### Lectures in

Physical Electrons University of NinevahDepartment of Electronics, Communication and Computer

in College of Electronic Engineering

by

Dr. Qais Thanon

2017-2018

### Chapter 1

Energy bands in Solids|||||||||||||||||||||||||||||||||{ All matters are made of atoms; and all atoms consist of electrons, pro-

tons, and neutrons. In this chapter, you will learn about the structure of

the atom, electron orbits and shells, valence electrons, ions, and the semi-

conductive materials. Semiconductive material is important because the

conguration of certain electrons in an atom is the key factor in determin-

ing how a given material conducts electrical current.

||||||||||||||||||||||||||||||||||

1.1 Atomic Structure An atom is the smallest particle of an element that retains the characteristics of that

element. Each atom consists of central nucleus surrounded by orbiting electrons. The

nucleus consists of positively charge particles called protonsand uncharged particles

called neutrons . Electrons are the basic particles of negative charge.

CHARGED PARTICLES Electric charges, positive or negative, occur in multiples of the electronic charge. The

electron is one of the fundamental particles constituting the atom. The charge of an

### 2 Energy bands in Solids

Figure 1.1: Atomic structureelectron is negative and is denoted by (e). The magnitude of eis 1 :6 10

19

Coulomb

and the mass of the electron is 9 :1 10 31

kg. The proton have the same charge of electron

but with opposite sign and the mass of proton equal to 1 :672 10

27

kg. The charge

of a positive ion is an integral multiple of the charge of the electron, although it is of

opposite sign. For the case of singly ionised particles, the charge is equal to that of the

electron. For the case of doubly ionised particles, the ionic charge is twice that of the

electron.

The electron in obit moving under the in

uence of the balance between the central

force and the electrostatic force between the electron and the positive nucleus.

e2 4

or 2 = mv

2 r

The electron velocity in orbit can be calculated by:

v= s e

2 4

omr (1.1)

Where mis the mass of electron, eis charge of electron and vis electron velocity and

r is the radius of orbit and

o is the permittivity of space = 8

:85 10

12

F/m .

According to equation (1.1) Bohr developed a model which contains three postulates:

1. Instead of innity of orbits which would be possible in classical mechanics is only possible for an electron to move in an orbit for which its orbital angular momentum

is an integer multiple by Plank's constant divided by 2 .

1.1

Atomic Structure 3

P

=mvr =nh 2

=

n~ (1.2)

where ~= h 2

h is Plank's constant = 6 :62 10

34

J s .

Equation (1.2) shows that the angular momentum of electron is quantized. From

equations (1.1) and (1.2) it can be written:

rn = n

2

h 2

o me

2 (1.3)

When n=1 equation (1.3) will gives r

1 = 5

:29 10

11

m . r

1

Bohr radius

Hydrogen radius ( a

o). Equation (1.3) can be written as:

rn =

a

on 2

(1.4)

where, ao = h

2

o me

2

2. Electromagnetic radiation is emitted if an electron initially moves from an orbit of total energy ( E

i) to lower orbit of energy (

E

f). The frequency of the emitted

radiation is equal to:

f= E

i

E

f h

(1.5)

3. An atom has a nite state of energy. These states are separated and the electron in which state is stationary and non-radiating.

### 4 Energy bands in Solids

1.2Electron Energy in Orbit The total energy of electron in orbit around the nucleus is the sum of kinetic energy

and electrostatic energy, that means:

E = electrostatic (work)+ kinetic energy

The potential energy of the electron at a distance rfrom the nucleus is

q2 4

or , and its

kinetic energy is mv

2 2

. Then, according to the conservation of energy and using equation

(1.1) we can write:

E= e

2 8

or +

e2 4

or (1.6)

The rst term in the above equation represents the kinetic energy of the electron

in the orbit whilst the second term represents the work, i.e., the electro-static energy.

The electrostatic energy can be dene as the work which required to move an electron

from innite to distance rfrom the positive nucleus. As a result the total energy of the

orbital electron can be written as:

E=

e2 8

or (1.7)

It is necessary to observe that equation (1.7) shows that the electron energy is negative

when it bounded to atom. If the total energy of electron is greater than zero,

then the electron has enough energy to separate from atom . Substituting the

value of rfrom (1.3) in the equation (1.7), the last equation would be as:

E=

me 4 8

h 2

2

o

1 n

2 (1.8)

The magnitude

me 4 8

h 2

2

o is constant and represents the energy of electron in the ground

state of the hydrogen atom. Then it can be re-write equation (1.8) as:

E=E

r 1 n

2 (1.9)

1.3

Atomic Energy Level 5

Where

E

r=

me 4 8

h 2

2

o and the magnitude of this constant is equal to

-13.6 eV.

This value is the energy which required to ionize the hydrogen atom .

It can be observed that it is used the electronvolt as a unit of the energy instead

of the 'Joule'. For energies involved in electron devices, 'Joule' is too large a unit.

Such small energies are conveniently measured in electronvolt , abbreviated eV. The

electron volt is dened as the kinetic energy gained by an electron, initially at rest,

in moving through a potential dierence of 1 volt. Since e= 1 :6 10

19

Coulomb, then

each 1 eVwill equal to 1 :6 10

19

Joule.

1 eV = 1 :6 10

19

Joule .

1.3 Atomic Energy Level According to equation (1.9) the energy of second state is -3.4ev. Then it is required

to 10.2ev to raise the electron from ground state to second state. Therefore there is

only discrete value of electron energy exist within atomic structures. This energy can

be calculate using the following equation:

E =E

R

1 n

2

i

1 n

2

f !

(1.10)

For each integral value of nin equation (1.9) a horizontal line is drawn. These lines

are arranged vertically in accordance with the numerical values calculated from equa-

tion (1.9). Such graph is called an energy-level diagram and is indicated in Fig. 1.2 for

hydrogen. The number to the left of each line gives the energy of this level in electron

volts. The number immediately to the right of a line is the value of n. Theoretically,

an innite number of levels exist for each atom, but only the rst ve and the level for

n = 1 are indicated in Fig. 1-2.

As the electron is given more and more energy, it moves into stationary states which

### 6 Energy bands in Solids

Figure 1.2: The lowest ve energy levels and the ionization level of hydrogen. Thespectral lines are in angstrom units.

are farther and farther away from the nucleus. When its energy is large enough to

move it completely out of the eld of in

uence of the ion, it becomes 'detached'from

it. The energy required for this process to occur is called the ionization potentialand is

represented as the highest state in the energy-level diagram.

1.3

Atomic Energy Level 7

Solved problems

1.

Calculate the electron velocity in the second orbit of the hydrogen atom?

Solution According to equation (1.1) the electron velocity in the orbit is given by:

v = q e

2 4

omr

Since r= a

o

n2

r = 5 :29 10

11

22

= 2 :11 10

10

m .

v = q (1

:6 10

19

)2 4

8:85 10

12

9:1 10

31

2:11 10

10

v = 1 :96 106

m:s

1

.....................................................

2. The total energy of electron in the second orbit is (-3.4ev). Find the

kinetic energy and work?

Solution As it is shown before in equation (1.6) that the total energy of the electron in the

orbit is the sum of kinetic energy and work. Then, E=K.E + W

but K:E =e

2 8

or and

W=

e2 4

or

then E= e

2 8

or +

e2 4

or =

e2 8

or =-K.E= W 2

K:E =E =( 3:4) = 3 :4 eV

W = 2 E = 2 3:4 = 6:8 eV

.....................................................

3. A photon with energy of 5eV is emitted for electron transition from or-

bit to another. Find the wavelength of the emitted electromagnetic wave?

Solution E

=h = h

c

= h

c E

=6

:62 10

34

3 10 8 5

1:6 10

19 =0.24

m.

.....................................................

### 8 Energy bands in Solids

4.What is the frequency of the electromagnetic wave which emitted from

the electron transition from the second orbit to the first orbit?

Solution

E =E

R

1 n

2

i

1 n

2

f

= 13 :6

1 2

2

1 1

2

= 10 :2 eV

1.4 De Broglie Hypothesis Since a photon is absorbed by only one atom, the photon acts as if it were con-

centrated in a very small volume of space, in contradiction to the concept of a wave

associated with radiation. De Broglie postulated that the dual character of wave and

particle is not limited to radiation, but is also exhibited by particles such as electrons,

atoms, or macroscopic masses. He postulated that a particle of momentum p= mv has

a wavelength associated with it given by:

= h P

(1.11)

We can make use of the wave properties of a moving electron to establish Bohr's postulate

that a stationary state is determined by the condition that the angular momentum

must be an integral multiple of ( h=2 ). It seems reasonable to assume that an orbit of

radius rwill correspond to a stationary state if it contains a standing-wave pattern. In

other words, a stable orbit is one whose circumference is exactly equal to the electronic

wavelength , or to ( n), where nis an integer (but not zero). Thus:

2 r =n =nh mv

(1.12)

Prove equation 1.12?

1.5

The energy-band theory of crystals 9

1.5

The energy-band theory of crystals As we have seen, all the electrons of a given atom having the same value of

nbelong

to the same prescribed bands (electronic shell). Each shell around the nucleus corre-

sponds to a certain energy band and is separated from adjacent shells by energy gaps,

in which no electron can exist.

A crystal is a solid consisting of a regular and repetitive arrangement of atoms or

molecules (strictly speaking, ions) in space. If the positions of the atoms in the crystal

are represented by points, called lattice points, we get a crystal lattice. The distance

between the atoms in a crystal is xed and is termed the lattice constant of the crystal.

To discuss the behaviour of electrons in a crystal, we consider an isolated atom of the

crystal. If Zis the atomic number, the atomic nucleus has a positive charge Z e. At a

distance rfrom the nucleus, the electrostatic potential due to the nuclear charge is (in

SI units).

V(r ) = Z e 4

or (1.13)

Since an electron carries a negative charge, the potential energy of an electron at a

distance rfrom the nucleus is:

Ep(

r ) = eV =

Z e 2 4

or (1.14)

V (r ) is positive while E

p(

r ) is negative. Both V(r ) and E

p(

r ) are zero at an innite

distance from the nucleus. Figs. 1.3(a) and (b) show the variation of V(r ) and E

p(

r ),

respectively with r.

Now, consider two identical atoms placed close together. The net potential energy of

an electron is obtained as the sum of the potential energies due to the two individual

nuclei. In the region between the two nuclei, the net potential energy is clearly smaller

than the potential energy for an isolated nucleus (see Fig. 1.3).

### 10 Energy bands in Solids

Figure 1.3: Variation of (a) Potential in the eld of a nucleus with distance, (b) Potentialenergy of an electron with its distance from the nucleus. Figure 1.4: Potential energy variation of an electron with distance between two identical

nuclei.

The potential energy along a line through a row of equispaced atomic nuclei, as in a

crystal, is diagrammatically shown in gure 1.5. The potential energy between the nuclei

is found to consist of a series of humps. The separation between the split-o energy levels

is very small. This large number of discrete and closely spaced energy levels forms an

energy band . Energy bands are represented schematically by shaded regions in gure

1.5(b).

The width of an energy band is determined by the parent energy level of the isolated atom

and the atomic spacing in the crystal. The lower energy levels are not greatly aected

by the interaction among the neighbouring atoms, and hence form narrow bands. The

higher energy levels are greatly aected by the interatomic interactions and produce

1.5

The energy-band theory of crystals 11

Figure 1.5: Splitting of energy levels of isolated atoms into energy bands as these atoms

are brought close together to produce a crystal.

wide bands.

The lower energy bands are normally completely lled by the electrons since the electrons

always tend to occupy the lowest available energy states. The higher energy bands may

be completely empty or may be partly lled by the electrons.

The interatomic spacing, although xed for a given crystal, is dierent for dierent

crystals. The width of an energy band thus depends on the type of the crystal, and is

larger for a crystal with a small interatomic spacing. The lower energy bands are

normally completely lled by the electrons since the electrons always tend to

occupy the lowest available energy states. The higher energy bands may be

completely empty or may be partly lled by the electrons. The lower energy

band calls the valence band and the rst, unlled or partially lled, band above the

valence is called conduction band. The energy gap between the valence and conduction

can be calculated as:

Eg=

E

c

E

v (1.15)

On the basis of the band structure, crystals can be classied into metals, insulators,

and semiconductors.

### 12 Energy bands in Solids

Figure 1.6: Energy band structure of (a) metal, (b) insulator, and (c) semiconductor.1.5.1 Metals A crystalline solid is called a metal if the uppermost energy band is partly

lled [gure 1.6(a)] or the uppermost lled band and the next unoccupied

band overlap in energy. The electrons in the uppermost band nd neighbouring va-

cant states to move in, and thus behave as free particles. In the presence of an applied

electric eld, these electrons gain energy from the eld and produce an electric current,

so that a metal is a good conductor of electricity. The partly lled band is called the

conduction band. The electrons in the conduction band are known as free electrons or

conduction electrons.

1.5.2 Insulators When the forbidden energy gap between the valence band and the conduction band,

is very large, only a few electrons can acquire enough thermal energy to move from the

valence band into the conduction band. Such solids are known as insulators. Since only

a few free electrons are available in the conduction band, an insulator is a bad conductor

of electricity. The energy band structure of an insulator is schematically shown in gure

1.6(b).

1.6

Fermi Dirac distribution function 13

1.5.3 Semiconductors

A material for which the width of the forbidden energy gap between the valence and

the conduction band is relatively small ( 1 eV) is referred to as a semiconductor. As the

forbidden gap is not very wide, some of the valence electrons acquire enough thermal

energy to go into the conduction band. These electrons then become free and can move

about under the action of an applied electric eld. The absence of an electron in the

valence band is referred to as a hole. The holes also serve as carriers of electricity. The

electrical conductivity of a semiconductor is less than that of a metal but greater than

that of an insulator. The band diagram of a semiconductor is given in gure 1.6(c).

At 0 o

K the semiconductor becomes insulator because the electrons do not has energy

to jump to conduction band.

1.6 FermiDirac distribution function Fermi Dirac distribution function describes the energies of single particles in a system

comprising many identical particles that obey the Pauli exclusion principle. Electrons

are Fermions, and thus follow Fermi Dirac distribution function.

At room temperature, the thermal energy of the atoms may allow a small number of

electrons to participate in the conduction process in semiconductor. The probability for

lling the band by electrons depends on temperature.

f(E ) = 1 1 +

exp

EE

f kT

(1.16)

where f(E ) is the probability of occupancy of the state with energy E,E

F is a char-

### 14 Energy bands in Solids

acteristic energy for a particular solid and is referred to as the Fermi level,Tis the

absolute temperature in o

K and kis Boltzmann's constant ( k= 1.38 10

23

JK

1

=

8.614 10

5

eV K

1

). Fermi energy can be dened as the energy at which there would

be a fty percent chance of nding an electron.

Figure 1.7 shows the probability f(E ) against E E

F for

T= 0, 300 and 2000 K. Figure 1.7:

f(E ) as a function of E =E

Ffor dierent values of

T

From Fermi Dirac function, equation 1.16 and gure 1.7, it can be conclude that:

1. At the absolute zero of temperature, i.e. at T = 0 K, equation 1.16 shows that

f (E ) = 1 for E < E

Fand

f(E ) = 0 for E > E

F. Thus all the energy states

below E

F are occupied by the electrons and all the energy states above

E

F are

completely empty. (The probability of nding electron above Fermi level at zero

o K is zero).

2. At temperatures greater than the absolute zero, f(E )> 0 for E > E

F, as shown

in gure 1.7. This means that at a nite temperature, some of the electrons in the

quantum states below EF acquire thermal energy to move into states above E

F.

The probability of electron above Fermi level at T >0o

K is given by:

f (E ) = exp

(E E

f) kT

(1.17)

1.6

Fermi Dirac distribution function 15

3. The probability of electron to ll a state below Fermi level at

T >0o

K is given by:

f (E ) = 1 exp

(jE E

fj

) kT

(1.18) Solved problems

1.

Find the probability of an electron to occupy a level (0.1 eV) above

Fermi level at 27 o

C ?

Solution f

(E ) = exp

(E

E

f) kT

= exp

(0 :1) 8

:614 10

5

300

= 0 :0209

2. The probability for an electron to occupy a level at 120 o

C is (210

6

).

Find the location of this level with respect to Fermi level?

Solution f

(E ) = 1 1+

exp

E E

f kT

= 1 1+

exp

E E

f 8

:614 10

5

300

= 2 10

6

E E

f = 0

:444 eV above Fermi level. Review Questions

1.

What is the orbital angular momentum of the electron in the third orbit? (Ans. 3 ~ =

1 :97 10

15

eV ).

2. What is the energy of the electron in the Hydrogen atom after absorption an electro- magnetic wave with frequency of 3 :284 1015

H z ?(Ans. 13.59 eV).

3. An electron, initially at rest, gains a speed of 10 7

m/s after being accelerated through

a potential dierence of V volt. Determine V. What is the nal kinetic energy of the

electron in J and eV? (Ans. 284.7 volt, 4 :555 10

17

J, 284.7 eV)

### 16 Energy bands in Solids

4. A particle carries a positive charge numerically equal to the electronic charge. It acquiresa velocity of 200 km=safter moving through a potential dierence of 210 V. Determine

the mass of the particle relative to the electronic rest mass. (Ans. 1844 m

o).

5. A particle of charge 1 :2 10

8

C and mass 5 gtravels a distance of 3 munder the action

of a potential dierence of 500 V. Calculate the nal velocity and the acceleration of the

particle if it starts from rest. (Ans. 4 :898 10

2

m=s , 410

4

m=s 2

).

6. A potential dierence of 400 volt is applied between two parallel metal plates 4 cm apart. An electron starts from rest from the negative plate. Obtain (i) the kinetic energy of

the electron when it reaches the positive plate and (ii) the time required by the electron

to reach the positive plate. (Ans. 400 eV, 6 :745 10

9

s ).

7. Find the probability for an electron to occupy a level at 50 o

C if this level is below Fermi

level by (0.2 eV)?

### Chapter 2

Transport Phenomena inSemiconductor

|||||||||||||||||||||||||||||||||{ The current is dened as the

ow of charge particles. In metal the

current results from the

ow of negative charges (electrons),whereas the

current in a semiconductor results from the motion of both electrons and

positive charges (holes). A pure semiconductor may be doped with impu-

rity atoms so that the current is due predominantly either to electrons or

to holes. The transport of the charges, i.e. conductivity, in a crystal under

the in

uence of an electric eld (a drift current), and also as a result of a

nonuniform concentration gradient (a diusion current), is investigated in

this chapter.

||||||||||||||||||||||||||||||||||

2.1 Mobility and Conductivity As it is observed in the preceding chapter, according to the energy band theory, the

materials can be classied into three types: insulators, conductors and semiconductors.

A conductor is a solid in which an electric current

ows under the in

uence of the electric

eld. By contrast, application of an electric led produces no current in an insulator.

### 18 Transport Phenomena in Semiconductor

The energy gap for an insulator is so wide that hardly any electrons acquired enoughto jump to conduction band. If a constant electric eld is applied to the metal, the

electrons will move with acceleration equal:

a= e E m

(2.1)

where, Eis the electric eld in unit ( V m

1

)

But the electron suering from collisions with other particles in metal and it speed

between two successive collisions is ( a t). t: is the relaxation time. The distance between

two successive collisions is called the mean free pathand equal to:

l = v

d t

(2.2)

Electrical mobility is the ability of charged particles (i.e. electrons) to move through

a medium in response to an electric eld that is pulling them. The external electrical

eld gives electron drift velocity and acceleration, therefore the drift velocity is equal

to:

vd =

E (2.3)

where is the electron mobility in unit ( m2

V

1

s

1

) and it is equal to ( et=m).

The minus sign means the drift velocity is in the direction opposite to that of the external

electric eld.

Electrical Conductivity: Electrical conductivity is a measure of a material's ability to

conduct an electric current. It is commonly represented by Greek letter . The following

gure shows a box of metal with length Land cross section area A. The voltage Vwas

applied on the ends of the box. According to Ohm's law: V=I R and;

R = L A

(2.4)

where is the resistivity in unit (

m).

Since R= V =I

Then 2.4 becomes:

( I =A ) = (1 =) (V =L ) =J(Current density)

but ( V =L) =Eand (1 =) = , then:

2.1

Mobility and Conductivity 19

Figure 2.1: Box of metal

J= E (2.5)

Where Jrepresents the drift current density.

Now, consider that the metal contains ( n) of free electrons per unit volume, then the

total free electrons inside the metal is:

q= n e A L (2.6)

but I= ( q=t) = q (v

d=L

), where v

d = (

L=t )

Since I= J A and from 2.5 we have J= E , then it can be write:

E =q v

d L

A =

n e v

d (2.7)

Substituting 2.7 in equation 2.3, we can obtain:

= n e (2.8)

This equation shows that the conductivity depends on the density of the free elec-

trons and the mobility of these electrons. Equation 2.8 can be written in new form as

shown below:

= n e

2

l m v

d (2.9)

### 20 Transport Phenomena in Semiconductor

2.2 Diusion CurrentIn addition to a conduction current, the transport of charges in a semiconductor

may be accounted for by a mechanism called diusion, not ordinarily encountered in

metals. The essential features of diusion are now discussed. It is possible to have a

nonuniform concentration of particles in a semiconductor. As indicated in gure 2.2, the

concentration nof electrons varies with distance xin the semiconductor, and there exists

a concentration gradient, dn=dx, in the density of carriers. The existence of a gradient

implies the density of electrons immediately on one side of the surface is larger than

the density on the other side. The electrons are in a random motion as a result of their

thermal energy. Accordingly, electrons will continue to move back. The net transport

of electrons across the surface constitutes a current in the positive Xdirection. Figure 2.2: A nonuniform concentration n(x) results in a diusion current

It should be noted that this net transport of charge is not the result of mutual re-

pulsion among charges of like sign, but is simply the result of a statistical phenomenon.

This diusion is exactly analogous to th at which occurs in a neutral gas if a concentra-

tion gradient exists in the gaseous container. The diusion electron-current density J

n

(amperes per square meter) is proportional to the concentration gradient, and is given

by: J

dif f /

(dn=dx )

also, J

dif f /

D

where Dis diusion constant ( m2

sec

1

)

J dif f =

e D dn dx

(2.10)

D and are related by Einstein relation:

D= kT e

(2.11)

2.3

Work Function 21

where

Tis the temperature in o

K

Total Current: It is possible for both a potential gradient and a concentration gradient

to exist simultaneously within a semiconductor. In such a situation the total hole current

is the sum of the drift current 2.5 and the diusion current 2.10;

Jtot =

eD dn dx

+

E (2.12)

2.3 Work Function Free electron moves in metal by random motion in absent external operator or in

additional to drift and /or diusion motion. The kinetic energy makes electrons reach

Fermi level. Then the energy required rising electron to state outside the metal is E

s

(surface energy), therefore the work function ( ) given as:

= E

s

E

f (2.13)

The work function is the minimum energy (usually measured in electron volts)

needed to remove an electron from a solid to a point immediately outside the solid

surface (or energy needed to move an electron from the Fermi level into vacuum). Here

"immediately" means that the nal electron position is far from the surface on the

atomic scale but still close to the solid on the macroscopic scale. The work function

is a characteristic property for any solid phase of a substance with a conduction band

(whether empty or partly lled). For a metal, the Fermi level is inside the conduction

band, indicating that the band is partly lled. For an insulator, the Fermi level lies

within the band gap, indicating an empty conduction band; in this case, the minimum

energy to remove an electron is about the sum of half the band gap and the electron

anity. When the electron absorbs energy Ethen the K:E :for this electron outside the

metal will be:

1 2

m v

2

= E (2.14)

This is called electronic emission . There is four type of electronic emission:

Thermionic Emission, Photo Emission, Field Emission and Secondary Emission . If

### 22 Transport Phenomena in Semiconductor

a thermal energy is supplied to the electrons in the metal, then the energy distributionof the electrons changes, because of the increasing in the temperature. The thermal

energy given to the charge carrier overcomes the binding potential (work function) and

can release it from the metal surface. This is called Thermionic emission.According

to the Richardson-Dushman equation the emitted electron current density, J(A:m

2

),

is related to the absolute temperature Tby the equation:

J = A

oT 2

exp (

kT

) (2.15)

where ( A

o) is Richardson-Dushman constant. As mentioned before, the work func-

tion is the minimum energy that must be given to an electron to liberate it from the sur-

face of a particular substance. In the photoelectric eect, electron excitation is achieved

by absorption of a photon. If the photon's energy is greater than the substance's work

function, photoelectric emission occurs and the electron is liberated from the surface.

Excess photon energy results in a liberated electron with non-zero kinetic energy. The

photo-electric work function is:

= hf

o

f o is the minimum (threshold) frequency of the photon required to produce photo-

electric emission. Field emission (F

E) (also known as

eld electron emission andelectron eld

emission ) is emission of electrons induced by an electrostatic eld. Field emission in

pure metals occurs in high electric elds and strongly dependent upon the work function.

The emission current density is given as:

J/ exp (

e E x

o) (2.16)

x o is the gap thickness.

Secondary electron emission is a phenomenon where primary incident electrons

of sucient energy, when hitting a surface of material, induce the emission of secondary

electrons. It was found experimentally the number of secondary electrons depend on

the following the number and the energy of primary electrons, the angle of incidence of

the particles on the material, the type of the material, and the physical condition of the

surface. The secondary emission ratio ( ) dened as:

2.3

Work Function 23

= no: of the secondary emitted electrons no: of the primary incident electrons

(2.17)

Solved problems 1.

A silicon crystal having across section area of ( 0:001 cm2

) and a length

of ( 10

3

cm ) is connected at its ends to ( 10V) battery at temperature

( 300 o

K ). Find the resistivity and the conductivity of the silicon crys-

tal if the current passing through the crystal is ( 100mA ).

Solution J

= E )= J E

=

I =A V =L

= 100

10

2

= 0:001 10

4 10

=(10

3

10

2

) = 10 (

m)

1

= 1

=

1 10

= 0

:1

m

.....................................................

2. Calculate the average drift velocity of hole in a bar of silicon with

across sectional area ( 10

4

cm 2

), containing holes concentration of ( 4:5

10 15

cm

3

) and carrying a current of ( 45mA )?

Solution v

d =

E ::::: (a )

= pe ::::: (b )

J = E )= J E

:::::

(c )

Therefore equation (b) becomes: J E

=

pe )= J pe

:::::

(d )

From equations (a and d),

vd =

J peE

E = J pe

=

I =A pe

=

45

10

3

= 10

4

10

4 4

:5 1015

106

1:6 10

19 = 6250

ms

1

### 24 Transport Phenomena in Semiconductor

3.A current of 1A passing through an intrinsic silicon bar has 3mm length

and 50100 m2

cross-section. The resistivity of the bar is 2:3 10 5

cm

at 300o

K . Find the voltage across the bar?

Solution J

d = I A

=

E

E = J

d

=

I A

1

=

I A

E = 10

6 50

10

6

100 10

6

2:3 105

10

2

= 4 :6 105

V :m

1

V bar =

EL= 4 :6 105

3 10

3

= 1380 V

4. The electron density variation along the x-axis is given as [ 1028

exp ( 10

6

x)].

Find the diffusion current at ( x= 0 ) and ( x= 10

5

m ) if the mobil-

ity of electron is ( 4 10

3

m 2

V

1

s

1

) at T= 300 o

K ?

Solution J

dif f =

eD dn dx

D = kT e

= 1

:38 10

23

300 1

:6 10

19

4 10

3

= 1 :035 10

4

m 2

sec

1

dn dx

=

10 28

10

6

exp

10

6

x

J dif f = 1

:6 10

19

1:035 10

4

10 28

10

6

exp

10

6

x

1 :J

dif f (at x =0) =

1:6 10

11

Am

2

2 :J

dif f (at x =10

5

) = 1

:035 10

4

1:6 10

19

10 28

10

6

exp

10

6

10

5

= 1 :76 1011

Am

2

5. A bar of copper of ( 2cm ) length and resistively of ( 1:8 10

8

### m ) is

connected to power supply of ( 10V). Find the mobility and drift ve-locity of the electrons if electron density in copper is ( 8:5 10 28

m

3

)?

Solution

= n e )= 1 n e

2.4

Generation and Recombination of Charges 25

= 1 8

:5 1028

1:6 10

19

1:8 10

8 = 4

:08 10

3

m 2

V

1

s

1

2.4 Generation and Recombination of Charges Generation = break up of covalent bond to form electron and hole pairs. A pure

silicon crystal at room temperature derives heat (thermal) energy from the surrounding

environment, causing some valence electrons to gain sucient energy to jump the gap

from valence band into the conduction band, becoming free electrons. When an electron

jumps to C.B., a vacancy is left in the valence band. This vacancy is called a hole.

If nand pis the free electron and hole concentration, respectively, per volume unit, Figure 2.3: Free charge carrier generation in semiconductor

at equilibrium status n= p= n

i. Where

n

i is the carrier concentration. Recombination

occurs when a conduction-band electron loses energy and falls back into a hole in the

valence band.

To summarise, a piece of intrinsic semiconductor at room temperature has, at any in-

stant, a number of conduction-band (free) electrons that are unattached to any atom and

are essentially drifting randomly throughout the material. There is also an equal number

of holes in the valence band created when these electrons jump into the conduction band.

Electron and Hole Current

When a voltage is applied across a piece of intrinsic silicon the thermally generated

free electrons in the conduction band, which are free to move randomly in the crystal

structure, are now easily attracted toward the positive end. This movement of free

electrons is one type of current in a semiconductor material and is called electron current.

### 26 Transport Phenomena in Semiconductor

Another type of current occurs in the valence band, where the holes created by the freeelectrons exist. Electrons remaining in the valence band are still attached to their atoms

and are not free to move randomly in the crystal structure as are the free electrons.

However, a valence electron can move into a nearby hole with little change in its energy

level. thus leaving another hole where it came from. Eectively the hole has moved

from one place to another in the crystal structure. as illustrated in Figure 2.4 This is

called hole current . Figure 2.4: Hole current in intrinsic silicon

2.4.1 Electrons and Holes Density in an Intrinsic Semiconductor In a pure (

intrinsic) semiconductor the number of holes is equal to the number of

free electrons and the electrical properties determined by host martial. In intrinsic

semiconductor the carrier concentration can be determined from Fermi-Dirac function

distribution:

n= N

cexp

E

c

E

f kT

(2.18)

p = N

vexp

E

f

E

v kT

(2.19)

2.4

Generation and Recombination of Charges 27

Where

nand pare the electron and hole concentration, respectively. N

cis the active

level density at C.B. and N

vis the active level density at V.B. and given by:

Nc=

2 m

n kT h

2

3=2

and Nv=

2 m

p kT h

2

3=2

m

n : eective mass of electron.

m

p : eective mass of hole.

( Eective mass : When we apply an external force to an electron in a crystal, it may

not respond as if it were a free electron. This is because of the interaction with the

crystal lattice). Then the number of carriers is:

ni= p n p

=p N

cN

vexp

E

c

E

v 2

kT

(2.20)

but E

g=

E

c

E

v, then;

ni= p n p

=p N

cN

vexp

E

g 2

kT

(2.21)

It can be observed that the concentration of electrons and holes in pure semiconductor

independent on the location of the Fermi level but it is depending on the temperature.

2.4.2 Electrons and Holes Density in an Extrinsic Semiconductor Semiconductor materials do not conduct current well and are of limited value in their

intrinsic state. This is because of the limited number of free electrons in the conduction

band and holes in the valence band. Intrinsic silicon (or germanium) must be modied

by increasing the number of free electrons or holes to increase its conductivity and

make it useful in electronic devices. This is done by adding impurities to the intrinsic

material as you will learn in this section. Two types of extrinsic (impure) semiconductor

materials, n-type and p-type, are the key building blocks for most types of electronic

devices. An extrinsic semiconductor can be formed from adding impurity atoms to

the intrinsic semiconductor in process known as doping. The electrical properties of

extrinsic semiconductor are determined by chemical impurities. For example, silicon

### 28 Transport Phenomena in Semiconductor

has four valence electrons. Doping silicon with Aluminum (Al) will produce a hole. The

dopant atoms have not enough number of electrons to share bonds with surrounding

silicon atoms. One of the silicon atoms has a vacancy for an electron. It creates hole

that contribute in conduction process and the semiconductor is called p-type as shown in

gure 2.5(a). The dopant atoms are called acceptors. While if the silicon is doped with

Phosphor ( P) or Arsenide ( As), which they have extra electron in valence bands, the

dopant atoms contributes an additional electron to the crystal and the semiconductor

is called n-type as shown in gure 2.5(b). The dopant atoms are called donors. Figure 2.5: Extrinsic s.c. (a) n-type s.c. (b) p-type s.c.

Determination of electrons density for n-type semiconductor

If, to intrinsic silicon, there is added a small percentage of phosphor ( P) atoms, a

doped, impure, or extrinsic, semiconductor is formed. The fth electron of the phosphor

( P ) will be released by energy 0 :05 eV , which is the smallest energy required an electron

of silicon atom by 20 times ( E

g= 1

:1 eV ).

Then the density of electron in host semiconductor which doped by N

D atoms is:

n = N

D

N D is the

concentration of donor atoms .

The increasing of the electron density in conduction band case shifting in Fermi-

level up word to C.B, then the dierence in energy between old and new position of

Fermi-level is:

E

n =

E

f n

E

f i

2.4

Generation and Recombination of Charges 29

Figure 2.6: Energy band structure in n-type s.c.

The concentration of the electrons in conduction band is: n= N

D =

N

cexp

E

c

E

f n kT

(2.22)

but, E

f n =

E

n +

E

f i

then 2.22 becomes;

n= N

D =

N

cexp

E

c

E

n

E

f i kT

n = N

D =

N

cexp

E

c

E

f i kT

exp

E

n kT

Since;

ni=

N

cexp

E

c

E

f i kT

then, n= N

D =

n

iexp

E

n kT

(2.23)

From the above equation the shift in Fermi level can be calculated as:

E

n =

kT ln

n n

i

= kT ln

ND n

i

(2.24)

### 30 Transport Phenomena in Semiconductor

Determination of holes density of n-type semiconductorThe shifting of Fermi

level by E

n up word in the n-type semiconductor, as shown in gure 2.6, means the

new Fermi level shift away from valance band, and this will cause a new concentration

of holes. Since the hole density given as:

p= N

vexp

E

f n

E

v kT

(2.25)

but, E

f n =

E

n +

E

f i then 2.25 becomes;

p = N

vexp

E

f i

E

v+

E

n kT

p = N

vexp

E

f i

E

v kT

exp

E

n kT

Since; ni=

N

vexp

E

f i

E

v kT

then, p= n

iexp

E

n kT

(2.26)

It can be observe that the hole density decrease with Fermi level shifting upward:

n p=n2

i

Since n= N

D, then;

p= n2

i =N

D

Determination of holes density for p-type semiconductor The doping solid

is the aluminum ( Al) or boron ( B), which have 3 valence electrons only, so the doping

atom need one additional electron to bonded with the silicon structure, in this case it's

can be lled from the nearest bond electrons and this will cause break up the bond near

vacancy. The energy which required is (0.05 eV) for the boron ( B) atoms, the number

of charge carriers is equal to the number of holes (doping atoms). The impurity atoms

in this case called the acceptors atoms and it's density is:

p= N

A

where N

A is the

concentration of acceptor atoms .

2.4

Generation and Recombination of Charges 31

The concentration of holes can be determined from Fermi-Dirac function as below:

p= N

vexp

E

f p

E

v kT

(2.27)

new position of Fermi-level is: E

p =

E

f i

E

f p Figure 2.7: Energy band structure in p-type s.c.

p= N

A =

N

vexp

E

f i

E

p

E

v kT

p = N

vexp

E

f i

E

v kT

exp

E

p kT

Since; ni=

N

vexp

E

f i

E

v kT

then, p= n

iexp

E

p kT

(2.28)

From the above equation the shift in Fermi level can be calculated as:

E

p=

kT ln

p n

i

= kT ln

NA n

i

(2.29)

and the number of electrons is equal to:

n= n

2

i N

A

### 32 Transport Phenomena in Semiconductor

:::::}}}}}}}::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::}}}}}}}:::::Solved problems ...........................................................................

1. The electron density in pure silicon is 1:45 10 16

m

3

at 300o

K . Find

the electron density when the temperature change to 350o

K , take E

g=

1 :1 eV ?

Solution n

i1 = p N

cN

vexp

E

g 2

kT

1

::::: (1

n i2 = p N

cN

vexp

E

g 2

kT

2

::::: (2

n i1 n

i2 = exp

E

g 2

kT

1 exp

E

g 2

kT

2

n i2 =

n

i1 exp

Eg 2

kT

1 exp

Eg 2

kT

2

n i2 =

n

i1 exp

Eg 2

k

1 T

1

1 T

2

= 1:45 10 16

exp

1:1 2

8:62 10

5

1 300

1 350

n i2 = 3

:03 1017

m

3

||||||||||||||||

2. Pure semiconductor with energy gap of 1:42 eV and charge carrier den-

sity of 1:79 10 12

m

3

at 300 o

K . Determine the position of the Fermi

level with respect of the mid of gap if N

c= 4

:7 1023

m3. What is

the value of N

v?

Solution Since the charge concentration in pure semiconductor is equal to electron concen-

tration, so:

n= N

cexp

E

c

E

f kT

2.4

Generation and Recombination of Charges 33

E

c

E

f =

kT ln N

c n

=

1

:38 10

23 1

:6 10

19

300 ln

4:7 1023 1

:79 1012

= 0 :68 eV

The Fermi level located at 0 :68 eV under the E

c, but the mid of the energy gap

at 0 :71 eV under the E

c. Therefore the position of Fermi level would be 0

:03 eV

above the mid of the gap.

** Home work )Find N

v**

||||||||||||||||

3. Pure silicon has electron concentration 1:45 10 16

m

3

at 300 o

K was doped

with 1022

m

3

phosphor ( P) atoms. Find the electron and hole densities

at 300 o

K and 500o

K ?

Solution n

i = 1

:45 1016

m

3

before doping at 500 o

K the doping solid is phosphor ( P)

which is donor atoms, then; n= N

D = 10 22

m

3

the density of the solid after

doping at 300 o

K

ni1 n

i2 = exp

E

g 2

kT

1 exp

E

g 2

kT

2

n i2 =

n

i1 exp

Eg 2

kT

1 exp

Eg 2

kT

2

n i2 =

n

i1

exp

Eg 2

k

1 T

1

1 T

2

= 1:45 10 16

exp

1:1 2

8:62 10

5

1 300

1 500

n i2 = 7

:2 1019

m

3

p 2 = n

2

i 2 N

D = 5

:2 1017

m

3

||||||||||||||||

4. The electron concentration in pure silicon is 1:5 10 16

m

3

at 300 o

K .

The silicon was doped with 1022

m

3

donor atoms. Find the electron and

hole densities after doping and calculate the position of the new Fermi

level with to the initial position?

### 34 Transport Phenomena in Semiconductor

SolutionAfter doping the density of electron is

n= N

D = 10 22

m

3

, while the hole density

is given as;

p= n

2

i N

D =

1

:5 1016

2 10

22 = 2

:25 1010

m

3

E

n =

kT ln

ND n

i

E

n = 1

:38 10

23 1

:6 10

19

300 ln

1022 1

:5 1016

= 0 :347 eV

The new position of Fermi level is above the initial position by 0 :347 eV

||||||||||||||||

5. If the position of Fermi level in impure semiconductor at 0:3 eV above

the mid of the energy gap at 300o

K , where the E

g = 1

:1 eV and n

i =

1 :45 1016

m

3

.

What is type of the impurities and what are its concentration?

Solution Since the Fermi level location is above the mid of the energy gap, therefore the

semiconductor would be from n-type. So the impurities are donor atoms.

ND =

n

i

exp

E

n kT

N D = 1

:45 1016

exp

0:3 8

:614 105

300

= 1 :58 1021

m

3

2.5 Electrical conduction in semiconductor As it was seen that the electron motion and the electrical conduction in metal depend

on several parameters which describe the electrical motion in metal, this description for

the electron motion and the electrical conduction is same to these in semiconductor but

with take care the ratio of doping.

2.5

Electrical conduction in semiconductor 35

2.5.1 Electrical conduction in intrinsic semiconductor

The electrical conduction in intrinsic semiconductor is same the general formula of

conductivity in metal:

= n e

Applied this formula on an intrinsic semiconductor, the electrons and holes contribute

in electrical conduction then:

i =

n e

n+

p e

p (2.30)

where; n =concentration of electrons ( m

3

)

n=electron mobility (

m2

V

1

s

1

)

p =concentration of holes ( m

3

)

p=hole mobility (

m2

V

1

s

1

)

In pure semiconductor the concentration of electrons equal to concentration of holes;

i.e., p= n= n

i, then equation 2.30 can be written as;

i =

n

ie

(

n +

n) (2.31)

and n

i is;

ni= p N

cN

vexp

E

g 2

kT

n i= s

2 m

n k T h

2

3=2

2 m

p k T h

2

3=2

exp

E

g 2

kT

n i=

2 k h

2

3= 2

m

n m

p

3=4

T3

=2

exp

E

g 2

kT

n i/

T3

=2

exp

E

g 2

kT

### 36 Transport Phenomena in Semiconductor

Since it is known that;n = e t

n m

n

and

p = e t

p m

p

where, t

n and

t

p are the relaxation time of electrons and holes respectively.

The mobility depends on the relation time and eective mass of moving charges.

Heating the semiconductor causes vibration of atoms and this will eect on electron

motion inside the crystal structure and hence the electrons collision with atoms will

increase due to vibration of atoms and therefore the mobility will decreases.

/ T

3= 2

Then it can be conclude that: / n

i=

) / T3

=2

exp

E

g 2

kT

and; / =) / T3

=2

As a result of that the semiconductor conductivity aected by temperature as:

i =

oexp

E

g 2

kT

(2.32)

where,

o is a constant and

independent to temperature . Equation 2.32 can also be

written as:

ln(

i) =

ln(

o)

E

g 2

kT

The semiconductor conductivity changes strongly with temperature variation.

1

id

i dT

=

E

g 2

kT (2.33)

2.6

Diusion and Drift currents density in semiconductor 37

2.5.2 Electrical conduction in extrinsic semiconductor

When the semiconductor is doped by impurities has N

D concentration (

n >> p):

(n ) =

n e

n+

p e

p (2.34)

In other words equation 2.34 can be written as;

(n ) =

n +

p

But n p= n2

i =

) N

D

p= n2

i =

) p= ( n2

i =N

D), then equation 2.34 can also be

written as;

(n ) =

N

D e

n+ n

2

i N

D e

p (2.35)

* N

D =

n >> n

ithat is meaning the

pconcentration has no eect and (

n >>

p), so:

(n ) =

N

D e

nIn the same manner if the semiconductor is doped by impurities have

N A acceptor atoms concentration (

N

A =

p >> n

i):

(p ) =

N

Ae

p

2.6 Diusion and Drift currents density in semiconductor There are two mechanisms by which holes and electrons move through a silicon

crystal diusion and drift. Diusion current density: As aforementioned the diusion current density is

given as:

Jdif f =

e D dn dx

(2.36)

Since the diusion current density in semiconductor is due to electrons and holes

motion, then:

Jdif f =

J

dif f (n ) +

J

dif f (p ) =

e

D ndn dx

+

D

pdp dx

(2.37)

Drift current density: As aforementioned the drift current density is given as:

Jd =

e n

n (2.38)

### 38 Transport Phenomena in Semiconductor

The free electrons will drift in the direction opposite to that of E.The total drift

current density is obtained by combining the two charge carriers:

Jdrif t =

J

d =

e E (n

n+

p

p) (2.39)

2.7 Photo-conductivity When the semiconductor exposure to electromagnetic wave has energy (

hf) then

this energy will cause a generation of a new charge carriers contribute in electrical

conduction process, this called the Photo-conductivity: If the energy of exposure photon

is: hf E

g. In other words the minimum wavelength of the absorbed electromagnetic

radiation which can produce a new charge carriers will given as:

1

:24 E

g (

m ) (2.40)

The ability of the semiconductor to absorb photons depend on its nature and fre-

quency. If the semiconductor surface exposure to the ray of the photons

n ph (o )

, so the

number of the photons will decrease with penetration depth ( x) of the surface and the

number of the photons which will arrive to depth ( x) would be:

n ph (x ) =

n

ph (o ):exp

( x ) (2.41)

: is the absorption constant. proportion to the absorption of solid ability to photons,

so if is large the solid has good ability to absorb.

2.7

Photo-conductivity 39

Solved problems

1.

Pure germanium has (41022

) atom: cm

3

doped by indium atoms, the

impurity is add to the extent of 1 part in (108

) germanium atoms, if

the intrinsic concentration of germanium 2:5 1013

cm

3

, note that

n = 3800

cm2

(V s )

1

and

p = 1800

cm2

(V s )

1

.

(a) Find the conductivity and the resistivity before the doping?

(b) Find the conductivity and the resistivity after the doping?

(c) What can you conclude from 1 and 2?

Solution 1. The conductivity of pure semiconductor (before doping) is given by:

= ne

n+

pe

p

since the semiconductor is intrinsic then, n= p= n

i

= n

ie

(

n +

p)

= 2 :5 1013

1:6 10

19

(3800 + 1800) = 0 :0224 S cm

1

= 1

= 1 0

:0224 = 44

:64

cm

2. Doping pure germanium with indium will produce increasing in hole density,

so:

NA = 4

1022 10

8 = 4

1014

cm

3

n p= n2

i )

n N

A =

n2

i )

n= n

2

i N

A

n = (2

:5 1013

)2 4

1014 = 1

:56 1012

cm

3

### 40 Transport Phenomena in Semiconductor

(2:5 103

)2 10

14

= ne

n+

pe

p

= 1 :6 10

19

1 :56 1012

3800 + 4 1014

1800

= 0 :116 S cm

1

= 1

=

1 0

:116 = 8

:62

cm

3. The ratio between the conductivity after and before doping is given as: (af ter doping )

(bef ore doping )=

0

:116 0

:0224 = 5

:18

The conductivity of the germanium increased more than 5 time after doping with

indium.

2. Pure silicon doped by antimony has concentration equal to

2 1015

atom: cm

3

, until N

D

N

A

2n

i, note that they represent

replacement of less than 10

5

% of the atoms in the silicon. Find

the conductivities

(n );

(p ) and

and the resistivity of the

silicon? note that

n = 1260

cm2

(V s )

1

and

p = 460

cm2

(V s )

1

.

Solution When the pure silicon doped with antimony atoms mean doping by donor atoms:

ND = 2

1015

atom cm

3

) 10

5

2 1015

0 2n

i

n i= 10 10

atom cm

3

n:p =n2

i

p N D=

n2

i =

) p= ( n2

i )

=N

D

p = (10 10

)2

= (2 1015

) = 5 104

atoms cm

3

n =

N

D e

n= 2

1015

1:6 10

19

1260 = 0 :403 S cm

1

p =

p e

p= 5

104

1:6 10

19

460 = 368 10

14

S cm

1

=

n +

p

2.8

Hall Eect 41

*

n

p

) =

n

= 0 :403 S cm

1

= 1

=

1 0

:403 = 2

:48

cm 2.8

Hall Eect Suppose that an electric current

J

x is

ow in a semiconductor in the x-direction,

and a magnetic eld B

z is applied normal to the s.c. in the z-direction. The current

J x will cause from the motion of holes (if the s.c. from p-type) with speed

v

Dx under

the in

uence the electric eld E

x. These holes will eect to force

F

L known as Lorentz

Force and in ( y) direction as shown in Figure below. Figure 2.8: Hall eect setup

FL =

e v

Dx

B

z (2.42)

This force will push the holes in the front surface direction; this will cause a high

density of holes on this surface while the back surface would be empty from the holes.

Since the

ow of hole current in y-direction ( E

y). This eld cause from the distribution

of holes, and the eld will cause a force on holes equal to Lorentz force, i.e.:

### 42 Transport Phenomena in Semiconductor

e Ey=

e v

Dx

B

z (2.43)

but Jx =

e v

Dx

p

E y= J

x B

z p e

(2.44)

or, 1 p e

=

E

y J

x B

z

The quantity 1 p e

called Hall Coecient ( R

H), then;

RH = 1 p e

=

E

y J

x B

z (2.45)

The induced potential between front and back surfaces can be measured and then;

VH =

EW

If the thickness of the slide is dthen,

I = J

x

W d

Substituting the value of J

x in equation 2.45, then:

RH = V

H d I B

z=

1 p e

(2.46)

Since V

H ; d I ;

andB

Z are a measurable values, then it can be calculated the Hall

coecient experimentally and hence calculated the hole concentration from equation

2.46. Similar analysis can be done for a semiconductor of n-type and nd:

RH = 1 n e

(2.47)

It is seen in last gure that there is a result of electrical eld ( E) in ( ) direction

about the direction of ( E

x), then,

2.8

Hall Eect 43

tan

=E

y E

x (2.48)

Substituting the value of ( E

y) from equation 2.46 in equation 2.48, then it can nd,

tan =J

x B

z p e

J

x =

pB

z (2.49)

or, p =

R

H

(2.50)

According to this simple analysis it can measured the mobility of hole from Hall coe-

cient and conductivity. Solved problems

1.

A current of ( 0:12 A) pass through n-type semiconductor have a width of

( w = 2 mm) and thickness of ( d= 1 mm). If the voltage along the

width of this sample of semiconductor is ( 3:4 mV ) and a normal magnetic

field of ( 500Gauss ) applied on this piece of semiconductor, find the

Hall coefficient and electron density?

Solution R

H = V

H

d I

B

z

R H = 3

:4 10

3

1 10

3 0

:12 500 10

4 = 5

:6 10

4

m 3

C

1

R H = 1 n e

5 :6 10

4

= 1 n

1:6 10

19 )

n= 2 1022

m

3

### 44 Transport Phenomena in Semiconductor

2.Find the Hall coefficient, electron density and the angle between the

field components for n-type semiconductor wire having thickness of ( d=

2 mm )? The normal applied magnetic field on this semiconductor is ( B=

0 :1 T ) and the current which passing through it is ( 10mA ), and ( V

H =

1 mV ), (

n = 0

:36 m2

s

1

).

Solution R

H = V

H

d I

B

z

R H = 1

10

3

2 10

3 10

10

3

0:1 = 2

10

3

m 3

C

1

R H = 1 n e

2 10

3

= 1 n

1:6 10

19 )

n= 3 :1 1021

m

3

tan =

pB

z= 0

:36 0:1 = 0 :036 )= tan

1

(0 :036) = 2 o

2.8

Hall Eect 45

Review Questions

1.

Dene mobility? Give its dimensions?

2. Indicate pictorially how a hole contributes to conduction?

3. (a) Dene intrinsic concentration of holes. (b) What is the relationship between this density and the intrinsic concentration for electrons? (c) What do these equal at 0 o

K ?

4. Given an intrinsic semiconductor specimen, state two physical processes for increasing its conductivity? Explain brie

y.

5. Explain physically the meaning of the following statement: An electron and a hole recombine and disappear?

6. Dene (a) donor, (b) acceptor impurities?

7. What properties of a semiconductor are determined from a Hall eect experiment?

8. A pure silicon contains 5 10 28

atom per cubic meter and the ratio of broken bonds are

one bond per 10 12

silicon atom at 38 o

C . What is the ratio of broken of covalent bonds

if the temperature raised to 75 o

C , where E

g= 1

:1 eV ?

9. A piece of pure semiconductor contains 5 10 18

donor atoms at 27 o

C . How far the Fermi

level will move and in which direction if an additional donor atoms of concentration of

10 22

will put in?

10. What is the wavelength of the electromagnetic waves which can release an electron from Germanium and Silicon? where E

g(

Ge ) = 0 :66 eV and E

g(

S i ) = 1 :1 eV .

11. The charge carriers concentration in pure silicon is 4 :5 10 16

m

3

at 300 o

K . Where the

Fermi level would be if the silicon doped by 1 10 21

donor atoms? Find the location of

Fermi level at 200 o

K and 900 o

K . Use E

g(

S i ) = 1 :1 eV .

12. If the conductivity of the pure Germanium change with temperature according the fol- lowing relation, so what is the energy gap of the Germanium?

exp

4350 T

### 46 Transport Phenomena in Semiconductor

13. A resistor of pure Silicon with a resistance of 2500at 20

o

C , if the resistance of this

resistor increased by 1% of its initial vaule when the temperature increased to 100 o

C ,

what is the energy gap of Silicon?

14. The resistivity of pure Silicon doped with impurities is 9 :27 10

7

### m and the Hall

coecient is 3 :84 104

m3

C

1

, what is the density and the mobility of impurities

atoms?

====== ||||||||||||||||||||||||||||||||||||| ======

### Chapter 3

P-N Junction (Diode)|||||||||||||||||||||||||||||||||{ If one side of a piece of silicon dope with a trivalent impurity and the

other side with a pentavalent impurity, a ( p n) junction will formed be-

tween the resulting p-type and n-type portions and a basic diode will cre-

ated. A diode is a device that conducts current in only one direction. In

this chapter we demonstrate the characteristics of the ( p n) junction re-

gion. The volt-ampere characteristics of the ( p n) junction is studied. The

capacitance across the junction is calculated.

||||||||||||||||||||||||||||||||||

3.1 The structure of p-n junction If a donor impurities are doped into one side and acceptors into the other side of a

single crystal of a pure semiconductor, a ( p n) junction will create. This two-terminal

device called a junction diode. 3.1 Fig.(1) shows the schematic symbol of the diode. The

key feature of this device that conducts current in only one direction. When the n-typeFigure 3.1: Diode schematic symbol.

### 48 P-N Junction (Diode)

semiconductor connect top-typesemiconductor the nregion loses free electrons as they

diuse across the junction. This creates a layer of positive ions near the junction. The

p region loses holes as electrons and holes combine. This creates a layer of negative ions

near the junction. These two layers of positive and negative form the depletion region,

as shown in gure3.2, with a built-in potential which is called the contact or barrier

potential (V) .The depletion region is completely free from the charges. Figure 3.2: Formation of the depletion region.

From the basic conception of the semiconductor, it is easy to determine the value

of the barrier orcontact potential . From the energy levels in gure 3.3 it can be

observed that:

Ecp

E

cn =

e V

o

In p-side, the electron concentration as a minor charge carrier is given as:

np =

N

c

exp

E

cp

E

f p kT

E cp =

E

f p

kT ln

np N

c

whilst the electron concentration as a ma jor charge carrier in n-side is given as: nn =

N

c

exp

E

cn

E

f n kT

E cn =

E

f n

kT ln

nn N

c

3.1

The structure of p-n junction 49

Figure 3.3: The P-N junction and its energy levels after contacted.

and eVo=

E

cp

E

cn

) eV

o=

E f p

kT ln

np N

c

E f n

kT ln

nn N

c

]

Since the junction is in equilibriumstatus, then: )E

f p =

E

f n , then it can be write:

V o = kT e

ln

nn n

p

(3.1)

Equation 3.1 can also be written as; nn =

n

p

exp

eV o kT

Since n

n =

N

D and

n

p =

n2

i =N

A, then the above equation becomes:

V o = kT e

ln

ND

N

A n

2

i

(3.2)

In same way, from n-side it could be found;

Vo = kT e

ln

pp p

n

or p p=

p

n

exp

eV o kT

### 50 P-N Junction (Diode)

Solved problems1.

A pn junction was formed from two pieces of silicon contain N

D = 10 24

m

3

and N

A = 10 20

m

3

at 300 o

K . Find the barrier potential? The carrier

concentration for pure silicon is: n

i= 1

:45 1016

m

3

.

Solution V

o = kT e

ln

ND

N

A n

2

i

= 1

:38 10

23

300 1

:6 10

19

ln

1024

1020 (1

:45 1016

)2

= 0 :7 volt

2. A pure silicon has n

i= 1

:45 10 16

m

3

, doped in its two ends by phos-

phor ( P) and Boron ( B) with carrier densities N

D = 10 22

m

3

and N

A =

10 20

m

3

, respectively to form a p-n junction at 300o

K . Find:

(a) The location of Fermi-level in n and p parts?

(b) Determine the barrier potential?

Solution (a) For n-side

ND =

n

i

exp

Ef n

E

f i kT

) E

f n

E

f i =

kT lnN

D n

i =

1

:38 10

23

300 1

:6 10

19

ln

1022 (1

:45 1016

)

= 0 :35 eV above F ermi level

(b) For p-side NA =

n

i

exp

Ef i

E

f p kT

) E

f i

E

f p =

kT lnN

A n

i =

1

:38 10

23

300 1

:6 10

19

ln

1020 (1

:45 1016

)

= 0 :27 eV below F ermi level

V o = kT e

ln

ND

N

A n

2

i

= 1

:38 10

23

300 1

:6 10

19

ln

1022

1020 (1

:45 1016

)2

= 0 :62 volt

3.2

PN junction characteristics 51

3.2 PN junction characteristics

If the external potential of V volt is applied across the P-N junction this will bias

the diode. There are two type of diode bias:

3.2.1 Forward bias Connecting an external voltage across the junction will cause a disturbance in the

junction equilibrium status and a net current will

ow through the P-N junction. Con-

necting the positive terminal of the external voltage source to the p-side and the negative

terminal to the n-type will cause a forward bias for the junction as shown in gure 3.4. Figure 3.4: Forward bias voltage (V) applied to p-n junction and the result energy band

structure.

The application of forward bias potential Vwill cause an injection of electrons from

n-side and hole from p-side in opposite direction across the junction region and some

of these carriers will recombine with the ions near the boundary region and reduce

the width of depletion region. On being injected across the junction, these carriers

immediately become minority carriers and the density of the minority carriers near the

junction rise to new values n

po and

p

no . as shown in gure 3.4.

n po =

n

n

exp

e

(V

o

V) kT

(3.3)

n po =

n

n

exp

e

(V

o) kT

exp

eV kT

### 52 P-N Junction (Diode)

Figure 3.5: The excess carriers in the terminals of junction.*n

n =

n

p

exp

e V o kT

S o n p=

n

n

exp

e V

o kT

) n

po =

n

p

exp

e V kT

(3.4)

where n

po is the electrons excess at the edge of the depletion region in

pside.

In same manner it can be prove that:

)p

no =

p

n

exp

e V kT

(3.5)

where p

no is the electrons excess at the edge of the depletion region in

nside.

As shown in gure 3.5, the density of carrier decreases far from the depletion region

to exist the forward bias.

n(x ) = n

p

(n

p

n

po )

exp

x L

n

(3.6)

L n :the electron diusion length in p-side.

p(x ) = p

n

(p

n

p

no )

exp

x L

p

(3.7)

L p :the electron diusion length in n-side.

The total current densities are due to holes and electrons diusion motion is dened

as:

JD =

J

n +

J

p =

e D

ndn dx

+

e D

pdp dx

3.2

PN junction characteristics 53

with caring to the direction of (

x), then the above equation can be written as:

J D =

e D nn

p L

n +

e D

pp

n L

p

exp

eV kT

1

(3.8)

Equation 3.8 can be written as; JF =

J

s

exp

eV kT

1

(3.9)

where; Js =

e D nn

p L

n +

e D

pp

n L

p

The value J

s represents the saturation current density, whilst

J

F the forward current

density. Saturation current

ows through the junction in equilibrium case between the

diusion currents under the eect of ( V

o).

3.2.2 Reverse bias If the positive terminal of the applied voltage connect to the

n type and the negative

terminal to the p type , as illustrated in gure 3.6, the junction will bias in reverse

direction. Because unlike charges attract, the positive side of the bias-voltage source

" pulls " the free electrons, which are the ma jority carriers in the nregion, away from

the pnjunction. As the electrons

ow toward the positive side of the voltage source,

additional positive ions are created. This results in a widening of the depletion region

and a depletion of ma jority carriers. In the pregion, electrons from the negative side Figure 3.6: A (p-n) junction under the reverse bias and the equivalent energy band

structure.

### 54 P-N Junction (Diode)

of the voltage source enter as valence electrons and move from hole to hole toward thedepletion region where they create additional negative ions. This results in a widening

of the depletion region and a depletion of ma jority carriers. The

ow of valence electrons

can be viewed as holes being "pulled"toward the positive side. The initial

ow of charge

carriers is transitional and lasts for only a very short time after the reverse-bias voltage

is applied. As the depletion region widens, the availability of ma jority carriers decreases

as shown in gure 3.7. As more of the nand pregions become depleted of ma jority

carriers, the electric eld between the positive and negative ions increases in strength

until the potential across the depletion region equals the bias voltage, V

bias . At this

point. The transition current essentially ceases (dies) except for a very small reverse

current that can usually be neglected. Reverse Current: The extremely small current Figure 3.7: The carrier densities in bulk of reverse bias p-n junction.

that exists in reverse bias after the transition current dies out is caused by the minority

carriers in the nand pregions that are produced by thermally generated electron-hole

pairs. The small number of free minority electrons in the p region are "pushed"toward

the pnjunction by the negative bias voltage. When these electrons reach the wide

depletion region, they "fall down the energy hill" and combine with the minority holes

in the nregion as valence electrons and

ow toward the positive bias voltage, creating

a small hole current. The conduction band in the pregion is at a higher energy level

than the conduction band in the nregion. Therefore, the minority electrons easily pass

through the depletion region because they require no additional energy. The current in

reverse-bias condition called Reverse Saturation Current ( I

s)

.

3.2

PN junction characteristics 55

The applied voltage lead to increase the height of barrier potential to a new value

equal to ( V

o +

V), thereby reducing the diusion current through the junction. The

current density at reverse bias is given as:

JR =

J

s

1 exp

eV kT

(3.10)

where J

R is the reverse current density.

The general formula of the current density in the p-n junction can be form as:

J= J

s

exp

eV kT

1

(3.11)

where (+):forward and ( ):reverse. If the junction cross-section area is equal to ( A)

then the current is:

I= A:J and I

s=

A:J

s

) I= I

s

exp

eV kT

1

(3.12)

Figure 3.8 shows the relation between the current and voltage in both forward and Figure 3.8:

I V characteristics of the p njunction.

reverse biases. For the large values of V

F , where (

V

F >

4kT ) the forward current is:

I F =

I

s exp

eV F kT

and for long time the reverse bias ( I

R =

I

S ).

### 56 P-N Junction (Diode)

3.3 Depletion region and depletion capacitancesThe two space charge density layers at the junction vary in width with the applied

voltage, and therefore in the amount of charges they contain as the bias voltage changes.

If ( w) is the width of the depletion region then:

w= d

n +

d

p (3.13)

where ( d

n) with in the n-side and (

d

p) with in the p-side.

It can be related between ( d

n and

d

p) from the charge density in these regions:

N

A A d

p=

N+

D A d

n or d

p d

n =

N

+

D N

A

Since the width of the depletion region eect by biasing voltage then:

w=

2 e

1 N

A +

1 N

+

D

(V

o

V)

1 2

(3.14)

The ve sign for the forward bias and + vesign for the reverse bias. Where =

o

r.

If the reverse bias at jV

R j

> V

oand

N

A

N

D or

N

D

N

A;

w =

2 V

R e N

1 2

(3.15)

N : the impurities concentration for much lesstype. The depletion region is a free of

charges (carriers), therefore it can be considered as insulator with capacitance.

Cj= w

=r e N

2

V

R in unit

F m

2

(3.16)

It can see that the capacitance of depletion region dependent on applied reverse voltage. Cj/

V

1=2

R (

variable capacitor f or abrupt j unction )

C j/

V

1= 2

R (

variable capacitor f or graded j unction )

The capacitance property for a pn junction in reverse bias is very useful for IC but

its value variable depend on applied voltage. variable capacitor!Tuning called

(varactor)

3.4

Diusion capacitors 57

3.4 Diusion capacitors

It is very important than the junction capacitor, where it is depend on the density of

minority carriers on the edge of the depletion region. These carriers eect with external

applied voltage ( V

F ), then the diusion capacity is given by:

CD = e

2 2

k T [

p

n L

p +

n

pL

n]

exp

e V F k T

(3.17)

Note that; CD /

exp

e V F k T

I F =

I

S exp

e V F k T

dI F dV

F=

e k T

I

S exp

e V F k T

) dI

F dV

F=

e I

F k T

=

1 r

d

Multiply equation 3.17 by I

F I

F

C D = e I

F k T

e 2

I

F [

p

n L

p +

n

pL

n]

exp

e V F k T

C D = 1 r

d

e 2

I

F [

p

n L

p +

n

pL

n]

exp

e V F k T

(3.18)

I S =

e

D nn

p L

n +

D

pp

n L

p

= saturation current

L = p D t

)D

n= L

2

n t

n and D

p= L

2

p t

p

t n and

t

p are the life times for the minority carriers which eect in

I

s.

I S =

e

L 2

n n

p t

n L

n +

L

2

p p

n t

p L

p

= e

L nn

p t

n +

L

pp

n t

p

I F =

e

L nn

p t

n +

L

pp

n t

p

exp

e V F k T

(3.19)

### 58 P-N Junction (Diode)

Substituting equation 3.19 in equation 3.18CD = 1 r

d

e 2

e h

Ln n

p t

n +L

pp

n t

p i

exp

e V F k T

[p

n L

p +

n

pL

n]

exp

e V F k T

C D = 1 2

r

d "

pn L

p +

n

pL

n L

n n

p t

n +L

pp

n t

p #

(3.20)

Since the p njunction current would be form from either( n

p or

p

n) then the equation

3.20 becomes:

CD r

d = 1 2

t

(3.21)

t :the life times of the minority carriers that have large eect in I

s.

r d:the dynamic resistance for the

p njunction.

C D is very important in digital circuit because it is limited the speed of the switching

(on/o ) which is given by:

tof f =

C

D k T e I

Ror t

of f=I

F I

R

L

2

p 2

D

n (3.22)

Solved problems 1.

Apn junction has a hole density in p-side 1024

m

3

and electron den-

sity in n-side 1022

m

3

, the cross-section area for the pnjunction is

10

6

m 2

, the mobility of the holes is 0:2 m 2

(V s )

1

and the mobility of

the electrons is 0:4 m 2

(V s )

1

. The diffusion length of the minorities

are ( L

n = 300

mand L

p = 200

m). If

r = 16

andn

i = 10 19

m

3

at

room temperature. Determine;

(a) The density of ma jority and minority carriers and the conductivity?

(b) The barrier potential? (c) The diusion constant for the both types of the carriers?

(d) Saturation current? (e) The junction current when V

F = 0

:25 V?

(f ) The junction current for the reverse bias, at high reverse voltage?

3.4

Diusion capacitors 59

(g) Width of the depletion region at

V

R = 10

V?

(h) Depletion capacity at V

R = 10

V?

(i) Ratio of holes current to electrons current across the junction?

Solution:

(a) At p-side:

np = n

2

i p

p =

(10

19

)2 10

24 = 10

14

m3electrons minority

N A = 10 24

m

3

holes maj ority

p =

e p

p

p = 1

:6 10

19

1024

0:2 = 3 :2 104

S m

1

(p side )

At n-side: pn = n

2

i n

n =

(10

19

)2 10

22 = 10

16

m3holes minority

N D = 10 22

m

3

electrons maj ority

n =

e n

n

n = 1

:6 10

19

1022

0:4 = 0 :64 104

S m

1

(n side )

(b) The barrier potential:

Vo = kT e

lnN

D

N

A (

n

i) 2

=1 40

ln

10 22

1024 (10

19

)2

= 0 :46 V olt

(c) The diusion constant given as: Dn= kT e

n = 1 40

0:4 = 0 :01 m2

s

1

D p= kT e

p = 1 40

0:2 = 0 :005 m2

s

1

(d) The saturation current given as:

Is =

J

s

A = A e

D nn

p L

n +

D

pp

n L

p

I s = 10

6

1:6 10

19

0:01 1014 300

10

6 + 0

:005 1016 200

10

6

= 0 :04 A

### 60 P-N Junction (Diode)

(e) The forward current:IF =

I

S

exp

e V F k T

1

= 0 :04 10

6

exp

1:6 10

19

0:25 8

:614 10

5

290

1

= 0 :88 mA

(f ) At high reverse voltage I

R =

I

s = 0

:04 A

(g) The depletion region width is:

w=

2 e

1 N

A +

1 N

+

D

(V

o +

V

R )

1 2

since =

o

r

w =

2 16 8:85 10

12 1

:6 10

19

1 10

24 + 1 10

22

(0:46 + 10)

1 2

= 1 :34 m

(h) The junction capacitor: Cj= w

A =

o

r w

A = 16

8:85 10

12 1

:34 106

10

6

= 100 pF

(i) The ratio of I

p to

I

n is:

Ip I

n = e D

pp

n L

p e D

nn

p L

n =

0

:005 10 16 200

10

6 0

:01 10 14 300

10

6 = 75

2. The conductivity of n-side in the Ge p njunction is 10 4

S m

1

and for the p-

side is 10 2

S m

1

. Find the barrier potential for the junction at 300 o

K ? where

n i= 2

:5 1019

m

3

,

n = 0

:36 m2

V s and

p = 0

:16 m2

V s .

Solution:

At n-side: (n ) =

n

n e

n+

p

n e

p=

N

D e

n+ n

2

i N

D e

p

10 4

= 1 :6 10

19

0:36 N

D + (2

:5 1019

)2 N

D

0:16 !

) N

D = 1

:7 1023

m

3

At p-side: (p ) =

p

p e

p+

n

pe

n=

N

Ae

p+ n

2

i N

A e

n

3.4

Diusion capacitors 61

10

2

= 1 :6 10

19

0:16 N

A + (2

:5 1019

)2 N

A

0:36 !

) N

A = 3

:9 1021

m

3

Then the barrier potential is:

V o = kT e

ln N

D

N

A (

n

i) 2

=8

:614 10

5

300 1

:6 10

19

ln 3

:9 1021

1:7 1023 (2

:5 1019

)2

= 0

:36 V olt

3. A forward p njunction is connected to 100

resistance and to power supply of

10 volt . If the applied voltage was reversed and t

of f = 0

:1 s . Find?

(a) The average reverse bias current during the period of inverting if D

n =

0 :0031 m2

s and L

p = 0

:5 m

(b) The diusion capacitor C

D.

Solution: (a) The reverse current: IF = V R

=

10 100

= 0

:1 A

t of f =I

F I

R

L

2

p 2

D

n )

I

R = I

F t

of f

L

2

p 2

D

n

I R = 0

:1 0

:1 10

6 (0

:5 10

6

)2 2

0:0031 = 40

A

(b) The diusion capacitor C

D:

C D =

t

of f I

R e kT

= 0

:1 10

6

40 10

6

40 = 160 pF

### 62 P-N Junction (Diode)

Review Questions1.

What is a pnjunction?

2. Describe the depletion region?

3. What is the eect of forward bias on the depletion region?

4. Which bias condition produces ma jority carrier current and how?

5. A cylindrical pnjunction with 200 m diameter and 10 m length for each side. In

n-side N

D = 10 22

m

3

,

n = 0

:13 m2

(V :s )

1

and in p-side N

A = 10 22

m

3

,

p =

0 :05 m2

(V :s )

1

at 20 o

C , if n

i= 1

:4 1016

m

3

. Find:

(a) The barrier potential?

(b) The bulk resistor? (c) The voltage which required to allow a current of 100mA passing through the junc- tion, if the saturation current is 1nA?

6. An abrupt Si pnjunction ( A= 10

4

cm 2

) has the following properties at 300 o

K :

p-side N

A = 10 17

cm

3

, t

n = 0

:1 s ,

p = 200

cm2

(V :s )

1

and

n = 700

cm2

(V :s )

1

n-side N

D = 5

10 22

cm

3

, t

p = 10

s,

p = 450

cm2

(V :s )

1

and

n = 1300

cm2

(V :s )

1

Find: (a) The depletion capacitance for V

R =100V?

(b) The total excess stored electric charge and the electric eld far from the depletion region on the p-side when the current =20mA?

====== |||||||||||||||||||||||||||||||||||||||||| ======

### Chapter 4

Diodes and their applications|||||||||||||||||||||||||||||||||{ If one side of a piece of silicon dope with a trivalent impurity and the

other side with a pentavalent impurity, a ( p n) junction will formed be-

tween the resulting p-type and n-type portions and a basic diode will cre-

ated. A diode is a device that conducts current in only one direction. In

this chapter we demonstrate the characteristics of the ( p n) junction re-

gion. The volt-ampere characteristics of the ( p n) junction is studied. The

capacitance across the junction is calculated.

||||||||||||||||||||||||||||||||||

4.1 Introduction Several common physical congurations of diodes are illustrated in Figure 4.1. The

anode and cathode are indicated on a diode in several ways, depending on the type of

package. The cathode is usually marked by a band, a tab, or some other feature. On

those packages where one lead is connected to the case, the case is the cathode. Summary of diode biasing:

Forward bias:

Bias voltage connections: positive to ( p) region: negative to ( n) region.

The bias voltage must be greater than the barrier potential.

### 64 Diodes and their applications

Figure 4.1: Typical diode packages with terminal identication.Ma jority carriers

ow toward the ( pn) junction.

Ma jority carriers provide the forward Current.

The depletion region narrows.

Reverse bias:

Bias voltage connections: positive to ( n) region; negative to ( p) region.

The bias voltage must be less than the breakdown voltage.

Ma jority carriers

ow away from the ( pn) junction during short transition time.

Minority carriers provide the extremely small reverse current.

There is no ma jority carrier current after transition time.

The depletion region widens.

4.2 The diode model There are three models of the diode:

4.2.1 The ideal model The ideal model of a diode is a simple switch. When the diode is forward biased,

it acts like closed ( on) switch, as shown in gure 4.2a. When the diode is reversed

biased. It acts like an open ( of f) switch, as shown gure 4.2b. The barrier potential,

the forward dynamic resistance, and the reverse current are all neglected.

4.2

The diode model 65

Figure 4.2: The ideal model of the diode (a) forward bias, (b) reverse bias and (c) ideal

characteristic curve.

In gure 4.2c, the ideal VIcharacteristic curve graphically depicts the ideal diode

operation.

In the ideal diode model: V

F = 0,

I

R = 0 and

V

R =

V

bias .

4.2.2 The practical model The practical model adds the barrier potential to the ideal switch model. When

the diode is forward biased, it is equivalent to a closed switch in series with a small

equivalent voltage source equal to the barrier potential with the positive side toward

the anode, as indicated in gure3a. This equivalent voltage source represents the xed

voltage drop ( V

F ) produced across the forward biased (

p n) junction of the diode and

is not an active source of voltage. This voltage ( V

F ) consists of the barrier potential

voltage ( V

o) plus the small voltage drop across dynamic resistance of the diode (

r

d), as

indicated by the portion of the curve to the right of the origin. The curve slopes because

the voltage drops due to dynamic ( r

d) as the current increases.

### 66 Diodes and their applications

Figure 4.3: The complete model of the diode (a) forward bias, (b) reverse bias and (c)ideal characteristic curve (silicon).

4.2.3 The complete model For the complete model of a silicon diode, the following formulas apply:

VF =

V

o +

I

F

r

d

I F = (

V

Bias

V

o) (

R

Limit +

r

d)

The reverse current is taken into account with the parallel resistance and is indicated

by the portion of the curve to the left of the origin.

Solved problem a

) Determine the forward voltage and forward current for the diode in gure (a) for

each of the diode models. Also nd the voltage across the limiting resistor in each case.

Assume r

d = 10

at the determined value of forward current.

b) Determine the reverse voltage and reverse current for the diode in gure (b) for

each of the diode models. Also nd the voltage across the limiting resistor in each case.

Assume I

R =1

A.

Solution:

4.2

The diode model 67

Figure 4.4:

Figure 4.5:

a ) Ideal model:

VF = 0

volt

I F = V

bias R

Limit =

10 1

103= 10

mA

V Rlimit =

I

F

R

Limit = 10

mA1k

= 10 volt

Practical model: VF = 0

:7 volt

I F = V

bias

V

F R

Limit +

r

d =

10

0:7 1

103

+ 10 = 9

:21 mA

V d = 0

:7 + I

F

r

d = 0

:7 + 9 :21 mA 10 = 792 mV

V Rlimit =

I

F

R

Limit = 9

:21 mA 1k

= 9 :21 volt

If we neglected r

d then;

IF = V

bias

V

F R

Limit =

10

0:7 1

103= 9

:3 mA

V Rlimit =

I

F

R

Limit = 9

:3 mA 1k

= 9 :3 volt

b ) Ideal model:

IR = 0

A

### 68 Diodes and their applications

VR =

V Bias = 5V

V Rlimit = 0

volt

Practical model IR = 1

A

V R(limit )=

I

R

R

(Limit

) = 1A1k

= 1 mV

V R =

V

(Bias

) V

R(limit

)=5V1mV =4:999 volt

4.3 Diode Applications Because of their ability to conduct current in one direction and block current in the

other direction, diodes are used in circuits called rectiers that convert acvoltage into

dc voltage. Rectiers are found in all dcpower supplies that operate from an acvoltage

source. A power supply is an essential part of each electronic system from the simplest

to the most complex. In this section, you will study the most basic type of rectiers,

the half-wave rectier, full-wave rectiers and power supply lters and regulators, and

the diode limiting and clamping circuits, and voltage multipliers.

4.3.1 Half-wave rectier Figure 4.6, illustrates the process called half wave rectication. The diode connection

to an ac source and to a load resistor R

L will form a half-wave rectier. When the input Figure 4.6: Half-wave rectier circuit with the input and output voltage waveform.

voltage ( V

in

) goes positive during the time duration between t

0 to

t

1 , as shown in Figure

4.6, the diode will biased forward and conducts current through the load resistor. The

current produces an output voltage across the load R

L which has the same shape as

4.3

Diode Applications 69

the positive half cycle of the input voltage. As the input voltage goes negative, during

the second half of the input voltage cycle ( t

1 to

t

2 ), the diode will biased reverse. As

a result there is no current will pass through the load R

L and the voltage across the

load resistor is 0 V. The net result is that only the positive half cycles of the ac input

voltage appear across the load as shown in Figure 5. Since the output does not change

polarity, it is a pulse dc voltage . When the practical diode model is used with the Figure 4.7: Half-wave output voltage for three input cycles.

barrier potential of ( V

o) taken into account. During the positive half cycle, the input

voltage must overcome the barrier potential before the diode becomes forward biased.

This result in a half wave output with a peak value that is less than the peak value of

the input by ( V

o), as shown in Figure 4.8.

Vp(out )=

V

p(in )

V

o

The mean value of the output voltage ( V

avg or

V

dc ) can be calculated mathematically Figure 4.8: The eect of the barrier potential on the half wave rectied output voltage

is to reduce the peak value of the input by V

o.

by the area under the curve over a full cycle, as shown in Figure 4.9 dividing by 2 , the

number of radians in a full cycle, where V

p is the peak value of the voltage.

V avg =V

p

### 70 Diodes and their applications

The root mean square value of the output voltage (V

rms ) is given as:

V rms =V

p p

2

Figure 4.9: Average value of the half wave rectied signal.

4.3.5 Half wave rectier with transformer coupled input voltage A transformer is often used to couple the ac input voltage from the source to the

rectier, as shown in gure 4.10. Transformer coupling provides two advantages: 1. It

allows the source voltage to be stepped up or stepped down as needed. 2. The ac source

is electrically isolated from the rectier, thus preventing a shock hazard in the secondary

circuit. The turn's ratio ( n) is equal to the ratio of secondary turns ( N

secondary ) to the

primary turns ( N

primary ):

n= N

secondary N

primary

N secondary =

n N

primary

If n > 1, the secondary voltage is greater than the primary voltage. If n <1, the sec-

ondary voltage is less than the primary voltage. The peak secondary voltage, V

p(secondary )

in a transformer coupled half wave rectier is equal to V p(in ). Therefore:

V p(out )=

V

p(secondary )

V

o

4.4

Full Wave Rectier 71

Figure 4.10: Half wave rectier with transformer coupled input voltage.

4.4 Full Wave Rectier A full-wave rectier allows unidirectional (one-way) current through the load during

the entire 360 o

of the input cycle, whereas a half-wave rectier allow current through

the load only during one-half of the cycle. The result of full-wave rectication is an

output voltage with a frequency twice the input frequency that pulsate every half-cycle

of the input, as shown in Figure 4.11. The average or dc value for a full wave rectied

sinusoidal voltage is twice that of the half wave, as shown in the following formula:

Vdc =

V

avg =2

V

p

V avg

is approximately 63 :7% of V

P for a full wave rectiered voltage. Figure 4.11: Full wave rectier.

4.6.1 The centre tapped full wave rectier A centre tapped rectier use two diodes connected to the secondary of a center

tapped transformer as shown in gure 4.12. The input voltage is coupled through the

transformer to the centre tapped secondary. Half of the total secondary voltage appears

between the center tap and each end of the secondary winding as shown. For a positive

half cycle of the input voltage, the polarities of the secondary voltages are as shown in

### 72 Diodes and their applications

gure 4.13(a). This condition forward biases diodeD

1and reverse biases diode

D

2.

The current path is through D

1and the load resistor

R

L as indicated. Figure 4.12: A centre-tapped full-wave rectier.

For a negative half cycle of the input voltage, the voltage polarities on the secondary

are as shown in gure 4.13(b). This condition reverses biases D

1and forward biases

D

2.

The current path is through D

2and

R

L as indicated. Because the output current during

through the load, the output voltage developed across the load resistor is a full wave

rectied dc voltage as shown. The output voltage of a center tapped full wave rectier Figure 4.13: Basic operation of a centre tapped full wave rectier. a) During positive

half cycles, D

1is forward biased and

D

2is reverse biased. b) During negative half cycles.

D 2is forward-biased and

D

1is reverse biased.

is always one half of the total secondary voltage less the diode drop, no matter what

is the turn's ratio.

Vout =V

sec 2

V

o

4.6.8 The Bridge Full wave Rectier The bridge rectier uses four diodes connected as shown in gure 4.14. When the

input cycle is positive, as shown in gure 4.14a, diodes D

1and

D

2are forward biased

4.4

Full Wave Rectier 73

and conduct current in the direction shown by arrow. A voltage is developed across

R

L

that looks like the positive half of the input cycle. During this time, diodes D

3and

D

4

are reverse biased. When the input cycle is negative as in gure 4.15b, diodes D

3and

D 4are forward biased and conduct current in the same direction through

R

L as during

the positive half cycle. During the negative half cycle, D

1and

D

2are reverse biased. A

full wave rectied output voltage appears across R

L as a result of this action.

BridgeFigure 4.14: Figure 12(a) During the positive half cycle of the input.

D

1and

D

2are

forward biased and conduct current D

3and

D

4are reverse biased. Figure 4.15: Figure 12(b) During the negative half cycle of the input.

D

3and

D

4are

forward biased and conduct current while D

1and

D

2are reverse biased.

output voltage: During the positive half cycle of the total secondary voltage, diodes D

1and

D

2are

forward biased. Neglecting the diode drops, the secondary voltage appears across the

load resistor. The same is true when D

3and

D

4are forward biased during the negative

half cycle.

Vp(out )=

V

p(sec )

It can be seen in gure 4.16. Two diodes are always in series with the load resistor

during both the positive and negative half cycles. If these diode drops are taken into

### 74 Diodes and their applications

account, the output voltage is:Vp(out )=

V

p(sec )

V

o Figure 4.16: Bridge operation during a positive half cycle of the primary and secondary

voltages.

4.7 Power Supply Filters and Regulators In the most power supply applications, the standard 50 Hz AC power line voltage

should be converted to an approximately constant DC voltage. The pulsating DC output

of a rectier must be ltered to reduce the large voltage variations.

4.9.1 Power Supply Filters and Regulators The lter is simply a capacitor connected from the rectier output to ground.

R

L

represents the equivalent resistance of a load.

1. During the positive rst quarter cycle of the input, the diode is forward biased, allowing the capacitor to charge to within V

o of the input peak, as shown in gure

4.17.

2. When the input begins to decrease below its peak. As shown in gure 4.18, the capacitor retains its charge and the diode becomes reverse biased because the

cathode is more positive than the anode. During the remaining part of the cycle,

4.7

Power Supply Filters and Regulators 75

Figure 4.17: Initial charging of the capacitor (diode is forward biased) happens only

once when power is turned on. Figure 4.18: The capacitor discharges through

R

L after peak of positive alternation

when the diode is reverse biased.

the capacitor can discharge only through the load resistance at a rate determined

by the R

LC

time constant.

3. During the rst quarter of the next cycle, as shown in gure 4.19, the diode will again become forward biased when the input voltage exceeds the capacitor voltage

by approximately V

o. Figure 4.19: The capacitor charges back to peak of input when the diode becomes

forward-biased.

### 76 Diodes and their applications

4.9.2 Ripple FactorThe variation in the capacitor voltage due to the charging and discharging is called

the ripple voltage. Generally, ripple is undesirable; thus, the smaller the ripple, the

better the ltering action, as shown in gure 4.20. When ltered, the full wave rectied

voltage has a smaller ripple than does a half wave voltage for the same load resistance

and capacitor values. The capacitor discharges less during the shorter interval between

full-wave pulses, as shown in gure 4.20a and b. The ripple factor is an indication of Figure 4.20: Comparison of ripple voltages for (

a) half wave and ( b) full wave rectied

voltages with the same lter capacitor and load.

the eectiveness of the lter and is dened as: r= V

rms V

dc

As shown in gure 4.21. The ripple factor can be lowered by increasing the value of

the lter capacitor or increasing the load resistance. For a full wave rectier with a

capacitor input lter, approximations for the peak to peak ripple voltage, V

rms and the

DC value of the lter output voltage is V

d:c , are given in the following expressions. The

variable V

p(rect )is the unltered peak rectied voltage.

Vdc =

1 0

:00417 R

LC

Vp(rect )

V rms =

0:0024 R

LC

Vp(rect )

The last formulas are for ripple ltered signal.

4.12

Diode Clipping Circuits 77

Figure 4.21:

V

rms and

V

dc determination for the ripple factor calculation

4.10.3 Inductor Input Filter When a choke is add to the lter input, as in gure 4.22, a reduction in the ripple

voltage V

rms is achieved. The choke has a high reactance at the ripple frequency. The

capacitive reactance is low compared to both R

L and

X

L (10 time at least). The

magnitude of the out ripple voltage of the lter is determined with voltage divider

equation: Figure 4.22: The

LClter as it look to the ACcomponent.

V rms (out )= X

C j

X

L

X

CjV

rms (in )

Since the choke presents a winding resistance R

w in series with load resistance

R

L.

This resistance produce undesirable reduction of DC value, therefore R

w must be small

compared of R

L.

Vdc (out )= R

L R

L

R

wV

dc (in )

4.12 Diode Clipping Circuits Diode circuits, called limiters or clippers, are sometimes used to clip o portions

of signal voltages above or below certain levels. Another type of diode circuit, called

### 78 Diodes and their applications

a clamper, is used to add or restore a DC level to an electrical signal. In the presentlecture we will discuss the clipping circuits only.

4.12.1 Positive Clipping Circuit Figure 4.23 shows a diode clipper that clips the positive part of the input voltage.

As the input voltage goes positive, the diode becomes forward biased and conducts the

current. So point Ais limited to (+ V

o), when the input go back below

V

o, the diode is

reverse biased and appears as an open the output voltage is look like the negative part

of the input voltage, but with magnitude determined by the voltage divider formed by

R 1and the load resistance

R

L, as follow: Figure 4.23: Circuit for the positive clipper.

Vout = R

L R

L +

R

1V

in

if R

1is much less than

R

L )

V

out

V

in

4.13.2 Negative Clipping Circuit To obtain negative biased clipper circuit, the diode and bias voltage must connect

as shown in gure 4.24. In this case, the voltage at point A must go below ( V

B

V

o)

to forward bias the diode an initiate limiting action as shown.

4.15 Voltage Divider Bias The bias voltage sources that have been used to explain the basic operation of diode

limiter can be replaced by a resistive voltage divider that derives the desired bias voltage

4.15

Voltage Divider Bias 79

Figure 4.24: Circuit for the negative clipper.

from the dcsupply voltage, as shown in 4.25. The bias voltage is set by the resistor

values according to the voltage divider formula. Figure 4.25: Circuit for the negative clipper.

VBias = R

3 R

2+

R

3V

supply

A positively biased limiter is shown in 4.25a, a negatively biased limiter is shown in part

(b), and a variable positive bias circuit using a potentiometer voltage divider is shown in

part (c). The bias resistors must be small compared to ( R), so that the forward current

through the diode will not aect the bias voltage.