Course Name: Electronics of Semiconductor Devices Duration: 45 hours

Course Code : SPH 1103

Course Level : Year One Semester: I

Credit Units : 3 CU

Course outline

• DC and AC circuit
• Semi – conductors and circuits
• Intrinsic and extrinsic semi-conductors
• Transistors
• Hybrid parameter of transistor
• Transistor Circuit

Introduction

Electronic devices such as diodes, transistors, and integrated circuits are made of

a semi conductive material. To understand how these devices work, you should have

a basic knowledge of the structure of atoms and the interaction of atomic particles.

An important concept introduced in this chapter is that of the p-n junction that is

formed when two different types of semi conductive material are joined. The pn

junction is fundamental to the operation of devices such as the solar cell, the diode,

and certain types of transistors.

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1.1 The Atom

All matter is composed of atoms; all atoms consist of electrons, protons, and

neutrons except normal hydrogen, which does not have a neutron. Each element in

the periodic table has a unique atomic structure, and all atoms for a given element

have the same number of protons. At first, the atom was thought to be a tiny

indivisible sphere. Later it was shown that the atom was not a single particle but was

made up of a small, dense nucleus around which electrons orbit at great distances

from the nucleus, similar to the way planets orbit the sun. Niels Bohr proposed that

the electrons in an atom circle the nucleus in different obits, similar to the way

planets orbit the sun in our solar system. The Bohr model is often referred to as the

planetary model. Another view of the atom called the quantum model is considered a

more accurate representation, but it is difficult to visualize. For most practical

purposes in electronics, the Bohr model suffices and is commonly used because it is

easy to visualize.

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❖ The Bohr Model

An atom* is the smallest particle of an element that retains the characteristics of

that element. Each of the known 118 elements has atoms that are different from

the atoms of all other elements. This gives each element a unique atomic structure.

According to the classical Bohr model, atoms have a planetary type of structure that

consists of a central nucleus surrounded by orbiting electrons, as illustrated in

Figure 1–1. The nucleus consists of positively charged particles called protons and

uncharged particles called neutrons. The basic particles of negative charge are

called electrons. Each type of atom has a certain number of electrons and protons

that distinguishes it from the atoms of all other elements. For example, the simplest

atom is that of hydrogen, which has one proton and one electron, as shown in Figure

1–2(a). As another example, the helium atom, shown in Figure 1–2(b), has two protons

and two neutrons in the nucleus and two electrons orbiting the nucleus.

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❖ Electrons and Shells

✓ Energy Levels

Electrons orbit the nucleus of an atom at certain distances from the nucleus.

Electrons near the nucleus have less energy than those in more distant orbits. Only

discrete (separate and distinct) values of electron energies exist within atomic

structures. Therefore, electrons must orbit only at discrete distances from the

nucleus. Each discrete distance (orbit) from the nucleus corresponds to a certain

energy level. In an atom, the orbits are grouped into energy levels known as shells.

A given atom has a fixed number of shells. Each shell has a fixed maximum number

of electrons. The shells (energy levels) are designated 1, 2, 3, and so on, with 1 being

closest to the nucleus. The Bohr model of the silicon atom is shown in Figure 1–4.

Notice that there are 14 electrons surrounding the nucleus with exactly 14 protons,

and usually 14 neutrons in the nucleus.

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✓ Valence Electrons

Electrons that are in orbits farther from the nucleus have higher energy and are

less tightly bound to the atom than those closer to the nucleus. This is because the

force of attraction between the positively charged nucleus and the negatively

charged electron decreases with increasing distance from the nucleus. Electrons

with the highest energy exist in the outermost shell of an atom and are relatively

loosely bound to the atom. This outermost shell is known as the valence shell, and

electrons in this shell are called valence electrons. These valence electrons

contribute to chemical reactions and bonding within the structure of a material and

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determine its electrical properties. When a valence electron gains sufficient energy

from an external source, it can break free from its atom. This is the basis for

conduction in materials.

✓ Ionization

When an atom absorbs energy, the valence electrons can easily jump to higher energy

shells. If a valence electron acquires a sufficient amount of energy, called ionization

energy, it can actually, escape from the outer shell and the atom’s influence. The

departure of a valence electron leaves a previously neutral atom with an excess of

positive charge (more protons than electrons). The process of losing a valence

electron is known as ionization, and the resulting positively charged atom is called a

positive ion. For example, the chemical symbol for hydrogen is H. When a neutral

hydrogen atom loses its valence electron and becomes a positive ion, it is designated

H+. The escaped valence electron is called a free electron. The reverse process can

occur in certain atoms when a free electron collides with the atom and is captured,

releasing energy. The atom that has acquired the extra electron is called a negative

ion. The ionization process is not restricted to single atoms. In many chemical

reactions, a group of atoms that are bonded together can lose or acquire one or more

electrons. For some nonmetallic materials such as chlorine, a free electron can be

captured by the neutral atom, forming a negative ion. In the case of chlorine, the ion

is more stable than the neutral atom because it has a filled outer shell. The chlorine

ion is designated as Cl-.

✓ The Quantum Model

Although the Bohr model of an atom is widely used because of its simplicity and ease

of visualization, it is not a complete model. The quantum model is considered to be

more accurate. The quantum model is a statistical model and very difficult to

understand or visualize. Like the Bohr model, the quantum model has a nucleus of

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protons and neutrons surrounded by electrons. Unlike the Bohr model, the electrons

in the quantum model do not exist in precise circular orbits as particles. Three

important principles underlie the quantum model: the wave-particle duality principle,

the uncertainty principle, and the superposition principle.

▪ Wave-particle duality. Just as light can be thought of as exhibiting both a wave

and a particle (photon), electrons are thought to exhibit a wave-particle duality.

The velocity of an orbiting electron is related to its wavelength, which interferes

with neighboring electron wavelengths by amplifying or canceling each other.

• Uncertainly principle. As you know, a wave is characterized by peaks and valleys;

therefore, electrons acting as waves cannot be precisely identified in terms of

their position. According to a principle ascribed to Heisenberg, it is impossible to

determine simultaneously both the position and velocity of an electron with any

degree of accuracy or certainty. The result of this principle produces a concept

of the atom with probability clouds, which are mathematical descriptions of

where electrons in an atom are most likely to be located.

• Superposition

A principle of quantum theory that describes a challenging concept about the

behavior of matter and forces at the subatomic level. Basically, the principle

states that although the state of any object is unknown, it is actually in all

possible states simultaneously as long as an observation is not attempted. An

analogy known as Schrodinger’s cat is often used to illustrate in an oversimplified

way quantum superposition.

The analogy goes as follows: A living cat is placed in a metal box with a vial of

hydrocyanic acid and a very small amount of a radioactive substance. Should even

a single atom of the radioactive substance decay during a test period, a relay

mechanism will be activated and will cause a hammer to break the vial and kill the

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cat. An observer cannot know whether or not this sequence has occurred.

According to quantum theory, the cat exists in a superposition of both the alive

and dead states simultaneously.

In the quantum model, each shell or energy level consists of up to four subshells

called orbitals, which are designated s, p, d, and f. Orbital s can hold a maximum

of two electrons, orbital p can hold six electrons, orbital d can hold 10 electrons,

and orbital f can hold 14 electrons. Each atom can be described by an electron

configuration table that shows the shells or energy levels, the orbitals, and the

number of electrons in each orbital. For example, the electron configuration table

for the nitrogen atom is given in Table 1–1. The first full-size number is the shell

or energy level, the letter is the orbital, and the exponent is the number of

electrons in the orbital.

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In a three-dimensional representation of the quantum model of an atom, the s-

orbitals are shaped like spheres with the nucleus in the center. For energy level 1,

the sphere is a single sphere, but for energy levels 2 or more, each single s-orbital

is composed of nested spherical shells. A p-orbital for shell 2 has the form of two

ellipsoidal lobes with a point of tangency at the nucleus (sometimes referred to as a

dumbbell shape.) The three p-orbitals in each energy level are oriented at right

angles to each other. One is oriented on the x-axis, one on the y-axis, and one on the

z-axis. For example, a view of the quantum model of a sodium atom (Na) that has 11

electrons as shown in Figure 1–5. The three axes are shown to give you a 3-D

perspective.

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1.2: Materials Used in Electronic Devices

In terms of their electrical properties, materials can be classified into three groups:

conductors, semiconductors, and insulators. When atoms combine to form a solid,

crystalline material, they arrange themselves in a symmetrical pattern. The atoms

within a semiconductor crystal structure are held together by covalent bonds, which

are created by the interaction of the valence electrons of the atoms. Silicon is a

crystalline material.

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❖ Insulators, Conductors and Semiconductors

All materials are made up of atoms. These atoms contribute to the electrical

properties of a material, including its ability to conduct electrical current. For

purposes of discussing electrical properties, an atom can be represented by the

valence shell and a core that consists of all the inner shells and the nucleus. This

concept is illustrated in Figure 1–6 for a carbon atom. Carbon is used in some types

of electrical resistors. Notice that the carbon atom has four electrons in the valence

shell and two electrons in the inner shell. The nucleus consists of six protons and six

neutrons, so the +6 indicates the positive charge of the six protons. The core has a

net charge of +4 (+6 for the nucleus and -2 for the two inner-shell electrons).

❖ Insulators

An insulator is a material that does not conduct electrical current under normal

conditions. Most good insulators are compounds rather than single-element materials

and have very high resistivities. Valence electrons are tightly bound to the atoms;

therefore, there are very few free electrons in an insulator. Examples of insulators

are rubber, plastics, glass, mica, and quartz.

❖ Conductors

A conductor is a material that easily conducts electrical current. Most metals are

good conductors. The best conductors are single-element materials, such as copper

(Cu), silver (Ag), gold (Au), and aluminum (Al), which are characterized by atoms with

only one valence electron very loosely bound to the atom. These loosely bound valence

electrons can become free electrons with the addition of a small amount of energy

to free them from the atom. Therefore, in a conductive material the free electrons

are available to carry current.

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❖ Semiconductors

A semi-conductor is a material that is between conductors and insulators in its

ability to conduct electrical current. A semiconductor in its pure (intrinsic) state is

neither a good conductor nor a good insulator. Single-element semiconductors are

antimony (Sb), arsenic (As), astatine (At), boron (B), polonium (Po), tellurium (Te),

silicon (Si), and germanium (Ge). Compound semiconductors such as gallium arsenide,

indium phosphide, gallium nitride, silicon carbide, and silicon germanium are also

commonly used. The single-element semiconductors are characterized by atoms with

four valence electrons. Silicon is the most commonly used semiconductor.

❖ Band Gap

✓ In solid materials, interactions between atoms “smear” the valence shell into a

band of energy levels called the valence band. Valence electrons are confined to

that band. When an electron acquires enough additional energy, it can leave the

valence shell, become a free electron, and exist in what is known as the conduction

band.

✓ The difference in energy between the valence band and the conduction band is

called an energy gap or band gap. This is the amount of energy that a valence

electron must have in order to jump from the valence band to the conduction

band. Once in the conduction band, the electron is free to move throughout the

material and is not tied to any given atom. Figure 1–7 shows energy diagrams for

insulators, semiconductors, and conductors. The energy gap or band gap is the

difference between two energy levels and electrons are “not allowed” in this

energy gap based on quantum theory. Although an electron may not exist in this

region, it can “jump” across it under certain conditions. For insulators, the gap

can be crossed only when breakdown conditions occur—as when a very high

voltage is applied across the material. The band gap is illustrated in Figure 1–7(a)

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 for insulators. In semi-conductors the band gap is smaller, allowing an electron in

the valence band to jump into the conduction band if it absorbs a photon. The

band gap depends on the semiconductor material. This is illustrated in Figure 1-

7(b). In conductors, the conduction band and valence band overlap, so there is no

gap, as shown in Figure 1–7(c). This means that electrons in the valence band move

freely into the conduction band, so there are always electrons available as free

electrons.

❖ Comparison of a Semiconductor Atom to a Conductor Atom

Silicon is a semiconductor and copper is a conductor. Bohr diagrams of the silicon

atom and the copper atom are shown in Figure 1–8. Notice that the core of the silicon

atom has a net charge of +4 (14 protons – 10 electrons) and the core of the copper

atom has a net charge of +1 (29 protons – 28 electrons). Recall that the core includes

everything except the valence electrons.

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The valence electron in the copper atom “feels” an attractive force of +1 compared

to a valence electron in the silicon atom which “feels” an attractive force of +4.

Therefore, there is more force trying to hold a valence electron to the atom in silicon

than in copper. The copper’s valence electron is in the fourth shell, which is a greater

distance from its nucleus than the silicon’s valence electron in the third shell. Recall

that, electrons farthest from the nucleus have the most energy. The valence

electron in copper has more energy than the valence electron in silicon. This means

that it is easier for valence electrons in copper to acquire enough additional energy

to escape from their atoms and become free electrons than it is in silicon. In fact,

large numbers of valence electrons in copper already have sufficient energy to be

free electrons at normal room temperature.

❖ Silicon and Germanium

The atomic structures of silicon and germanium are compared in Figure 1–9. Silicon

is used in diodes, transistors, integrated circuits, and other semiconductor devices.

Notice that both silicon and germanium have the characteristic four valence

electrons.

The valence electrons in germanium are in the fourth shell while those in silicon are

in the third shell, closer to the nucleus. This means that the germanium valence

electrons are at higher energy levels than those in silicon and, therefore, require a

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smaller amount of additional energy to escape from the atom. This property makes

germanium more unstable at high temperatures and results in excessive reverse

current. This is why silicon is a more widely used semi-conductive material.

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Covalent Bonds Figure 1–10 shows how each silicon atom positions itself with four

adjacent silicon atoms to form a silicon crystal, which is a three-dimensional

symmetrical arrangement of atoms. A silicon (Si) atom with its four valence electrons

shares an electron with each of its four neighbors. This effectively creates eight

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shared valence electrons for each atom and produces a state of chemical stability.

Also, this sharing of valence electrons produces a strong covalent bond that hold

the atoms together; each valence electron is attracted equally by the two adjacent

atoms which share it. Covalent bonding in an intrinsic silicon crystal is shown in Figure

1–11. An intrinsic crystal is one that has no impurities. Covalent bonding for

germanium is similar because it also has four valence electrons.

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Page 18 of 117
❖ Kirchhoff ‘s rules in circuit analysis

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 DEPARTMENT OF HUMANITIES AND BASIC ELECTRICAL AND ELECTRONICS
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 M is the number of branches that are connected to a node.

The above statement of KCL can also be expressed as "the algebraic sum of currents entering a node is
equal to the algebraic sum of currents leaving a node". Let us verify this statement through the following
example.
Example
Write KCL equation at node P of the following figure.

 In the above figure, the branch currents I1, I2 and I3 are entering at node P. So, consider negative
signs for these three currents.
 In the above figure, the branch currents I4 and I5 are leaving from node P. So, consider positive signs
for these two currents.
The KCL equation at node P will be
-I1-I2-I3+I4+I5=0
I1+I2+I3=I4+I5

In the above equation, the left-hand side represents the sum of entering currents, whereas the right-hand
side represents the sum of leaving currents.
In this tutorial, we will consider positive sign when the current leaves a node and negative sign when it
enters a node. Similarly, you can consider negative sign when the current leaves a node and positive sign
when it enters a node. In both cases, the result will be same.
Note − KCL is independent of the nature of network elements that are connected to a node.
Kirchhoff’s Voltage Law
Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of voltages around a loop or
mesh is equal to zero.
A Loop is a path that terminates at the same node where it started from. In contrast, a Mesh is a loop that
doesn’t contain any other loops inside it.

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 DEPARTMENT OF HUMANITIES AND BASIC ELECTRICAL AND ELECTRONICS
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Mathematically, KVL can be represented as

∑ 𝑉𝑛 = 0
𝑛=1

Where,
 Vn is the nth element’s voltage in a loop (mesh).
 N is the number of network elements in the loop (mesh).
The above statement of KVL can also be expressed as "the algebraic sum of voltage sources is equal to the
algebraic sum of voltage drops that are present in a loop." Let us verify this statement with the help of the
following example.
Example
Write KVL equation around the loop of the following circuit.

The above circuit diagram consists of a voltage source, V S in series with two resistors R1 and R2. The
voltage drops across the resistors R1 and R2 are V1 and V2 respectively.
Apply KVL around the loop.
VS-V1-V2=0
VS=V1+V2
In the above equation, the left-hand side term represents single voltage source VS. Whereas, the right-hand
side represents the sum of voltage drops. In this example, we considered only one voltage source. That’s
why the left-hand side contains only one term. If we consider multiple voltage sources, then the left side
contains sum of voltage sources.
In this tutorial, we consider the sign of each element’s voltage as the polarity of the second terminal that is
present while travelling around the loop. Similarly, you can consider the sign of each voltage as the polarity
of the first terminal that is present while travelling around the loop. In both cases, the result will be same.
Note − KVL is independent of the nature of network elements that are present in a loop.
In this chapter, let us discuss about the following two division principles of electrical quantities.

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 Current Division Principle

 Voltage Division Principle
Current Division Principle
When two or more passive elements are connected in parallel, the amount of current that
flows through each element gets divided(shared) among themselves from the current that is entering the
node.
Consider the following circuit diagram.

The above circuit diagram consists of an input current source IS in parallel with two resistors R1 and R2.
The voltage across each element is VS. The currents flowing through the
resistors R1 andR2 are I1 and I2 respectively.
The KCL equation at node P will be
IS = I1+I2
 Substitute I1= 𝑉𝑠 and I2=𝑉𝑠 in the above equation
𝑅1 𝑅2
𝑉𝑠 𝑉𝑠 1 1
IS = + = 𝑉 (1 + )
𝑅1 𝑅2 𝑅2
𝑅1𝑅2
VS=𝐼 𝑠( )
𝑅1+𝑅2
 Substitute the values of V in I =𝑉𝑠
S 1
𝑅1
𝐼 𝑅1𝑅2
1I = 𝑆( )
𝑅1 1+𝑅2
𝑅2
I1=𝐼𝑆 ( )
𝑅1+𝑅2
 Substitute the values of V in I =𝑉𝑠
S 2
𝑅2
𝐼 𝑅1𝑅2
I2 = 𝑆( )
𝑅2 1+𝑅2
𝑅1
I2=𝐼𝑆 ( )
𝑅1+𝑅2

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 DEPARTMENT OF HUMANITIES AND BASIC ELECTRICAL AND ELECTRONICS
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From equations of I1 and I2, we can generalize that the current flowing through any passive element can be
found by using the following formula.
IN=IS Z1 ⃦ Z2 ⃦………. ⃦ ZN-1
Z1 +Z2 +……+ ZN

This is known as current division principle and it is applicable, when two or more passive elements are
connected in parallel and only one current enters the node.
Where,
 IN is the current flowing through the passive element of Nth branch.
 IS is the input current, which enters the node.
 Z1, Z2, …,ZN are the impedances of 1st branch, 2ndbranch, …, Nth branch respectively. 
Voltage Division Principle
When two or more passive elements are connected in series, the amount of voltage present
across each element gets divided (shared) among themselves from the voltage that is available across that
entire combination.
Consider the following circuit diagram.

The above circuit diagram consists of a voltage source, VS in series with two resistors R1 and R2. The
current flowing through these elements is IS. The voltage drops across the resistors R1and R2 are V1 and
V2 respectively.
The KVL equation around the loop will be
VS=V1+V2
 Substitute V1 = IS R1 and V2 = IS R2 in the above equation

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VS= IS R1 + IS R2 = IS (R1 +R2)

𝑉𝑆
IS =
𝑅1+𝑅2

 Substitute the value of IS in V1 = IS R1.

V =𝑅 𝑉𝑆
1
1 ( +𝑅 )
𝑅1 2
𝑅1
V =𝑉 ( )
1 𝑆 𝑅1 +𝑅2

 Substitute the value of IS in V2 = IS R2.

V =𝑅 ( 𝑉𝑆 )
2
2
𝑅1 +𝑅2
𝑅2
V =𝑉 ( )
2 𝑆 𝑅1 +𝑅2

From equations of V1 and V2, we can generalize that the voltage across any passive element can be found by
using the following formula.

vN = vS ( ZN )
Z1 +Z2 +……+ ZN

This is known as voltage division principle and it is applicable, when two or more passive elements are
connected in series and only one voltage available across the entire combination.
Where,
 VN is the voltage across Nth passive element.
 VS is the input voltage, which is present across the entire combination of series passive elements.
 Z1,Z2, …,Z3 are the impedances of 1 st passive element, 2nd passive element, …, Nth passive element
respectively.

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 Find the current flowing through 10 ohms resistor using KVL and KCL.

Method 1: LOOP ANALYSIS (BY CONSIDERING BRANCH CURRENTS)

INDICATE CURRENTS IN EACH BRANCH

APPLY KVL AT LOOP 1

20-5I1-5I2=0
I1+I2=4 1)

APPLY KVL AT LOOP 2

-10I3-5I4+5I2=0
I2-2I3-I4=0 2)
APPLY KVL AT LOOP 3
-5I5-5I5+5I4=0
I4-2I5=0 3)

FROM KCL I1=I2+I3, I3=I4+I5

SUB I5=I3-I4 I EQUATION 3, WE GET,

I4=2/3I3 4)

SUB I4 VALUE IN EQUATION 2, WE GET,

-8I1+11I2=0 5)
SOLVE 5 AND 1, WE WILL GET
I2=1.684A

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 DEPARTMENT OF HUMANITIES AND BASIC ELECTRICAL AND ELECTRONICS
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FROM EQUATION 1, SUB I2 VALUE WE GET I1= 2.31A

FROM KCL I3=I1-I2

I3 =2.31-1.64=0.63A
Current flowing through 10 ohm resistor Is I3 and the Value Is 0.63 A.

EX: Find the current flowing through R2 resistor

APPLY KCL AT NODE

I1+I2+I3=0
𝑉 − 20 𝑉
+ + 2𝑚𝐴 = 0
5𝐾 7.5𝐾
(𝑉 − 20)15 10𝑉
+ + 2𝑚𝐴 = 0
75𝐾 75𝐾
15𝑉 − 300 + 10𝑉
+ 2𝑚𝐴 = 0
75𝐾
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DEPARTMENT OF HUMANITIES AND BASIC ELECTRICAL AND ELECTRONICS
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25𝑉 − 300
+ 2𝑚𝐴 = 0
75𝐾
25𝑉 − 300 + 150
=0
75𝐾
25𝑉 − 150
=0
75𝐾
150
𝑉= = 6𝑉
25

Current Flowing Through R2 IS 𝑉 = 6

7.5𝐾 7.5𝐾

=0.8mA

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 DEPARTMENT OF ELECTRICAL AND BASIC ELECTRICAL AND ELECTRONICS
ELECTRONICS ENGINEERING ENGINEERING

Network Theorems:
Introduction:
Any complicated network i.e. several sources, multiple resistors are present if the single element response is
desired then use the network theorems. Network theorems are also can be termed as network reduction
techniques. Each and every theorem got its importance of solving network. Let us see some important
theorems with DC and AC excitation with detailed procedures.
Thevenin’s Theorem and Norton’s theorem (Introduction) :
Thevenin’s Theorem and Norton’s theorem are two important theorems in solving Network problems having
many active and passive elements. Using these theorems the networks can be reduced to simple equivalent
circuits with one active source and one element. In circuit analysis many a times the current through a
branch is required to be found when it’s value is changed with all other element values remaining same. In
such cases finding out every time the branch current using the conventional mesh and node analysis methods
is quite awkward and time consuming. But with the simple equivalent circuits (with one active source and
one element) obtained using these two theorems the calculations become very simple. Thevenin’s and
Norton’s theorems are dual theorems.
Thevenin’s Theorem Statement:
Any linear, bilateral two terminal network consisting of sources and resistors(Impedance),can
be replaced by an equivalent circuit consisting of a voltage source in series with a resistance
(Impedance).The equivalent voltage source VTh is the open circuit voltage looking into the terminals(with
concerned branch element removed) and the equivalent resistance RTh while all sources are replaced by their
internal resistors at ideal condition i.e. voltage source is short circuit and current source is open circuit.

(a) (b)
Figure (a) shows a simple block representation of a network with several active / passive elements with the
load resistance RL connected across the terminals ‘a & b’ and figure (b) shows the Thevenin's equivalent
circuit with VTh connected across RTh & RL .

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Main steps to find out VTh and RTh :

1. The terminals of the branch/element through which the current is to be found out are marked as say a & b
after removing the concerned branch/element
2. Open circuit voltage VOC across these two terminals is found out using the conventional network
mesh/node analysis methods and this would be VTh .
3. Thevenin's resistance RTh is found out by the method depending upon whether the network contains
dependent sources or not.
a. With dependent sources: RTh = Voc / Isc
b. Without dependent sources : RTh = Equivalent resistance looking into the concerned terminals with all
voltage & current sources replaced by their internal impedances (i.e. ideal voltage sources short
circuited and ideal current sources open circuited)
4. Replace the network with VTh in series with RTh and the concerned branch resistance (or) load resistance
across the load terminals (A&B) as shown in below fig.

Fig.(a)
Example: Find VTH, RTH and the load current and load voltage flowing through RL resistor as shown in fig.
by using Thevenin’s Theorem?

Solution:
The resistance RL is removed and the terminals of the resistance RL are marked as A & B as shown in the
fig. (1)

Fig.(1)

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Calculate / measure the Open Circuit Voltage. This is the Thevenin Voltage (VTH). We have already
removed the load resistor from fig.(a), so the circuit became an open circuit as shown in fig (1). Now we
have to calculate the Thevenin’s Voltage. Since 3mA Current flows in both 12kΩ and 4kΩ resistors as this
is a series circuit because current will not flow in the 8kΩ resistor as it is open. So 12V (3mA x 4kΩ) will
appear across the 4kΩ resistor. We also know that current is not flowing through the 8kΩ resistor as it is
open circuit, but the 8kΩ resistor is in parallel with 4k resistor. So the same voltage (i.e. 12V) will appear
across the 8kΩ resistor as 4kΩ resistor. Therefore 12V will appear across the AB terminals.
So, VTH = 12V

Fig (2)
All voltage & current sources replaced by their internal impedances (i.e. ideal voltage sources short circuited
and ideal current sources open circuited) as shown in fig.(3)

Fig(3)
Calculate /measure the Open Circuit Resistance. This is the Thevenin's Resistance (RTH)We have Reduced
the 48V DC source to zero is equivalent to replace it with a short circuit as shown in figure (3) We can see
that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor. i.e.:
8kΩ + (4k Ω || 12kΩ) ….. (|| = in parallel with)
RTH = 8kΩ + [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)]
RTH = 8kΩ + 3kΩ
RTH = 11kΩ

Fig(4)

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Connect the RTH in series with Voltage Source VTH and re-connect the load resistor across the load
terminals(A&B) as shown in fig (5) i.e. Thevenin's circuit with load resistor. This is the Thevenin’s
equivalent circuit.

VTH

Fig (5)
Now apply Ohm’s law and calculate the load current from fig 5.
IL = VTH/ (RTH + RL)= 12V / (11kΩ + 5kΩ) = 12/16kΩ
IL= 0.75mA
And VL = ILx RL= 0.75mA x 5kΩ
VL= 3.75V
Norton’s Theorem Statement:
Any linear, bilateral two terminal network consisting of sources and resistors(Impedance),can
be replaced by an equivalent circuit consisting of a current source in parallel with a resistance
(Impedance),the current source being the short circuited current across the load terminals and the resistance
being the internal resistance of the source network looking through the open circuited load terminals.

(a) (b)
Figure (a) shows a simple block representation of a network with several active / passive elements with the
load resistance RL connected across the terminals ‘a & b’ and figure (b) shows the Norton equivalent
circuit with IN connected across RN & RL .
Main steps to find out IN and RN:
 The terminals of the branch/element through which the current is to be found out are marked as say a
& b after removing the concerned branch/element.

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 Open circuit voltage VOC across these two terminals and ISC through these two terminals are found
out using the conventional network mesh/node analysis methods and they are same as what we
obtained in Thevenin’s equivalent circuit.
 Next Norton resistance RN is found out depending upon whether the network contains dependent
sources or not.
a) With dependent sources: RN = Voc / Isc
b) Without dependent sources : RN = Equivalent resistance looking into the concerned terminals
with all voltage & current sources replaced by their internal impedances (i.e. ideal voltage
sources short circuited and ideal current sources open circuited)
 Replace the network with IN in parallel with RN and the concerned branch resistance across the load
terminals(A&B) as shown in below fig

Example: Find the current through the resistance R L (1.5 Ω) of the circuit shown in the figure (a)
below using Norton’s equivalent circuit.

Fig(a)
Solution: To find out the Norton’s equivalent ckt we have to find out IN = Isc ,RN=Voc/ Isc. Short the 1.5Ω
load resistor as shown in (Fig 2), and Calculate / measure the Short Circuit Current. This is the Norton
Current (IN).

Fig(2)

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We have shorted the AB terminals to determine the Norton current, I N. The 6Ω and 3Ω are then in parallel
and this parallel combination of 6Ω and 3Ω are then in series with 2Ω.So the Total Resistance of the circuit
to the Source is:-
2Ω + (6Ω || 3Ω) ….. (|| = in parallel with)
RT = 2Ω + [(3Ω x 6Ω) / (3Ω + 6Ω)]
RT = 2Ω + 2Ω
RT = 4Ω
IT = V / RT
IT = 12V / 4Ω= 3A..
Now we have to find ISC = IN… Apply CDR… (Current Divider Rule)…
ISC = IN = 3A x [(6Ω / (3Ω + 6Ω)] = 2A.
ISC= IN = 2A.

Fig(3)
All voltage & current sources replaced by their internal impedances (i.e. ideal voltage sources short circuited
and ideal current sources open circuited) and Open Load Resistor. as shown in fig.(4)

Fig(4)
Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (R N) We have Reduced the
12V DC source to zero is equivalent to replace it with a short circuit as shown in fig(4), We can see that 3Ω
resistor is in series with a parallel combination of 6Ω resistor and 2Ω resistor. i.e.:
3Ω + (6Ω || 2Ω) ….. (|| = in parallel with)
RN = 3Ω + [(6Ω x 2Ω) / (6Ω + 2Ω)]
RN = 3Ω + 1.5Ω
RN = 4.5Ω

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Fig(5)
Connect the RN in Parallel with Current Source IN and re-connect the load resistor. This is shown in fig (6)
i.e. Norton Equivalent circuit with load resistor.

Fig(6)
Now apply the Ohm’s Law and calculate the load current through Load resistance across the terminals
A&B. Load Current through Load Resistor is
IL = IN x [RN / (RN+ RL)]
IL= 2A x (4.5Ω /4.5Ω +1.5kΩ)
IL = 1.5A IL = 1. 5A
Superposition Theorem:
The principle of superposition helps us to analyze a linear circuit with more than one current
or voltage sources sometimes it is easier to find out the voltage across or current in a branch of the circuit by
considering the effect of one source at a time by replacing the other sources with their ideal internal
resistances.
Superposition Theorem Statement:
Any linear, bilateral two terminal network consisting of more than one sources, The total
current or voltage in any part of a network is equal to the algebraic sum of the currents or voltages in the
required branch with each source acting individually while other sources are replaced by their ideal internal
resistances. (i.e. Voltage sources by a short circuit and current sources by open circuit)
Steps to Apply Super position Principle:
1. Replace all independent sources with their internal resistances except one source. Find the output
(voltage or current) due to that active source using nodal or mesh analysis.

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2. Repeat step 1 for each of the other independent sources.

3. Find the total contribution by adding algebraically all the contributions due to the independent
sources.
Example: By Using the superposition theorem find I in the circuit shown in figure?

Fig.(a)
Solution: Applying the superposition theorem, the current I2 in the resistance of 3 Ω due to the voltage
source of 20V alone, with current source of 5A open circuited [ as shown in the figure.1 below ] is given by
:

Fig.1
I2 = 20/(5+3) = 2.5A
Similarly the current I5 in the resistance of 3 Ω due to the current source of 5A alone with voltage source of
20V short circuited [ as shown in the figure.2 below ] is given by :

Fig.2
I5= 5 x 5/(3+5) = 3.125 A
The total current passing through the resistance of 3Ω is then = I2 + I5= 2.5 + 3.125 = 5.625 A
Let us verify the solution using the basic nodal analysis referring to the node marked with V in fig.(a).Then
we get :

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𝑉 − 20 𝑉
+ =5
5 3
3V-60+5V=15× 5
8V-60=75
8V=135
V=16.875
The current I passing through the resistance of 3Ω =V/3 = 16.875/3 = 5.625 A.

EX:Using KCL,KVL Find the currents in all the sources of the circuit of the following figure[10 M]

Method 1: LOOP ANALYSIS (BY CONSIDERING BRANCH CURRENTS)

INDICATE CURRENTS IN EACH BRANCH

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APPLY KVL AT LOOP 1

5 -3I1-3I2-3I1=0
-6I1-3I2-5=0 1)
APPLY KVL AT LOOP 2
-2I3-2(I3+ I4) -5-2(I3+ I4) +3 I2=0
3I2-6I3-4I4-5=0 2)
APPLY KVL AT LOOP 3
-I4-I4-I4+2I3+5 =0
2I3-3I4+5=0 3)
FROM KCL I1=I2+I3 +I4 ------------------------------------------------------------------------------------------------------ 4)
SUB EQUATION 4 IN 1, WE WILL GET
-6(I2+I3 +I4) - 3I2+5=0
-9I2-6I3-6I4+5=0 5)
SOLVE 5 AND 2, WEWILL GET
-6I2-I4+5=0 6)
SOLVE 5 AND 3, WE WILL GET
-9I2-15I4 +20=0 7)
SOLVE 6 AND 7, WE WILL GET
75
I4 = = 0.925 𝐴
81
Sub I4 value in equation 6 we will get,
5−0.925
I2 = = 0.679 𝐴
6
Sub I2 value in equation 1 we will get,
5 − 3(0.679 )
I1 = = 0.493 𝐴
6
Sub I4, I2, I3 values in equation 4 we will get
I3=−0.493 − 0,679 -0.925
I3=−1.11 A
currents in all the sources are I1, I3 +I4, & I4
I1= 0.493 A
I4= 0.925 A
I4+ I3 = 0.925 + (-1.11) = -0.185A

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EX: Find the current flowing through R4 resistor [10 M]

Method 1: NODAL ANALYSIS

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Assume considered node voltage is superior, so all currents are always leaving at node

Apply KCL at node

I1+I2+I3+I4=0

𝑉 − 50 𝑉 𝑉 𝑉
+ + + =0
25 50 200 100

8𝑉−400+4𝑉+𝑉+2𝑉
=0
200

V=26.6V
𝑉
` Current through R4 Resistor Is I4=
100

I4=26.6/100=0.266 A

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 BET-102 Basic Electronics Engineering.
Chapter-1
1. ELECTRONIC DEVICES
1.1 Define Electronics & its application.

The world's reliance on electronics is so great that commentators claim people live in an
"electronic age." People are surrounded by electronics—televisions, radios, computers, mobiles,
Laptop and DVD players, along with products with major electric components, such as
microwave ovens, refrigerators, and other kitchen appliances, automatic vehicles, Robotics, as
well as hearing aids and medical instruments and numerous applications in industry.

Definition: The branch of engineering which deals with current conduction through a Vacuum
or Gas or Semiconductor is known as Electronics. An electronic device is that in which current
flows through a vacuum or gas or semiconductor. This control of electrons is accomplished by
devices that resist, carry, select, steer, switch, store, manipulate, and exploit the electron.

Or
Electronics deals with electrical circuits that involve active electrical components such as
vacuum tubes, transistors, diodes and integrated circuits, and associated passive
interconnection technologies. Commonly, electronic devices contain circuitry consisting
primarily or exclusively of active semiconductors supplemented with passive elements; such a
circuit is described as an electronic circuit.

Pre Knowledge (Some of the basic definitions):

Passive Components: Capable of operating without an external power source. Typical passive
components are resistors, capacitors, inductors.
Active components: Requiring a source of power to operate. Includes transistors (all types), integrated
circuits (all types), TRIACs, SCRs, LEDs, etc.

APPLICATIONS of Electronics:

Electronic components: capacitor (C), cathode ray tube (CTR), diode (D), digital signal
processor (DSP, field effect transistor (FET), integrated circuit (IC), junction gate field effect
transistor (JFET), inductor (L), Liquid crystal display (LCD), light dependent resistor (LDR, light
emitting diode (LED), Metal oxide semiconductor field effect transistor (MOSFET), transistor
(Q), resistor (R), relay (RLA, RY), switch (SW), transformer (T), thermistor (TH), transistor
(Tr), integrated circuit (U, IC), variable capacitor (VC), variable resistor (VR) and more.

Consumer Electronics include products like – Audio Systems, Video Systems, TV (Television),
Computer, Laptop, Digital Camera, DVD Players, Home and Kitchen Appliances, GPS, Mobiles
Phones etc.

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Communication. Electronic communication systems connect people around the world. Using
telephones, Internet and computers, people in different countries communicate almost
instantly. Radios transmit sounds and televisions transmit sounds and pictures great distances.
Cellular telephones enable a person to call another person. Within seconds, fax machines send
and receive copies of documents over telephone lines/Satellite.

Information processing. Scientists, artists, students, government and business workers, and
hobbyists at home all rely on computers, Internet to handle huge amounts of information
quickly and accurately. Computers solve difficult mathematical problems, maintain vast
amounts of data, create complex simulations, and perform a multitude of other tasks that help
people in their everyday lives.

Medicine and research. Include product like X-ray machines ECG (Electrocardiogram) use
radiation to take images of bones and internal organs. Radiation therapy, or radiotherapy, uses
X-rays and other forms of radiation to fight cancer. Many hearing-impaired people depend on
hearing aids to electrically amplify sound waves.

Computers and other electronic instruments provide scientists and other researchers with
powerful tools to better understand their area of study. Computers, for example, help scientists
design new drug molecules, track weather systems, and test theories about how galaxies and
stars develop. Electron microscopes use electrons rather than visible light to magnify specimens
1 million times or more.

Automation. Electronic components enable many common home appliances, such as

refrigerators, washing machines, and toasters, to function smoothly and efficiently. People can
electronically program coffeemakers, lawn sprinklers, and many other products to turn on and
off automatically. Microwave ovens heat food quickly by penetrating it with short radio waves
produced by a vacuum tube.

Instrumentation. Measuring Instruments like CRO, Multimeter, ph-meter, strain gauge,

VTVM, Frequency Counter are used in different Laboratory/organisations.

Many automobiles have electronic controls in their engines and fuel systems. Electronic devices
also control air bags, which inflate to protect a driver and passengers in a collision.

1.2 Define Electronic Emission & different types of Emission.

The Electronics devices depends the movements of free Electrons in an evacuated space.
The liberation of electrons from the surface of a metal is known as Electron Emission.

 For electron emission, metals are used because they have many free electrons.
 The electrons are free only to transfer from one atom to another within the metal but
they cannot leave the metal surface to provide electron emission.
 Thus at the surface of the metal, a free electron encounters forces that prevent it to
leave the metal.
 In other words, the metallic surface offer a barrier to free electrons, their kinetic
energy increases and is known as surface barrier.
 However, if sufficient energy is given to the free electrons, their kinetic energy
increases and thus the electrons will cross over the surface barrier to leave the metal.
 This additional energy required by an electron to overcome the surface barrier of the
metal is called work function of the metal.

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The metallic surface offers a barrier to free electrons and is known as surface barrier.

Work function (W0): The amount of additional energy(such as heat energy, energy stored in
electric field, light energy or kinetic energy of the electric charges bombarding the metal
surface) required to emit an electron from a metallic surface is known as work function of that
metal. The minimum energy required by an electron to just escape (i.e. with zero velocity) from
metal's surface is called Work function (W0) of the metal. The work function of pure metals
varies (roughly) from 2eV to 6eV. Its value depends upon the nature of the metal, its purity
and the conditions of the surface.

Different types of Emission:

There are following four principal method of obtaining electron emission from the surface of a
metal:
1. Thermionic Emission - (Due to Thermal energy)
2. Field Emission - (Due to application of strong electric field)
3. Secondary Emission – ( due to bombardment of high-speed electrons)
4. Photo Electric Emission – ( by the application of light)

1. Thermionic Emission

The process of electron emission from a metal surface by supplying thermal energy to it is
known as Thermionic emission.

In this type of emission the electron emission is achieved by heating the electrode to a
sufficient temperature (about 2500oC) to enable the free electrons to leave the metal surface.
Due to heating the electrons get enough energy that they emit from the surface of that material
heat energy is converted into kinetic energy, causing accelerated motion of free electrons and
electrons acquire additional energy equal to the work function of the metal. An electron emitted
from a hot cathode comes out with a velocity that presents different between the kinetic energy
possessed by electron just before emission usually used in cathode of diode, triode, pentode,
CRT and many other. The higher the temperature, the greater is the emission of electrons. The
commonly used materials for electron emission are tungsten, thoriated tungsten and metallic
oxides of barium and strontium.

S.No. Emitter Work Function Operating Temperature

1 Tungsten 4.52 eV 2327ºC

2 Thoriated tungsten 2.63 eV 1700ºC
3 Oxide-coated 1.1 eV 750ºC

2. Field Emission
The process of electron emission by the application of strong electric field at the surface of a
metal is known as field emission.

When metal surface is placed in an electric field, the electron rotating in their orbits
experience a force due to electrostatic field. Hence the process of electron emission by
application of strong electric field at the surface of a metal is called field emission. It is also
called cold cathode emission or auto- electronic emission.

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 3. Secondary Emission
Electron emission from a metallic surface by the bombardment of high-speed electrons
or other particles is known as secondary emission.

When high-speed electrons suddenly strike a metallic surface, they may give some or all
of their kinetic energy to the free electrons in the metal. If the energy of the striking
electrons is sufficient, it may cause free electrons to escape from the metal surface. This
phenomenon is called secondary emission. The electrons that strike the metal are called
primary electrons while the emitted electrons are known as secondary electrons. The
intensity of secondary emission depends upon the emitter material, mass and energy of the
bombarding particles.

4. Photo Electric Emission

Electron emission from a metallic surface by the application of light is known as photo
electric emission.

When a beam of light strikes the surface of cathode normally made of potassium, Sodium the
energy of photons of light is transfer to the free electrons of cathode. In this method, the
energy of light falling upon the metal surface is transferred to the free electrons within the
metal to enable them to leave the surface. The greater the intensity of light beam falling on the
metal surface, the greater is the photoelectric emission. The emitted electrons are known as
photo electrons and the phenomenon is known as photoelectric emission. Photo-electric
emission is utilised in photo tubes which form the basis of television and sound films.

1.3 Classification of solid according to electrical conductivity (Conductor, Semiconductor &

Insulator) with respect to energy band diagram only.

Pre-Knowledge:
(i) Valence band. The range of energies (i.e. band) possessed by valence electrons is
known as valence band. The electrons in the outermost orbit of an atom are known as
valence electrons. This band may be completely or partially filled.

(ii) Conduction band.

The range of energies (i.e. band) possessed by conduction band electrons is known as
conduction band. Generally, insulators have empty conduction band. On the other
hand, it is partially filled for conductors. The free electrons which are responsible for the
conduction of current in a conductor are called conduction electrons.

(iii) Forbidden energy gap. The separation between conduction band and valence
band on the energy level diagram is known as forbidden energy gap.

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 Fig 1. Energy band diagram

Classification:

Solid-state materials According to electrical conductivity can be classified into three

groups. Such as:

1. Insulators - Insulators are materials having an electrical conductivity

(like diamond: 10-14S/cm);
2. Semiconductors - semiconductors have a conductivity (for
silicon it can range from 10-5S/cm to 103S/cm)
3. Conductors - at last conductors are materials with high conductivities:
(like silver: 106S/cm.)

Figure 2 : Representation of energy bands

(i) Insulators. Insulators (e.g. wood, glass, plastics,rubber etc.) are those substances which do
not allow the passage of electric current through them. In terms of energy band, the
valence band is full while the conduction band is empty as shown in Fig 2. Further, the
energy gap between valence and conduction bands is very large (15 eV). Therefore, a very
high electric field is required to push the valence electrons to the conduction band. For this
reason, the electrical conductivity of such materials is extremely small. At room

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 temperature, the valence electrons of the insulators do not have enough energy to cross
over to the conduction band. However, when the temperature is raised, some of the
valence electrons may acquire enough energy to cross over to the conduction band. Hence,
the resistance of an insulator decreases with the increase in temperature i.e. an insulator
has negative temperature coefficient of resistance.

(ii) Conductors. Conductors (e.g. copper, aluminum) are those substances which easily allow
the passage of electric current through them. It is because there are a large number of free
electrons available in a conductor. In terms of energy band as in Fig 2, the valence and
conduction bands overlap each other due to this overlapping; a slight potential difference
across a conductor causes the free electrons to constitute electric current.

(iii) Semicondutors. Semiconductors (e.g. germanium, silicon etc.) are those substances
whose electrical conductivity lies in between conductors and insulators. In terms of energy
band, the valence band is almost filled and conduction band is almost empty in fig 2.
Further, the energy gap between valence and conduction bands is very small. The
semiconductor has :
(a) Filled valence band
(b) Empty conduction band
(c) Small energy gap or forbidden gap ( 1 eV) between valence and conduction bands.
(d) Semiconductor virtually behaves as an insulator at low temperatures. However, even at
room temperature, some electrons cross over to the conduction band, imparting little
conductivity (i.e. conductor).

1.3 Discuss Intrinsic & Extrinsic Semiconductor.

Intrinsic Semiconductor
A semiconductor in an extremely pure form is known as an intrinsic semiconductor.

In this case the holes in the valence band are vacancies created by electrons that have been
thermally excited to the conduction band and hole-electron pairs are created. When electric
field is applied across an intrinsic semiconductor, the current conduction takes place by two
processes, namely; by free electrons and holes as shown in Fig 3. The free electrons are
produced due to the breaking up of some covalent bonds by thermal energy. At the same time,
holes are created in the covalent bonds. Under the influence of electric field, conduction
through the semiconductor is by both free electrons and holes. Therefore, the total current
inside the semiconductor is the sum of currents due to free electrons and holes. This creates
new holes near the positive terminal which again drift towards the negative terminal.

Figure 3 : Diagram showing the electronic bonds in an intrinsic semiconductor (Si)

Extrinsic Semiconductor
An extrinsic semiconductor is a semiconductor doped by addition of small amount
impurity which is able to change its electrical properties (conduction), making it suitable for
electronic applications (diodes, transistors, etc.) or optoelectronic applications (light emitters
and detectors). This is achieved by adding a small amount of suitable impurity (having 3 or 5
valence electron) to a semiconductor (having 4 valence electron). It is then called impurity or
extrinsic semiconductor.

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  The process of adding impurities to a semiconductor is known as doping. The purpose of
adding impurity is to increase either the number of free electrons or holes in the
semiconductor crystal.
 If a penta valent impurity (having 5 valence electrons) is added to the semiconductor, a
large number of free electrons are produced in the semiconductor.
 If a trivalent impurity (having 3 valence electrons) is added to the semiconductor, large
number of holes are produced in the semiconductor crystal.
 Depending upon the type of impurity added, extrinsic semiconductors are classified into:
(i) n-type semiconductor
(ii) p-type semiconductor
(i) n-type Semiconductor
When a small amount of pentavalent impurity is added to a pure semiconductor, it is known as n-type
semiconductor.

The addition of pentavalent impurity pro-vides a large number of free electrons in the
semiconductor crystal. Typical examples of pentavalent impurities are arsenic ,
antimony,Bismuth and Phosphorous etc. Such impurities which produce n-type semiconductor
are known as donor impurities because they donate or provide free electrons to the
semiconductor crystal.

Figure 4 : Schematic representation of electronic bonds in a Silicon crystal doped with Arsenic As (n
doping)

 Electrons are said to be the majority carriers whereas holes are the minority carriers.

(ii) p-type Semiconductor

When a small amount of trivalent impurity is added to a pure semiconductor, it is called p-type
Semiconductor.

The addition of trivalent impurity provides a large number of holes in the semiconductor.
Typical examples of trivalent impurities are gallium , indium, boron etc. Such impurities which
produce p-type semiconductor are known as acceptor impurities because the holes created can
accept the electrons fig 5.

Figure 5 : schematic representation of a Si crystal doped with boron (B)

Electrons are said to be the minority carriers whereas holes are the majority carriers.

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Basic Electronics [18EC14/24]

Basic Electronics Notes (Old Syllabus)

Subject Code: 18EC14/24

UNIT-I

Chapter-1: Semiconductor Diodes.

Introduction:

Semiconductor is a chemical Element in which the conductivity lies between conductor

and insulator, Hence the movement of electrons (Current Conduction) can be controlled easily by
means of an external voltage (Biasing).

Two types of Semiconductors

1. Intrinsic Semiconductor (Pure Semiconductor) and
2. Extrinsic Semiconductor (Impure semiconductor/Added impurity)

Extrinsic semiconductor is further classified into two types

a. P-Type Extrinsic Semiconductor (Doping Trivalent element) and
b. N-Type Extrinsic Semiconductor (Doping Pentavalent element).

The holes are majority charge carriers and electrons are minority charge carriers in P-Type and
electrons are majority and holes are minority charge carriers in case of N-Type Semiconductor.

P-N Junction(Semiconductor Diode/ Diode):

Construction: +

Starting with a piece of intrinsic semiconductor and divide it into two halves, one half is
doped with any tri-valent element such as Boron, Aluminum etc., to form P-Type semiconductor,
in which the holes are majority charge carriers and electrons are minority charge carriers. Other
half is doped with any penta-valent element such as phosphorus, arsenic etc., to form N-Type
Semiconductor, in which the electrons are majority charge carriers and holes are minority charge
carriers.
The Junction or a line dividing the P-Type and N-Type is called P-N Junction. Metallic
contact is connected to P-Type and N-Type material to get terminals for the device called
Electrodes such as Anode and Cathode, this device is called P-N Junction Diode or Semiconductor
diode or simply Diode as shown in figure(1).

Figure(1): P-N Junction Diode/Diode

Working:

The working principle can be studied in three different operations or Biasing arrangements
as follows.

Case (1): Zero Biasing.

Without any external supply and at normal room temperature, at the time of contact with
P-Type and N-Type material, it has a tendency to move or diffusing the electrons from N side and

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occupies holes from the P side. Similarly holes in the P side attract electrons in the N side. This
results forming a thin layer near the P-N Junction due to loosing electrons near the junction from
the N side and holes near the junction from the P side. This layer or region is called depletion layer
and it acts as an intrinsic semiconductor as shown in figure (2).

Figure (2): P-N Junction Diode/Diode with depletion layer/region.

Case (2): Reverse Biasing.

External supply with Positive terminal of the battery is connected to the cathode and
negative terminal of the battery is connected to the anode is called Reverse biasing.
With this biasing the negative terminal of the battery sucks out or attracts the holes from
P- Type material and positive terminal of the battery sucks out or attract electrons from N-Type
material, this results wider depletion region and the resistance is very high, and the current that
flows through the device only due to minority charge carriers as shown in figure (3).
The magnitude of current under reverse biasing is in terms of nano amperes for silicon
diodes.

Figure (3): P-N Junction Diode is under Reverse Biasing.

Case (3): Forward Biasing.

External Supply with Positive terminal of the battery is connected to the anode and negative
terminal of the battery is connected to the cathode is called forward biasing.
With this biasing the negative terminal of the battery pushes or pumps more electrons to
the N-Type material and positive terminal of the battery pushes or pumps more holes to the P-
Type material. By go on increasing the biasing voltage the width of the depletion region decreases,
Resistance decreases and the current flowing through the device is increases( not proportional to
voltage). If the Biasing voltage VS is greater than or equal to Vγ (Thresold Voltage) the depletion
layer completely vanishes and easily current will flow as shown in figure (4).The cut in voltage or
threshold voltage (Vγ) for silicon diodes is 0.7 V and for Germanium diodes is 0.3V.

Figure (4): P-N Junction Diode is under Forward Biasing.

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V-I Characteristics:

Figure (5): V-I Characteristics of P-N Junction Diode.

Figure (5) shows the V-I Characteristics of P-N Junction diode, VS is the biasing voltage,
ID is the Diode Current and VBD is the Break down voltage. The leakage current flows through the
device under reverse biasing due to minority charge carriers. Under forward biasing and biasing
voltage is greater than or equal to the threshold voltage, the device then acts as a conducting
material.

Diode Characteristic Parameters:

1. Reverse Resistance (Rr):
The ratio of Reverse biasing voltage to the reverse current is called Reverse resistance of the PN
Junction Diode.
𝑉
i.e.,𝑅𝑟 = 𝐼 𝑅 ; where, VR is the Biasing voltage under reverse biasing, called as reverse voltage and
𝑜
Io is the reverse leakage current.
2. Forward Resistance (Rf):
The ratio of Forward biasing voltage to the Forward current is called Forward resistance of the PN
Junction Diode.
𝑉
i.e.,𝑅𝑟 = 𝐼 𝐹 ; where, VF is the Biasing voltage under Forward biasing, called as Forward voltage
𝐹
and ID is the Forward Current.

Diode Current Equation:

The diode current equation is given by,
𝑉𝐹
𝐼𝐷 = 𝐼𝑂 (𝑒𝜂𝑉𝑇
-1)
Where,
ID is the diode current,
IO is the reverse saturation or leakage current,
VF is the applied forward voltage,
η is a constant 1 for Ge and 2 for Si and
𝒌𝑻
VT is the volt equivalent temperature (Thermal Voltage) is given by, 𝑽𝑻 = 𝒒 ;
Where,
k is the boltzmann’s constant,
T is the temperature in Kelvin and
q is the charge of an electron.
𝒌 = 𝟏. 𝟑𝟖𝟎𝟔𝟒𝟖𝟓𝟐 × 𝟏𝟎−𝟐𝟑 𝑱/𝑲
𝒒 = 𝟏. 𝟔𝒙𝟏𝟎−𝟏𝟗 𝑪𝒐𝒍𝒐𝒖𝒎𝒃𝒔
At room Temperature, i.e., at 300oK,
𝑽𝑻 = 𝟐𝟓. 𝟖𝟓𝒎𝑽 ≈ 𝟐𝟔𝒎𝑽

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Or
𝐓
𝑽𝑻 =
𝟏𝟏𝟔𝟎𝟎
Equivalent Circuit of diode:
1. DC Equivalent Circuit.
The DC equivalent circuit of a diode under reverse biasing is an open circuit or Reverse
Resistance Rr(typically in terms of MΩ) shown in figure (6a), and under forward biasing as shown
in figure (6). Where Rf is the forward resistance of the diode, VON is the voltage drop across the
diode under Conduction State (VON=0.7V for Silicon diodes and VON=0.3V for Germanium
Diodes).

Figure (6a): Diode DC Equivalent Circuit under reverse biasing.

Figure (6b): Diode DC Equivalent Circuit under forward biasing.

2. AC Equivalent Circuit.
The AC equivalent circuit of a diode under reverse biasing and for forward biasing is the
parallel connection of a Resistor and a Capacitor as shown in figure (7a) and figure (7b)
respectively.
Under reverse biasing the depletion layer width increases and acts as a parallel plate
capacitor with dielectric, hence the diode will be considered as a capacitor called Transition
Capacitance/ Junction Capacitance/ Space charge Capacitance.

Figure (7a): AC Equivalent circuit under Reverse Biasing.

Under forward biasing the rate of change of charge carriers increases with respect to the
applied forward voltage, hence the diode under forward biasing acts as a capacitor called
diffusion Capacitance.

Figure (7b): AC equivalent Circuit under forward biasing.

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Effect of Temperature on Diodes:

The number of charge carriers will vary depending on the temperature. i.e., if temperature
increases the number of charge carriers also increases and due to this the conduction current (ID)
also increases. The relation between current, voltage and temperature is as follows.

 The reverse saturation current doubles for every 10oC rise in Temperature.
𝒕𝟐 −𝒕𝟏
i.e., 𝑰’𝑶 = 𝑰𝑶 𝒙 𝟐( 𝟏𝟎
)
;
where,
IO’ is the reverse saturation current at temperature t2 and
IO is the reverse saturation current at temperature t1.

 The forward voltage drop across the diode reduces 2.56mV for every 1oC rise in
temperature.
i.e., 𝑽′𝑭 = VF − 𝟐. 𝟓𝟔𝒎(𝒕𝟐 − 𝒕𝟏 ).
Where,
VF’ is the voltage drop across the diode at t2 and
VF is the voltage drop across the diode at t1

Rectifiers:
Rectifiers are the electronics circuits that convert AC quantity into to DC quantity. This
can be achieved by using unidirectional conduction devices like diode.
Depending on the conduction angle the rectifier circuits are classified into two types, they
are,
1. Half wave Rectifier and
2. Full wave Rectifier.
The Full wave Rectifiers are further classified (based on number of diodes using) into two
types, they are,
a. Center Tap Transformer (Two Diodes) full wave rectifier and
b. Bridge Type (Four Diodes) full wave rectifier.

1. Half wave Rectifier:

The half rectifier is an electronic circuit, which converts AC quantity into pulsating DC,
by using a single diode with conduction angle only 180o that is only half cycle.

Circuit Diagram:

Figure (8): Half wave Rectifier circuit.

Figure (8) shows the circuit diagram of a half wave rectifier, where D is a diode (Assume Diode
is ideal), RL is the load resistor, input is an AC signal and output is the Pulsating DC Signal.

Explanation:

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During every Positive half cycle diode D conducts and acts as a short circuit, hence the
current flows through the Load resistor and is proportional to the input voltage according to Ohm’s
law, therefore the voltage across RL is same as input signal.
i.e., 𝑽𝒐 = 𝑽𝒊

During every negative half cycle diode D does not conducts and acts as an open circuit and
no current flows through the load element, hence the voltage across RL is zero.
i.e., 𝑽𝒐 = 𝟎

Waveforms:
Figure (9) shows the waveforms of an half wave rectifier circuit, and it can be observed that the
output is only half cycle for every complete cycle input and also pulsating DC (Ripples/ some AC
Components also present), i.e., not a pure DC.

Figure (9): Input and Output Waveforms of a Half wave Rectifier circuit.

Mathematical expressions:
The output of half wave rectifier circuit is irregular in nature and hence, need to analyze the circuit
for average DC and AC voltage or current along with the efficiency and ripple factor.
Transformer voltage and current is given by,
𝒗(𝒕) = 𝒗𝒎 𝒔𝒊𝒏𝝎𝒕.
𝒊(𝒕) = 𝒊𝒎 𝒔𝒊𝒏𝝎𝒕
Therefore
1. Average DC Voltage.
𝟏 𝑻
𝑽𝒅𝒄 = ∫ 𝒗(𝒕)𝒅𝝎𝒕.
𝑻 𝟎

𝟏 𝟐𝝅
𝑽𝒅𝒄 = ∫ 𝒗 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕.
𝟐𝝅 𝟎 𝒎
𝟏 𝝅 𝟏 𝟐𝝅
𝑽𝒅𝒄 = ∫ 𝒗 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕 + ∫ 𝒗 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕 .
𝟐𝝅 𝟎 𝒎 𝟐𝝅 𝝅 𝒎
𝟏 𝝅 𝟏 𝟐𝝅
𝑽𝒅𝒄 = ∫ 𝒗𝒎 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕 + ∫ 𝟎 𝒅𝝎𝒕 .
𝟐𝝅 𝟎 𝟐𝝅 𝝅
𝒗𝒎
𝑽𝒅𝒄 = [−𝒄𝒐𝒔𝝎𝒕]𝝅𝟎
𝟐𝝅
𝒗𝒎
𝑽𝒅𝒄 = 𝒙𝟐
𝟐𝝅
𝒗𝒎
𝑽𝒅𝒄 =
𝝅

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2. Average DC Current.

𝟏 𝑻
𝑰𝒅𝒄 = ∫ 𝑰(𝒕)𝒅𝝎𝒕.
𝑻 𝟎

𝟏 𝟐𝝅
𝑰𝒅𝒄 = ∫ 𝒊 𝒔𝒊𝒏𝝎𝒕 𝑑𝝎𝒕.
𝟐𝝅 𝟎 𝒎
𝟏 𝝅 𝟏 𝟐𝝅
𝑰𝒅𝒄 = ∫ 𝒊 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕 + ∫ 𝒊 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕 .
𝟐𝝅 𝟎 𝒎 𝟐𝝅 𝝅 𝒎
𝟏 𝝅 𝟏 𝟐𝝅
𝑰𝒅𝒄 = ∫ 𝒊 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕 + ∫ 𝟎 𝒅𝝎𝒕 .
𝟐𝝅 𝟎 𝒎 𝟐𝝅 𝝅
𝒊𝒎
𝑰𝒅𝒄 = [−𝒄𝒐𝒔𝝎𝒕]𝝅𝟎
𝟐𝝅
𝒊𝒎
𝑰𝒅𝒄 = 𝒙𝟐
𝟐𝝅
𝒊𝒎
𝑰𝒅𝒄 =
𝝅
Or
𝑽𝒅𝒄
𝑰𝒅𝒄 =
𝑹𝑳

3. Root Mean Square value of the output voltage.

𝟏 𝑻
𝑽𝒓𝒎𝒔 = √ ∫ 𝒗𝟐 (𝒕)𝒅𝝎𝒕.
𝑻 𝟎

𝟏 𝝅
𝑽𝒓𝒎𝒔 = √ ∫ (𝒗𝒎 𝒔𝒊𝒏𝝎𝒕)𝟐 𝒅𝝎𝒕.
𝟐𝝅 𝟎

𝟏 𝝅 𝟐
𝑽𝒓𝒎𝒔 = √ ∫ 𝒗𝒎 𝒔𝒊𝒏𝟐 𝝎𝒕 𝒅𝝎𝒕.
𝟐𝝅 𝟎

𝟏 𝝅 𝟐
𝑽𝒓𝒎𝒔 = √ ∫ 𝒗 (𝟏 − 𝒄𝒐𝒔 𝟐𝝎𝒕)/𝟐 𝒅𝝎𝒕.
𝟐𝝅 𝟎 𝒎
𝒗𝒎
𝑽𝒓𝒎𝒔 =
𝟐
4. Root mean square value of the output current.
𝟏 𝑻𝟐
𝑰𝒓𝒎𝒔 = √ ∫ 𝒊 (𝒕)𝒅𝝎𝒕.
𝑻 𝟎

𝟏 𝟐𝝅
𝑰𝒓𝒎𝒔 = √ ∫ (𝒊𝒎 𝒔𝒊𝒏𝝎𝒕)𝟐 𝒅𝝎𝒕.
𝟐𝝅 𝟎

𝟏 𝟐𝝅 𝟐
𝑰𝒓𝒎𝒔 = √ ∫ 𝒊𝒎 𝒔𝒊𝒏𝟐 𝝎𝒕 𝒅𝝎𝒕.
𝟐𝝅 𝟎

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𝟏 𝝅𝟐
𝑰𝒓𝒎𝒔 = √ ∫ 𝒊 (𝟏 − 𝒄𝒐𝒔 𝟐𝝎𝒕)/𝟐 𝒅𝝎𝒕.
𝟐𝝅 𝟎 𝒎
𝒊𝒎
𝑰𝒓𝒎𝒔 =
𝟐
Or

𝑽𝒓𝒎𝒔
𝑰𝒓𝒎𝒔 =
𝑹𝑳
5. Ripple factor.
Ripple factor is the ratio of rms value of the ac component to the dc value of the component.
𝑽𝒓𝒓𝒎𝒔
𝒓𝒊𝒑𝒑𝒍𝒆 𝒇𝒂𝒄𝒕𝒐𝒓 (𝜸) =
𝑽𝒅𝒄
√𝑽𝟐𝒓𝒎𝒔 − 𝑽𝟐𝒅𝒄
𝜸=
𝑽𝒅𝒄
𝟐 𝟐
√(𝑽𝒎 ) − (𝑽𝒎 )
𝟐 𝝅
𝜸=
𝑽𝒎
𝝅
𝜸 = 𝟏. 𝟐𝟏𝟏𝟒.

6. Efficiency.
It is the ratio of dc output power present in the output to the ac power input.
𝑷𝒅𝒄
𝜼= 𝒙𝟏𝟎𝟎
𝑷𝒂𝒄
𝑽𝟐𝒅𝒄
𝑹
𝜼 = 𝟐 𝑳 𝒙𝟏𝟎𝟎
𝑽𝒓𝒎𝒔
𝑹𝑳
𝜼 = 𝟎. 𝟒𝟎𝟔𝒙𝟏𝟎𝟎
𝜼 = 𝟒𝟎. 𝟔%.

7. Peak Inverse Voltage.

The maximum voltage applied across the diode under reverse biasing, in other words maximum
reverse voltage, which occurs at the peak of the input signal when the diode is reverse biased is
called Peak Inverse Voltage or Peak Reverse Voltage.

The Maximum reverse voltage applied across the diode is Vm.

i.e., 𝑷𝑰𝑽 = 𝑽𝒎
Note: The above expressions holds good only for the ideal diode (Rf=0 and VON=0) and ideal
transformer (Rs=0).
Case (i): For a diode with Rf.
𝑉𝑚
𝐼𝑚 =
𝑅𝑓 + 𝑅𝐿
𝐼𝑚
𝐼𝐷𝐶 =
𝜋

𝑉𝐷𝐶 = 𝐼𝐷𝐶 𝑋 𝑅𝐿

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𝐼𝑚
𝐼𝑟𝑚𝑠 =
2
𝑉𝑟𝑚𝑠 = 𝐼𝑟𝑚𝑠 𝑋 𝑅𝐿
2 2
√𝐼𝑟𝑚𝑠 − 𝐼𝐷𝐶
𝛾=
𝐼𝐷𝐶
2
𝐼𝐷𝐶 𝑅𝐿
%𝜂 = 2 𝑥100
𝐼𝑟𝑚𝑠 (𝑅𝑓 + 𝑅𝐿 )

Case (ii): For a diode with Rf and Transformer with Rs

𝑉𝑚
𝐼𝑚 =
𝑅𝑓 + 𝑅𝑠 + 𝑅𝐿
𝐼𝑚
𝐼𝐷𝐶 =
𝜋

𝑉𝐷𝐶 = 𝐼𝐷𝐶 𝑋 𝑅𝐿
𝐼𝑚
𝐼𝑟𝑚𝑠 =
2
𝑉𝑟𝑚𝑠 = 𝐼𝑟𝑚𝑠 𝑋 𝑅𝐿
2 2
√𝐼𝑟𝑚𝑠 − 𝐼𝐷𝐶
𝛾=
𝐼𝐷𝐶
2
𝐼𝐷𝐶 𝑅𝐿
%𝜂 = 2 (𝑅 + 𝑅 + 𝑅 )
𝑥100
𝐼𝑟𝑚𝑠 𝑓 𝑠 𝐿

Rectifier with C Filter:

From the above discussion it was observed that the output of the rectifier is not a pure dc.
i.e., some ac components (ripples) also present, to remove the ripples we need to connect a filter.
Filter can be capacitor or inductor, in the following context the capacitor filter will be discussed.
Filter is an electronic circuit which converts pulsating dc into pure dc.
The property of the capacitor is that it allows the ac component (High frequency component and
low resistance) and blocks the dc component (low frequency component and high resistance).

Figure (10), shows the circuit diagram of an half-wave rectifier with capacitor filter, i.e., a
capacitor is connected across the load resistor RL. When AC voltage is applied, during the positive
half cycle, the diode D is forward biased and allows electric current through it and capacitor starts
charging instantly to its maximum level. During negative half cycle diode does not conducts and
no current flow through the capacitor and hence starts discharging slowly. The discharging time
constant is given by,
𝑻𝒅 = 𝑪𝑹𝑳

Note: If C or RL or both increases the discharging time also increases and we will get pure dc. For
very high Td the capacitor never discharges and it acts as a voltage source of value is equal to load
voltage.

As we already know that, the capacitor provides high resistive path to dc components (low-
frequency signal) and low resistive path to ac components (high-frequency signal).

Electric current always prefers to flow through a low resistance path. So when the electric
current reaches the filter, the dc components experience a high resistance from the capacitor and
ac components experience a low resistance from the capacitor.

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The dc component does not like to flow through the capacitor (high resistance path). So
they find an alternative path (low resistance path) and flows to the load resistor (RL) through that
path shown in figure (11).

Figure (10): Half wave Rectifier circuit with Capacitor Filter.

Figure (11): Working of Capacitor filter.

Electric current always prefers to flow through a low resistance path. So the AC
components will flow through the capacitor whereas the DC components are blocked by the
capacitor. Therefore, they find an alternate path and reach the output load resistor R L. The flow of
AC components through the capacitor is nothing but the charging of a capacitor.

Waveforms:

Figure (12): waveform of a Half wave Rectifier with C-filter.

Figure (12) shows the waveform of a half-wave rectifier filtered output. If discharging time
increases then the output become pure dc.

The Ripple factor of a filtered output signal is given by

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1
𝛾=
2√3𝑓𝑅𝐿 𝐶
Advantage:
 Simple and easy to construct.
 PIV is only Vm.

Disadvantages:
 Conducts only half cycle, due to this more power will be wasted.
 More ripples occur in the output.

2. Full wave Rectifier using Center tap transformer:

A full wave rectifier is a type of rectifier which converts both half cycles of the AC signal
into pulsating DC signal.
Circuit Diagram:

Figure (13): Full-wave Rectifier circuit.

As shown in the figure (13), the full wave rectifier converts both positive and negative half cycles
of the input AC signal into output pulsating DC signal.

Note: The center tapped transformer shown in figure (14), works almost similar to a normal
transformer. Like a normal transformer, the center tapped transformer also increases or reduces
the AC voltage. However, a center tapped transformer has another important feature. That is the
secondary winding of the center tapped transformer divides the input AC current or AC signal (VP)
into two parts.

Figure (14): Working of Center Tapped Transformer.

The upper part of the secondary winding produces a positive voltage V1 and the lower part of the
secondary winding produces a negative voltage V2. When we combine these two voltages at output
load, we get a complete AC signal.
i.e. 𝑽𝑻𝒐𝒕𝒂𝒍 = 𝑽𝟏 + 𝑽𝟐

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The voltages V1 and V2 are equal in magnitude but opposite in direction. That is the voltages
(V1 and V2) produced by the upper part and lower part of the secondary winding are 180 degrees
out of phase with each other. However, by using a full wave rectifier with center tapped
transformer, we can produce the voltages that are in phase with each other. In simple words, by
using a full wave rectifier with center tapped transformer, we can produce a current that flows only
in single direction.

The AC source is connected to the primary winding of the center tapped transformer. A
center tap (additional wire) connected at the exact middle of the the secondary winding divides the
input voltage into two parts.
The upper part of the secondary winding is connected to the diode D1 and the lower part of the
secondary winding is connected to the diode D2. Both diode D1 and diode D2 are connected to a
common load RL with the help of a center tap transformer. The center tap is generally considered
as the ground point or the zero voltage reference point.

Explanation:
The center tapped full wave rectifier uses a center tapped transformer and two diodes to
convert the input AC voltage into output DC voltage.
When input AC voltage is applied, the secondary winding of the center tapped transformer
divides this input AC voltage into two parts: positive and negative.
During every positive half cycle of the input AC signal, terminal A become positive,
terminal B become negative and center tap is grounded (zero volts). The positive terminal A is
connected to the p-side of the diode D1 and the negative terminal B is connected to the n-side of
the diode D1. So the diode D1 is forward biased during the positive half cycle and allows electric
current through it.

Figure (15): Full-wave rectifier circuit during every positive half cycle.

On the other hand, the negative terminal B is connected to the p-side of the diode D2 and
the positive terminal A is connected to the n-side of the diode D2. So the diode D2 is reverse biased
during every positive half cycle and does not allow electric current through it.
The diode D1 supplies DC current to the load RL. The DC current produced at the load RL will
return to the secondary winding through a center tap.
i.e., 𝑽𝒐 = 𝑽𝒊
During the positive half cycle, current flows only in the upper part of the circuit while the
lower part of the circuit carry no current to the load because the diode D2 is reverse biased. Thus,
during the positive half cycle of the input AC signal, only diode D1 allows electric current while
diode D2 does not allow electric current as shown in figure (15).
During every negative half cycle of the input AC signal, terminal A become negative,
terminal B become positive and center tap is grounded (zero volts). The negative terminal A is
connected to the p-side of the diode D1 and the positive terminal B is connected to the n-side of
the diode D1. So the diode D1 is reverse biased during the negative half cycle and does not allow
electric current through it.

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Figure (16): Full-Wave Rectifier during every negative half cycle.

On the other hand, the positive terminal B is connected to the p-side of the diode D2 and
the negative terminal A is connected to the n-side of the diode D2. So the diode D2 is forward
biased during the negative half cycle and allows electric current through it.
The diode D2 supplies DC current to the load RL. The DC current produced at the load RL will
return to the secondary winding through a center tap as shown in figure (16).
During the negative half cycle, current flows only in the lower part of the circuit while the
upper part of the circuit carry no current to the load because the diode D1 is reverse biased. Thus,
during the negative half cycle of the input AC signal, only diode D2 allows electric current while
diode D1 does not allow electric current.
Thus, the diode D1 allows electric current during the positive half cycle and diode
D2 allows electric current during the negative half cycle of the input AC signal. As a result, both
half cycles (positive and negative) of the input AC signal are allowed. So the output DC voltage
is almost equal to the input AC voltage as shown in figure (17).

Figure (17): Full-wave rectifier with total output.

The diodes D1 and D2 are commonly connected to the load RL. So the load current is the sum of
individual diode currents.
We know that a diode allows electric current in only one direction. From the figure (17), we can
see that both the diodes D1 and D2 are allowing current in the same direction.
We know that a current that flows in only single direction is called a direct current. So the
resultant current at the output (load) is a direct current (DC). However, the direct current appeared
at the output is not a pure direct current but a pulsating direct current.
The value of the pulsating direct current changes with respect to time. This is due to the
ripples in the output signal. These ripples can be reduced by using filters such as capacitor and
inductor.
The average output DC voltage across the load resistor is double that of the single half wave
rectifier circuit.

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Waveforms:
The output waveforms of the full wave rectifier is shown in figure (18).

Figure (18): waveforms of full wave rectifier.

The first waveform represents an input AC signal. The second waveform and third
waveform represents the DC signals or DC current produced by diode D1 and diode D2. The last
waveform represents the total output DC current produced by diodes D1and D2. From the above
waveforms, we can conclude that the output current produced at the load resistor is not a pure DC
but a pulsating DC.

Mathematical Expressions:
The output of half wave rectifier circuit is irregular in nature and output would be equal to the
average voltage or current.
Transformer voltage and current is given by,
𝒗(𝒕) = 𝒗𝒎 𝒔𝒊𝒏𝝎𝒕.
𝒊(𝒕) = 𝒊𝒎 𝒔𝒊𝒏𝝎𝒕
Therefore
1. Average or DC Voltage.
𝟏 𝑻
𝑽𝒅𝒄 𝒐𝒓 𝑽𝒂𝒗𝒈 = ∫ 𝒗(𝒕)𝒅𝝎𝒕.
𝑻 𝟎

𝟏 𝟐𝝅
𝑽𝒅𝒄 = ∫ 𝒗 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕.
𝟐𝝅 𝟎 𝒎
𝟏 𝝅 𝟏 𝟐𝝅
𝑽𝒅𝒄 = ∫ 𝒗𝒎 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕 + ∫ 𝒗 𝒔𝒊𝒏𝝎𝝎𝒕 𝒅𝒕 .
𝟐𝝅 𝟎 𝟐𝝅 𝝅 𝒎
𝟏 𝝅 𝟏 𝟐𝝅
𝑽𝒅𝒄 = ∫ 𝒗 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕 + ∫ 𝒗 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕 .
𝟐𝝅 𝟎 𝒎 𝟐𝝅 𝝅 𝒎
Component Between π to 2𝝅 is same as the 0 to π component.
Therefore,
𝟏 𝝅
𝑽𝒅𝒄 = 𝟐𝒙 ∫ 𝒗 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕.
𝟐𝝅 𝟎 𝒎
𝒗𝒎
𝑽𝒅𝒄 = 𝟐𝒙 [−𝒄𝒐𝒔𝝎𝒕]𝝅𝟎
𝟐𝝅
𝒗𝒎
𝑽𝒅𝒄 = 𝟐𝒙 𝒙𝟐
𝟐𝝅
𝟐𝒗𝒎
𝑽𝒅𝒄 =
𝝅

2. Average or DC Current.

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𝟏 𝑻
𝑰𝒅𝒄 𝒐𝒓 𝑰𝒂𝒗𝒈 = ∫ 𝑰(𝒕)𝒅𝝎𝒕.
𝑻 𝟎

𝟏 𝟐𝝅
𝑰𝒅𝒄 = ∫ 𝒊 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕.
𝟐𝝅 𝟎 𝒎
𝟏 𝝅 𝟏 𝟐𝝅
𝑰𝒅𝒄 = ∫ 𝒊𝒎 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕 + ∫ 𝒊 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕 .
𝟐𝝅 𝟎 𝟐𝝅 𝝅 𝒎
𝟏 𝝅
𝑰𝒅𝒄 = 𝟐𝒙 ∫ 𝒊 𝒔𝒊𝒏𝝎𝒕 𝒅𝝎𝒕
𝟐𝝅 𝟎 𝒎
𝒊𝒎
𝑰𝒅𝒄 = 𝟐𝒙 [−𝒄𝒐𝒔𝝎𝒕]𝝅𝟎
𝟐𝝅
𝒊𝒎
𝑰𝒅𝒄 = 𝟐𝒙 𝒙𝟐
𝟐𝝅
𝟐𝒊𝒎
𝑰𝒅𝒄 =
𝝅
Or
𝑽𝒅𝒄
𝑰𝒅𝒄 =
𝑹𝑳
3. Root Mean Square value of the output voltage.

𝟏 𝑻 𝟐
𝑽𝒓𝒎𝒔 = √ ∫ 𝒗 (𝒕)𝒅𝝎𝒕.
𝑻 𝟎

𝟏 𝟐𝝅
𝑽𝒓𝒎𝒔 = √ ∫ (𝒗𝒎 𝒔𝒊𝒏𝝎𝒕)𝟐 𝒅𝝎𝒕.
𝟐𝝅 𝟎

𝟏 𝟐𝝅 𝟐
𝑽𝒓𝒎𝒔 = √ ∫ 𝒗 𝒔𝒊𝒏𝟐 𝝎𝒕 𝒅𝝎𝒕.
𝟐𝝅 𝟎 𝒎

𝟏 𝝅 𝟐
𝑽𝒓𝒎𝒔 = √𝟐𝒙 ∫ 𝒗 (𝟏 − 𝒄𝒐𝒔 𝟐𝝎𝒕)/𝟐 𝒅𝝎𝒕.
𝟐𝝅 𝟎 𝒎
𝒗𝒎
𝑽𝒓𝒎𝒔 =
√𝟐
4. Root mean square value of the output current.
𝟏 𝑻𝟐
𝑰𝒓𝒎𝒔 = √ ∫ 𝒊 (𝒕)𝒅𝝎𝒕.
𝑻 𝟎

𝟏 𝟐𝝅
𝑰𝒓𝒎𝒔 = √ ∫ (𝒊𝒎 𝒔𝒊𝒏𝝎𝒕)𝟐 𝒅𝝎𝒕.
𝟐𝝅 𝟎

𝟏 𝝅𝟐
𝑰𝒓𝒎𝒔 = √𝟐𝒙 ∫ 𝒊 𝒔𝒊𝒏𝟐 𝝎𝒕 𝒅𝝎𝒕.
𝟐𝝅 𝟎 𝒎

𝟏 𝟐𝝅 𝟐
𝑰𝒓𝒎𝒔 = √𝟐𝒙 ∫ 𝒊 (𝟏 − 𝒄𝒐𝒔 𝟐𝝎𝒕)/𝟐 𝒅𝝎𝒕.
𝟐𝝅 𝟎 𝒎
𝒊𝒎
𝑰𝒓𝒎𝒔 =
√𝟐
Or
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𝑽𝒓𝒎𝒔
𝑰𝒓𝒎𝒔 =
𝑹𝑳
5. Ripple factor.
Ripple factor is the ratio of rms value of the ac component to the dc value of the component.
𝐕𝐫𝐫𝐦𝐬
𝐫𝐢𝐩𝐩𝐥𝐞 𝐟𝐚𝐜𝐭𝐨𝐫 (𝛄) =
𝐕𝐝𝐜
𝟐 − 𝐕𝟐
√𝐕𝐫𝐦𝐬 𝐝𝐜
𝛄=
𝐕𝐝𝐜
𝟐 𝟐
√(𝐕𝐦 ) − (𝟐𝐕𝐦 )
√𝟐 𝛑
𝛄=
𝟐𝐕𝐦
𝛑
𝛄 = 𝟎. 𝟒𝟖.

6. Efficiency.
It is the ratio of dc output power present in the output to the ac power input.
𝐏𝐝𝐜
𝛈= 𝐱𝟏𝟎𝟎
𝐏𝐫𝐦𝐬
𝟐
𝐕𝐝𝐜
𝐑
𝛈 = 𝟐𝐋 𝐱𝟏𝟎𝟎
𝐕𝐫𝐦𝐬
𝐑𝐋
𝛈 = 𝟎. 𝟖𝟏𝟐𝐱𝟏𝟎𝟎
𝛈 = 𝟖𝟏. 𝟐%.

7. Peak Inverse Voltage.

The maximum voltage applied across the diode under reverse biasing, in other words maximum
reverse voltage, which occurs at the peak of the input signal when the diode is reverse biased is
called Peak Inverse Voltage or Peak Reverse Voltage.

The Maximum reverse voltage applied across the diode is 2Vm.

i.e., 𝑷𝑰𝑽 = 𝟐𝑽𝒎

Note: The above expressions holds good only for the ideal diode (Rf=0 and VON=0) and ideal
transformer (Rs=0).
Case (i): For a diode with Rf.
𝑉𝑚
𝐼𝑚 =
𝑅𝑓 + 𝑅𝐿
2𝐼𝑚
𝐼𝐷𝐶 =
𝜋

𝑉𝐷𝐶 = 𝐼𝐷𝐶 𝑋 𝑅𝐿
𝐼𝑚
𝐼𝑟𝑚𝑠 =
√2
𝑉𝑟𝑚𝑠 = 𝐼𝑟𝑚𝑠 𝑋 𝑅𝐿

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2 2
√𝐼𝑟𝑚𝑠 − 𝐼𝐷𝐶
𝛾=
𝐼𝐷𝐶
2
𝐼𝐷𝐶 𝑅𝐿
%𝜂 = 2 𝑥100
𝐼𝑟𝑚𝑠 (𝑅𝑓 + 𝑅𝐿 )

Case (ii): For a diode with Rf and Transformer with Rs

𝑉𝑚
𝐼𝑚 =
𝑅𝑓 + 𝑅𝑠 + 𝑅𝐿
2𝐼𝑚
𝐼𝐷𝐶 =
𝜋

𝑉𝐷𝐶 = 𝐼𝐷𝐶 𝑋 𝑅𝐿
𝐼𝑚
𝐼𝑟𝑚𝑠 =
√2
𝑉𝑟𝑚𝑠 = 𝐼𝑟𝑚𝑠 𝑋 𝑅𝐿
2 2
√𝐼𝑟𝑚𝑠 − 𝐼𝐷𝐶
𝛾=
𝐼𝐷𝐶
2
𝐼𝐷𝐶 𝑅𝐿
%𝜂 = 2 (𝑅 + 𝑅 + 𝑅 )
𝑥100
𝐼𝑟𝑚𝑠 𝑓 𝑠 𝐿

Full-wave Rectifier with C Filter:

The filter is an electronic device that converts the pulsating Direct Current into pure Direct
Current.
In the circuit diagram, the capacitor C is placed across the load resistor RL.

Note: The working of the full-wave rectifier with filter is almost similar to that of the half wave
rectifier with filter. The only difference is that in the half wave rectifier only one half cycles (either
positive or negative) of the input AC current will charge the capacitor but the remaining half cycle
will not charge the capacitor. But in full wave rectifier, both positive and negative half cycles of
the input AC current will charge the capacitor.

The main duty of the capacitor filter is to short the ripples to the ground and blocks the
pure DC (DC components), so that it flows through the alternate path and reaches output load
resistor RL.

Figure (19) shows the circuit diagram of a full-wave rectifier with Capacitor filter, when input AC
voltage is applied, during the positive half cycle, the diode D1 is forward biased and allows electric
current whereas the diode D2 is reverse biased and blocks electric current. On the other hand,
during the negative half cycle the diode D2 is forward biased (allows electric current) and the diode
D1 is reverse biased (blocks electric current).

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Figure (19): Full-wave Rectifier circuit with Capacitor filter.

During the positive half cycle, the diode (D1) current reaches the filter and charges the
capacitor. However, the charging of the capacitor happens only when the applied AC voltage is
greater than the capacitor voltage.
Initially, the capacitor is uncharged. That means no voltage exists between the plates of the
capacitor. So when the voltage is turned on, the charging of the capacitor happens immediately.
During this conduction period, the capacitor charges to the maximum value of the input
supply voltage. The capacitor stores a maximum charge exactly at the quarter positive half cycle
in the waveform. At this point, the supply voltage is equal to the capacitor voltage.

Figure (20): Full-wave Rectifier filtered output.

When the AC voltage starts decreasing and becomes less than the capacitor voltage, then
the capacitor starts slowly discharging as shown in figure (20).
The discharging of the capacitor is very slow as compared to the charging of the capacitor.
So the capacitor does not get enough time to completely discharge. Before the complete discharge
of the capacitor happens, the charging again takes place. So only half or more than half of the
capacitor charge get discharged.
When the input AC supply voltage reaches the negative half cycle, the diode D1 is reverse
biased (blocks electric current) whereas the diode D2 is forward biased (allows electric current).
During the negative half cycle, the diode (D2) current reaches the filter and charges the capacitor.
However, the charging of the capacitor happens only when the applied AC voltage is greater than
the capacitor voltage.
The capacitor is not completely uncharged, so the charging of the capacitor does not happen
immediately. When the supply voltage becomes greater than the capacitor voltage, the capacitor
again starts charging.
In both positive and negative half cycles, the current flows in the same direction across the
load resistor RL. So we get either complete positive half cycles or negative half cycles. In our case,
they are complete positive half cycles.

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The Ripple factor of a filtered output signal is given by

1
𝛾=
4√3𝑓𝑅𝐿 𝐶

Advantages:
 Conducts both the half cycles.
 Efficiency is improved
 Ripples factor is reduced.

Disadvantages:
 Center tapped transformer more expensive and bulky.
 PIV is 2Vm.

3. Bridge type Full wave Rectifies:

The Bridge Type Full-wave rectifier is an electronic circuit, which converts AC quantity
into pulsating DC, by using the four diodes with conduction angle only 360o that is complete cycle.

Circuit Diagram:
The Circuit diagram of a bridge rectifier is shown in figure (21). The bridge rectifier is
made up of four diodes namely D1, D2, D3, D4 and load resistor RL. The four diodes are connected
in a closed loop (Bridge) configuration to efficiently convert the Alternating Current (AC) into
Direct Current (DC).

Figure (21): Full-wave Bridge Rectifier Circuit diagram.

Explanation:
The input AC signal is applied across two terminals A and B and the output DC signal is
obtained across the load resistor RL which is connected between the terminals C and D.
The four diodes D1, D2, D3, D4 are arranged in series with only two diodes allowing electric
current during each half cycle. For example, diodes D1 and D2 are considered as one pair which
allows electric current during the positive half cycle whereas diodes D3 and D4 are considered as
another pair which allows electric current during the negative half cycle of the input AC signal.
When input AC signal is applied across the bridge rectifier, during the positive half cycle
diodes D1 and D2 are forward biased and allows electric current while the diodes D3 and D4 are
reverse biased and blocks electric current. On the other hand, during the negative half cycle diodes
D3 and D4 are forward biased and allow electric current while diodes D1 and D2 are reverse biased
and blocks electric current.
During the positive half cycle, the terminal A becomes positive while the terminal B becomes
negative. This causes the diodes D1 and D2 forward biased and at the same time, it causes the
diodes D3 and D4 reverse biased.
The current flow direction during the positive half cycle is shown in the figure (22) (i.e. A to D to
C to B).
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Figure (22): Bridge Rectifier during every positive half cycle.

During the negative half cycle, the terminal B becomes positive while the terminal A becomes
negative. This causes the diodes D3 and D4 forward biased and at the same time, it causes the
diodes D1 and D2reverse biased.
The current flow direction during negative half cycle is shown in figure (23) (I.e. B to D to C to
A).

Figure (23): Bridge Rectifier during every negative half cycle.

From the two figures (22 and 23), we can observe that the direction of current flow across
load resistor RL is same during the positive half cycle and negative half cycle. Therefore, the
polarity of the output DC signal is same for both positive and negative half cycles. The output DC
signal polarity may be either completely positive or negative. In our case, it is completely
positive. If the direction of diodes is reversed then we get a complete negative DC voltage.
Thus, a bridge rectifier allows electric current during both positive and negative half cycles of the
input AC signal.
The output waveforms of the bridge rectifier is shown in figure (24).

Waveforms:

Figure (24): Bridge Rectifier waveforms.

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Mathematical Expressions:
Same as Full wave Rectifier except PIV.
PIV of Bridge type full wave rectifier is only Vm.
i.e., 𝑷𝑰𝑽 = 𝑽𝒎

Note: The above expressions holds good only for the ideal diode (Rf=0 and VON=0) and ideal
transformer (Rs=0).
Case (i): For a diode with Rf.
𝑉𝑚
𝐼𝑚 =
2𝑅𝑓 + 𝑅𝐿
2𝐼𝑚
𝐼𝐷𝐶 =
𝜋

𝑉𝐷𝐶 = 𝐼𝐷𝐶 𝑋 𝑅𝐿
𝐼𝑚
𝐼𝑟𝑚𝑠 =
√2
𝑉𝑟𝑚𝑠 = 𝐼𝑟𝑚𝑠 𝑋 𝑅𝐿
2 2
√𝐼𝑟𝑚𝑠 − 𝐼𝐷𝐶
𝛾=
𝐼𝐷𝐶
2
𝐼𝐷𝐶 𝑅𝐿
%𝜂 = 2 𝑥100
𝐼𝑟𝑚𝑠 (2𝑅𝑓 + 𝑅𝐿 )

Case (ii): For a diode with Rf and Transformer with Rs

𝑉𝑚
𝐼𝑚 =
2𝑅𝑓 + 𝑅𝑠 + 𝑅𝐿
2𝐼𝑚
𝐼𝐷𝐶 =
𝜋

𝑉𝐷𝐶 = 𝐼𝐷𝐶 𝑋 𝑅𝐿
𝐼𝑚
𝐼𝑟𝑚𝑠 =
√2
𝑉𝑟𝑚𝑠 = 𝐼𝑟𝑚𝑠 𝑋 𝑅𝐿
2 2
√𝐼𝑟𝑚𝑠 − 𝐼𝐷𝐶
𝛾=
𝐼𝐷𝐶
2
𝐼𝐷𝐶 𝑅𝐿
%𝜂 = 2 (2𝑅 + 𝑅 + 𝑅 )
𝑥100
𝐼𝑟𝑚𝑠 𝑓 𝑠 𝐿

Bridge Rectifier with C Filter:

Same as Full wave Rectifier with C Filter.

Advantages:
 Center tapped Transformer is not necessary hence circuit becomes simple and cheap.
 PIV is only Vm.

Disadvantages:
 Requires four diodes.

Choke Filter:

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Choke filter is a circuit consists of an inductor connected in series with load and capacitor
connected across the load. The choke filter is also called L-section filter as shown in figure (25).

Figure (25): Choke Filter.

The choke filter is used to remove the ripples present in the rectified output. The Pulsating DC
signal pass through the inductor or choke will blocks ac and allows dc and further the ripples i.e.,
some of the residual ripples will get bypassed through the capacitor (capacitor blocks dc and allows
ac). Hence from the choke filtered output is pure dc.
The Ripple factor of a filtered output signal is given by

1.19
𝛾=
𝐿𝐶

Significance of Choke filters:

The series inductor filter decreases the current, and shunt capacitor filter increases the current.

Zener Diode:

The reverse current through the normal diode is in terms of microamperes and it is almost
constant until the reverse voltage is less than break down voltage, if the reverse voltage is greater
than or equal to the break down voltage the junction breaks and high current will flow through the
device and more power will be dissipated then the device may be destroyed or damaged.

If we limit the current through the device by means of connecting a resistor in series with
the device, the power dissipation reduces and the device may not be destroyed even under
breakdown region. By using this principle the special type of diode is designed by Clearance Zener
called as Zener diode.
There are two types of breakdown occurs in Zener diode depending on the break down
voltage levels.
i) Zener Break down:
This type of breakdown occurs in the device if the breakdown voltage is less than or equal
to 6V(typically), this strong electric field at the junction becomes very large and breaks the
covalent bonds to release free electrons, due to this very high current will flow through the device.
This mechanism or process is called ionization by Electric field.

ii) Avalanche Breakdown:

This type of breakdown occurs in the device if the breakdown voltage is greater than 6V
(Typically), this high potential forces minority charge carriers to move quickly means kinetic
energy increases, due to this the minority charge carriers collide with atoms to break covalent
bonds which increase the free electrons and hence the current increases sharply. This process or
mechanism is called, Impact Ionization or Ionization by collision. In this mechanism, the free
electrons increase in multiples and hence called avalanche breakdown.

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VI Characteristics of Zener Diode:

Figure (26): VI Characteristics of Zener Diode.

Zener Diode Voltage Regulator:

Zener diode provides constant voltage if the Zener Current is between IZmin to IZmax under
reverse biasing, this feature of Zener diode will be utilized to design a voltage regulator.
Voltage regulator is a system which produces the constant voltage irrespective of variations
in line (Input) and load.
Figure (27) shows the circuit diagram of a Zener Diode Voltage Regulator. Where, R is the
Series Resistor used to limit or control the sharp current flowing through the Zener diode under
breakdown condition, RL is the Load resistance, Vi is the unregulated power supply, IZ is the
current through the Zener diode, IL is the load current, VR is the voltage drop across series resistor.
VZ is the Zener Voltage, Vo is the voltage across the load resistor called output voltage and I is the
input current.

Figure (27): Zener Diode Voltage Regulator Circuit.

From the above figure, the input unregulated power supply Vi, Positive terminal is
connected to the cathode terminal and negative terminal is connected to the anode terminal, hence
the Zener diode is operating in the reverse biasing mode.
And the above circuit provides a constant voltage even by varying the input voltage and
varying load, i.e., provides regulation for both line and load.
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Case(i): Line Regulation; Variable input and constant load:

If the input voltage Vi is less than the Zener Voltage VZ ( Zener Break Down Voltage), the
output voltage is same as the input voltage Vi, because the Zener diode is in off state, if the input
voltage is greater than the Zener voltage VZ, the diode in ON state and hence it acts as a voltage
source of VZ Volts.

If Vi increases, the input current I also increases, and IZ increases to maintain IL constant,
but IZ should be between IZmin to IZmax.
If Vi decreases, the input current I also decreases, and IZ decreases to maintain IL constant,
but IZ should be between IZmin to IZmax.

Therefore the voltage across the load resistor constant and is given by
Vo=VZ

Case(ii): Load Regulation; Fixed input and variable load:

If RL increases, IL decreases and to keep input current I constant IZ increases, but IZ should
be between IZmin to IZmax.
If RL decreases, IL increases and to keep input current I constant IZ decreases, but
IZ should be between IZmin to IZmax.
Therefore the voltage across the load resistor constant and is given by
Vo=VZ

Special types of diodes

The PN junctions with additional features are called special types of diodes, which have
special form of PN junction compared to normal diodes. Most of the special diodes converts one
form of energy into another form, few of them are discussed in the following section.

1. Photodiodes:

Photo diodes are special type of diode or special form of PN junction, which conducts or
produces current when the junction is exposed to light radiation under reverse biasing condition.
I.e., photo diodes converts light radiation into electrical energy. If the radiation increases, the
current produced by the device also increases.

Circuit symbol and realistic view of photodiode:

The pair of inward arrow indicates that, the device starts conducting, when the device receives
light radiation.

Internal structure with biasing arrangement:

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Figure shows the internal structure of the photodiode, these types of diodes are surrounded by a
glass surface to allow the radiation into the junction. V is the voltage supply for biasing the device
in reverse biasing mode. Ammeter connected in series with the device to read the current produced
by the device. Photo diodes will work in two modes
i) Photovoltaic (Solar cells) : Under un-biasing condition, when the junction exposed to radiation,
the device will produces current is called photovoltaic mode of operation of photodiode. Example:
Solar cells.
ii) Photoconductive (Photo diode): The device is under reverse biasing and the junction exposed
to radiation, produces current. This mode of operation is called photoconductive mode.

Working principle:

Under reverse biasing the depletion region increases and immobile ions accumulates near
the junction on both sides, which acts as a barrier and avoids further movement of charge carriers
from one region to another region. When the junction is exposed to light radiation, the junction
takes sufficient energy to break the covalent bonds. The covalent bonds breaks and generates free
electrons, due to the movement of these free electrons, current will be flowing through the device.

VI Characteristics:

Figure shows the VI characteristics of a photodiode, which is a plot of reverse voltage vs

current through the device with constant and different light intensities. If the device under reverse
biasing and does not exposed to radiation, only small current will flow through the device due to
minority charge carriers, this current is called dark current. If the light intensity increases, the
current through the device increases.

Applications:
• Theft detection
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• Automatic door opening and closing

• Obstacle detection

2. Light emitting diode (LED):

Light emitting diode is a special type of diode or special form of PN junction, which converts
electrical energy into light energy under forward biasing. The reverse bias operation of LEDs is
avoided due to very low breakdown voltage.

Circuit symbol and realistic view of LED:

The pair of outward arrows, indicates when the device conducts, which emits visible light.

Internal structure with biasing arrangement and working:

Figure, shows the internal structure and biasing arrangement of light emitting diode, these
types of diodes are surrounded by a transparent plastic hard epoxy hemisphere cell to protect the
device from the external shock and to emit the visible light in different colors.

V is the voltage supply for biasing the device in forward biasing mode. Energy gap must be greater
than 1.8eV in order to emit the visible light. The device needs some energy to move the free
electrons from valence band to conduction band and during the recombination, electrons releases
approximately same amount of energy, if the released energy in the form of photon.
Energy of photon is the product of plank’s constant and frequency of radiated electromagnetic
wave.
i.e., 𝐸𝑔 = ℎ𝑓 − − − (1)
𝑐
𝑤. 𝑘. 𝑡. , 𝑓 = − − − (2)
𝜆
Where, c is the velocity of light and 𝜆 is the wavelength of electromagnetic wave.

From the equations (1) and (2), we get,

1
𝐸𝑔 𝛼 − − − (3)
𝜆
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i.e., energy of a photon is directly proportional to frequency and inversely proportional to

wavelength, if the energy gap increases, LEDs emit electromagnetic waves of different
wavelength(different colors).

If the energy gap is less than 1.8eV, electrons releases energy in the form of infra-red radiation,
which is not visible to human eyes. If the energy gap is greater than or equal to 1.8eV, during the
recombination process, which releases energy in the form of light, which are visible to human eye.
Ge and Si are not suitable elements to produce visible light signals, so compounded semiconductor
materials like gallium arsenide, gallium phosphide, gallium nitride, gallium arsenide phosphide
etc…are used to develop light emitting diodes. The cut-in voltage of LEDs ranges from 1.2V to
4V.

VI Characteristics:

Figure shows the VI characteristics of an LED, which is a plot of forward voltage vs current
through the device.

Applications:
• Traffic signal lamps
• Medical devices
• Camera flash lights etc.

3. Photocoupler:

Also called as optocoupler or optoisolator, acts as an interface between two different circuits
with different voltage levels. These devices supplies electrical isolation between input and output
source.

Photocouplers are composed of two semiconductor devices, namely light emitting diode and
photo-transistor. The photocouplers receives electrical signal as an input from the source and
emits light, photo-transistor receives the light signal and converts back into electrical signal.
Figure shows the working principle of optocouplers.

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Circuit symbol:

Applications:
• Switching the DC circuits
• PC communication
• AC power control
• Microprocessor input/output switching etc.
•
4. Voltage Regulators (78XX series)

78XX series regulators are linear voltage regulators, which provide stabilized output
voltage from a potentially unstable power supply source. These regulators comes in IC package
and these are most commonly used voltage regulators to provide a stable output.

Voltage Regulators 78xx Family Models

Model VOUT Typical

7805 5

7806 6

7808 8

7809 9

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7810 10

7812 12

7815 15

7818 18

7824 24

Advantages:
• Very easy to use - just select the required 78XX series regulator and place it in circuit for
it to work.
• Very few additional electronic components are required - using the basic circuit only
capacitors are required for the input and output.
• Low cost - these linear voltage regulators can be obtained for a very low cost.

NOTE: 79XX series voltage regulators are similar to 78XX series regulators, but 79XX series
regulators provides negative voltage.

******

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Basic Electronics Notes

Subject Code: 18EC14/24
UNIT-II
Chapter-1: Bipolar Junction Transistor.
Introduction:
Electronic systems are interconnection or combination of electronic devices and electrical
elements. Electrical elements in the electronic systems are used to bias the electronic devices.
Semiconductor devices such as diodes and transistors are called electronic devices. Electronic
devices are used for switching and amplification purpose. Diodes are used for switching
applications and they are not having the capability of amplifying the signals. Therefore diodes are
limited to switching applications.
The transistor is a three-terminal device that can amplify the signals as well as performing
switching action.
The term transistor is derived from the statement transfer of resistor that is a transistor
will transfers the resistance from one junction to another junction through proper biasing. Hence,
transistors perform amplification by changing their resistance.

Transistor
Transfer of Resistor

The first transistor was demonstrated on 23rd Dec 1947, by Dr. Willian Shockley and his
team at the Bell Telephone Laboratories, USA
The important features of transistors compared to vacuum tubes are listed as follows.
1. Three terminal solid-state device
2. Smaller and lightweight
3. Rugged construction
4. No heater requirement
5. Requires less power
6. Lower operating voltage and
7. More efficient.
With these advantages, transistors are developed and used in all electronic systems as a
switch and/or amplifier.

Classification:
Figure (1) shows the classification of transistors.

Transistor

BJT FET

NPN PNP JFET MOSFET

P- Enhancement
N-Channel Depletion mode
Channel mode

P-
N-Channel N-Channel P-Channel
Channel
Figure 1: Classification of transistors.

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In the following section, BJTs construction, working principle, characteristics and

applications are discussed.
BJT, stands for bipolar junction transistor, figure (2) shows the structure of BJT and
terminologies.

Figure 2: Structure of transistor.

 Bipolar: Both electrons and holes are involved in current flow.

 Junction: has two p-n junctions.
 Transistor: Transfer + Resistor.
• BJTs are current-controlled devices
• Have three regions with three terminals labeled as
i. Emitter (E)
ii. Base (B) and
iii. Collector (C)

Figure (3) shows the structure and circuit symbol of NPN and PNP transistors.

 NPN- Bipolar Junction Transistor

 PNP- Bipolar Junction Transistor



Figure 3: Structure and circuit symbols of NPN and PNP transistor.

The arrow in the circuit symbols indicates the direction of the current flow. In the NPN
transistor, current will flow from collector to emitter terminal and in the PNP transistor current
will flow from emitter to collector.
Construction:

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In the following section, the step-by-step process of constructing NPN – BJT is explained,
the following specifications are to be considered for constructing BJTs.

Emitter Base Collector

Width Medium Thin High

Doping
High Low Moderate
Concentration

Step1: start with a piece of intrinsic semiconductor and divided it into three regions

Figure 4: Piece of intrinsic semiconductor

Step2: Three regions are doped to form NPN regions

Figure 5: Doped intrinsic semiconductor

Doping with pentavalent element gives N-type material and doping with trivalent element gives
P-type semiconductor material.

Step3: Metallic contacts are deposited at each layer to connect the electrodes to form terminals

Figure 6: Electrodes connected to each layer

Step4: Terminals are named as Emitter, Base, and Collector

Figure 7: Terminals are named as emitter, base, and collector

The emitter terminal emits more electrons and hence, the emitter region is doped heavily,
Collector terminal collects the emitted electrons, hence the width of the collector region is high.

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The base terminal is a lightly doped and thin region, which controls the electrons flow from
emitter to collector.

Working Principle:
The BJTs are working under three different modes, such as cut-off, saturation and active modes
and the working principle of NPN-BJT is explained with these three modes.
Case (i): Cut-off mode
• Both Emitter-Base and Collector-Base junctions are reverse biased
• The depletion region widens at both the junctions and no current will flow through the
device.
• Acts as an OFF switch

Figure 8: Cut-off mode operation of BJT

Case (ii): Saturation Mode

• Both Emitter-Base and Collector-Base junctions are forward biased
• The depletion region reduces at both the junctions and maximum current will flow
through the device.
• Acts as an ON switch.

Figure 8: Saturation mode operation of BJT

Case (iii): Active Mode

• Emitter-Base Junction is forward biased and Collector-Base junction is reverse biased
• The width of the depletion region decreases at EB Junction.
• Width of the depletion region Increases at CB Junction.
• Electrons move through the base region, but the base region is lightly doped and only a
few electrons recombine with the holes present in the base region, remaining electrons are
drifted towards the collector region and constitute a collector current.
• Only 2-5% of electrons recombine at the base region, remaining 95 to 98% electrons move
to the collector region
• Transfer of resistance takes place, hence it is called the transistor.
• Acts as an amplifier

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Figure 8: Active mode operation of BJT

Characteristics of BJT:

Bipolar junction transistor is a three-terminal device and BJTs are need to be modeled as
a two-port network. A pair of terminals is called a port, the two-port network means, the network
has two pairs of terminals. One pair of terminals used to apply the input and another pair of
terminals is used to take the corresponding output. Figure (9) shows the block diagram of two-
port model.

Figure 9: Two-port model

One terminal of BJT can be connected to the ground or made common to both input and
output to form a two-port model, which leads to three different configurations they are.

1. Common Base (CB) Configuration

2. Common Emitter (CE) Configuration and
3. Common Collector (CC) Configuration.

Each of these configurations is having its advantages and disadvantages. To study the behavior of
these configurations, VI characteristics need to be obtained, VI Characteristics of BJTs are divided
into two types they are.

1. Input characteristics and

2. Output characteristics

1. Common Base (CB) Configuration:

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Figure 10: CB Configuration of NPN-BJT

• The base terminal(grounded) is common for both input and output
• Input is applied between E and B
• Output is taken across C and B
• IE is the input current and IC is the output current
• VEB is the input voltage and VCB is the output voltage
Where,
• IE - Emitter current
• IC - Collector current
• VEB - Emitter to base voltage
• VCB - collector to base voltage.

Figure (11) shows the circuit arrangement to obtain the VI characteristics

Figure 11: Circuit arrangement to obtain the VI Characteristics

Input Characteristics:

• VEB vs IE with zero or constant VCB

• Initially, the collector-base voltage VCB is kept at zero, and emitter current IE is increased
from zero by increasing VEB.
• The EB junction depletion region reduces by increasing VEB and if VEB is greater than or
equal to cut-in voltage, current starts increasing.
• This characteristic is similar to the forward bias characteristics of the diode.
• Effect of VCB: If VCB is increased, the depletion region of the CB junction increases and
the depletion region penetrates deeper into the base region, which leads to a decrease in
the width of the based region, so emitter current increases. This effect is called early
effect or based width modulation.

Figure 12: Input characteristics of CB configuration

Output Characteristics:

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• VCB vs IC with zero or constant IE

• If IE is zero, and CB junction reverse-biased, a small current will flow through the device.
This current is called reverse leakage current, denoted as ICBO.
• If IE increases by increasing VEB, the emitter terminal injects more electrons and less
recombination takes place at the base region due to early effect, and hence collector
current increases proportionally to emitter current.
• i.e., 𝐼𝐶 = 𝛼𝐼𝐸 + 𝐼𝐶𝐵𝑂; ICBO is the leakage current and small
• 𝐼𝐶 = 𝛼𝐼𝐸 ; α is the current amplification factor of CB configuration typically 0.95 to 0.98
• 𝐼𝐶 ≈ 𝐼𝐸 ; Collector current is approximately equal to emitter current.

Figure 13 Output characteristics of CB configuration

Mathematical Expressions:

𝐼𝐸 = 𝐼𝐶 + 𝐼𝐵 − − − (1)
𝐼𝐶 = 𝛼𝐼𝐸 + 𝐼𝐶𝐵𝑂 − − − (2)
𝛼𝐼𝐸 ≫ 𝐼𝐶𝐵𝑂
𝐼𝐶 = 𝛼𝐼𝐸
Where,
𝑰𝑪
𝜶=
𝑰𝑬
𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑎𝑚𝑙𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝐶𝐵 𝑐𝑜𝑛𝑓𝑖𝑔𝑢𝑟𝑎𝑡𝑖𝑜𝑛
Typically
𝛼 = 0.95 𝑡𝑜 0.98 ;
𝑰𝑪 ≈ 𝑰𝑬

2. Common Emitter (CE) Configuration

Figure 14: CE configuration

• Emitter terminal is common (grounded) for both input and output

• Input is applied between B and E
• Output is taken across C and E
• IB is the input current and IC is the output current
• VBE is the input voltage and VCE is the output voltage
Where,
• IB – Base current
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• IC - Collector current
• VBE – Base to Emitter voltage
• VCE - collector to Emitter voltage

Figure (15) shows the circuit arrangement for obtaining the VI Characteristics

Figure 15: Circuit arrangement to obtain the characteristics.

Input characteristics

• VBE vs IB with zero or constant VCE

• Initially, the collector-emitter voltage VCE is kept at zero, and emitter current IB is
increased from zero by increasing VBE.
• The EB junction depletion region reduces by increasing VBE and if VBE is greater than or
equal to cut-in voltage, current starts increasing.
• This characteristic is similar to the forward bias characteristics of the diode.
• Effect of VCE: If VCE is increased, the depletion region of the CB junction increases and
the depletion region penetrates deeper into the base region, which leads to a decrease in
the width of the base region, so emitter current increases and base current decreases
(IE=IB+IC). This effect is called early effect or based width modulation.

Figure 16: Input characteristics of CE configuration

Output characteristics
• VCE vs IC with zero or constant IB
• If IB is zero, and CB junction reverse-biased, a small current will flow through the device.
This current is called reverse leakage current, denoted as ICEO.
• 𝐼𝐶 = 𝛽𝐼𝐵 + (1 + 𝛽)𝐼𝐶𝐵𝑂 − − − (1)
• 𝐼𝐶 ≈ 𝛽𝐼𝐵 − − − (2); β is the current amplification factor in CE configuration
• If IB increases IC also increases.
• Effect of VCE: If VCE increases base current decreases and collector current increases
due to early effect.

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Figure 17: Output characteristics of CE configuration

Mathematical Expressions:
𝐼𝐸 = 𝐼𝐶 + 𝐼𝐵 − − − (1)
𝐼𝐶 = 𝛼𝐼𝐸 + 𝐼𝐶𝐵𝑂 − −(2)
𝐼𝐶 = 𝛼(𝐼𝐶 + 𝐼𝐵 ) + 𝐼𝐶𝐵𝑂
𝐼𝐶 = 𝛼𝐼𝐶 + 𝛼𝐼𝐵 + 𝐼𝐶𝐵𝑂
(1 − 𝛼)𝐼𝐶 = 𝛼𝐼𝐵 + 𝐼𝐶𝐵𝑂
Divide (1 − 𝛼) on both sides
𝛼 1
𝐼𝐶 = 𝐼𝐵 + 𝐼
(1 − 𝛼) (1 − 𝛼) 𝐶𝐵𝑂
𝛼 1
𝐿𝑒𝑡 𝛽 = ; (1 + 𝛽) =
(1 − 𝛼) (1 − 𝛼)
(1
𝐼𝐶 = 𝛽𝐼𝐵 + + 𝛽)𝐼𝐶𝐵𝑂
𝐼𝐶 = 𝛽𝐼𝐵 + 𝐼𝐶𝐸𝑂 − − − (3);
Where,
𝐼𝐶𝐸𝑂 = (1 + 𝛽)𝐼𝐶𝐵𝑂
𝑰𝑪
𝜷= = 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑔𝑎𝑖𝑛 𝑜𝑟 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝐶𝐸
𝑰𝑩

3. Common Collector (CC) Configuration

Figure 18: CC configuration

• Collector terminal is common (grounded) for both input and output

• Input is applied between B and C
• Output is taken across E and C
• IB is the input current and IE is the output current
• VBC is the input voltage and VEC is the output voltage
Where,
• IB – Base current

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• IE – Emitter current
• VBC – Base to Collector voltage
• VEC - Emitter to collector voltage.

Figure () shows the circuit arrangement to obtain the VI Characteristics

Figure 19: Circuit arrangement to obtain the VI characteristics

Input characteristics
• VBC vs IB with zero or constant VEC
• Initially, the emitter to collector voltage VEC is kept at zero.
• If VCB increases the CB junction depletion region increases and the width of the base
region decreases, thereby decreasing the base current shown in figure.
• Effect of VEC: If VEC is increased, the depletion region of CB junction increases and
depletion region penetrates deeper into the base region, which leads to further decrease
in the width of the base region, so emitter current increases and base current further
decreases.

Figure 20: Input characteristics of CC configuration

Output characteristics
• VEC vs IE with zero or constant IB
• If IB is zero, and CB junction reverse-biased, a small current will flow through the device.
This current is called reverse leakage current, denoted as ICBO.
• 𝐼𝐸 = 𝛾𝐼𝐵 + 𝛾𝐼𝐶𝐵𝑂 − −(1);
𝛾 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑖𝑛 𝐶𝐶 𝑐𝑜𝑛𝑓𝑖𝑔𝑢𝑟𝑎𝑡𝑖𝑜𝑛.
• The emitter current is proportional to the input current (Base current), the
proportionality constant is the current amplification factor

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Figure 21: Output characteristics of CC configuration

Mathematical Expressions:

𝐼𝐶 = 𝛼𝐼𝐸 + 𝐼𝐶𝐵𝑂 − − − (1)

𝐼𝐸 = 𝐼𝐵 + 𝐼𝐶 − − − (2)
𝑓𝑟𝑜𝑚, (2)
𝐼𝐶 = 𝐼𝐸 − 𝐼𝐵 − − − (3)
𝐼𝐸 − 𝐼𝐵 = 𝛼𝐼𝐸 + 𝐼𝐶𝐵𝑂
𝐼𝐸 − 𝛼𝐼𝐸 = 𝐼𝐵 + 𝐼𝐶𝐵𝑂
(1 − 𝛼)𝐼𝐸 = 𝐼𝐵 + 𝐼𝐶𝐵𝑂
𝐼𝐵 𝐼𝐶𝐵𝑂
𝐼𝐸 = + − − − (4)
(1 − 𝛼) (1 − 𝛼)
1
Let, (1−𝛼) = 𝛾
𝐼𝐸 = 𝛾𝐼𝐵 + 𝛾𝐼𝐶𝐵𝑂
𝐼𝐸 = 𝛾𝐼𝐵 + 𝐼𝐶𝐶𝑂 − − − (5)
𝐼𝐶𝐶𝑂 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑒𝑎𝑘𝑎𝑔𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑛 𝐶𝐶 𝑐𝑜𝑛𝑓𝑖𝑔𝑢𝑟𝑎𝑡𝑖𝑜𝑛
𝐼𝐶𝐶𝑂 ≪ 𝛾𝐼𝐵
𝐼𝐸 = 𝛾𝐼𝐵 − − − (6)
𝑰𝑬
𝜸= ⇒ 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝐶𝐶 𝑐𝑜𝑛𝑓𝑖𝑔𝑢𝑟𝑎𝑡𝑖𝑜𝑛
𝑰𝑩

Relation between 𝜶, 𝜷 𝒂𝒏𝒅 𝜸

𝐼𝐸 = 𝐼𝐶 + 𝐼𝐵 − − − (1)
𝐼 𝐼 𝐼
We know that, 𝛼 = 𝐼𝐶 ; 𝛽 = 𝐼𝐶 𝑎𝑛𝑑 𝛾 = 𝐼𝐸
𝐸 𝐵 𝐵
Divide 𝐼𝐵 on both sides
𝐼𝐸 𝐼𝐶
= +1
𝐼𝐵 𝐼𝐵
𝜸=𝜷+𝟏
Divide IE, both sides
𝐼𝐶 𝐼𝐵
1= +
𝐼𝐸 𝐼𝐸
1
1=𝛼+
𝛾
1 𝜸−𝟏
𝛼 =1− ⇒𝜶=
𝛾 𝜸
Divide IC, both sides
𝐼𝐸 𝐼𝐵 1 1 𝜷
=1+ ⇒ =1+ ⇒ 𝜶 =
𝐼𝐶 𝐼𝐶 𝛼 𝛽 𝟏+𝜷
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Β in terms of α
1 1
=1+
𝛼 𝛽
1 1
−1=
𝛼 𝛽
1−𝛼 1
=
𝛼 𝛽
𝛼
𝛽=
1−𝛼

Comparison of three configurations

Characteristic Common Base Common Emitter Common Collector

Input Impedance Low Medium High

Output Impedance Very High High Low

Phase Shift 0o 180o 0o

Voltage Gain High Medium Low

Current Gain Low Medium High

Power Gain Low Very High Medium

Transistor DC Load line analysis

Applying DC voltage of proper magnitude and polarity at the two junctions without AC
signal is called DC biasing. When the BJT is biased, it will establish certain current and voltage
conditions, these conditions are called operating conditions of the transistor.
In Common emitter configuration, collector to emitter voltage (VCE) and collector current(IC)
are Operating conditions, these operating conditions are also called as DC operating point or
quiescent point or simply Q-point. This operating point must be stable for the proper operation
of the transistor.

Q point will be written in the following format: Q (VCE, IC)

In the following section, finding the Q point for CE configuration is explained.

Consider a BJT in common emitter configuration, Base-emitter junction is biased for the forward
bias and Collector-base junction is biased for reverse bias. Resistors are connected through the
base and collector terminals for biasing the device with proper magnitudes.

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Figure 22: DC biasing of CE configuration NPN-BJT

The value of the collector current and emitter current at any time will satisfy
𝑉𝐶𝐶 = 𝐼𝐶 𝑅𝐶 + 𝑉𝐶𝐸 − − − (1)
If 𝑉𝐶𝐸 = 0,
𝐕
𝐈𝐂(𝐦𝐚𝐱) = 𝐑𝐂𝐂 − − − (𝟐)
𝐂
i.e., the collector current is maximum at zero collector-emitter voltage, denote this point as A.
If 𝐼𝐶 = 0,
𝐕𝐂𝐄(𝐦𝐚𝐱) = 𝐕𝐂𝐂 − − − (𝟑)
i.e., the voltage across collector and emitter terminals is maximum, if collector current is zero,
denote this point as B.

By locating the points A and B on output characteristics of common emitter configuration BJT,
and joining the points gives DC load line. The intersection of the output characteristic curve and
DC load line gives the operating point of the transistor. As shown in figure (23).

Figure 23: DC load line

The Q point must be at the middle of the active region for perfect amplification if the Q point
moves towards the x-axis, the transistor will operate in cur-off mode and if the Q point moves
towards the Y-axis, then the transistor will operate in saturation mode. The slope of the DC load
1
line is𝑅 .
𝐶

NOTE:

1. The leakage current, ICBO greatly affected by temperature variations. The collector to base
leakage current doubles for every 10oC rise in temperature.

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𝑡2 −𝑡1
𝐼𝐶𝐵𝑂 𝑎𝑡 𝑡2 = 𝐼𝐶𝐵𝑂 𝑎𝑡 𝑡1 ∗ 2 10 − − − (4)
Also,
𝐼𝐶𝐸𝑂 = (1 + 𝛽)𝐼𝐶𝐵𝑂 − − − (5)
And,
𝐼𝐶𝐶𝑂 = 𝛾𝐼𝐶𝐵𝑂 − − − (6)
Collector current depends on ICBO which is the temperature-dependent quantity and hence, the Q
point may shift towards saturation or cut-off region, leads to unexpected behavior and response
from the system. This process is termed as a thermal run-away, i.e., destruction of Q point due
to temperature variations is called thermal run-away. So, proper biasing is necessary to maintain
the location of the Q point constantly.
Also, VBE is a temperature-dependent variable, which decreases by 2.5mV for every 1oC rise in
temperature.

2. The Q point also depends on the value of β, β is a highly sensitive factor, and small changes in
the input current lead to very high changes at the output (Collector current and collector to emitter
voltage). So, proper biasing is necessary to maintain the location of Q point constantly.
i.e.,
∆𝐼𝐶 = 𝛽∆𝐼𝐵 − − − (7)

3. Location of the Q point and its corresponding response.

i. If the Q point is at the middle of the active region

Figure 24: Q point is located in the middle of the active region.

ii. If the Q point is nearer to the cut-off region

Figure 25: Q point is located the on x-axis

iii. If the Q point is nearer to the saturation region

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Figure 26: Q point is located the on y-axis

Transistor as an amplifier:

Amplifier is an electronic circuit, which increases the strength of the weak signal, in
otherward amplifier is an electronic circuit, which increases the amplitude of the input (current
or voltage or both) signal.
Common emitter configuration of BJT is best suitable for amplification purpose, because
common collector configuration can amplify both voltage as wells as current.
Consider a common emitter configured BJT with simple biasing method, shown in figure
(27). RB is the base resistor and RC is the collector resistor which is to be properly chosen to
maintain the location of the Q point in the middle of the active region for zero signal.

Figure 27: Transistor amplifier circuit.

When an ac signal is applied across the base to emitter terminal, that signal superimposed
with the DC base current. The small change in base current ∆𝐼𝑏 provides a very high change in
collector current∆𝐼𝑐 , because of large𝛽.
i.e.,
∆𝑰𝒃
= 𝜷 − − − (𝟏)
∆𝑰𝒄
Current amplification
Assume, 𝑟𝑒 is the internal emitter resistance of the PN junction (EB Junction) and
𝑉𝑏 = 𝐼𝑏𝑅𝐵 − − − (2)
The output voltage is measured across Rl.
𝑉𝑐 = 𝐼𝑐 𝑅𝐿 − − − (3)
𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡, 𝐼𝑐 ≈ 𝐼𝑒
𝑽𝒄 𝑹
𝑽
= 𝑹 𝑳 𝜷 − − − (𝟒)
𝒃 𝑩
Voltage amplification
Transistor as a switch:

Figure shows the circuit arrangement to utilize the transistor as a switch. Collector
terminal is biased with VCC through RC and base terminal is biased with VBB through RB. RB and

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RC decides the transistor to work in saturation and cu-off region with respect to VBB. The switching
action of transistor is discussed as follows.

Case (i): Cut-off

If VBB=0, based current is zero and hence collector current through the transistor is zero lead to
collector to emitter voltage is equal to the applied voltage VCC, hence the terminals collector and
emitter acts as an open circuit called OFF switch.
i.e., 𝑉𝐶𝐸(𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛) = 𝑉𝐶𝐶 − − − (1)

Figure 27: Transistor .

Case (ii): Saturation
When positive voltage of sufficient magnitude applied across the base terminal, base to emitter
junction conducts and sufficient base current will drive the transistor to saturation region and
maximum collector current flows through the device. Hence, the collector to emitter terminal acts
as a short circuit and called ON switch.
𝑉𝐶𝐶 − 𝑉𝐶𝐸(𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛)
𝑖. 𝑒. , 𝐼𝐶 (𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛) = − − − (2)
𝑅𝐶
𝑉𝐶𝐸 (𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛) = 0
𝑉𝐶𝐶
𝐼𝐶(𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛) = − − − (3)
𝑅𝐶

Example to show the amplifier and switch action of transistor.

Consider 𝑉𝐶𝐶 = 5𝑉, 𝑅𝐶 = 1𝐾Ω, 𝛽 = 100

We know that, 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶

If IB = 0, VCE = VCC => 5V (𝐜𝐮𝐭 − 𝐨𝐟𝐟)

If IB = 10μA, VCE = 4V
If IB = 20μA, VCE = 3V
} (Active region)
If IB = 30μA, VCE = 2V
If IB = 40μA, VCE = 1V

If IB = 50μA, VCE = 0V (𝐒𝐚𝐭𝐮𝐫𝐚𝐭𝐢𝐨𝐧)

10 < 𝐼𝐵 < 50𝜇𝐴 => 𝐴𝑐𝑡𝑖𝑣𝑒 𝑟𝑒𝑔𝑖𝑜𝑛

𝐼𝐵 > 50𝜇𝐴 => 𝑠𝑤𝑖𝑡𝑐ℎ

Design of RB to operate the transistor as an amplifier

𝐼𝐵 𝑚𝑖𝑛 = 10𝜇𝐴

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𝐼𝐵 𝑚𝑎𝑥 = 50𝜇𝐴
Assume 𝑉𝐵𝐵 = 5𝑉

𝑉𝐵𝐵 − 𝑉𝐵𝐸
𝑅𝐵 = − − − (1)
𝐼𝐵
5 − 0.7
𝑅𝐵 𝑚𝑖𝑛 = => 86𝐾Ω
50𝜇
5 − 0.7
𝑅𝐵 𝑚𝑎𝑥 = => 430𝐾Ω
10𝜇
𝐑 𝐁 𝐬𝐡𝐨𝐮𝐥𝐝 𝐛𝐞 𝐜𝐡𝐨𝐨𝐬𝐞𝐧 𝐛𝐞𝐭𝐰𝐞𝐞𝐧 𝟖𝟔 𝐊 𝐭𝐨 𝟒𝟑𝟎𝐊

Design of RB to operate the transistor as a switch

If 𝑽𝑩𝑩 = 𝟎; 𝑰𝑩 = 𝟎, 𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕 𝒐𝒇 𝑹𝑩 and 𝑽𝑪𝑬 = 𝑽𝑪𝑪 (𝒄𝒖𝒕 − 𝒐𝒇𝒇)
If 𝑉𝐵𝐵 = 5𝑉; 𝑓𝑜𝑟 𝐼𝐵 > 50𝜇𝐴
𝑅𝐵 < 86𝐾Ω

Verification:
If 𝑅𝐵 = 80𝐾Ω; 𝐼𝐵 = 53.75𝜇𝐴, 𝐼𝐶 = 5.375𝑚
𝑽𝑪𝑬 = 𝑽𝑪𝑪 − 𝑰𝑪 𝑹𝑪 => 𝟎𝑽𝒐𝒍𝒕𝒔(𝑺𝒂𝒕𝒖𝒓𝒂𝒕𝒊𝒐𝒏)

******

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Basic Electronics Notes

Subject Code: 18EC14/24
UNIT-II

Chapter-2: Feedback amplifier.

Feedback is a process/technique where a portion of the output signal of a system is fed back
and recombined with the input.
• Feedback is used in all amplifier circuits.
• Some circuits virtually acts as a feedback in electronic systems and also can be introduced
externally for some amplifiers.
• Invented in the year 1928 by Harold black, engineer, Western electric company.

Figure (1) shows the canonical form representation of a feedback system, block AOL is the
amplifier with gain A or AOL and block β is the feedback network with gain β.

Figure 1: Canonical for representation of feedback system

Based on the way of combining the signals at input, feedback systems are classified into:
• Positive feedback
• Negative feedback

Positive Feedback:
• Positive feedback is a process of adding the portion of the output with the input and
amplifying. This process is continuous and regenerative action takes place shown in
figure (2).

Figure 2: Canonical for representation of positive feedback system

Though the positive feedback increases the gain of the amplifier, it has the disadvantages such
as
• Increasing distortion
• Instability

NOTE: Positive feedback is mainly used in the design of oscillators.

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Negative Feedback:
 Negative feedback is a process of subtracting the portion of the output with the input and
amplifying. This process is continuous and degenerative action takes place shown in figure (3).

Figure 3: Canonical for representation of negative feedback system

NOTE: Negative feedback is used in the design of all types of amplifiers.

Advantages of Negative Feedback/Properties of negative feedback:
 Stabilizes the gain
 Increases the input impedance
 Decreases the output impedance
 Bandwidth increases
 Increases upper cut-off frequency
 Decreases the lower cut-off frequency
 Decreases the distortions
 Reduction in noise

Disadvantage:
 Decreases the gain

Basic structure of feedback amplifiers

Figure (4) shows the basic structure of a feedback system, which consisting of input unit,
output unit, summing network, sampling network, basic amplifier and feedback network. The
function each of these blocks are explained as follows.

Figure 4: Basic block diagram of neagative feedback system

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1. Input signal:
• It is a source wither modelled by a practical voltage source or practical current source.
• An ideal voltage source in series with a resistor is called practical voltage source
• An ideal current source in parallel with a resistor is called practical current source.
Practical voltage source Practical Current source

2. Output signal:
• The output of the system is either voltage across the load or current through the load.
• The output of the feedback amplifiers must be independent of the load variations and parameter
variations in the amplifier.

3. Sampling Network :
• A network is used to mesure and sending the output signal to the feedback network is called
sampling network.
• For measuring the voltage a parallel(shunt) connection is required
• For measuring the current a series connection is required

Voltage measurement Current measuremen

4. Comparison or summing network:

• Comparing the output signal with the input signal.
• Comparing voltage signals – series network
• Comparing the currents signals – parallel(Shunt) network

Voltage summing Current summing

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5. Amplifier:
• Amplifies the compared or combined signal
• Electronic circuits

6. Feedback network:
• Combination of electronic/electrical/both elements
• Provides the feedback signal in proportional to the output signal.

Based on the type of summing network and sampling network, feedback systems are classified
into four types, they are.
1. Voltage - Series negative feedback
2. Voltage - Shunt negative feedback
3. Current - Series negative feedback and
4. Current - Shunt negative feedback

1. Voltage Series negative feedback

Summing network is a series network and sampling network is a shunt network, output
voltage is fed to the feedback network and feedback voltage and input voltage are combined at the
summing netwrk. Actual inpt to the amplifier circuit is the difference of the feedback voltage and
supply voltage. Figure (5) shows the block diagram of voltage series negative feedback system.

Figure 5: Block diagram of voltage series negative feedback system

Property-1: Reduces the overall gain and stabilizes the gain

Proof:
𝑉𝑜 = 𝐴𝑉𝑖 − − − (1)
𝑉𝑖 = 𝑉𝑠 − 𝑉𝑓 − − − (2)
𝑉𝑓 = 𝛽𝑉𝑜 − − − (3)
𝑉𝑜 = 𝐴[𝑉𝑠 − 𝑉𝑓 ]
𝑉𝑜 = 𝐴[𝑉𝑠 − 𝛽𝑉𝑜 ]
𝑉𝑜 = 𝐴[𝑉𝑠 − 𝛽𝑉𝑜 ]
𝑉𝑜 = 𝐴𝑉𝑠 − 𝐴𝛽𝑉𝑜
𝑉𝑜 [1 + 𝐴𝛽] = 𝐴𝑉𝑠
𝑽𝒐 𝑨
= 𝟏+𝑨𝜷 − − − (𝟒) Closed loop gain
𝑽 𝒔

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Example:

P1: If gain of the amplifier is 105, what is the overall gain after introducing the feedback with
feedback gain 0.01? Also compare the overall gain if the gain of the amplifier is changed by ±50%
Given data:
𝐴 = 105 𝑎𝑛𝑑 ± 50%
𝛽 = 0.01
To find:
𝐴𝐶𝐿 𝑓𝑜𝑟 𝐴 𝑎𝑛𝑑 𝐴 ± 50%
Solution:
𝐴 105
𝐴𝐶𝐿1 = ⇒ ⇒ 99.9
1 + 𝐴𝛽 1 + 105 ∗ 0.01
𝐴 + 50% 150000
𝐴𝐶𝐿2 = ⇒ ⇒ 99.93
1 + 𝐴𝛽 1 + 150000 ∗ 0.01
𝐴 − 50% 50000
𝐴𝐶𝐿3 = ⇒ ⇒ 99.8
1 + 𝐴𝛽 1 + 50000 ∗ 0.01

Property-2: Increases the input impedance

Proof:
𝑉𝑖
𝑍𝑖𝑛 = 𝑖𝑛𝑝𝑢𝑡 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘 ⇒
𝐼𝑖
𝑉𝑠
𝑍𝑖𝑓 = 𝑖𝑛𝑝𝑢𝑡 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑤𝑖𝑡ℎ 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘 ⇒
𝐼𝑖
𝑉𝑠 𝑉𝑠 𝑉𝑖 𝑽𝒔
𝑍𝑖𝑓 = ⇒ ∗ ⇒ ∗ 𝒁𝒊𝒏 − − − (𝟏)
𝐼𝑖 𝑉𝑖 𝐼𝑖 𝑽𝒊
𝑉𝑖 = 𝑉𝑠 − 𝑉𝑓
𝑉𝑖 = 𝑉𝑠 − 𝛽𝑉𝑜
𝑉𝑖 = 𝑉𝑠 − 𝛽𝐴𝑉𝑖
𝑉𝑖 [1 + 𝐴𝛽] = 𝑉𝑠
𝑉𝑠
= 1 + 𝐴𝛽 − − − (2)
𝑉
𝑖
Substitute (2) in (1) we get
𝒁𝒊𝒇 = (𝟏 + 𝑨𝜷)𝒁𝒊𝒏 − − − (𝟑)
Property-3: Decreases the output impedance

Proof:
𝑍𝑜𝑢𝑡 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘
𝑉𝑜
𝑍𝑜𝑓 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑤𝑖𝑡ℎ 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘 =
𝐼𝑜
𝐴𝑉𝑓 + 𝑉𝑜 = 𝐼𝑜 𝑍𝑜𝑢𝑡
𝐴[𝛽]𝑉𝑜 + 𝑉𝑜 = 𝑍𝑜𝑢𝑡 𝐼𝑜

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Basic Electronics [18EC14/24]

𝑉𝑜 [1 + 𝐴𝛽] = 𝐼𝑜 𝑍𝑜𝑢𝑡
𝑽𝒐 𝒁𝒐𝒖𝒕
𝒁𝒐𝒇 = ⇒ − − − (𝟏)
𝑰𝒐 [𝟏 + 𝑨𝜷]
Property-4: Bandwidth increases

𝑩𝑾 = 𝒇𝟐 − 𝒇𝟏 − − − (𝟏)

𝑩𝑾𝒇 = 𝑩𝑾(𝟏 + 𝑨𝜷) − − − (𝟐)

𝒇𝟏
𝒇𝟏𝒇 =
(𝟏 + 𝑨𝜷)

𝒇𝟐𝒇 = 𝒇𝟐 (𝟏 + 𝑨𝜷)

Property-5: Reduction in distortion

𝑫
𝑫𝒇 =
(𝟏 + 𝑨𝜷)

Property-6: Reduction in Noise

𝑵
𝑵𝒇 =
(𝟏 + 𝑨𝜷)

Summary:

Voltage Gain Decreases and stabilizes

Bandwidth Increases

Input resistance Increases

Output resistance Decreases

distortion Decreases

Noise Decreases

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Basic Electronics [18EC14/24]

2. Voltage shunt negative feedback

Properties
Block diagram Voltage Gain Decreases and
stabilizes the gain

Bandwidth Increases

Input resistance Decreases

Output Decreases
resistance
distortion Decreases

Noise Decreases
3. Current series negative feedback

Properties
Voltage Gain Decreases and
Block diagram stabilizes the gain

Bandwidth Increases

Input resistance Increases

Output Increases
resistance
distortion Decreases

Noise Decreases
5. Current shunt negative feedback system

Properties
Voltage Gain Decreases and
Block diagram stabilizes the gain

Bandwidth Increases

Input resistance Decreases

Output Increases
resistance
distortion Decreases

Noise Decreases
By comparing the properties of all the four topologies of negative feedback systems,
voltage series negative feedback has more advantages, and hence, voltage series negative feedback
will be used for the design of amplifiers.

******
Prepared by Mohankumar V., Assistant Professor, Dept. of ECE, Dr.AIT Page
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Chapter 4

Bipolar Junction Transistors.

Problem Solutions

4.1 Problem 4.37

It is required to design the circuit in Figure (4.1) so that a current of 1 mA is established in
the emitter and a voltage of +5 V appears at the collector. The transistor type used has a
nominal β of 100. However, the β value can be as low as 50 and as high as 150. Your design
should ensure that the specified emitter current is obtained when β = 100 and that at the
extreme values of β the emitter current does not change by more than 10% of its nominal
value. Also, design for as large value for RB as possible. Give the values of RB , RE , and
RC to the nearest kilo-ohm. What is the expected range of collector current and collector
voltage corresponding to the full range of β values?

+15 V

RC

RB

RE

-15 V

Figure 4.1:

Page 100 of 117

2 CHAPTER 4. BIPOLAR JUNCTION TRANSISTORS. PROBLEM SOLUTIONS

Solution

Nominal β = 100, so nominal α = β/(1 + β) = 0.99, nominal IE = 1 mA, nominal IC = αIE

= 0.99 mA, nominal VC = +5 V. RC can then be calculated as:
VCC − VC
RC =
IC
15 − 5
=
0.99
= 10.1 kΩ
= 10 kΩ

Applying Kirchhoff’s voltage rule on the base-emitter loop we get:

IE RE + IB RB = VEE − VBE

Using IB = IE /(β + 1), we then get:

VEE − VBE
IE = RB
(4.1)
RE + 1+β
15 − 0.7
= RB
RE + 101
14.3
1 = RB
RE + 101
RB
RE + = 14.3 (4.2)
101
The collector current depends only on VBE , while IB and IE depends also on β. Note that
for the same collector current, changing β from 100 to 50 changes the base current by a
factor of 2, while changing it from 100 to 150, changes the base current by a factor 2/3. This
means that reducing β will have more effect on the emitter current then increasing it. So
we design the circuit to limit the maximum change in the emitter current at β = 50. Since
decreasing β decreases the emitter current we then use the lower limit of IE of 0.9 mA and
β = 50 in Equation (4.1):
14.3
0.9 =
RE + R51B
RB
RE + = 15.89 (4.3)
51

Page 101 of 117

4.1. PROBLEM 4.37 3

Using Equation (4.2) and Equation (4.3) we get:

RB = 164 kΩ
RE = 13 kΩ

to find the range of IC and VC for the full range of β values we use:

IC = αIE
β VEE − VBE
= × RB
(4.4)
1+β RE + 1+β
VC = VCC − IC RC (4.5)

Using Equation (4.4) and Equation (4.5) we get for β = 50:

50 15 − 0.7
IC = ×
1 + 50 13 + 164 51
= 0.864 mA
VC = 15 − 0.864 × 10
= 6.36 V

and for β = 150:

150 15 − 0.7
IC = ×
1 + 150 13 + 164151
= q.008 mA
VC = 15 − 1.008 × 10
= 4.92 V

Page 102 of 117

4 CHAPTER 4. BIPOLAR JUNCTION TRANSISTORS. PROBLEM SOLUTIONS

4.2 Problem 4.49

We wish to design the amplifier circuit of Figure (4.2) under the constraint that VCC is fixed.
Let the input signal vbe = V̂be sin ωt where V̂be is the maximum value for acceptable linearity.
Show for the design that results in the largest signal at the collector without the BJT leaving
the active region, that
VCC − VBE − V̂be
RC IC =
1 + V̂Vbe
T

and find an expression for the voltage gain obtained. For VCC = 10 V, VBE = 0.7 V, and
V̂be = 5 mV, find the dc voltage at the collector, the amplitude of the output voltage signal,
and the voltage gain.

iC
RC

iB VCC
vbe
vCE

VBE vBE
iE

Figure 4.2:

Solution
The total collector current (ac and dc) iC is given by:

iC = IC + gm vbe
= IC + gm V̂be sin ωt

The total collector voltage vC is similarly given by:

vC = VCC − IC RC − gm V̂be sin ωt

Page 103 of 117

4.2. PROBLEM 4.49 5

To maintain the BJT in the active region vc ≥ vbe then:

VCC − IC RC − gm V̂be ≥ VBE + V̂be

To maximize vC we should use the equal sign in the last equation, i.e.

VCC − IC RC − gm RC V̂be = VBE + V̂be

Using the expression for gm :

IC
gm =
VT
the last equation then becomes:

V̂be
VCC − IC RC = VBE + V̂be
VT
!
V̂be
IC RC 1+ = VCC − VBE − V̂be
VT
VCC − VBE − V̂be
IC RC =
V̂be
1+ VT

The voltage gain Av = −gm RC , using the last equation we get:

Av = gm RC
IC
= − RC
VT
VCC − VBE − V̂be
= − (4.6)
VT + V̂be
Substituting with the given numerical values we get:

VCC − VBE − V̂be

IC RC =
1 + V̂Vbe
T
10 − 0.7 − 0.005
= 5
1 + 25
= 7.75 V
VC = VCC − IC RC
= 10 − 7.75
= 2.25 V

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6 CHAPTER 4. BIPOLAR JUNCTION TRANSISTORS. PROBLEM SOLUTIONS

The ac output voltage vc is given by:

vc = VC − vbe
= VC − (VBE + V̂be sin ωt)

The amplitude of vc will determined by the amplitude of vbe , i.e.

V̂c = VC − (VBE + V̂be )

= 2.25 − (0.7 + 0.005)
= 1.55 V

The voltage gain can be calculated from −V̂c /V̂be and from Equation (4.6):

V̂c
Av = −
V̂be
1.55
= −
0.005
= −310
VCC − VBE − V̂be
= −
VT + V̂be
10 − 0.7 − 0.005
= −
0.025 + 0.005
9.295
= −
0.03
= −309.8
= −310

Page 105 of 117

4.5. PROBLEM 4.92 15

4.5 Problem 4.92

In the emitter follower in Figure (4.7), the signal source is directly coupled to the transistor
base. If the dc component of vs is zero, find the dc emitter current. Asume β = 120.
Neglecting r◦ , find Ri , the voltage gain v◦ /vs , the current gain i◦ /is and the output resistance
R◦ .

+5 V

3.3 kΩ

vo
100 kΩ ii io

1 kΩ
vs

Ri Ro

-5 V

Figure 4.7: The capacitor is a blocking capacitor of very large capacitance.

Solution

The T-model equivalent of the given circuit is shown in Figure (4.8)

Given that α ≈ 1, the emitter current IE is given by:
VCC − VBE
IE = RB
RC + 1+β
5.0 − 0.7
=
3.3 + 100
121
= 1.042 mA

Page 106 of 117

16 CHAPTER 4. BIPOLAR JUNCTION TRANSISTORS. PROBLEM SOLUTIONS

αie
Rs B ib C ix
vo
vb
vx
Vs
re RC RL

Ri Ro

Figure 4.8:

We can calculate re and Ri from:

VT
re =
IE
25
=
1.042
= 24 Ω

The input resistance Ri is the resistance that the source will see looking into the base. It is
clear from Figure (4.8) that Ri is composed of re , RC , and RL . The last two resistors are
connected in parallel and obviously the RCL = RC //RL is in series with re since they both
carry the same current. This situation is similar to that where a resistor Re is connected to
the emitter and is in series with re , in this case Ri = (1 + β)(re + Re ). In the case at hand
Ri is then given by:

Ri = (1 + β)(re + RCL )
= (1 + β)(re + RC //RL )
 
3.3 × 1
= 121 × 24 +
3.3 + 1
= 121 × (24 + 767)
= 95.8 kΩ

vb , re , and RCL form a voltage divider. The output voltage vo is the voltage across RCL we
then have:
vo RCL
=
vb re + RCL
while vs , Rs , and Ri form another voltage divider where vb is the voltage across Ri , we then
have:
vb Ri
=
vs Rs + Ri

Page 107 of 117

4.5. PROBLEM 4.92 17

Using the last two equations, the overall voltage gain vo /vs is:

vo vb vo
= ×
vs vs vb
Ri RCL
= ×
Rs + Ri re + RCL
95.8 0.767
= ×
100 + 95.8 0.024 + 0.767
= 0.474

The input current ii is the current produced by the input voltage vs in the series combination
of Rs and Ri , while the output current io is produced by the output voltage through the
load resistor RL , so the overall current gain io /ii is given by:

io vo vs
= /
ii RL Rs + Ri
v o Rs + Ri
= ×
vs RL
100 + 95.8
= 0.474 ×
1
= 92.8

To find the output resistance Ro we set vs to zero and insert a virtual voltage source vx at
the point where the load device looks back at the circuit. Let us assume that vx produces
a virtual current ix , as shown by the dashed part of the circuit in Figure (4.8). Taking vx
across the input part of the circuit (vs = 0), we get:

v x = ie r e + i b R s
= ie re + (1 − α)ie Rs
Rs
= ie r e +
1+β
 
Rs
= ie r e +
1+β

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18 CHAPTER 4. BIPOLAR JUNCTION TRANSISTORS. PROBLEM SOLUTIONS

The virtual current ix is given by:

vx
ix = + ie
RC
vx vx
= + Rs
RC re + 1+β
ix 1
=
vx Ro
1 1
= + Rs
RC re + 1+β
 
Rs
Ro = RC // re +
1+β
 
100
= 3.3// 0.024 +
121
= 3.3//0.85 kΩ
3.3 × 0.85
=
3.3 + 0.85
= 0.676 kΩ

Page 109 of 117

4.6. PROBLEM 4.96 19

4.6 Problem 4.96

For the follower circuit in Figure (4.9) let transistor Q1 have β = 20 and transistor Q2 have
β = 200, and neglect the effect of r◦ . Use VBE = 0.7 V.

(a) Find the dc emitter current of Q1 and Q2 . Also find the dc voltages VB1 and VB2 .

(b) If a load resistance RL = 1 kΩ, is connected to the output terminal, find the voltage gain
from the base to the emitter of Q2 , v◦ /vb2 , and find the input resistance Rib2 looking into
base of Q2 . (Hint: Consider Q2 as an emitter follower fed by a voltage vb2 at its base.)

(c) Replacing Q2 with its input resistance Rib2 found in (b), analyze the circuit of emitter
follower Q1 to determine its input resistance Ri , and the gain from its base to its emitter,
ve1 /vb1 .

(d) If the circuit is fed with a source having a 100-kΩ resistance, find the transmission to
the base of Q1 , vb1 /vs .

(e) Find the overall voltage gain v◦ /vs .

+9 V

R1 = 1
MΩ
Q1

R2 = 1 Q2
MΩ
vo
20 µA

2 mA

Ri

Figure 4.9: The capacitors are a blocking capacitors of very large capacitance.

Page 110 of 117

20 CHAPTER 4. BIPOLAR JUNCTION TRANSISTORS. PROBLEM SOLUTIONS

Solution

(a) In the base circuit of Q1 , one can replace VCC , R1 = 1M Ω, R2 = 1M Ω by their the
Thevenin’s equivalent of RBB and VBB , such that:
R1 R2
RBB =
R1 + R2
1×1
=
1+1
= 0.5 M Ω
R1
VBB = VCC ×
R1 + R2
= 9.0 × 0.5
= 4.5 V

The emitter currents of Q1 and Q2 are given by:

IE1 = 2 mA
IE2 = 20 µA + IB2
IE2
= 20 µA +
1 + β2
2000(µA)
= 20 µA +
201
= 30 µA

The base voltages of Q1 and Q2 , are:

VB1 = VBB − IB1 RBB

IE1 1 + β1
= VBB − RBB
×
30(µA)
= 4.5 − × 0.5(M Ω)
21
= 4.5 − 1.43(µA) × 0.5(M Ω)
= 3.79 V
VB2 = VB1 − VBE
= 3.79 − 0.7
= 3.09 V

Page 111 of 117

4.6. PROBLEM 4.96 21

gm1 vbe1
Rs RBB ib1
B1
vb1
Vs
VBB re1

gm2 vbe2
vb2

E1 B2 re2

Vo

RL

Figure 4.10:

(b) the T-model equivalent of the whole circuit is shown in Figure (4.10). It is clear from
the figure that:

RL
vo = × vb2
RL + re2
VT
re2 =
IE2
25
=
2
= 12.5 Ω
vo RL
=
vb2 RL + re2
1000
=
1000 + 12.5
= 0.988
Rib2 = (1 + β2 )(re2 + RL )
= 201 × (1000 + 12.5)
= 203.5 kΩ

Page 112 of 117

22 CHAPTER 4. BIPOLAR JUNCTION TRANSISTORS. PROBLEM SOLUTIONS

(c) Replacing the second transistor Q2 by its input resistance in Figure (4.10)we get:

VT
re1 =
IE1
25000(µV )
=
30(µA)
= 833 Ω
= 0.833 kΩ
Rib2
ve1 = × vb1
Rib2 + re1
ve1 Rib2
=
vb1 Rib2 + re1
203.5
=
203.5 + 0.833
= 0.996
Ri = RBB //(1 + β1 )(re1 + Rib2 )
= 500//[21 × (.833 + 203.5)] kΩ
= 0.5//4.29 M Ω
= 0.448 M Ω
= 448 kΩ

(d) In Figure (4.10) let us connect vs with it internal resistance Rs = 100 kΩ, and replaceing
Q1 by its internal resistance Ri we get:
vb1 Ri
=
vs Ri + Rs
448
=
448 + 100
= 0.818

(e) finally the overall gain is (note that ve1 = vb2 ):

vo vb1 ve1 vo
= × ×
vs vs vb1 vb2
= 0.818 × 0.996 × 0.988
= 0.805

Page 113 of 117

Basic Electronics [18EC14/24]

Basic Electronics Notes

Subject Code: 18EC14/24

Unit-III: Field Effect Transistors(FETs) and Silicon Controlled

Rectifier(SCR).

1. FETs:

FET is an acronym for the field-effect transistor. The FETs are three-terminal unipolar
devices and conduction will be controlled by the electric field. Hence FETs are also called Field
controlled/Voltage controlled devices.

1.1. Similarities between BJTs and FETs.

 Both are transistors.

 Both have two junctions and three terminals.
 Both are controlling the conduction of current and
 Both are semiconductor devices.

1.2. Difference between BJTs and FETs.

Sl.
BJTs FETs
No.

Two Types Two Types

2  NPN(Never Points iN)  N-Channel(points iN)
 PNP(Points iN Permanently)  P-Channel(Points out)

𝐼𝐶 = 𝑓(𝐼𝐵 ) 𝐼𝐷 = 𝑓(𝑉𝐺𝑆 )
3 𝑉𝐺𝑆 2
i.e., 𝐼𝐶 = 𝛽𝐼𝐵 i.e., 𝐼𝐷 = 𝐼𝐷𝑆𝑆 (1 − )
𝑉𝑃

Collector Current is approximately equal

Drain Current is equal to Source Current.
4 to Emitter Current.
𝐼𝐷 = 𝐼𝑆
𝐼𝐶 ≅ 𝐼𝐸

5 Current Controlled Device Voltage Controlled Device

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Basic Electronics [18EC14/24]

Current conduction takes place by both Current conduction takes place by only the
6 holes and electrons. Hence Called Bipolar majority charge carriers, either electrons or
Transistors. holes. Hence Called Unipolar Transistors.

7 Poor Thermal Stability Good Thermal Stability

8 Low Input impedance Very High input impedance

9 High Power Consumption Low Power Consumption

Occupies less space, hence can be fabricated

10 Occupies More Space in a circuit
easily in Integrated Chips.

11 High Gain* Low Gain*

12 Large Bandwidth* Bandwidth is low*

*Disadvantages of FETs over BJTs.

1.3. Classification of FETs.

FET

JFET MOSFET

Enhancement Depletion
N-Channel P-Channel Mode Mode

N-Channel P-Channel N-Channel P-Channel

Figure 1.1: Classification of FET

N-channel and P-channel JFETs, Construction, Working principles, and Characteristics are
discussed in the present context.

Analogy for discussing the working principle of JFETs:

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11561 of 117
ASSIGNMENT (TO BE SUBMITTED 18/10/2024)

TASK 1

Read and make a thorough study on;

(a) Hybrid parameter of transistors

(b) Working Principle of different Field effect transistors (FETs)

TASK 2

Page 116 of 117

TASK 3

Page 117 of 117