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Lectures in

Physical Electrons University of Ninevah
Department 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
con guration 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 structure
electron 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 in nity 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

1.2
Electron 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 de ne as the work which required to move an electron
from in nite 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 de ned as the kinetic energy gained by an electron, initially at rest,
in moving through a potential di erence 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 in nite 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. The
spectral 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 in nite
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) Potential
energy 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 a ected
by the interaction among the neighbouring atoms, and hence form narrow bands. The
higher energy levels are greatly a ected 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 di erent for di erent
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, un lled 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 classi ed 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 de ned 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 di erent 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 di erence 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 acquires
a velocity of 200 km=safter moving through a potential di erence 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 di erence 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 di erence 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 in
Semiconductor
|||||||||||||||||||||||||||||||||{ The current is de ned 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 di usion 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 classi ed 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 enough
to 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 su ering 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 Di usion Current
In addition to a conduction current, the transport of charges in a semiconductor
may be accounted for by a mechanism called di usion, not ordinarily encountered in
metals. The essential features of di usion 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 di usion 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 di usion is exactly analogous to th at which occurs in a neutral gas if a concentra-
tion gradient exists in the gaseous container. The di usion 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 di usion 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 di usion 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 di usion 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 distribution
of 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 e ect, 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 ( ) de ned 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 free
electrons 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. E ectively 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 : e ective mass of electron.
m
p : e ective mass of hole.
( E ective 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 modi ed
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 di erence 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 semiconductor
The 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

Solution
After 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 e ective mass of moving charges.
Heating the semiconductor causes vibration of atoms and this will e ect 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 a ected 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
Di usion 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 e ect 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 Di usion and Drift currents density in semiconductor There are two mechanisms by which holes and electrons move through a silicon
crystal di usion and drift. Di usion current density: As aforementioned the di usion current density is
given as:
Jdif f =
e D dn dx
(2.36)
Since the di usion 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 E ect 41
*

n 

p
) = 
n
 = 0 :403 S cm
1
 = 1 
=
1 0
:403 = 2
:48
cm 2.8
Hall E ect 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 e ect to force
F
L known as Lorentz
Force and in ( y) direction as shown in Figure below. Figure 2.8: Hall e ect 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 E
y=
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 E ect 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 E ect 45
Review Questions
1.
De ne mobility? Give its dimensions?
2. Indicate pictorially how a hole contributes to conduction?
3. (a) De ne 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. De ne (a) donor, (b) acceptor impurities?
7. What properties of a semiconductor are determined from a Hall e ect 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 2500
at 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 10
4
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 to
p-typesemiconductor the nregion loses free electrons as they
di use 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 problems
1.
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 di usion length in p-side.
p(x ) = p
n
(p
n
p
no )
exp 
x L
p
(3.7)
L p :the electron di usion length in n-side.
The total current densities are due to holes and electrons di usion motion is de ned
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
di usion currents under the e ect 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 the
depletion 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 di usion 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 capacitances
The 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 e ect 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
Di usion capacitors 57
3.4 Di usion 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 e ect with external
applied voltage ( V
F ), then the di usion 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 e ect 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 3.18
CD = 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 e ect 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 di usion 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
Di usion 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 di usion 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
Di usion 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 di usion 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 di usion 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 Questions
1.
What is a pnjunction?
2. Describe the depletion region?
3. What is the e ect 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 con gurations 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 identi cation.
 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

V
R =
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 recti ers that convert acvoltage into
dc voltage. Recti ers 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 recti ers,
the half-wave recti er, full-wave recti ers and power supply lters and regulators, and
the diode limiting and clamping circuits, and voltage multipliers.
4.3.1 Half-wave recti er Figure 4.6, illustrates the process called half wave recti cation. The diode connection
to an ac source and to a load resistor R
L will form a half-wave recti er. When the input Figure 4.6: Half-wave recti er 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 e ect of the barrier potential on the half wave recti ed 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 recti ed signal.
4.3.5 Half wave recti er with transformer coupled input voltage A transformer is often used to couple the ac input voltage from the source to the
recti er, 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 recti er, 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 recti er is equal to V p(in ). Therefore:
V p(out )=
V
p(secondary )
V
o

4.4
Full Wave Recti er 71
Figure 4.10: Half wave recti er with transformer coupled input voltage.
4.4 Full Wave Recti er A full-wave recti er allows unidirectional (one-way) current through the load during
the entire 360 o
of the input cycle, whereas a half-wave recti er allow current through
the load only during one-half of the cycle. The result of full-wave recti cation 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 recti ed
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 recti ered voltage. Figure 4.11: Full wave recti er.
4.6.1 The centre tapped full wave recti er A centre tapped recti er 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 diode
D
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 recti er.
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
recti ed dc voltage as shown. The output voltage of a center tapped full wave recti er Figure 4.13: Basic operation of a centre tapped full wave recti er. 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 Recti er The bridge recti er 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 Recti er 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 recti ed 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 recti er 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 recti er 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 Factor
The 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 recti ed
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 recti ed
voltages with the same lter capacitor and load.
the e ectiveness of the lter and is de ned 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 recti er 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 un ltered peak recti ed 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
LC lter 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 present
lecture 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 a ect the bias voltage.


رفعت المحاضرة من قبل: Mohammed Aldhanuna
المشاهدات: لقد قام 4 أعضاء و 54 زائراً بقراءة هذه المحاضرة






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