1- SEMICONDUCTOR BAND STRUCTURE



1- SEMICONDUCTOR CRYSTAL STRUCTURE

1.1 INTRODUCTION

While studying semiconductors we are interested in behaviour of a large number of electrons moving through positively charged ions.

To appreciate the complexity of the problem, we can examine the number density of electrons involved.

"Free" valance electrons

Electrons in a solid

valance electrons (-eZc)

core electrons (-e(Za - Zc))

nucleus with charge (eZa)

where Za = atomic number

Fig: 1.1 Electrons in a solid

Consider now an element with density, ( , having a unit volume.

Then mass = ( , . Let the atomic mass be A , and use Avagadro's Number (6.022 1023 atoms per mole),

Then the number of moles = ( /A, and

n = (6.022 1023 atoms per mole)* (( /A)*Zc ---------------1.1

where,

Zc = number of electrons in the outermost shell, and

n = electron density for conduction electrons.

For most metals n ( 1023 electrons.

1.2 BASIC LATTICE TYPES

Crystal is formed by identical building blocks that consist of an atom or a group of atoms.

Natural crystals have a fixed symmetry, BUT artificial crystals are produced with a modified crystalline structure. These lead to "superlattices". Altering crystal structure also changes the electronic properties.

The building block which is repeated infinitely to produce the crystal may be :

very simple (single atom for many metals, such as copper, gold, aluminium etc.) OR

complex ( as in some proteins)

In superlattices the number of the building block can be arbitrary, but in natural crystals it is mainly two.

Now we shall have some definitions:

Lattice is a set of points in space which form a periodic structure. It is a mathematical abstraction.

Basis is a building block of atoms attached to each lattice point.

Lattice point (R', can be obtained from any other lattice point (R by a translation:

(R' = (R +m1 (a1 + m2 (a2 + m3 (a3 ----------------1.2

where (a1, (a2,( a3 are vectors and m1, m2, m3, are integers. Such a lattice is called Bravais Lattice and the lattice structure together with the basis form the Crystal Structure, as indicated below in Fig. 1.2.

* * * * * *

* * * * * *

a- Space lattice

0o

b- Basis with two different ions

0o 0o 0o 0o 0o 0o

0o 0o 0o 0o 0o 0o

c- Crystal structure

Fig. 1.2 Crystal structure

The translation vectors, (a1, (a2,( a3 , are called primitive if the volume of the cell formed is smallest possible.

One choice would be to take,

(a1 as the shortest period of the lattice,

(a2 as the shortest period of the lattice, and not parallel to (a1

( a3 as the shortest period of the lattice, and not coplanar with( a1 and (a2

The basic building block of the crystal is called the unit cell.

The lattice separation is very small, so in practical terms one deals with almost infinite lattices. Boundaries, interfaces play an important role, as will be seen later.

Basic lattice types: Lattice types are important in the sense that the structure of lattice also describes the symmetry which effects the electronic properties. We will only consider cubic lattices, which is structure taken by all semiconductors. Basically there are THREE types:

Simple cubic

Body centred cubic ( abbreviated to bcc)

Face centred cubic ( abbreviated to fcc)

Simple cubic lattice

This is generated by the primitive vectors

a (x, a (y, a(z --------------------------------------1.3

where (x, (y, (z are unit vectors, and a is the cube side.

Fig. 1.3 Simple cubic structure.

Body centred cubic ( abbreviated to bcc)

This is generated from the simple cubic structure by placing an extra atom at the Centre of the cube

Fig. 1.4. Body centred cubic structure

Face centred cubic ( abbreviated to fcc)

This is generated from the simple cubic structure by adding atoms in the Centre of each square face.

Fig. 1.5 Face centred cubic structure.

"fcc" is the most important lattice for semiconductors. The primitive vector set is given by:

a1 = a/2 ((y + (z ), a2 = a/2 ((z + (x ), a3 = a/2 ((x + (y )------1.4

where a is the cube edge and is called lattice constant of the semiconductor.

The diamond and zinc-blende structures:

Essentially all semiconductors of interest for electronics and optoelectronics have "fcc" structure with two atoms per basis.

The two basis atoms have coordinates relative to each other as, (0,0,0), and (a/4, a/4, a/4) where "a" is the lattice constant.

Fig 1.6 Two basis atoms of a diamond structure, and two diamond structures interpenetrating.

Each atom lies on its own "fcc" , and the structure may be thought of two interpenetrating "fcc" structures, where one is displaced by (a/4, a/4, a/4) along the body diagonal of a cube with sides "a".

Now, if the two atoms of the basis are identical the structure is called DIAMOND (eg. as in Si, Ge, C, etc.)

But, if the two atoms are different, it is called ZINC BLENDE (GaAs, CdS, AlAs, etc.)

Usually, semiconductors with a diamond structure is called ELEMENTAL semiconductors, whereas those having Zinc Blende structure are called COMPOUND semiconductors. As an example, in a compound semiconductor of GaAs there are 4 Ga and 4 As atoms in each cube of volume a3.

Example 1.1

Lattice constant of silicon is 5.43 A(. Calculate the number of silicone atoms in a cubic centimetre. Also calculate the number density of Ga atoms in GaAs which has a lattice constant of 5.65 A(.

Solution

Since we have silicone we should already know that we have a n "fcc" diamond structure where we also have two identical atoms per basis.

"fcc" gives ( 8 lattice points at the cube edges

"fcc" gives ( 6 points at the face centres

Lattice points at the edges are shared by 8 other cubes, while the faces are shared by 6 other cubes, as indicated below.

Hence the number of point per cube of volume a3 is:

N(a3) = 8*(1/8) + 6*(1/2) = 1 + 3 = 4 points

In silicone there are two atoms per lattice point (basis), hence the number density is then, (Angstrom = 10-10 meters, 10-8 cm):

N Si = (4*2 ) / a3 = 8 / a3 = 8 /(5.43 10-8 cm)3 = 4.997 10 22 atoms/cm3.

In GaAs there are only one Ga atom and only one As atom at each lattice point. Hence the number density of Ga atoms is given by:

N Ga = 4 / a3 = 4 / (5.65 10-8 cm)3 = 2.22 1022 atoms / cm3

Also there are equal number of As atoms.

Example 1. 2

A Si device on a VLSI chip represents one of the smallest devices and GaAs laser one of the largest. If Si device has dimensions of (5*2*1)((m)3 and GaAs has (200*10*5)( (m)3 , calculate number of atoms in each device.

Solution

Using the results obtained in Example 1, for the number of atoms per cm3 , Si (5 1022 atoms / cm3 ) and GaAs (2.22 1022 atoms /cm3 ) .

N Si = (5 1022 atoms / cm3 ) * 10 (10-6 102 cm)3 = 5 1011 atoms.

N Ga = (2.22 1022 atoms /cm3 ) * 104 (10-6 102 cm)3 = 2.22 1014 atoms.

Also,

N As = (2.22 1022 atoms /cm3 ) * 104 (10-6 102 cm)3 = 2.22 1014 atoms.

1.3 MILLER INDICES

Miller Indices are used to simplify description of lattice planes.

There is a simple procedure to follow when a drown plane is to be indexed. This is:

Define the x, y, z, axes.

Take the intercepts of the plane along the axes in units of lattice constant.

Take reciprocals of the intercepts and reduce them to smallest integers.

A family of parallel planes are indicated by a notation ( h k l ( .

Fig. 1.7 Miller Indices for various planes

Example 1.3

Calculate the surface density of Ga atoms on a Ga terminated (001) GaAs surface.

Solution

The atoms at the top of (001) surface are either Ga or As. If it is Ga terminated then the atom at the top are Ga. The (001) plane has an area of a 2 , 4 corner atoms and 1 Centre atom. Each atom of the corner atoms are shared by 4 similar planes, so that the contribution is 1/4, but the Centre atom is contributed wholly.

Hence the number of atoms per a2 = 4*(1/4) + 1 = 2 atoms

The surface density is then, S Ga = 2 / a2 = 2 / (5.65 10-8 cm)2 = 6.26 1014 atoms cm-2.

Example 1. 4

Plot the following planes for "bcc" and "fcc" structures.

(010),(110),(200),(123).

2- QUANTUM MECHANICS AND STATISTICAL PHYSICS OF ELECTRONS

2.1 INTRODUCTION

The physical world can be viewed in two ways using:

Classical physics

Quantum mechanics

The interaction of experimental observations and theoretical descriptions in 20th century defied the classical explanations. Experimental observations such as ;

black-body radiation, with

Energy (E) = hv -------------------------------------------2.1

where "v" is the frequency and "h" is the Planck's constant;

photoelectric effect with,

Energy (E) = h' w -----------------------------------------2.2

where h' = h / 2(, and w = 2(v;

and atomic spectra are all explained by the quantum mechanics approach.

The quantum mechanics approach can be described by either,

Heisenberg uncertainty relation between momentum "p" and position "x", and between energy "E" and time "t". That is

(p*(x ( h' /2 and (E*(t ( h' /2

Schrodinger Equation ( This is what we will choose to study).

2.2 SCHRODINGER EQUATION

The quantities that are physically seen are represented by mathematical operators in Schrodinger approach. Some important operators are represented by following operations.

Momentum : px ( -i h' (( / (x)

py ( -i h' (( / (y) ---------------------------------2.3

Energy E ( i h' (( / (t)

Now consider kinetic energy (KE) and potential energy (PE) of an electron with mass mo. Then,

KE + PE = E or,

p2 /2mo + U(r,t) = E -----------------------------------------------------2.4

where U(r,t) is the potential in which the electron moves. If we represent observables "p" and "E" by their operators, and operate on a general state ((r,t), we get Schrodinger equation;

(-h'2/2mo) (2 ((r,t) + U(r,t) ((r,t) = i h' (( / (t) ((r,t) = E ((r,t) --2.5

where "t" is the time and "r" all the spatial variables (x,y,z), and ((r,t) is called the wave function. The wave function has the meaning that;

(((r,t)(2 dr ------------------------------------------------------------2.6

gives the probability that the electron can be found at time "t" in the volume element "dr".

As an example take a hypothetical (not true) case and plot (((x)(2

in one dimension as in Fig. 2.1 and assume that it is time independent.

Fig. 2.1 The electron is always between "xo" and "x4" , (i.e. probability is zero for xo ( x ( x4 ).

Since, in Fig. 2.1, the electron must be somewhere, the probability of finding it between xo" and "x4" must be unity. That is;

[pic](((x)(2 dx = 1 -------------------------------------------------------2.7

This equation only helps to illustrate meaning of (((x)(2 .

If the potential energy term U(r,t) has no time dependence, then we can rewrite Eq. 2.5 with i) time dependent and ii) time independent parts.

We may also rewrite the wave function with the time dependent and time independent components as;

((r,t) = ((r) f(t) -----------------------------------------------------2.8

If we substitute Eq. 2.8 into Eq. 2.5, divide by "((r) f(t)" and equate each side to "E" , we get ;

one time independent equation as,

(-h'2/2mo) (2 ((r) + U(r) ((r) = E ((r) -----------------------------2.9

one time dependent equation as;

i h' (( / (t) f(t) = E f(t) -------------------------------------------------2.10

The solutions for the two cases are;

i) Time dependent case;

The time dependent part of the solution has a general form,

f(t) = exp((iEt / h')[pic] --------------------------------------------2.11a[pic]

and if we set E = h' ( then we get;

f(t) = exp((i(t )[pic] ----------------------------------------------2.11b[pic]

where "E" is the energy of the electron that has to be solved for by solving Eq. 2.9.

(In Eq.2.9 the unknowns are both "((r)" and "E". In general, there will be a series of solutions "((r)" with associated energies En.

It is clear that the associated wave functions are complex, and the physical interpretation according to which the probability to find the electron in a region of space Vo is given by;

P(Vo) = [pic]d3r ((*(r )((r )( ------------------------------------------2.12

The allowed wave functions may be normalised to a particular volume "V" (ie. unit volume) signifying that there is one electron in the volume "V". (

ii) time independent case;

The time independent case can be considered under two different conditions;

condition 1, when there is no potential, U(r) = 0

condition 2, when the potential changes uniformly in space

Condition 1

This is the case when we set U(r) = 0, in Eq. 2.9.

This concept deals with quantum mechanics of "free" electrons moving in absence of any potential.

So the Schrodinger (time independent ) equation for free electrons is ; ( Eq. 2.9 with no potential energy term),

(-h'2/2mo) ((2 / (x2 + (2 / (y2 + (2 / (z2 )((r) = E ((r) ------2.14

with a general solution;

((r) = (1 / (V) exp((ik.r) ------------------------------------------2.15

and the corresponding energy (only the kinetic energy since we already had no potential energy) is given by;

E = (h'2 k2 )/2mo -----------------------------------------------------2.16

where "k" is a wave vector and the factor, "(1 / (V)" is used such that "V" is a volume having one electron and obeys the relation,

[pic]d3r ( ((r) (2 = 1 ---------------------------------------------------2.17

The energy of the electron is,

E = (h'2 k2 )/2mo ( p2 /2mo --------------------------------------------2.18

and the momentum is;

-ih' (( / (r ) ( h' k -------------------------------------------------------2.19

while the velocity is;

v = h' k / mo --------------------------------------------------------------2.20

Assume now that the volume "V" is a cube of sides "L". To correlate with physical conditions, two boundary conditions are imposed on the wave function.

Firstly, for electrons confined in a finite volume; the wave function goes to zero at the boundaries of the volume. The wave solutions are of the form sin(kxx) or cos(kxx) and k-values are restricted to the positive values;

kx = ( / L, 2( / L, 3( / L, nx( / L ----------------------------------2.21

where "nx" is an integer, and the allowed values of energy are;

E = (h'2 k2 )/2mo = (h2 / 8m0L2 )(nx2 + ny2 + nz2 ) -----------------2.22

This standing wave solution is used to describe stationary electrons (e.g. electrons in quantum wells) as shown in Fig. 2.2 below.

Secondly, for moving electrons; the boundary conditions is known as periodic boundary conditions. The wave can be spread in all space (although finite in volume, V,) made up of cubes of side "L".

Then,

((x, y, z+L)= ((x,y,z)

((x, y+L, z) = ((x,y,z) ------------------------------------------------2.23

((x+L, y, z) = ((x,y,z)

Fig.2.2 a) Standing waves(stationary boundary).

b) Exponential wave function with electron probability

equal in all regions (periodic boundary).

Because of boundary conditions, the allowed values of "k" are;

kx = 2( nx / L, ky = 2( ny / L, kz = 2( nz / L, -------------------2.24

where "n" are integers, positive or negative.

If "L" is large, spacing between "k" values is very small.

Now consider volume in k-space that each electron occupies. As shown in Fig. 2.3 this volume in 3-dimensions is;

(2( / L)3 = 8(3 / V -----------------------------------------------------2.25

where the volume in real space V= L3

Fig.2.3 k-space , the allowed values are separated by 2( / L.

Now, if "(" is any volume in k-space, the number of electronic states in this volume are;

(() / (8(3 / V) = (V / 8(3 --------------------------------------------2.26

Condition 2

This is the case when the electrons are in a periodic potential as in a crystal lattice.

Normally, the equation of motion for "free" electrons may be written as;

mo dv / dt = Fint + Fext ---------------------------------------------2.27

where " Fint " and " Fext " are the internal and external forces respectively.

The internal forces, " Fint " may be very complex. However, the periodic nature of the crystal potential allow an equation like, "Eq. 2.27" to be written except that the effect of the internal forces are accounted for by using "m*" (effective mass) rather than the free electron mass "mO".

Hence; dpeff / dt = Fext ----------------------------------- 2.28

The Schrodinger equation for an electron in a semiconductor is;

(-h'2/2mo) (2 ((r) + U(r) ((r) = E ((r) -------------------------2.29

where U(r) is the background potential. Due to crystalline nature of material, U(r) has the same periodicity "R" as the crystal lattice.

Hence; U(r) = U( r + R ) --------------------------------------2.30

and the electron probability is the same in all unit cell.

If the potential was random, this would not be the case as shown in Fig. 2.4.

Fig. 2.4 : a) non-periodic potential. b) periodic potential.

Through the Bloch's Theorem, Fig. 2.4b, shows that the probability (not the wave function) is periodic through the crystal.

2.3 FROM ATOMIC LEVELS TO BANDS

There are different models representing the back-ground potential in the crystal. One of them, the Kronig-Penny model represents the back-ground potential (periodic) seen by electrons in the crystal as a simple potential as shown in Fig. 2.4b above.

Graphical solution of Eq.2.29 to obtain allowed energy levels is shown in Fig. 2.5 below, where f(E) is a function of energy that must lie between the values of 1.0 and -1.0.

Fig. 2.5 a) f(E) against electron energy. b) electron energy, E, against wave vector kx.

A simpler representation of energy states is shown in Fig. 2.6, where the allowed bands are infinitesimally narrow and the forbidden gaps are relatively wide.

Vacuum Energy Level Electrons in free space

Electrons are bound in the material

Crystal Isolated atoms

(allowed bands) (atomic levels)

spacing between atoms----------------------------------------------------(

spacing ( 1-2 A( spacing ( 1000 A(

Fig. 2.6 Allowed and forbidden bands.

If the atoms are brought together to form a crystal, the discrete energy levels broaden. The core electrons are tightly bound and not effected.

Hence, it is the outermost electrons forming the allowed and forbidden bands, and the properties of the semiconductors are essentially determined by these valance electrons.

The wavelengths that are allowed in a crystal is governed by the k-vector such that;

k = 2 ( / ( ------------------------------------------------------------2.31

Hence, the shortest wavelength that is allowed in a crystal depends on spacing between the two adjacent lattice points.

[pic][pic][pic]

3- THE HYDROGEN ATOMAND QUANTUM NUMBERS

3.1 THE HYDROGEN ATOM

Hydrogen atom forms a relatively simple system with one electron and one proton.

Electron can stay in any stable energy level and can move to a higher or a lower energy state by absorbing or radiating a well defined quantum of energy, as shown in Fig 3.1 below.

--------------------------------

electron in orbit at certain

energy level

-a-

Free electrons

Vacuum level

Bound

electrons

(300 ((310) ((320) or 3s (3p) (3d)

13.6 eV

-b- (200 ((210)or 2s (2p)

(100 or 1s

Fig. 3.1 a- Hydrogen atom b- Energy levels.

Mathematical model of Hydrogen atom can be used, as a simplification, while studying the doped semiconductors. Here an electron is bound to an ionised impurity.

The Schrodinger Equation for the electron-proton system is;

((-h'2/2mo (e2 )-( h'2/2mp(h2 )- e2/(4((o(re-rp(2)(((r) = Etot ((r) -3.1

where the first and second terms on the left-hand side of the equation represent the kinetic energy of an electron and a proton respectively, and the third term represent the potential energy of the system (e.g. coulomb energy between electron and proton at points re and rp respectively.)

If we now consider the relative motion of the electron and proton we have;

((-h'2/2mr( 2 ) - e2 / (4((o / r)( ((r) = E((r) ------------------3.2

where r = re – rp , is the relative coordinate and mr is the reduced mass such that;

1 / mr = 1 / mo + 1 / mp ( 1/ mo , since mp (( mo .

The energy values are;

En = -(mo e4 / (4((o )2 h'2)((1 / n2( = 13.6 / n2 eV-------------------3.3

Where “n” is an integer.

The general wavefunction has the form,

(nml ( r,(,(), where “ n,m,l “ are quantum numbers.

The ground state wavefunction has the form;

(100 ( r,(,() = (1 / ((a3 )exp (-r / a )-----------------------------------3.4

where “a” is called Bohr radius, given as;

a = h'2 / mo2 = 0.53 A(------------------------------------------------- 3.5

Some energy levels with corresponding quantum numbers are

shown in Fig 3.1b. For quantum number l = 0, 1, 2, 3,… the

representation used is s, p, d, f,….respectively.

Hence;

(100 is called 1s, and

(210 is called 2p states etc…

3.2 QUANTUM NUMBERS

Symbol “n” Principal quantum number. It determines the

total energy and can have integers, 1,2,3,4, ..(.

Symbol “l” Azimuthal quantum number. It determines the

Orbital momentum and can have integers, 0,1,2,3

4, ……(n-1).

Symbol “m” Magnetic quantum number. It determines

magnetic moments. It can have values in the

range -l,……..0………..,+l

Spectrographic designation of Azimuthal quantum numbers are;

Value of “l” Designation

0. s (sharp)

1. p (principal)

2. d (diffuse)

3. f (fundamental)

n l m Spectrographic designations

1 0 0 1s 2 electrons

---------------------------------------------------------------------

2 0 0 2s 2 electrons

1 -1,0,+1 2p 6 electrons

----------------------------------------------------------------------

3 0 0 3s 2 electrons

1 -1,0,+1 3p 6 electrons

2 -2,-1, 0,+1,+2 3d 10 electrons

----------------------------------------------------------------------

4 0 0 4s 2 electrons

1 -1,0,+1 4p 6 electrons

2 -2,-1, 0,+1,+2 4d 10 electrons

3 -3,-2,-1,0,+1,+2,+3 4f 14 electrons

Subscripts or superscripts on s,p,d,…. Levels ( eg. s2 or s2 means 2 electrons) represent the number of electrons in that level. Hence;

Group 4 Semiconductors

C (Carbon) 1s22s22p2

Si (Silicone) 1s22s22p63s23p2

Ge (Germanium) 1s22s22p63s23p63d104s24p2

Group 3-5 Semiconductors

Ga (Gallium) 1s22s22p63s23p63d104s24p1

As (Arsenic) 1s22s22p63s23p63d104s24p3

Now, as the “p” level have ( -1, 0, +1 ) but all have the same energy, they are called 3-fold degenerate, eg. in a magnetic field they split into 3 levels.

Similarly, “d” levels are 5-fold degenerate.

Example 3.1

Calculate the wavelength associated with a 1 eV a) photon, b) electron, c) neutron. (h = 6.6 10 –34 Js; c = 3 10 8 m/s;

e = 1.6 10 –19 J; mo = 0.91 10 –30 kg; mn = 1824 mo) .

Solution

a) Using the equation; E = hv or E = hc / ( or (ph = hc / E

Hence, (ph = (6.6 10 –34 Js)(3 10 8 m/s) / 1.6 10-19 J = 1.24 (m.

b) From Eq.2.31; we have k = 2( / ( and for an electron the energy is given by;

E = (h'2 k2 )/2mo which gives for “k”, k = ( (2moE) / h'2(1/2 . Since we have, ( = 2( / k, now substitute for “k” and h'= h / 2( to get;

( = ( (h / 2() 2(( / (2moE)1/2 = h / (2moE)1/2 ( = (6.6 10 –34 Js) / (2(0.91 10 –30 kg )( 1.6 10-19 J )(1/2 = 12.3 A(

c) As we see in part b) we have a relationship ( ( 1 / (m)1/2 and we are given that, mn (neutron mass) = 1824 mo , then,

(n /(e = (mo)1/2 / (mn)1/2 = 1 /(1824)1/2 and (n = 12.3 A((1 / 1824)1/2

(n = 0.28 A(

Example 3.2

Calculate the wavelength associated with an electron having 1eV energy, if the electron is inside GaAs crystal where its effective mass is 0.067 mo. ( mo = 0.91 10 –30 kg)

Solution

From the previous example we have , ( ( 1 / (m)1/2, hence, if the effective mass is represented by m* then we can write,

(GaAs / (free = (mo)1/2 / (m*)1/2 which gives for;

(GaAs = 12.3 A( (mo / 0.067 mo )1/2 = 47.5 A(

4- DENSITY OF STATES AND THE DISTRIBUTION FUNCTION

4.1 DENSITY OF STATES

Density of states (DOS), is the number of available states per unit volume per unit energy around an energy “E”. Now, represent

The number of states in a unit volume = N(E) dE

Where N(E) = density of states

dE = energy interval

To find N(E) we need to know “E-k” relation of an electron.

For the free electron case;

E = (h'2 k2 )/2mo --------------------------------------------------4.1

Suppose now that we represent energies E and E + dE as surfaces of spheres with radii k and k + dk as in Fig. 4.1 .

3-Dimensional case 2-Dimensional case 1-Dimensional case

Shaded volume = 4(k2dk Shaded area = 2(kdk Shaded space = 2dk

Fig. 4.1 Representation of E and E+dE in 3-2-and 1 dimensional cases.

From Eq.2.25, we see that k-space volume per electron = (2(/L)3

Now considering the 3-dimensional case only, the number of electronic states between k and k + dk = (4(k2dk) / (2(/L)3

= (k2 dk) V / 2(2 -------------------------------------------------4.2

Hence, N(E) dE = (k2 dk) V / 2(2V = (k2 dk) / 2(2 -------4.3

Here in this equation we will simplify for k and dk, and use equations,

E = (h'2 k2 )/2mo which gives dE / dk = (2h'2 k )/2mo)---4.4

So that,

k = (2moE / h'2)1/2 and k dk = 2modE / 2h'2

Multiplying the last two equations give;

k2 dk = ((2 mo 3/2 E1/2 dE( / h'3 ---------------------------4.5

Hence,

N(E) dE = (k2 dk) / (2(2) = ((2 mo 3/2 E1/2 dE( / h'3 (2(2)

= ( mo 3/2 E1/2 dE( / h'3(2 (2 -------------------4.6

or, (multiplying by two for two spin values of the electron) we have,

N(E) = ((2 mo 3/2 E1/2 ( / h'3 (2 ---------------------------4.7

Hence a plot of N(E) against the energy E will have the shape shown in Fig. 4.2 below.

Fig. 4.2 Density of states against energy(3-D case).

4.2 THE DISTRIBUTION FUNCTIONS

Distribution function f(E) gives the probability that states with energy E are occupied. The three distribution functions (Classical, Fermions, and Bosons ) are summarised below.

• CLASSICAL f(E) = exp –(E – EF )/kBT-----------------4.8

(Maxwell-Boltzmann)

• FERMI-DIRAC f(E) = 1 / (1 + exp (E – EF )/kBT (-------4.9

• BOSE-EINSTEIN f(E) = 1 / (( exp E /kBT) - 1(-------------4.10

Where kB is Boltzmann constant = 1.38 10-23 J/K or 8.62 10-5 eV/K

The Fermi- Dirac expression shows that f(E) is always less than 1, as shown in Fig. 4.3 below.

Fig. 4.3 f(E) against E at various temperatures. At T = 0 K, step-like behaviour. At higher temperatures, function f(E) becomes mainly exponential (Boltzmann distribution).

The EF is the chemical potential or Fermi Energy. It represents the energy where f(E) becomes ½. It can be determined once the density of states N(E) and the electron density n are known. That is;

n = [pic](N(E) dE( (f(E)(

n = [pic](N(E) dE( (1 / (1 + exp (E – EF )/kBT (( ------------4.11

Case When T = 0 K,

we have;

f(E) = 1 / (1 + exp (E – EF )/kBT ( = 1 if E ( EF

or

f(E) = 1 / (1 + exp (E – EF )/kBT ( = 0 if E ( EF ----4.12

Hence for values of energy between 0 and EF , we have;

n = [pic](N(E) dE( --------------------------------------------------4.13

So for the 3-Dimensional system at T = 0 K, using Eq.4.7 for N(E) we get;

n = ((2 mo 3/2 ( / h'3 (2 [pic]E 1/2 dE

n = (2(2 mo 3/2 / 3h'3 (2 ((EF 3/2( -------------------------------4.14

or,

EF =( h'2 / 2mo ) (3 (2 n)2/3 -----------------------------------------4.15

Eq. 4.15 is used for metals such as copper, gold, etc., and not for semiconductors. Fermi energy EF is the highest occupied energy state at T = 0 K,.

Now define corresponding wavevector kF (Fermi vector) and velocity vF (Fermi velocity) for the same case when T = 0 K.

Hence, comparing Eq 4.15 with Eq. 2.16 we have;

kF 2 = (3(2 n )2/3, or kF = (3(2 n )1/3 -----------4.16

Using Eq. 2.23, (v = h' k/ mo) and substituting kF for k, we have;

Fermi -Velocity, vF = h' k/ mo = ( h' / mo) (3(2 n)1/3 ---------4.17

Hence, we now see that the velocity at the highest occupied state is vF at T = 0 K, and not zero as in the case of Boltzmann statistics.

Case When T ( 0 K,

At finite temperatures, n, is not easy to determine. But, if n is small so that f(E) is always small, Fermi Function can be written without the “1” in the denominator, and this leads to Boltzmann Function;

f(E) = exp –(E – EF )/kBT---------------------------------------4.18

In this case when f(E) is small, it is called non-degenerate statistics, and ;

n = [pic](N(E) dE( (f(E)( = Nc exp ( EF )/kBT ------------------4.19

where Nc = effective density of states, and it is given by;

Nc = 2 (mo / 2 ( h'2 (3/2 ( kBT )3/2 -------------------------------4.20

But if n is HIGH so that f(E) approaches unity (the degenerate condition) the Joyce-Dixon approximation is used, eg;

EF = kB T (ln (n / Nc) + (1 /( 8 )(n / Nc)( ------------------------4.21

And,

Etot = = [pic] (N(E) dE( E (f(E)( -----------------------------------4.22

Example 4.1

A particular metal has 10 28 electrons per cubic meter. Calculate the Fermi Energy and Fermi velocity at T = 0 K.

Solution

From Eq. 4.15 we have for the EF ;

EF =( h'2 / 2mo ) (3 (2 n)2/3

EF = (1.05 10 –34 J s(2 ( 3 (2 10 28( 2/3 / 2 ( 0.91 10 –30 kg (

= 2.69 10 –19 J

Also; fro Eq. 4.17 we have ;

vF = h' k/ mo = ( h' / mo) (3(2 n)1/3

= (1.05 10 –34 J s( ( 3 (2 10 28( 1/3 / 0.91 10 –30 kg

= 7.69 107 cm/s

Example 4.2

Schematically Plot the distribution function :

f(E) = 1 / (1 + exp (E - EF ) / kBT (

as a function of energy “E” for temperatures T = 0 K, and T2 (( T1 ((T. Clearly indicate values of important points. At what value of energy does the distribution value equal 50% ?

Solution

5- METALS – SEMICONDUCTORS – INSULATORS

5.1 ENERGY BANDS

Considering the allowed and forbidden energy bands, the allowed bands may be completely or partially filled or completely empty, as shown in Fig. 5.1 below.

VACUUM ENERGY LEVEL

----------------------------------------------------------------------------------

e(

conduction

band

e(

valance

band

core

level

core

level

METAL SEMICONDUCTOR

Highest occupied band is Highest occupied band is

Partially filled. Completely filled.

Fig. 5.1 Electron occupation at 0 degrees Kelvin is described schemattically. e( = work function; e( = electron affinity.

Concept : If an allowed band is completely filled with electrons, there is no conduction of current

COMPLETELY FILLED PARTIALLY FILLED

Hence, a material with a partially filled band is called a METAL; while those with the highest occupied level completely filled are called INSULATORS.

An insulator having a band-gap ( 3 eV is called a SEMICONDUCTOR.

The band which is normally filled at 0 degrees Kelvin in a semiconductor is called VALANCE BAND while the upper unfilled band is called the CONDUCTION BAND.

Metal Workfunction; is the energy difference between the vacuum level and the highest occupied state.

Electron Affinity; is the energy difference between the vacuum level and the bottom of the conduction band.

5.2 DIRECT AND INDIRECT SEMICONDUCTORS

For most semiconductors, top of the valance band occurs at wave vector value k = 0.

A typical structure is shown in Fig.5.2 below.

Direct conduction band indirect conduction band

Direct bandgap indirect bandgap

Valance band

Fig. 5.2 Direct and indirect bandgaps.

If the bottom of the conduction band is at k = 0, we have a Direct bandgap material; ( eg GaAs, InP, InGaAs, etc,) BUT, if k ( 0 we have an indirect bandgap material (eg. Si, Ge, AlAs, etc.). These indirect bandgap materials involve a momentum change (k ( 0) for transitions between the valance and the conduction bands and therefore are less suitable for use in optical devices.

In most cases for a direct bandgap semiconductors;

E(k) = Ec + (h'2 k2 )/2m* -------------------------------------------5.1

Where “Ec “ is the conduction band edge, and the band structure is a simple parabola.

For indirect bandgap materials;

E(k) = Ec + (h'2 kl2 )/2ml* + (h'2 kt2 )/2mt* --------------------5.2

Where kl = longitudinal part of “k”

And kt = transverse part of “k” and the constant energy surface of the band structure is an ellipsoid.

Near the band edges, electrons in semiconductors have effective mass represented by “m*” which is directly proportional to the materials energy gap Eg.

Fig.5.3 below shows how the effective mass normalised to the free mass (m* / m0) changes with Eg .

Fig. 5.3 Effective mass against energy gap Eg.

Masses of valance band electrons are usually heavier and also negative. In general effective mass is defined as;

1 / m* = (1 / h'2)* (d2E / dk2) ----------------------------------------5.3

5.3 HOLES IN SEMICONDUCTORS

Considering a semiconductor, at finite temperatures, some electrons leave the valance band and go into the conduction band, as shown in Fig. 5.4 below.

Conduction band

Wavevector associated with

the missing electron.

Electron removed Valance band

Missing electron

Means a hole

Fig. 5.4 Holes in semiconductor.

As the electron moves to conduction band then the missing state can be represented with a hole having wavevector –ke = kh .

The hole responds as if it has a positive charge. Holes and electrons move in opposite directions as shown in Fig. 5.5 below. So, when discussing conduction band one is dealing with electrons; whereas for valance band holes have the major role.

Fig. 5.5 Holes in valance band; when a) t = 0 ; b) t = t1 ; c) t = t2 and t2 ( t1 ( t , d) motion of electrons and holes under an electric field of “F”.

6- DOPING OF SEMICONDUCTORS

6.1 INTRINSIC CARRIER CONCENTRATION

We already know that when T = 0 Kelvin, the valance band is completely filled, while the conduction band is completely empty. This means that there is a high resistance to current transport.

When T( 0 Kelvin however, there will be some electrons that will go into the conduction band , leaving behind holes in the valance band. These carriers are called “intrinsic carriers”. They are nuisance in semiconductor devices, as there is no easy and practical control of their populations. We shall later deal with controllable carriers introduced by a process called ‘Doping”.

The intrinsic carrier concentration depends on mainly;

a- bandgap, Eg

b- Temperature in Kelvin, T

c- Band edge masses, m*.

Fig. 6.1 below shows the conduction and valance band density of states.

Fig. 6.1 Density of states, Ne, Nh, and electron and hole concentrations.

The electron concentration “n” is given (using Boltzmann approximation, for low “n” ( 1016 cm-3 ) by; ( using Eq. 4.19 with limits from Ec to ( , and not from 0 to ( as in Eq. 4.19.)

n = Nc exp ( (EF – EC )/ kB T (

where Nc = 2 ((m*e kB T)/(2 ( h'2 )(3/2 ------------------------------6.1

and represent the effective density of states at the conduction band edge.

To find hole concentration, p, we use hole distribution function;

fh = 1 – fe = 1 – 1 / ( exp ( (E – EF )/ kB T (( + 1

( exp - ( ( EF - E)/ kB T ( --------------------------------------------6.2

when EF – E (( kBT

This results for p;

P = Nv exp ( (EV – EF )/ kB T ( ----------------------------------------6.3

Where Nv = 2 ((m*h kB T)/(2 ( h'2 )(3/2 and represents the effective density of states for the valance band edge.

For intrinsic semiconductors; n = p = ni = pi

And the product;

np = 4(( kB T)/(2 ( h'2 )( 3 (me*mh*) 3/2 ( exp (- Eg ) / kB T( ---6.4

and if we take square root of the product we get;

ni = pi =2(( kB T)/(2 ( h'2 )( 3/2 (me*mh*) 3/4 (exp (- Eg )/2 kB T(--6.5

From Eq. 6.1 and Eq. 6.3, equating n = p (this will be the intrinsic case so we may use EFi now for the Fermi energy), we get;

exp (2 EFi ) / kB T = (mh*/ me*) 3/2 (exp (Eg )/kB T( ------------6.6

and

EFi = (Ec – Ev)/ 2 + ¾ kBT ln (mh*/ me*) ----------------------6.7

Where EFi is the intrinsic Fermi level which is seen in Eq. 6.7 to be very close to the midgap.

As we can see in Eq. 6.5 the carrier concentration increases exponentially as the bandgap decreases. A logarithmic plot in Fig. 6.2 below shows this dependence for intrinsic carriers.

Fig. 6.2 Intrinsic concentration against reciprocal of temperature.

The intrinsic carriers are detrimental to device performance and once its concentration increase to ( 1015 cm –3, the material becomes unsuitable for electronic devices.

We can see in Fig.6.2 that, for a given temperature, the intrinsic concentrations ni = pi decrease as Eg increases.

Hence, high Eg materials are required for high temperature operations ( eg ni = pi will remain low).

For a system in equilibrium, the product pn is independent of EF, and is CONSTANT at a given temperature ( the law of mass action).

Now that we have some idea of concentrations in a semiconductor (eg intrinsic carriers , order of 1011 cm –3 ), and a metal (eg order of 1021 cm –3 ), it is clear that conduction in semiconductors need enhancement. Such enhancement can be done by a process called DOPING.

6.2 DONORS AND ACCEPTORS

Doping involves placing an impurity into a pure crystal lattice. Since it is placed intentionally to serve a purpose then it is better to name it as a dopant rather then an impurity, which is the general name used for unintentional placements in a lattice.

When a dopant is placed in a lattice, it disturbs the symmetry and changes the background potential (this location is sometimes called a defect centre, or an impurity or dopant centre, or a dopant defect). If the potential change is weak and long ranged, then the dopant centre can be described, approximately, by the Hydrogen model, except now the effective mass at the bandedge is used. Dopants placed in a host lattice can effect properties of the material mainly in two ways. If the effect is to enrich electron concentration, then the dopant is called a donor; but if the effect is to enrich hole concentration, then the dopant is called an acceptor.

We can define the lattice and the dopants as;

• Host lattice is a lattice into which a dopant is placed.

• Donor is an element which has one-more valance electron than the host material.

• Acceptor is an element which has one-less valance electron than the host material.

DONOR

Consider now a semiconductor Si (group 4 element) which has 4 valance electrons. If Si is a host material then, Arsenic, As, (group 5 element) is a donor is Si host lattice ( eg As has one-more valance electron than the Si). Such a donor defect is shown in Fig. 6.3 below.

all 4 valance electrons

go into the valance band

Silicone host atom Band

-a-

Pentavalent Silicone like Electron-ion

donor coulombic attraction

-b-

Fig. 6.3 a) silicone host. b) donor defect.

It is clear from Fig. 6.3 that the donor defect is neutral when occupied (eg when the excess electron is there) or positively charged when not occupied (eg when the excess electron leaves the defect).

The energy equation for the electron – ion combination is similar to that of the Hydrogen atom, except that electron mass is m* and the coulombic potential is reduced by, (0 / ( ;

where, (, is the dielectric constant of the semiconductor, and

(0 is the permittivity of vacuum ( 8.85 10 -12 Fm-1 ).

That is (see Eq. 3.3 );

Ed = Ec – (e 4 m*e ) / 2 ( 4 ( ( )2 h'2 n2 where n = 1,2,3,…---------6.8

and Ed - Ec is the donor defect energy with respect to conduction band edge.

The ground -state energy level is obtained by setting n = 1, which gives;

Ed = Ec – (e 4 m*e )/2( 4 ( ( )2 h'2 = Ec –13.6(m*/m0)((0/()2 eV--6.9

Fig. 6.4 below shows the donor defect and the donor energy level.

Fig. 6.4 Donor defect and the energy level.

The wave function is similar to Hydrogen atom case and is;

(d(r) = (exp –r/a) / (( a3 )1/2 -----------------------------------------6.10

and Bohr radius a=( 4 ( ( )2 h'2 /m*e e2 = 0.53(( / (0) /(m*e/m0) 6.11

For most semiconductors; Ed ( few meV; and a ( 100 (A.

Although Ed seems to be dependant only on the host material (through ( and m* in Eq. 6.9) more accurate theories show some dependence on the dopant atom itself.

ACCEPTOR

Considering Si again as the host material, then, group 3 elements can be used as acceptors . Fig. 6.5 below shows Boron (B) acting as an acceptor in Si, and the corresponding energy level of the acceptor.

Fig. 6.5 Acceptor defect and the energy level.

An electron from valance band go to Boron to complete bonding, leaving behind a hole. So we see from Fig. 6.5 that the acceptor defect is neutral when it is not occupied, but it is negatively charged when occupied.

In compound semiconductors, the dopants can be “amphoteric” . (eg Si (group 4) can act as a donor in GaAs if it replaces Ga (group 3) atom, or as an acceptor if it replaces As(group 5) atom.)

6.3 EXTRINSIC CARRIER DENSITY

In the lowest energy state of the donor atom, the extra electron is trapped at the donor site as showing Fig. 6.6a and cannot carry current; eg it is not useful for changing electronic properties. This state is called carrier “freeze out”, and also occurs at very low temperatures.

At higher temperatures, the donor electron is “ionised” and resides in conduction band as a free electron as shown in Fig. 6.6b. Ionised donor is positively charged and offers scattering centres for free electrons. So the donor site can be represented as;

Donor = atom core + 4 bound electrons + 1 “loosely” bound electron ------------------------------------------------------------------6.12

localised electron “free” electron

Ground state excited state

Donor is neutral donor is positively charged

Fig. 6.6 a) Donor site neutral b)donor ionised (positively charged)

When the material is doped, we may have donor density, N.D., and acceptor density, An, and no equality between them, eg electrons and holes; so

n – p = (n ( 0 ----------------------------------------------------------6.13

However the law of mass action still holds at normal doping levels and we can still write;

np = constant = ni2 = pi2 = ni pi -------------------------------------6.14

From equations Eq. 6.13 and Eq. 6.14 one can show that;

n = ½ (n + ½ ( (n2 + 4ni2 )1/2 --------------------------------------6.15

p = -½ (n + ½ ( (n2 + 4ni2 )1/2 --------------------------------------6.16

Also using equation Eq. 6.1 with approximation for the distribution function, eg use Boltzmann like function, f(E) = exp –(E – EF )/kBT

we get by setting first n = n, and then n = ni

n = ni exp ( (EF – EFi )/ kB T ( ---------------------------------------6.17

( use, n = Nc exp ( (EF – Ec )/ kB T (, ni = Nc exp ( (EFi – Ec )/ kB T ( )

Similarly;

p = pi exp ( -(EF – EFi )/ kB T ( -------------------------------------6.18

where EF is the Fermi level corresponding to the doped semiconductor. We also have from Eq, 6.17 and Eq. 6.18;

(n – p) / ni = (n / 2 = 2 sinh ( (EF – EFi )/ kB T ( ----------------6.19

• The relative position of the Fermi changes depending on the doping level and type of doping. Hence,

• when the semiconductor is doped n-type, the Fermi level moves towards the conduction band edge.

• when the semiconductor is doped p-type, the Fermi level moves towards the valance band edge.

In such cases, (eg when doped) the Boltzmann approximation is not valid any more and the simple expression relating the carrier concentration and the Fermi level are not very accurate.

A more accurate relation is given by;

[pic] ----------------------------------------------------------6.20

where, [pic] = [pic]and is called Fermi Dirac Integral or (just Fermi half integral), and,

[pic] ; [pic] -----------------------------------------6.21

and Nc is the effective density. The Fermi half integral is plotted in Fig. 6.7 below.

Fig. 6.7 Fermi half integral.

Using the Joyce Dixon approximation we have;

[pic] or

[pic] -------------------------------------------6.22

If the last term (eg [pic] ; and [pic] ) is omitted the result corresponds to Boltzmann approximation.

We shall assume in our studies that the Boltzmann approximation is valid and all the dopant centres (donors and acceptors) are ionised. Hence, the electron concentration in n-type materials will be the same as the donor density, Nd and the hole concentration in p-type materials will be the same as the acceptor density Na

7- TRANSPORT PROPERTIES IN SEMICONDUCTORS

7.1 SCATTERING

Impurities and imperfections, either intentional or unintentional, cause scattering of electrons.

Some important sources of scattering may be summarised as below;

Ionised impurities (caused by dopants)

Phonons (lattice vibrations)

Alloys (causes potential variation)

Interface (surface roughness in heterostructures)

Chemical impurities (causes potential variations).

Scattering causes loss of coherence between the initial and final states of momentum and energy of an electron, as represented in Fig. 7.1 below.

Fig. 7.1 Scattering of electrons

We shall now define relaxation time ([pic]) as the time taken (Average) to loose coherence, or memory of the initial state.

Conceptual picture of scattering is given in Fig. 7.2 below.

Fig. 7.2 Conceptual picture of scattering.

Scattering rate is proportional to;

• How strong the scattering potential (Vr) couples to the initial and final states.

• How many final states there are to scatter to.

7.2 VELOCITY-ELECTRIC FIELD RELATIONS IN

SEMICONDUCTORS

If an electric field is applied, electrons move in the field direction (in opposite sense) scatter but in steady-state gain some net drift velocity.

Electron distribution changes now and it can be determined by velocity-field relationship at low and moderately high electric fields.

LOW FIELD RESPONSE – MOBILITY

For this case assume that;

• Electrons do not interact with each other,

• [pic], is mean time between successive collisions from scattering sources,

• electron move like free electrons (eg [pic]) in between collisions.

Electron gain velocity in between collisions, ie., only during the time [pic]. Hence, we can write ;

[pic] (drift velocity) -----------------------------------------7.1

where F is the electric field.

The current density J = -nevd and after substituting for vd (Eq. 7.1)we have;

[pic] ----------------------------------------------------------------7.2

If we now compare Eq. 7.2 with that of Ohm’s Law ;

[pic] --------------------------------------------------------------------7.3

where [pic] is the conductivity, then we can write;

[pic] -----------------------------------------------------------------7.4

From the definition of mobility ([pic] );

[pic]---------------------------------------------------------------------7.5

we have from Eq. 7.1 and 7.5 ;

[pic] -------------------------------------------------------------------7.6

Mobility has strong dependence on the reciprocal effective mass, m*( which is the conductivity effective mass).

HIGH FIELD RESPONSE

Under strong electric fields (1 – 100 kV/cm), electrons get hot (gain more energy) and the carrier drift velocity increases.

In Si, typical of indirect bandgap materials, this velocity initially increases and then saturates, as shown in Fig. 7.3 below, at about 1.2 10 7 cm s –1.

In GaAs, typical of direct bandgap materials, the drift velocity shows a peak and then decreases (shows a negative differential resistance) and saturates at about 10 7 cm s –1.

Fig. 7.3 Drift velocity against the electric field.

7.3 CARRIER TRANSPORT BY DIFFUSION

Diffusion takes place, always from higher to lower concentrations.

The extend of diffusion is effected by the collision of electrons (holes). The random collision of electrons (holes) can be described by “mean free path” (l), and the “mean collision time” ([pic]sc[pic]) and represented as below.

collision collision

* *

l = average distance

[pic]sc[pic]= time

Now, consider a concentration gradient profile n(x,t), of electrons as in Fig. 7.4 , at time t .

Fig. 7.4 Electron concentration profile n(x), against distance x, at time t.

Now, we are going to calculate electron flux [pic](x,t) across the plane x = xo at any instant of time.

Electrons in the L (left) and R(right) regions move randomly and half of them in the L region will go across x = xo to the right. Similarly, half of those in the region R will go across x = xo to the left in time [pic]sc .

The flux to the right can be written as;

[pic] --------------------------------------------------------7.7

where nL and nR are the average carrier densities. As the two regions are separated by the distance l, we can write;

[pic] or [pic] -------------------------------7.8

Hence, the net flux is;

[pic] --------------------------------------7.9

where Dn is called the Diffusion coefficient of the electron system, and depend on l and [pic].

We also have;

[pic];

where [pic] is the mean thermal speed, and shows that Dn is also temperature dependant.

A similar argument for holes gives;

[pic]--------------------------------------7.10

Hence total current flow (in the absence of any applied electric field) is given by;

Jtot (diff) = Jn (diff) + Jp (diff) = [pic] --------7.11

Example 7.1

For n-type GaAs at 300 K, electron concentration varies as;

[pic] for x[pic]0; where L = 1 micron(micrometer = 10-6 meter).

Calculate diffusion current density at x = 0, if the electron diffusion coefficient is 220 cm2/s at 300 K.

Solution

The diffusion current density is given by;

Jn (diff) = eDn [pic][pic] = - (1.6 10–19C)(220 cm2/s) (10 16/ 10-4)

= - 3.5 kA/cm2.

7.4 TRANSPORT BY DRIFT AND DIFFUSION

(EINSTEIN RELATION)

In many electronic devices charge moves under the influence of both;

• electric field

• concentration gradients

The current is then given by; (using Eq. 7.2, 7.5 and 7.11)

Jn (x) =-e(nn(x)F(x) + [pic]

-----------------------------7.12

Jp (x) = e(pp(x)F(x) - [pic]

The mobility (n and (p are only constant at low fields, but decrease at high fields. The product “(F” reaches a constant value at even higher fields, when the saturation velocity is reached.

Thus the Drift current saturates and becomes independent of the Field, (eg as we have, J = nevd, and vd = (F ).

The effect of an electric field on the energy bands can be examined through Fig. 7.5 below.

Fig. 7.5 Bending of energy bands due to an electric field.

The force on the electron is “ –eF(x)” and it is related to potential energy “U(x)” by;

F(Force) = - (U(x) -----------------------------------------------7.13

For a uniform electric field, then;

U(x) = eF(x) x ------------------------------------------------------7.14

Then,

Ec(x) = Ec (F = 0) + eF(x) x -------------------------------------7.15

and since the electron charge is negative, bands bend as shown in Fig. 7.5, and electrons drift downhill in the energy band picture (eg current flow in opposite sense).

At equilibrium, the total electron and hole currents are individually (each one) zero and we have from Eq. 7.12;

[pic]-----------------------------------------------------7.16

For n(x) we may use Boltzmann distribution (from Eq. 6.17);

[pic]------------------------------------------------------7.17

where EFi and EF(x) are the intrinsic Fermi level and the Fermi level in the presence of extrinsic charge. Tis gives;

[pic]--------------------------------------------------7.18

But at equilibrium, the Fermi level can NOT vary spatially;

Hence, [pic]--------------------------------------------------------7.19

So now Eq. 7.19 becomes;

[pic] and now differentiate Eq.7.14 to get [pic],

also use Eq. 7.16 for [pic] and substitute both to get;

[pic] , and this gives;

[pic]------------------------------------------------------------7.20

which is the Einstein Relation satisfied for electrons. A similar relation also holds for holes, using the hole parameters.

7.5 QUASI FERMI LEVELS

We already know that at equilibrium, distribution of n and p are given by Fermi function, if of course we know the Fermi level and the product pn = constant.

If however, excess n and p are injected, the same Fermi function will not describe the occupancy. So, quasi-Fermi levels are used now, assuming that excess n and p are in thermal equilibrium in conduction and valance bands respectively.

In such cases, the quasi–equilibrium n and p can be represented by an electron Fermi function fe and a hole Fermi function fh respectively. Hence;

[pic][pic]------------------------------------------------------7.21

[pic]-------------------------------------------------------7.22

where [pic]-------------------------------------------------7.23

[pic]----------------------------7.24

where EFn and EFp are electron and hole Fermi levels.

At equilibrium, EFn = EFP . But, when excess carriers are injected

EFn moves towards the conduction band

And

EFP moves towards the valance band.

The carrier density relationships are as follows;

[pic]----------------------------------------7.25

The Joyce Dixon approximation is then,

[pic]

-----------------------------------------7.26

[pic]

Example 7.2

Use Boltzmann statistics to calculate the position of the electron and hole Fermi levels when an e-h density of 1017 cm-3 is injected (by field or optical) into pure (undoped ) silicone at RT (300K).

( Nc = 2.8 1019 cm-3, Nv = 1.04 1019cm-3, Eg = 1.1 eV.)

Solution

Given; n = p = 1017 cm-3 , kBT = 0.026 eV

Hence, from; [pic] and [pic], we have

[pic]

[pic]

For Si, Eg = 1.1 eV = Ec – Ev , hence;

EFn – EFp = (Ec – Ev ) – ( 0.146 + 0.121 ) = 1.1 – 0.267 = 0.833 eV

If the injected e-h density was 10 15 cm-3, then

EFn – EFp = 1.1 – 0.506 = 0.59 eV, hence

As the injected carrier density increase the separation between the Fermi levels widen.

7.6 THE CONTINUITY EQUATION – DIFFUSION LENGTH

In our discussion up to now we did not consider a process called recombination ( loss of carriers, radiative or non-radiative) of electrons and holes.

Continuity equation is developed to account for the recombination process in carrier transport.

Hence;

Particle flow rate = Particle flow rate due to current – Particle loss rate due to Recombination.

This is simply a statement of conservation of particles. Consider now a volume of space as in Fig.7.6 below.

Fig. 7.6 Recombination rate R in a volume element of A(x.

If [pic] is the excess carrier density in the region, the recombination rate, R, in volume A(x may be written as (losses);

[pic]-----------------------------------------------------------------7.27

where, [pic] is the electron recombination time per excess particle (radiative or non-radiative process together).

Also, particle flow rate into the same volume due to the current Jn is (gains);

[pic]------------------------------7.28

Using Eq.7.27-28, total rate of electron build-up can be written as;

[pic]-------------------------7.29

where (n is the only part which changes with time.

Hence, from Eq. 7.29 the continuity equation for electrons and holes can be written as;

[pic] ------------------------------------------------7.30

[pic] ------------------------------------------------7.31

In the case where we only have charge transport by diffusion, the diffusion currents are (from Eq.7.11);

[pic] ---------------------------------------------------7.32

[pic] ---------------------------------------------------7.33

we have from Eq. 7.30 and Eq. 7.31, and differentiating Eq.7.32 and Eq.7.33;

[pic] -------------------------------------------7.34

[pic] ---------------------------------------------7.35

In steady state, (used to study steady state charge profile), we have ;

[pic] and now, Eq. 7.34 and Eq.7.35 become,

[pic] and [pic] - --7.36

where Ln Lp are defined as;

[pic] and [pic] and are called Diffusion Lengths.

The meaning of Diffusion Length is better understood if we study the spatial variation in carrier density as shown below in Fig. 7.7.

Fig. 7.7 Spatial variation of injected carrier density in x-direction.

Due to some external injection, an excess electron density (n(0) is maintained at the semiconductor edge at x = 0.

If no is the equilibrium density, then we are interested in finding out how (n changes with x.

The general solution of Eq. 7.36 is given by;

[pic] ------------------------------------------7.37

However, when x is large (n(x) must go to zero, therefore, it is required that A1 = 0.This condition is valid only if the sample is large (large x).

Since (n(0) is known and fixed by injection condition, we have;

A2 = (n(0) -----------------------------------------------------------7.38

Hence the solution is;

[pic] ---------------------------------------------7.39

The diffusion length Ln (or Lp)represents the distance over which the injected carrier density falls to 1/e of its original value.

It also represents the average distance an electron will diffuse before it recombines with a hole.

Example 7.3

In p-type GaAs sample, electrons are injected from a contact. If the minority carrier mobility is 4000cm 2/V-s at 300K, calculate the diffusion length of the electrons.(recombination time ( = 0.6 ns)

Solution

At 300 K, kBT/e = 0.026 Volts, ( = 4000 cm2 /V-s = 0.4 m2 /V-s

Hence, using Einstein relation, Dn = ( kBT/e

Dn = (0.4 m2 / V-s )(0.026 V) = 1.04 10 –2 m2 /s .

The diffusion length = ((Dn() = (((1.04 10 –2 m2 /s )(0.6 10-9 s)( = 2.5 10 –6 m = 2.5 micrometer = 2.5 micron = 2.5 (m.

( In pure silicone ( can be much longer in the range 10-3 to 10-6 seconds and Ln can be much longer.)

Example 7.4

The electron mobility of Silicone at 300 K is 1400 cm2 /V-s . Calculate the mean free path and the energy gained in a mean free path at an electric field of 1 kV /cm . Assume that the mean free path l = vth(sc and vth =2 107 cm/s, m* = 0.26 mo, and eV = 1.6 10-19 Joule.)

Solution

Based on the data given we can first find the collision time, (sc , which is given by;

(sc = ( m*/e = (1400 10-4 m2/V-s)(0.26 )(0.91 10-30kg)/ 1.6 10-19 C

= 2.07 10-13 s

Now we can find the mean free path as;

l = vth(sc = (2 107 cm/s)( 2.07 10-13 s) = 4.14 10-6 cm = 414 Angstrom.

Electrons gain drift velocity, vd, and the energy gained between the collisions is kinetic energy , given by;

(E = m*(vd)2 /2 = m*((F)2 /2 = (0.5)(0.26)(0.91 10-30 kg)( ( 1400

10-4 m2/V-s )(105 V/m)(2 = 2.32 10-23 J = 0.145 meV .

Example 7.5

In a Silicone sample at 300 K the electron concentration drops linearly from 10 18 cm –3 to 10 16 cm –3 over a length of 2 (m. Calculate the current density due to the electron concentration.( Dn = 35 cm 2 /s; e = 1.6 10 –19 C).

Solution

The diffusion current density due to electron concentration is given by;

[pic]

here the only unknown is the concentration gradient [pic] , which can be found from the data given.

So, using the data,

[pic]

Hence the diffusion current density is then;

[pic]Jn = (1.6 10 –19 C)( 35 cm 2 /s)(4.95 10 21 cm -4)

= 2.77 10 4 A/cm 2.

Example 7.6

In GaAs sample it is known that the electron concentration varies linearly. The diffusion current density at 300 K is found to be 100 A/cm2 . Calculate the slope of the electron concentration profile. (Dn = 220 cm 2/s; e = 1.6 10 –19 C).

Solution

From,

[pic]

we have

[pic]

Example 7.7

Calculate the density of electrons in Silicone if the Fermi level coincides with the conduction band edge at 300K. Compare the results using Boltzmann approximation and the Fermi Half Integral. (Nc = 2.78 1019 cm-3).

Solution

The Boltzmann approximation gives;

[pic]

Since we have EC – EF = 0 , then n = NC = 2.78 1019 cm –3.

The Fermi Half Integral gives for, [pic], F 1/2 (0) = 0.65

So, from [pic] we have ; [pic]

n= 0.74 NC which is more accurate than the Boltzmann approximation. However the Boltzmann approximation is good enough when the term (EC – EF/kBT ) is large.

Example 7.8

The electron density in Silicone sample at 300 K is 1016 cm-3. Calculate EC – EF and the hole density using Boltzmann approximation. (NC = 2,78 1019 cm-3, np = 2.25 1020 cm-6 at 300 K).

Solution

Using Boltzmann approximation we have

[pic]

which gives for;

EC – EF = -kBT ln(n/NC) = (-0.026)ln(1016/2.781019) eV = 0.206 eV

Since the product pn is given then we can calculate hole concentration as;

P = 2.25 1020 / 1016 = 2.25 104 cm-3.

Example 7.9

In a GaAs sample electrons are moving under the influence of an electric field of 10 kV/cm, and the drift velocity is saturated at 107 cm/s. If the electron concentration is uniform at 1016 cm-3, calculate the drift current density.( e = 1.6 10 –19 C)

Solution

Since the drift velocity is saturated we do not need the field and mobility values and can directly use the drift velocity value.

The drift current density is given by;

Jn = -nevd

Jn = (1016 cm-3)(1.6 1019 C)(107cm/s)

Jn = -1.6 104 A/cm2.

Example 7.10

A GaAs sample is doped n-type at 5 1017 cm-3. Assume that all the donors are ionised. What is the position of the Fermi level at 300 K. (NC = 4.45 1017 cm-3.)

Solution

Since all the donors are ionised, the electron density in the conduction band is ;

N = 5 1017 cm-3 = ND , since the intrinsic density is much less than the donors.

Using Boltzmann approximation;

EC – EF = -kBT ln(n/NC) = (0.026)ln(5 1017 cm-3 / 4.45 1017 cm-3 )

= -3meV

As EF is very close to EC, we may use Joyce-Dixon approximation for more accurate result.

EF = Ec + kBT(ln(n/NC) + n/(8 NC( eV

EF = Ec + 0.026(ln(5 1017 /4.45 1017 ) + 5 1017 /(8 5 1017 ( eV

Ec - EF = -13.36 meV.

8- THE P-N JUNCTION

8.1 UNBIASED P-N JUNCTION

There are many different procedures for doping semiconductors n- or p-type. Some of these are, a) epitaxial layers doped during crystal growth, b) ion implantation is used to implant dopant ions, c) diffusion of dopants, etc..

We will now consider a junction between n-type material and a p-type material, and assume that the junction is abrupt. Fig. 8.1 below shows what happens when the p-type and n-type materials are made to form a junction and there is no externally applied field.

Fig. 8.1 a) p-type and n-type material, before forming a junction.

b) pn junction formation.

In Fig. 8.1 we can identify 3 distinct regions. These are;

• p-type region at the far left-hand side. Material is neutral, bands are flat. Acceptor and hole densities are equal.

• n-type region at the far right-hand side. Material is neutral, bands are flat. Donor and free electron densities are equal.

• The depletion region in the middle. Bands are bent. Field exists which swept mobile charge carriers(depleted), leaving behind negatively charged acceptors in the p-region and positively charged donors in the n-region.

Electric field exist in the depletion region which extends Wp into the p-region and Wn into the n-region.

Hence , both currents exist, but they counterbalance each other in the absence of an externally applied field.

In analysing the depletion layer, we will assume, as shown in Fig. 8.2 , that;

• The junction is abrupt and each side is uniformly doped.

• Mobile charge density in the depletion region is zero (depletion approximation). This approximation will be used to solve Poisson’s equation but not in calculating current flow.

• The transition between the bulk neutral n- type or p- type regions and the depletion region is abrupt.

Fig. 8.2 pn junction and the bulk neutral regions.

The electron and hole drift and diffusion currents may be summarised as below in Fig. 8.3 below.

Fig. 8.3 Drift and diffusion currents and charge movement in a pn junction.

These current cancel each other individually (eg hole currents add up to zero as does the electron currents). That is the hole current density given by;

Jp(x) = e((pp(x)F(x) - Dp[pic]( = 0------------------------------------8.1

This gives; [pic] ------------------------------------8.2

Now remember that (from Eq. 7.20) Einstein relation gives ;

[pic] --------------------------------------------------------------8.3

and in terms of potential gradient the electric field can be written as;

[pic] ----------------------------------------------------------8.4

Substitution of Eq. 8.3 and 8.4 into Eq. 8.2 gives;

[pic] -------------------------------------------8.5

If Vp and Vn denote the potential in the neutral p-region and n-regions (see Fig. 8.4a below), then Eq. 8.5 can be written in integral form as;

[pic] ------------------------------------------------8.6

where pp = hole density in p-region, and

pn = hole density in n-region.

Fig. 8.4a Structure, Potential profile, and the band profile of a p-n junction(not biased).

Now, using Eq. 8.6 and after integration we get;

[pic] ------------------------------------------8.7

Thus the contact potential or, build-in potential Vbi = Vn – Vp is given by;

[pic]--------------------------------------------------8.8

Similar start with electron drift current and diffusion currents will yield;

[pic] -------------------------------------------------8.9

where nn and np are the electron densities in n-region and p-region respectively, and the law of mass action still gives;

nnpn =nppp = n2i --------------------------------------------------------8.10

Therefore, we can now write from Eq. 8.8 and 8.9 ;

[pic] -------------------------------------------------8.11

If now an external bias (V) is applied, a similar relationship holds except Vbi is now replaced by (Vbi + V).

The electric field and potential in the depletion layer and the width of the depletion layer can be calculated using Poisson equations (see appendix ….).Considering that the charge on p- and n- regions has the same magnitude, then we can write;

AWpNa = AWnNd ----------------------------------------------------8.12

where A is the cross-sectional of the p-n structure.

Poisson equations can now be used which will have solutions for the electric field in the depletion region as ;

• [pic] -Wp ( x ( 0 ---- ----8.13

for the p-side of the junction.

• [pic] 0 ( x ( Wn ---------8.14

for the n-side of the junction.

• [pic]

for the p-side, and Wp ( x ( 0 ------------------------------------8.15

• [pic]

for the n-side and, 0 ( x ( Wn -------------------------------------8.16

The charge neutrality (Eq. 8.12) gives;

Nd Wn = NaWp and we can obtain the relationship;

[pic] ------------------------------8.17

[pic] ------------------------------8.18

and the depletion region width (total width) is given by;

W(Vbi) = Wp(Vbi) + Wn(Vbi) or;

[pic] ---------------------------------8.19

Fig, 8.4b below shows the Structure, Charge density, and Electric field profile of the region.

Fig. 8.4b Structure, charge density, and electric field profile in the depletion region.

Here it should be noticed that, in Na (( Nd (eg p-side is heavily doped, and shown as p+), then Wp (( Wn and a very strong field exists over a very narrow region in heavily doped side of the junction. In such cases, (p+n, n+p), depletion region is primarily on the lightly doped side of the junction(eg for p+n Wn (( Wp).

The maximum field, [pic] ---------------8.20

(this is obtained from Eq. 8.13 and 8.14 by setting x = 0).

8.2 P-N JUNCTION BIASED

When an external bias (voltage ) is applied across the junction, we can no longer assume that there is a balance between the Drift and Diffusion currents. A charge may be injected, and now a net current will flow. However we will still assume the following;

• injected minority carrier density (( majority carrier density,,

• Boltzmann distribution is valid in the depletion region,

• Mobile charge density is low in the depletion region and the applied voltage drops mainly across this region.

Fig. 8.5 below shows some profiles for forward and reverse biased junction cases.

Equilibrium Forward Bias Reverse Bias

Fig. 8.5 Biasing, voltage profile, and band profile for equilibrium, forward bias, and reverse bias cases of a p-n junction.

In forward bias (FB);

Vtot = Vbi - Vf -------------------------------------------------8.21

In reverse bias(RB);

Vtot = Vbi + Vr -------------------------------------------------8.22

Based on the approximation we used, equations for depletion region width, electric field profile, and potential profile are all applicable except that Vbi (used in Section 8.1 for the unbiased junction) is now replaced by Vtot as described by Eq. 821 and 8.22 above.

Now we are equipped with knowledge to study charge injection and current flow due to minority and majority carriers.

Charge injection and current flow

We will assume that under moderate external bias the electric field in the depletion region is always higher than the field for carrier velocity saturation (see Fig. 7.3, where the drift velocity is saturated). This means that the drift part of the current is not changing with bias, once the saturation level of field is reached.

From Eq. 8.10 we have;

[pic]---------------------------------------------8.23

Now in the presence of bias, we may write;

[pic]-----------------------------------8.24

where injection is much lower than majority carrier density, so that majority carrier densities are essentially unchanged because of injection. This is to say that;

p(-Wp) = pp and the ratio of Eq. 8.23 to Eq. 8.24 (for a forward bias with +V) gives ;

[pic]------------------------------------------------8.25

Here we see that the hole minority density p(Wn) at the edge of the n-side of depletion region will increase exponentially with bias V.

Similarly, for electrons we have;

[pic]------------------------------------------------8.26

n(-Wp) - np = (np (pn = p(Wn) - pn

The excess carriers injected across the depletion region are (from Eq. 8.25 and 8.26);

[pic]---------------8.27

[pic]------------8.28

At this moment we are going back to our discussion on diffusion and recombination of excess carriers (see Fig.7.7 ). Considering the p-n junction now, we have excess minority carriers just outside the depletion region and similarly we can write for minority carrier density (for holes or electrons);

[pic]

for x (Wn ------------------------------------------------------- 8.29

[pic]

for x ( -Wp --------------------------------------------------------8.30

as shown in Fig 8.6 below.

Fig. 8.6 Excess minority carrier density outside the depletion region.

Now we can write an equation that will describe net current flow with the applied bias voltage.

Only the diffusion currents will be considered, since the drift current will not change with bias ( eg. high field values within the depletion region and the drift velocity is saturated).

For net current we will consider both the excess electron and holes injected as given by Eq. 8.29 and 8.30.

Hence the diffusion current for holes (minority carriers in n-type) is given by;

[pic]

for x (Wn ------------------------------------------------------- 8.31

The total current injected (when x = Wn in equation 8.29) is given by;

[pic] -------------------------------8.32

Similarly the total electron current injected into p-side is given by;

[pic] --------------------------------8.33

The overall total current I(V) is then given by;

[pic][pic]

or

[pic] --------------------------------------------8.34

where

[pic] ---------------------------------------8.35

and Eq. 8.34 is known as the diode equation for forward bias (when V ( 0 ) or reverse bias (when V( 0 ). Hence, it is clear that for forward bias current increases exponentially, but for reverse bias, it reaches –Io value.

Minority and majority currents

The minority diffusion currents decrease exponentially as given by Eq. 8.31. If we now substitute Eq.8.29 into Eq. 8.31 we get for the hole diffusion current;

[pic] for x (Wn ----------------8.36

Now, as the diffusion current decreases due to recombination of holes with majority carriers (eg. electrons), then an equal number of electrons are injected (eg. used up). This causes drift current due to electrons which then exactly balance the hole current lost in recombination.(This is why x = Wn was taken for the total diffusion current in Eq. 8.32).

Hence,

Electron drift current = total current – hole diffusion current

Indr(x) = I(V) – Ip(x) and substituting for equations Eq. 8.36 and 8.34;

[pic]-8.37

Similarly on the p-side, hole drift current compensates the loss in electron diffusion current, as shown in Fig. 8.7 below. Hence,

[pic]-8.38

Fig. 8.7 Excess charge profile and the currents

The behaviour of current with bias is shown in Fig. 8.8 below.

Fig. 8.8 Diode current “I” with bias voltage “V”.

The forward bias voltage at which the diode current becomes significant ( order of mA), is called cut-in voltage. This is about 0.8 V for Si and about 1.2 V for GaAs.

The case of a narrow diode

Consider the geometry in Fig. 8.9 below;

Fig. 8.9 Structure of a narrow diode

We will assume that;

• there is no recombination in the depletion region, that is to say, Wn (( Lp and Wp (( Ln .

• injected charge goes to zero linearly at the contacts.

The hole current injected across Wn is then given by;

[pic]

and using Eq. 8.29 for (pn(Wn), and setting (pn(Wln) = 0 ;

[pic]---------------------8.38

A similar expression can be developed for the electron current, using the Eq. 8.28.

The diode current in this case would be ( compare with Eq. 8.32 and 8.33),

[pic]-------------------------------------8.39

similar to Eq. 8.34 and where;

[pic]---------------------8.40

This can be compared with Eq. 8.35 for Io where the difference is that;

Lp is replaced by (Wln – Wn), and

Ln is replaced by (Wlp – Wp).

8.3 SMALL SIGNAL EQUIVALENT CIRCUIT

Diode capacitance occurs as a combination of mainly two capacitive effects. These are;

• junction capacitance Cj, (dominates under reverse bias)

• diffusion capacitance Cdiff. (dominates under forward bias)

In general form the capacitance is given by;

[pic]-----------------------------------------------------8.41

The depletion width is (from Eq. 8.19) given for a forward bias V,

[pic]------------------------8.42

and the depletion charge is;

[pic]-------------------------------8.43

Also from Eq. 8.17 and 8.19 we have;

[pic], and [pic] -------------------8.44

Hence, substituting Eq. 8.44 into Eq. 8.43 we get;

[pic]---------8.45

The junction capacitance is then obtained by differentiating Eq. 8.45 as;

[pic]----------------8.46

Notice that Cj changes with bias V, and such a device may be called varactor.

Regarding the diffusion capacitance, Cdiff, consider now the injected holes. The charge injection is I(p ((p is the hole recombination time) and use the equation (pDp = Lp2 , and ignore the 1 in Eq. 8.32 ([pic]), then we get;

[pic]-----------8.47

Hence, we can now differentiate Eq. 8.47 to get;

[pic]----------8.48

For small signal ac response, the conductance Gs of a diode is given by;

[pic] ----------------------------------------8.49

where rs is the diode resistance.

Consider now a forward-biased p-n junction having an ac signal and an equivalent circuit as shown in Fig. 8.10 below;

Fig. 8.10 Forward biased p-n junction I-V characteristic and equivalent circuit.

In forward bias only the Cdiff will dominate and the small signal current is and voltage vs are related as;

[pic] ----------------------------------------8.50

or

is = Gsvs + jwCdiffvs ------------------------------------------------8.51

where w is the angular frequency. The admittance of the diode is then;

y = is/vs = Gs + jwCdiff --------------------------------------------8.52

9- SEMICONDUCTOR JUNCTIONS WITH METALS

9.1 INTRODUCTION

Before we say anything about the metal-semiconductor (MS) junctions, let recall how metals are usually used in electrical engineering. They are used mainly to serve as;

• Interconnection between different parts of a circuit,

• A part of a junction in Schottky barrier diode, etc.

• Ohmic contact

Regarding resistivities, we can compare some as;

• Aluminium (Al) 2.7 10 -6 ohm-cm.

• Copper (CU) 1.77 10 -6 ohm-cm.

• Silver (AG) 1.59 10 –6 ohm-cm

Example 9.1

Resistivity of Al at RT is 2.7 ((-cm. If 10 22 electrons/cm 3 participate in current flow, calculate the mobility of electrons.

Solution

From Eq. 7.3 we have, J = (F = nev, and substitute for v = (F, then ;

J = ne(F . Now comparing this equation with Eq. 7.3 we can write for conductivity;

( = ne( and therefore the resistivity ( is 1/( = 1/ ne(. This gives;

( =1/ne( = 1/(10 22 electrons cm-3)(1.6 10 –19 C)( 2.7 10 -6(-cm)

= 240.4 cm2/Vs (eg. dimensionally, V=IR = (C/s)( () or C( = Vs )

9.2 THE SCHOTTKY BARRIER DIODE

The Schottky barrier diode ;

• is similar to p-n diode (the p-n diode is a, semiconductor(p)- semiconductor(n) diode).

• has a metal-semiconductor junction.

• has a faster response than the normal p-n diode.

Now consider the energy band profiles shown in Fig. 9.1 below.

Fig. 9.1 Metal-Semiconductor (n) junction at equilibrium.

The height of the barrier (e(b) = Ec(semiconductor) – EFm . Hence from Fig. 9.1 we have;

e(b = e(m - e(s + (Ec – EFs) = e(m - e(s -----------------------9.1

The build-in potential for electron transport is;

eVbi = e(m - e(s -----------------------------------------------------9.2

And will change with bias.

In the case of p-type semiconductor the profile is shown in Fig. 9.2 below.

Fig. 9.2 Metal-Semiconductor (p) junction at equilibrium.

The build-in potential for hole transport is,

eVbi = e(s - e(m -----------------------------------------------------9.3

and will change with bias.

The barrier height, e(b , is essentially dependent on the properties of materials used and does not much change with bias.

Capacitance – Voltage characteristics

Using Poisson’s Equations and a similar approach to p+-n junction (as there is no depletion region on the metal side), we obtain, in the presence of a bias V (see Eq. 8.13 and 8.19, with Na (( Nd );

[pic] and [pic] --9.4

[pic] ---------------------------------------9.5

Using the maximum field value Fm = F(0),when x = 0, then;

[pic] -------------------------------------9.6

where;

[pic] ------------------------------------------9.7

and now substituting for W we get;

[pic] ------------------------------9.8

The depletion region charge is given by;

Qd = AeNdW = A(2e(Nd(Vbi – V)(1/2 -----------------------------9.9

The capacitance is then given by;

[pic] ------------------------------9.10

where A is the device area. Rearrange Eq. 9.10 to get;

[pic] ---9.11

where a and b are constants for the system, and a = bVbi .

Hence, a plot of reciprocal of capacitance square (1/C2) against Voltage (V) is a straight line and provides information on both the Vbi and Nd.

Example 9.2

A metal-silicone Schottky barrier with a 100 (m diameter has

“1/C2 (V” plot with a slope measured as “ –3 1024 F-1 V-1”. Calculate the doping density in the silicone.

Solution

Regarding the “1/C2 (V” plot the slope is given as (see Eq.9.11);

Slope = -2/A2e(Nd , and Nd = -2/ A2e((slope), where

permittivity(() = relative dielectric constant of Si (11.9 ) x permittivity of vacuum, (o (8.85 10-12 F m-1 ) = 105 .3 F m-1.

Nd = 2/(3 1024)(((50 10-6)2(2(1.6 10-19)(105.3 F m-1) = 6.42 1020 m-3 = 6.42 1014 cm-3 .

9.3 CURRENT FLOW IN A SCHOTTKY BARRIER

The most important mechanism of current flow is the thermionic emission, where electrons with energy greater than the barrier height (Vbi – V) pass across the junction, as shown in Fig. 9.3 below.

Fig. 9.3 Schottky barrier under forward and reverse bias, and I-V characteristic

The current is limited to junction barrier and only that fraction of electrons, nb, with energy greater than the barrier height “e(Vbi – V)” will pass over the barrier. That is;

[pic]--------------------------------------------9.12

where n0 is the electron density in the neutral region of the semiconductor, and is given by;

[pic] -------------------------------------------9.13

Hence;

[pic]--------------------------------------------9.14

(since e(b = eVbi + (Ec – EFs).

If the average flux of electrons towards the barrier is; [pic], where vav is the average speed of the electrons; then current, Ism, (semiconductor to metal) is given by, for diode area A;

[pic] and using Eq. 9.14 we have;

[pic] -----------------------------------------9.15

Now when the bias V = 0, the currents semiconductor –to- metal Ism and metal -to –semiconductor Ims are equal and given by;

[pic] -----------------------------9.16

Even with bias, the Ims(=Is) remains constant as it mainly depends on the barrier which essentially remains unchanged. Hence the net current, from Eq. 9.15 and 9.16, can be written as;

[pic] --------------------------------------9.17

Eq. 9.17 is very similar to the diode equation (for p-n diode) we have derived before (see Eq. 8.34). The most important differences between the normal p-n diode and the Schottky diode may be summarised as below;

|p-n diode |Schottky diode |

|Forward bias current is due to minority carrier injection |Forward bias current due to majority injection from the |

| |semiconductor |

|Reverse current due to minority carrier diffusion, and strongly |Reverse current due to majority carriers, and weakly temperature|

|temperature dependent |dependent |

|Cut-in voltage is high |Cut-in voltage is low |

|Switching speed is slow |Switching speed is fast |

|Some recombination in the depletion region |Essentially no recombination in the depletion region |

9.4 SMALL SIGNAL EQUIVALENT CIRCUIT

The small signal equivalent circuit of a Schottky diode is printed below in Fig. 9.4.

Fig. 9.4 Small signal equivalent circuit of a Schottky diode.

For depletion region resistance Rd, and differential capacitance Cd, we have;

[pic] -----------------------------------------------------9.18

[pic](from Eq. 9.10) -----------------------9.19

Rs is the contact resistance together with resistance of the neutral region of the semiconductor, and Ls, is parasitic inductance.

The capacitance due to geometry of the device is given by;

[pic] -------------------------------------------------------9.20

where L is the length of the device.

The absence of any diffusion capacitance makes these devices much faster than the normal p-n diodes.

9.5 OHMIC CONTACTS

A metal-semiconductor junction which has a linear, non-rectifying I-V characteristics, as shown in Fig. 9.5 below are called Ohmic contacts.

Fig. 9.5 I -V characteristic of an Ohmic contact

For a build-in potential, Vbi, the depletion width on the semiconductor side is;

[pic] ----------------------------------------------------9.21

which is much dependent on the level of doping.

Hence, W will decrease with increasing Nd. Using this dependence, the semiconductor interface can be heavily doped to reduce W, so that electrons can tunnel through the barrier as shown in Fig. 9.6 below.

Fig. 9.6 Band structure of an Ohmic contact

The resistance, R, of the contact of area A, defines the quality of an ohmic contact.

The specific contact resistance rc is given by;

rc = RA -----------------------------------------------------------9.22

and under heavy doping has the dependence that;

[pic] --------------9.23

where the field is;

[pic] -------------------------------9.24

Hence, from Eq. 9.23 and 9.24 we have;

[pic] -------------------------------------9.25

[pic]

Example 9.3

A silicone diode is fabricated by starting with an n-type (Nd = 10 16 cm –3 ) substrate, into which indium is diffused to form as a p-type region doped at 10 18 cm –3. Assuming that an abrupt p+n junction is formed by the diffusion process, calculate;

• Fermi level position in the p- type and n-type regions

• The build-in potential in the diode

• Depletion widths Wp and Wn.

( Nc = 2.8 10 19 cm –3 , Nv = 1.0 10 19 cm –3 at 300 K, Eg = 1.1 eV,

( = 11.9 x 8.84 10 –12 F m –1 ).

Solution

First we will assume that all the dopants are ionised, that is;

nn = Nd ; pp = Na .

a) We have from;

[pic] and,

[pic] and hence;

[pic]and

[pic]

b) The build-in potential eVbi is given by;

eVbi = Eg – 0.06 – 0.206 = EFn – EFp = 1.1 – 0.06 – 0.206 = 0.834 eV

c) The depletion width Wp(Vbi) is given by(see Eq. 8.17);

[pic]

= 3.2 10 –9 m = 32 Angstrom.

And;

[pic]

= 0.32 10-6 m = 3200 Angstrom. Showing that most of the

depletion region is on the lightly doped side of the junction, Wn .

Example 9.4

A silicone p-n diode has a diameter of 20 microns. Donor and acceptor densities are 10 16 cm –3 and 1018 cm –3 respectively. Calculate;

• Wp and Wn and the electric field profile under a reverse bias of 10 Volts and a forward bias of 0.5 Volts.

• The charge in the depletion region.

( ni = 1.45 1010 cm –3, T = 300 K, ( = 11.9 x 8.84 10 –12 F m –1)

Solution

To calculate the depletion width , first we need to know Vbi, which is give by;

[pic] where we are given, pp = 1018 cm –3 and using the mass action law we can calculate pn = ni2/nn = (1.451010)2/1016

= 2.1 10 4 cm –3 .

Hence;

[pic]

a) Eq. 8.17 gives the depletion widths, and for the Reverse bias case we have (Vtot = Vbi + Vr) where Vr = 10 V;

[pic]

= 115.2 Angstrom.

And

[pic]

= 1.151 micron.

And for the Forward bias case (Vtot = Vbi – Vf ) where Vf = 0.5 V;

[pic]

= 20.3 Angstrom.

And

[pic]

= 0.203 micron.

The electric field profile is a straight line on both sides of the junction and decrease to zero value at “-Wp” and “Wn” .

The maximum value of the electric field is given by;

[pic] and depends on values of the

depletion widths ( or the bias ).So the values of the maximum field

at reverse and forward bias cases are;

Fmax (Vr = 10 V) = -1.78 105 V/cm.

Fmax (Vf = 0.5 V) = -3.14 104 V/cm

b) The charge in the depletion region depends on the bias

condition and is given by;

Qr (Vtot=10.819V) = eNdWnA = eNaWpA = (1.6 10 -19 C )(10 22 m-3

)(1.151 10 -6 m)( ( )(1010 -6 m)2 = 5.78 10 –13 C.

Qf (Vtot=0.319V) = eNdWnA = eNaWpA = (1.6 10 -19 C )(10 22 m-3

)(0.203 10 -6 m)( ( )(1010 -6 m)2 = 1.02 10 –13 C.

Example 9.5

A metal-silicone Schottky junction device is disc shaped with radius 100 microns. The device has a capacitance(C) - voltage(V) relationship of C -2 = 2 (Vbi - V ) ( A2 e ( Nd )-1 where A is the barrier area. This relationship is plotted below. Calculate :

a- the build-in potential Vbi of the barrier.

b- the donor density Nd . ( e = 1.6 10-19 c , ( = 105.3 10-12 Fm-1 )

[pic]

Solution

a) From Eq. 9.11, [pic], we see that when V = Vbi,

then C –2 = 0. So from the plot we have Vbi = 0.6 Volt.

b) The slope from the plot is, 1.8 1024 / 0.6 = 3 10 24

But the slope is also given by Eq. 9.11 as;

Slope = [pic] where the area of the device, A, is given by;

A = (r 2 = ((100 10 –6 m)2 = ( 10 –8 m2 and, A2 = ( 2 10 –16 m 4.

Hence,

Nd = 2 / (3 10 24 F –2 V -1)( ( 2 10 –16 m 4)(1.6 10 –19 C)(105.3 10 –12 F m –1 ) = 0.4 10 20 m –3 = 0.4 10 14 cm –3 .

Example 9.6

a- Draw a small signal equivalent circuit of a Schottky diode and clearly label the parameters.

b- When a forward bias of Vf = kBT/e was applied it was found that the net current flow was 10.95 10-8 A. What is the net current flow when Vf = 0.3 volts and T = 300 degrees Kelvin.

Solution

a)

b) The Schottky diode current is given by Eq. 9.17 as ;

[pic]

Here, when we set V = Vf = kBT / e , then, from the diode equation

the corresponding current, say Ix, will be given as;

Ix = Is (e – 1 ) . But this current is also given as 10.95 10 –8 amp.

Hence;

Ix = Is (e – 1 ) = 10.95 10 –8 amp.

So, Is = Ix / (e – 1) =10.95 10 –8amp. / (e – 1) = 6.37 10 –8 amp.

Is = 6.37 10 –8 amp.

When Vf = 0.3 Volt, then

I(0.3) = 6.37 10 –8 amp.(e(0.3 / 0.026) –1) = 6.37 10 –8 amp(e 11.53 – 1)

= 6.53 10 –3 amp. = 6.53 mA.

Example 9.7

Schematically show the current – voltage (I-V) characteristics and Energy band diagram of an Ohmic contact. Clearly indicate energy levels and all regions of importance.

Solution

BIPOLAR JUNCTION TRANSISTOR

10.1 INTRODUCTION

The bipolar junction transistor (BJT) is a simple device, consisting of two junction regions; one having a forward bias, and the other having a reverse bias.

Consider now the reverse current Io (as in Equation 8.35) in a reverse biased p-n junction.

[pic] 10.1a

Io is dependent on the minority charge, pn and np.

In other words, the reverse biased p-n junction current is determined by the minority carrier that diffuse into the depletion region edge and then swept away by the field to contribute to current.

If the minority carrier density can be altered the current can be changed as shown in Fig. 10.1.

Remembering now that a forward biased p-n junction involves injection of minority carriers, then it is possible to use such an additional junction to supply (change) the minority carrier concentration in the reverse biased junction.

Consider now a forward biased junction with doping profile as, n+-p so that np >>pn, then Eq. 10.1a can be rewritten as,

[pic] 10.1b

so that any change in minority carrier concentration np (due to forward bias injection) will be reflected in Io, as shown in Fig. 10.1.

The two junctions together is shown in Fig. 10.2, where the forward biased junction is emitting minority carriers (Emitter part) and the reverse biased junction is collecting them (Collector part). The region in between is called Base and it is made so thin that most electron will diffuse through before recombination, e.g. base-width 0

cut-off RB, VBE < 0 RB, VCB> 0

saturation FB, VBE > 0 FB, VCB < 0

10.3 CURRENT FLOW IN A BJT

Using the notation adopted in the p-n diode study, Fig.10.5 shows the structure and concentration profiles of a BJT, where equilibrium minority concentrations are noted as, peo, nbo, nco in the emitter, base and collector regions respectively. Also;

δpe ( xe = 0 ) = excess hole density at the emitter side of the EBJ (emitter-base junction)

= [pic] 10.10

δnb ( xb = 0 ) = excess electron density at the base side of the EBJ (emitter-base junction)

= [pic] 10.11

δnb ( xb = Wb ) = excess electron density at the base side of the CBJ (collector-base junction)

= [pic] 10.12

δpc ( xc = 0 ) = excess hole density at the collector side of the CBJ(collector-base junction)

= [pic] 10.13

As in the case of a p-n diode, the continuity equation can be used to get;

[pic] 10.14

with solution;

[pic] 10.15

The boundary conditions as described by equations Eq. 10.11 and Eq. 10.12, can be used to solve for A1 and A2 .

Hence, the emitter current can be calculated through the following operations;

[pic] 10.16

[pic] 10.17

Similarly, the collector currents;

[pic] 10.18

[pic] 10.19

The solution for currents can be obtained readily if some simplifying assumptions are made. For example, assuming that the γe = 1, and Wbn ................
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