The -functions can be expressed in terms of the constant D

Figure 4.8 Square well with finite potential barriers

+=

i

D

A 12

=

i

D

B 12


32/93


4.0 Electron in a Periodic Field of a Crystal (the Solid State)Atoms in a crystal are arrange periodically except for amorphous orglassy cases. We need to find the potential distribution and thebehaviour of an electron in a crystal.

0 X

Figure 4.9: One-dimensional periodic potential distribution.

V

V0

I IIIIII I

-b a

Potential Wells of length a Region I, separated byPotential barriers of widthb and height V0 Region II

V0 >> electron energy, E

b


33/93


4.0 Electron in a Periodic Field of a Crystal - continuedModel is over simplified. Neglects:

Inner electrons are more strongly bound to the core.

Individual potential from each lattice site overlap.

Figure 4.10: One-dimensional periodic potential distribution for a real crystal.

Too complicated to use for a simple calculation !


34/93


4.0 Electron in a Periodic Field of a Crystal - continuedThe Schrdinger Equations for Regions I and II are:

(I) Region I:

(II) Region II:

(4.44)02

22

2

=+

E

m

x

( ) 02

022

2

=+

VE

m

x (4.45)

(4.46)

Need to solve equations (4.44) & (4.45) simultaneously tough.

(4.47)

For abbreviation we write: Em22 2

=

and ( )EVm

= 022 2

ikxeukik

dx

du

dx

ud

x

+=

22

2

2

2

2

(4.49)Use later


35/93


4.0 Electron in a Periodic Field of a Crystal - continued

Bloch showed that the solution has the form:

( ) ( ) ikxexux .= (4.48)

Bloch Function, where u(x) is a periodic function mimics thePeriodicity of the lattice in the x-direction. Therefore, u(x) is no

longer constant (e.g., amplitude A) but, changes periodically withincreasing x.

Also, u(x) is dependent on the various directions of the crystal lattice.


36/93



Differentiating the Bloch function twice with respect to x gives:

ikxeukikdx

du

dx

ud

x

+=

22

2

2

2

2

Inserting (4.49) into (4.44) & (4.45) and using (4.46) & (4.47):

(4.49)

( ) 02 222

2

=+

uk

dx

duik

x

I (4.50)

( ) 02 222

2

=++

uk

dx

duik

x

II (4.51)

Equations (4.50) & (4.51) that of damped vibration.


37/93


4.0 Electron in a Periodic Field of a Crystal - continuedEquations (4.50) & (4.51) that of damped vibration & solution isa:

(4.55)

II (4.56)

xixiikx BeAeeu

+=I

( )xxikx DeCeeu +=

a Differential equation of a damped vibration for spatial periodicity:

02

2

=++

Cu

dx

duD

x

(4.52)

Solution is:xixixD BeAeeu

+=)2/(

where

4

2D

C=

(4.54)

(4.53)


38/93



The four constants, A, B, C and D need to eliminate usingboundary conditions:

The functions and d/dx pass from Region I to II continuouslyat x=0. Here, Equation I = Equation II and thus:

A + B = C + D

Also, at x=0, du/dx for Equation I = du/dx for Equation II, thus:

A(i-ik) + B(-i-ik) = C(--ik) + D(-ik)

At the distance (a+b), the function and therefore u, is continuous.

(4.57)

(4.58)


39/93



At the distance (a + b), the function and therefore u, is continuous.

Thus Equation I at x = 0 must equal Equation II at x = (a + b); or,Equation I at x = a must equal Equation II at x = - b (Fig 4.9).

(4.57)

(4.60)

( ) ( ) ( ) ( )bikbikaikiaiki

DeCeBeAe +

+=+

Finally, (du/dx) is periodic in a + b

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )bikbikkiakia eikDeikCekBiekAi ++ ++=+


40/93



The constants can be obtained by solving equations (4.57)through (4.60).

Inserting these constants into equations (4.55) and (4.56) provides u,which in turn gives solutions for the function not what we really want.

We are looking for a condition that tells us where solutions to theSchrdinger Equations (4.44) and (4.45) exist.

We use equations (4.57) through (4.60) and eliminate the fourconstants A-D and using Eulers equations, we finally have:

( ) ( ) ( ) ( ) ( )bakabab +=+

coscoscoshsin.sinh

2

22

(4.61)


41/93



For simplification, let us imagine that b is very small and V0is very large. Also, the product V0b (Potential BarrierStrength), i.e., area of potential barrier is finite. That is, if V

0grows, b diminishes.

If V0 is very large, then E in (4.7) can be considered small

compared to V0 and can be neglected so that:

( ) ( ) ( ) ( ) ( )bakabab +=+

coscoscoshsin.sinh2

22

(4.61)

02

2V

m

=

(4.62)

S i d ti & M ti M t i l


42/93



Multiplying both sides by b gives:

( )bVbmb 022

=(4.63)

Since V0b is finite and b0, we have b becoming very small.

For small b, cosh(b) 1 and sinh(b) b

Neglecting 2 compared to 2 and b compared to a(equations 4.46, 4.47 & 4.49), equation (4.61) becomes:

(4.64)

kaaabVm

coscossin02

=+

(4.65)



43/93



(4.65)kaaabVm

coscossin02 =+

Substituting2

0

bmaVP=

(4.67)

(4.66)

We finally get:kaaa

a

P coscos

sin=+

This is the desired solution to the Schrdinger equations:

Only certain values of are possible and thus, only certain values of

energy E are possible.

02 22

2

=+ Emx

(4.45)(4.44) ( ) 02 0222

=+ VEmx



44/93


4.0 Electron in a Periodic Field of a Crystal - continuedPlotting the function [P(sin a/ a)+cos a] versus a for P=(3/2)we get the following:

Figure 4.11: Function [P(sin a/ a)+cos a] versus a for P =(3/2)RHS of (4.67) allows only certain values of this function because cos kais only defined between +1 and -1. Allowed values of the function

[P(sin a/ a)+cos a] are marked by heavy lines on the a axis.



45/93


4.0 Electron in a Periodic Field of a Crystal - continuedBecause a is a function of the energy, we see that an electronthat moves in a periodically varying potential field can onlyoccupy certain allowed energy zones. Energies outside of thesezones or bands are not allowed. With increasing values ofa(i.e. with increasing energy), the forbidden bands becomenarrower.

The size of the allowed & forbidden bands varies with P.

We have 4 special cases:

(a) If the potential barrier strength V0b is large, then accordingto (4.66), P is also large & the curve in figure 4.11 becomes steep the allowed bands are narrow.

Semi conducting & Magnetic Materials


46/93


4.0 Electron in a Periodic Field of a Crystal - continued(b) If the potential barrier strength V0b is small, then accordingto (4.66), P is also small & the allowed bands are wider.

Figure 4.12: Function [P(sin a/ a)+cos a] versus a for P = /10



47/93



(c) If the potential barrier strength V0b becomes smaller &

smaller, then P 0, then from (4.67),cos a = cos ka

Or, = k. Combining this with equation (4.46) we have

(4.68)

m

kE

2

22

=

This is the well-known equation (4.8) for free electrons



48/93



(d) If the potential barrier strength V0b is very large, then

P , however the LHS of (4.67) has to stay with the limits 1,

Combining (4.46) & (4.69) gives:

0sin

a

a

i.e., sin a 0. This is only possible ifa = n or,

2

222

a

n = for n = 1, 2, 3. (4.69)

2

2

22

2

n

ma

E

=



49/93


4.0 Electron in a Periodic Field of a Crystal - continuedTo summarize:If the electrons are strongly bound, i.e. potential barrier is verylarge, one obtains sharp energy levels.If the electrons are not bound, one obtains a continuous energyregion (free electrons).If the electron moves in a periodic potential field, one receives

energy bands (solid).

Figure 4.13: Allowed energy levels for (a) bound electrons

(b) free electrons and (c) electrons in a solid.



50/93


The energy levels get wide and

become energy bands transitioninto quasi-continuous energyregions.

Occurs due to interactions of atomsas their separation distancedecreases.

Figure 4.13: Allowed energy levels for (a) bound electrons(b) free electrons and (c) electrons in a solid.




51/93


Energy Bands in Crystals

In equation (4.8) we saw that:

22

2

k

m

E

=

In the case of free electrons, it is simple:

2/1

.Econstkx =

We plot the energy versus momentum (or wave vector k)of the electrons.

Fig 5.1: Electron energy versus wave vector k for free electrons- parabola.

(5.1)



52/93


kaaa

aP coscos

sin=+

Equation (4.67) was:

When P=0, we found in equation (4.68) that cos a = cos ka

Now, since the cosine function is periodic in 2,

equation (4.68) can be written in general form:

cos a = cos kxa = cos (kxa + n2) (5.2)

Where, n = 0, 1, 2, , therefore:

a = kxa + n2 (5.3)

Energy Bands in Crystalscontd.



53/93


We saw in equation (4.7) that: 22

mE=

Putting this value of in (5.3):

2/1

2

22

E

m

ankx =+

(5.4)

In equation (5.4) we see that the parabola of Fig 5.1 is repeatedin intervals of n . 2/a as shown below:

Fig 5.2: Family of free electronparabolas with periodicity of 2/a




54/93


Energy Bands in Crystalscontd.We have seen that when an electron propagates in a periodic potentialwe always observe discontinuities of the energies when coskxa has amaximum or a minimum, i.e. when coskxa = 1

This happens when, kxa = n., n = 1, 2, (5.5)

Or,

a

nkx

.=(5.6)

At these points, a deviation from the parabolic E vs kx curveoccurs & the branches of the individual parabolas merge intothe neighboring ones.

Leads to an important point:

The electrons in a crystal, behave, for most kx values, like free electrons,except when k

x n. / a



55/93



Fig 5.2: Family of free electron

parabolas with periodicity of 2/a

Fig 5.3: Periodic zone scheme.

Allowed bands

Band gap



56/93

Semi conducting & Magnetic MaterialsEnergy Bands in Crystalscontd.

Besides the periodic zone scheme, there a re 2 others.

One is the reduced zone scheme, which is common

it is the section of Fig 5.3, between the limits / a

Fig 5.4: Reduced zone scheme. Fig 5.5: Extended zone scheme.

Allowed bands



57/93

g gEnergy Bands in Crystalscontd.

Important question:

What do the E versus |k| curves mean?

They relate the energy of an electron with its k-vector, i.e. its momentum

Note Allowed Bands (Valence & Conduction bands) & Band Gaps.

The wave vectorkis inversely proportional to the wavelength

of the electrons.

Thus, khas the units of reciprocal length.



58/93


Brillouin Zones



59/93


As you move them closer, they start to pile up on each other-height increases.



60/93


Electrical conductivity of a material represents how easily chargeswill flow through the material.Materials with high conductivity are called conductors.Materials that do not readily conduct electricity are called insulators.

Semiconductors form a third category of material with conductivities

somewhere between conductors and insulators, but that is not exactlythe case. Semiconductors, despite the name, form a subgroup ofinsulators and have properties that differ greatly from the propertiesof conductors.

Pure crystalline silicon, in fact, is a rather poor conductor.

To understand how the term semiconductor arose, we return to theconcepts of electron states and energy bands.



61/93

Energy Bands in Crystalscontd.Electric current is generally due to the motion of valence electrons.An electron can move through a material only by moving from oneallowed energy state to another. But most materials are formed by

bonds that completely fill a valence band, as shown below.Electrons in this filled valence band have no empty states to moveinto, unless they somehow gain enough energy to jump across theforbidden band gap into the empty conduction band above.

Conduction is therefore very difficult. As you might imagine,this energy band diagram represents an insulator.

Conduction Band

Band Gap

Valence Band

Electrons in an insulator fill all

available states in the valence

band. Electrons must jump tothe next higher band before they

can move freely. This band

where electron motion occurs is

called the "conduction band"



62/93


Metals, insulators and semiconductors

Once we know the band structure of a given material we still need to

find out which energy levels are actually occupied and whetherspecific bands are empty, partially filled or completely filled.

Empty bands do not contain electrons and therefore are not expectedto contribute to the electrical conductivity of the material.

Partially filled bands do contain electrons as well as unoccupied

energy levels which have a slightly higher energy. These unoccupiedenergy levels enable carriers to gain energy when moving in anapplied electric field. Electrons in a partially filled band therefore docontribute to the electrical conductivity of the material.



63/93



Completely filled bands do contain plenty of electrons but do not

contribute to the conductivity of the material. This is due to the factthat the electrons can not gain energy since all energy levels arealready filled.

In order to find the filled and empty bands we must find out howmany electrons can be fit in one band and how many electrons areavailable: Since one band is due to one ore more atomic energy levels

we can conclude that the minimum number of states in a band equalstwice the number of atoms in the material. The reason for the factorof two is that even a single energy level can contain two electronswith opposite spin.



64/93



To further simplify the analysis we assume that only the valence

electrons (the electrons in the outer shell) are of interest, while thecore electrons are assumed to be tightly bound to the atom and arenot allowed to wander around in the material.Four different possible scenarios are shown in the figure below:



65/93



(a)

Half filled bandOne valence electron

High conductivity

(b)

Filled band overlappingEmpty band.

Two valence electron

Good conductivity

(d)

Full band separatesEmpty band.

No conductivity

Insulator

(c)

Almost full band

separates

Almost Empty

band.

Semiconductor



66/93

Energy Bands in Crystalscontd.Energy bands of semiconductors

Ev is valence band edge,Ec is the conduction band edge.Evacuum is the energy of the free electron =q.

q is the electronic charge

is the electron affinity

Semi-conducting & Magnetic Materialsd i C l d


67/93


Shown is theE-kdiagram for Silicon within the firstBrillouin zone and along the (100) direction. Theenergy is chosen to be to zero at the edge of the

valence band.

Conduction Band

Valence Band

3.2 eV

1.12 eV

Semi-conducting & Magnetic MaterialsE B d i C l d


68/93


Effect of Temperature on Semiconductors

At absolute Zero:

All electrons are tightly held.Inner orbit electrons are bound.Valence electrons covalent bonds strong no free electrons.

Behaves like insulators.

Above absolute Zero:

Covalent bonds break free electrons semiconductor if potential applied.

Generation of holes missing electrons.

Semi-conducting & Magnetic MaterialsE B d i C t l td


69/93


Intrinsic Semiconductors

Above absolute Zero: Generation of electron-hole pairs.

Extrinsic Semiconductors

Doping with impurities!

N-type and p-type

One impurity atomper 108 atoms of semiconductor

N-type

Si add Group V - electrons

P-type

Si add Group III deficit of electrons or holes

Majority and minority carriers Charge neutrality

Semi-conducting & Magnetic MaterialsE B d i C t l td


70/93


Fermi Energy-defined as the highest energy the electrons attain at T=0 K.

Probability that a certain energy level is occupied by electronsis given by the Fermi Function, f(E):

TkEE BfeEf /)(

11)(

+=

If an energy level is completely occupied by electrons, f(E)=1For an empty energy level f(E)=0When E=Ef, exponential is always 1 and f(E)=1/2Fermi Energy is that energy for which f(Ef)=1/2

Semi-conducting & Magnetic MaterialsDirect & Indirect semiconductors.


71/93

Minimum energy of the conduction band in Indirect semiconductors is at a different

momentum than that of the maximum energy of the valence band.

Electrons in the conduction band rapidly relax to the minimum band energy andlikewise, holes rapidly rise to the maximum energy of the valence band.

Therefore, electrons and holes do not have the same momentum in an indirect one.

In direct semiconductors these momenta are equal.

Semi-conducting & Magnetic MaterialsDirect & Indirect semiconductors


72/93

Direct & Indirect semiconductors.

Indirect semiconductors three-body interaction.

Such interaction is 1000 times less likely than simple electron-photon interactions.

Electrons & holes of different momenta do not recombine rapidly (of the order ofmicro-seconds) while in direct semiconductors it is nano-seconds.

Important effect for light-emitting devices direct semiconductors

Phonons = Heat



73/93

Energy Bands in Crystalscontd impurity ionization levels.

Conduction Band

Valence Band

Donor Level

Acceptor Level

~0.01eV

EgEf

Impurity ionization levels

Distance between the Donor

level & Conduction band =

energy needed to transfer the

extra electrons.

Distance between the Acceptor

level & Valence band =

(implied) energy needed to

transfer the extra holes.



74/93

Zinc-blende Crystal structure



75/93

Energy Bands in Crystalscontd impurity ionization levels.

~0.01eV

Semi-conducting & Magnetic MaterialsEnergy Bands in Crystalscontd.


76/93


Lattice constant ()



77/93




78/93

gy y

Table 1. Chemical Composition of Commercial LEDs.

Color Wavelength (nm) Composition

Blue 470 In0.06Ga0.94N

Green 556 GaPYellow 578 GaP0.85As0.15Orange 635 GaP0.65As0.35Red 660 GaP0.40As0.60

or Al0.25Ga0.75AsInfrared >700 GaAs



79/93

gy y

pn junction

Depletion layer electron and

hole combination nearjunction. depletion of charge carriers. acts as a barrier for further

movement of free electrons. Potential difference acrossthe depletion layer called Barrier Potential (V

0

). depends on type of sc,amount of doping, temperatureSi: 0.7V, Ge: 0.3V

Potential distribution curve

Semi-conducting & Magnetic MaterialsBiasing a pn junction


80/93

p n

No External

Field

External

Field

+ - - +External

FieldNo External

Field

np

Forward Bias Reverse Bias

Applied potential against barrier

Potential barrier is reduced eliminated

Junction offers low resistance

Current flows magnitude depends on voltage

Potential barrier is increased

Junction offers high resistance

- (reverse resistance, Rr)

No Current flow

g p j


Current flow in a Forward Biased pn junction


81/93


Forward Bias - Negative terminal of battery is connected to n-type repels the free electrons in n-type to the junction-leaving behind positively charged atoms-More electrons arrive from the -ve battery terminal & enter the n-region.

-As the free electrons reach the junction, they become-valence electrons. how & why? (Hole is in the covalent bond free electron combines with a hole, it becomes a valence electron)




82/93

p j

-As valence electrons they moves through the holes in the p-region.-Effectively, they move to the left in the p-region same as holes moving

to the right.-Valence electrons move to the extreme left of the crystal

-flow into the +ve terminal of the battery.-Therefore, in n-type region, current is carried by free electrons-In p-type region, current is carried by holes.-In the external connecting wires, current is carried by free electrons



83/93



84/93

(1)Zero external voltage.Circuit is open at KPotential barrier prevents current flow.Circuit current is zero, shown by point O.



85/93

(2)Forward Bias

N-type to ve terminal, p-type to +ve terminalPotential barrier -reduced - then eliminated (0.7 V for Si, 0.3 V for Ge)- current starts flowing.As voltage is increased, current initially increases slowly (overcome

potential barrier) and then rapidly.



86/93

(3)Reverse Bias

P-type to ve terminal, n-type to +ve terminalPotential barrier at junction -increased- Junction resistance becomes very high practically no current flows.However, in practice, a very small current flows called reverse

saturation (why?) current (Is) due to minority carriers.



87/93

(3)Reverse BiasMinority carriers few free electrons in p-type & few holes in n-type.

Applied reverse voltage appears to the minority carriers as forward bias!As reverse voltage is increased, the KE of electrons (minority carriers) high knock out electrons from semiconductor atoms Breakdownoccurs.

Semi-conducting & Magnetic MaterialsBreakdown Voltage.


88/93

Minimum reverse voltage at which pn junction breaks downwith sudden rise in reverse current

Electron-hole pairs in depletion layer minority carriers.

Reverse bias electrons move towards +ve terminal.At large reverse voltage, these electrons have high enough velocity todislodge valence electrons from semiconductor atoms

Avalanche effect

Semi-conducting & Magnetic MaterialsKnee Voltage.


89/93

Forward voltage at which current through the junction increases rapidly

Forward bias overcome potential barrier.

To get useful current through a pn junction,the applied Voltage > Knee Voltage.


Zener Diode.


90/93

Breakdown Voltage = Zener Voltage

Sudden increase in current is called zener current.

Breakdown Voltage depends on the amount of doping.

Why?

If pn-junction is heavily doped, the depletion layer will be thin.

Lightly doped pn-junction will have higher breakdown voltage.

With proper doping, a pn-junction diode with a sharp breakdown voltageis called a zener diode.

A zener diode is always reverse biased.

Semi-conducting & Magnetic Materialspn junction


91/93

Limitation in operating conditions of pn junction

Maximum forward current

exceeding this can destroy due to overheating.

Peak Inverse Voltage (PIV) maximum reverse voltagewithout damaging the pn junction important for rectifier circuit.

Maximum power rating.

A pn junction is known as a Semiconductor or Crystal Diode.


A pn junction is a Semiconductor Diode.


92/93

Alternating Current to Direct Current.

Maximum forward current exceeding this can destroy due to overheating.Peak Inverse Voltage (PIV) maximum reverse voltage

during ve half cycle important for rectifier circuit.Reverse current or Leakage current current flow during reverse bias due to minority carriers


Light Emitting Diodes forward biasing of pn-junctionInteraction of Photons with Matter or semiconductors


93/93

Interaction of Photons with Matter, or semiconductors

Absorption of light Emission of light

Electrons from n-type crosspnjunction recombine with holes in p-type.Free electrons in conduction band higher energy than holes in valence bandRecombining electrons release energy in the form of heat & light.With GaAs intense light.

semi-conducting magnetic materials-week 2-jan 16-2012

Documents