BASIC ELECTRONICS

MAHESH PRASANNA K., AIET, MOODBIDRI

10 ELN

– 15/25

1

BASIC ELECTRONICS

Sub Code: 10 ELN – 15/25 IA Marks: 25

Hrs/Week: 04 Exam Hours: 03

Total Hrs: 52 Exam Marks: 100

UNIT – 1

SEMICONDUCTOR DIODE AND APPLICATIONS

The branch of engineering which deals with the flow of

Electrons through vacuum, gas or semiconductor is called

Electronics.

Electronics essentially deals with electronic devices and their utilization.

The flow of electrons is generally termed as electric current.

Based upon ability to conduct current, all substances can be broadly classified as

follows:

Classification of materials on the basis of electrical conductivity:

1. Good conductors – materials which offers very little resistance to the flow of

current.

E.g.:- Copper, Aluminum, Silver, etc.

2. Insulator – resistance offered to the flow of current is very large.

E.g.:- Wood, Glass, Paper, Mica, Plastic, Etc.

3. Semiconductors – exhibit the properties of both conductors & insulators. At low

temperature, they behave like perfect insulators, but at higher temperature, they

behave like conductors.

E.g.:- Silicon & Germanium.

Properties of Semiconductors:-

1. The conductivity of a semiconductor lies between the conductivities of conductors

& insulators.

2. The conductivity of a semiconductor increases with temperature. Hence, a

semiconductor has a negative temperature coefficient of resistance.

3. The conductivity of a semiconductor increases, when suitable metallic impurity is

added to it.

NOTE: - If the temperature of a good conductor is increased, its resistance increases. This

is known as positive temperature coefficient. But, the resistance of a semiconductor

decreases with increase of temperature.

BASIC ELECTRONICS

MAHESH PRASANNA K., AIET, MOODBIDRI

10 ELN

– 15/25

2

Energy Band Diagrams:-

The difference in the electrical behavior of good conductors, insulators &

semiconductors can be illustrated on the basis of energy band diagrams.

All matter is made up of atoms.

In an individual atom there is a central nucleus & a definite number of planetary

electrons revolving around the nucleus in different orbits.

The nucleus is composed of protons & neutrons.

Protons are positively charged particles & they posses mass.

Neutrons have only mass but no charge.

Electrons are negatively charged particles. Each electron is associated with

definite quantum of negative charge, i.e., 1.602 * 10-19

Coulomb.

The entire mass of the atom is practically concentrated in the nucleus. Electrons

possess negligibly small mass, and hence they have high mobility.

Electrostatic force

+

Centrifugal force

Atomic structure

In a normal atom, which is electrically neutral, the number of protons in the

nucleus is equal to the number of electrons.

Hence, the total positive charge of the nucleus is balanced by the total negative

charge of the electrons.

The total number of electrons in an atom depends up on the atomic number.

atomic weight = no. of protons + no. of neutrons

atomic number = no. of protons or electrons in an atom These electrons are distributed in several orbits or shells around the nucleus.

The number of electrons which can be accommodated in a shell is fixed; and is

given by 2n2

,where n = the shell number.

Thus, the first orbit can accommodate 2*12

= 2 electrons, the second orbit can

accommodate 2*22

= 8 electrons, the third orbit can accommodate 2*32

= 18

electrons, & so on.

As the atomic number increases, the total number of electrons in the atom

increases; and hence, there would be more number of orbits.

In any atom, the number of electrons in the last (outermost) orbit is limited to 8.

Fundamental

Particle

Nature of

Charge

Mass in Kg.

Neutron No charge 1675 * 10-27

Proton Positive 1672 * 10-27

Electron Negative 9.107 * 10-31

BASIC ELECTRONICS

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The outermost orbit is called valence orbit & the electrons in this orbit are termed

as valence electrons.

An electron in any orbit of an atom possess energy. This energy increases as the

distance of the orbit from the nucleus increases. Hence, the valence electrons

possess maximum energy.

In a given mass of a material, there will be millions of atoms & the valence

electrons of different atoms in the mass have slightly different energies; since,

each atom is surrounded by several other atoms.

The range of energies possessed by the electrons of any one orbit of all the atoms

is referred as energy band.

The energy band in relation to valence electrons is termed as valence band.

The valence electrons possess the greatest energy & they are least attracted

towards the positively charged nucleus.

Hence, the valence electrons are loosely held to the parent atom; and if sufficient

energy is provided to them, they get detached from the parent atoms.

Electrons which are thus removed from the valence orbit of atoms are termed as

free electrons.

These free electrons constitute flow of current on application of voltage.

The range of energies possessed by the free electrons is termed as conduction

band.

An electron in the last orbit of an atom is in the valence band.

When it gets detached from the atom, it becomes a free electron & hence, it enters

the conduction band.

There is a void (or gap) separating conduction band & valence band, and no

electron can exist in this gap. This is termed as forbidden band.

NOTE:-

The range of energies possessed by valence electrons is called

valence band.

The range of energies possessed by free electrons is called

conduction band.

Valence band and conduction band are separated by an energy

gap in which no electrons normally exist this gap is called

forbidden gap.

BASIC ELECTRONICS

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In a metal, which a good conductor, there are large number of free electrons; and

hence, the conduction band is practically full.

Also, the energy required to push the electron from the valence band to the

conduction band is quite small, due to the fact that the two bands over lap as

shown in the fig. above.

In a metal, the number of valence electrons is less than 4.

In an insulator, the conduction band is empty, whereas the valence band is full.

i.e., valence band is saturated with all 8 electrons it can accommodate.

Also, enormous energy is needed to remove the valence electrons from the atoms

& push them into the conduction band; since, the forbidden energy gap is quite

large (about 15eV).

In a semiconductor, the number of valence electrons is 4, and this is less than 8;

and hence, the valence orbit is not saturated.

The conduction band is almost empty at low temperatures, and the valence band is

partially filled.

The forbidden energy gap is quite small (about 1eV).

Hence, at low temperatures, the material behaves like an insulator.

With increase of temperature, more & more valence electrons are pushed into the

conduction band; and hence, the material behaves like a good conductor.

BASIC ELECTRONICS

MAHESH PRASANNA K., AIET, MOODBIDRI

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Covalent Bonds:-

We know that, the valence orbit of any atom can accommodate 8 electrons.

There is a natural tendency for the atoms to complete their valence orbits.

– In an insulator, the valence orbit is full; having all the 8 electrons.

– In a metal, the number of valence electrons is less than 4, and these

electrons can move freely with the material.

– In semiconductors like Germanium (atomic number is 32) and Silicon

(atomic number is 14), the number of valence electrons is 4.

– These are arranged as follows (The number of electrons in any orbit is

2n2):

Germanium (32) Silicon (14)

First orbit 2 x 12 = 2 First orbit 2 x 1

2 = 2

Second orbit 2 x 22 = 8 Second orbit 2 x 2

2 = 8

Third orbit 2 x 32 = 18 Third orbit

(Valance orbit)

4

Fourth orbit

(Valance orbit)

4

Both germanium & silicon are crystalline in structure; i.e., their atoms are

arranged in an orderly way.

It is found that, in a given mass of Ge or Si, the atoms so arrange themselves that

each atom is surrounded by four other atoms, and it completes its valence orbit by

taking one electron from these neighboring atoms, as shown in the fig. below.

BASIC ELECTRONICS

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10 ELN

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Each atom shares 4 valence electrons with 4 other atoms by forming bonds, so

that all atoms are enabled to complete their valence orbits.

These bonds are termed as covalent bonds.

The covalent bonds are the characteristic of all semiconductor materials.

Effect of Temperature:-

At very low temperatures; the covalent bonds in semiconductors are very strong

& hence, the valence electrons cannot leave the valence band and enter the

conduction band. Thus, the material cannot conduct current.

At room temperature; due to the thermal energy imparted to them, a few covalent

bonds break & few free electrons are released & can contribute to the current

flow.

If the temperature is further increased; many more covalent bonds break down,

and a large number of free electrons would become available for conduction of

current. The material would then behave like a good conductor.

Effect of Light:-

Since light is a form of energy, when it incident on a semiconductor material, it

imparts energy to it, resulting in breaking of many covalent bonds & electrons are

set free. Hence, the conductivity is increased.

Free Electrons & Holes:-

When thermal energy is supplied to a semiconductor material, some of the

covalent bonds break.

When a covalent bond breaks, a valence electron is set free; i.e., it escapes from

the valence band into the conduction band.

At the same time, a vacancy for an electron is created in the covalent bond.

This is termed as hole.

A hole represents a missing electron. Holes can be looked upon as positive

charges, and just like free electrons, they also contribute to current flow, in a

semiconductor.

Such a current is usually termed as hole current.

The current which results in semiconductor material, due to the movement of

holes, is termed as hole current.

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MAHESH PRASANNA K., AIET, MOODBIDRI

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Consider a mass of Germanium as shown in the above Fig. If thermal energy is supplied,

a few covalent bonds break, resulting free electrons & equal number of holes.

Assume that, a hole is created at 1 due to the escape of valence electron from this

position.

Since, hole are positively charged, they attract negatively charged electrons.

Thus, an electron at 2 escapes, after breaking the covalent bond, and it fills the hole at 1.

When a free electron jumps into a hole, the hole disappears & the free electrons becomes

a valence electron. Also, a hole is created at 2.

This attracts an electron at position 3. Another covalent bond breaks & the electron at 3

escapes, and it fills the hole at 2. At the same time, a hole is created at 3.

Thus it is seen that holes are created successively at 1,2,3,….

Also, the hole at 1 disappears when a hole is created at 2. This hole at 2 disappears when

a hole is created at 3, and so on.

Hence, hole at 1 is moved to 2,3,4,….

This motion of hole contributes positive current flow in a direction opposite to the

negative current flow due to the motion of electrons.

Conduction in solids

Conduction in any given material occurs when a voltage of suitable magnitude is applied to

it, which causes the charge carriers within the material to move in a desired direction.

This may be due to electron motion or hole transfer or both.

Electron motion

Free electrons in the conduction band are moved under the influence of the applied electric field.

Since electrons have negative charge they are repelled by the negative terminal of the applied

voltage and attracted towards the positive terminal.

Hole transfer

Hole transfer involves the movement of holes.

Holes may be thought of positive charged particles and as such they move through an

electric field in a direction opposite to that of electrons.

I I

+ +

V V

(a) Conductor (b) Semiconductor

Flow of electrons Flow of electrons

Flow of current Flow of holes

Flow of current

BASIC ELECTRONICS

MAHESH PRASANNA K., AIET, MOODBIDRI

10 ELN

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In a good conductor (metal) as shown in fig (a) the current flow is due to free electrons

only.

In a semiconductor as shown in fig (b). The current flow is due to both holes and

electrons moving in opposite directions.

The unit of electric current is Ampere (A) and since the flow of electric current is

constituted by the movement of electrons in conduction band and holes in valence band,

electrons and holes are referred as charge carriers.

Intrinsic & Extrinsic Semiconductors:-

A semiconductor, in its pure form, is termed as „intrinsic semiconductor‟

The conduction which takes place, on application of voltage, is not of much

practical utility.

In order to achieve significance increase of conductivity, even at room

temperature, the normal practice is to add a small quantity of suitable metallic

impurity to the semiconductor material. This process of impurity addition is called

doping.

It is usual practice to add one impurity atom to 108 atoms of silicon or germanium.

A semiconductor to which an impurity is added with a view to increase its

conductivity is termed as „extrinsic semiconductor‟.

n – type Semiconductor:-

If a pentavalent impurity like arsenic or antimony or phosphorus is added to pure

germanium or silicon, an n – type semiconductor results.

When arsenic is added to germanium, each arsenic atom positions itself inside the

crystal lattice of germanium that it shares four of its five valence electrons with

four germanium atoms, as shown in the fig. below.

The fifth valence electron is free, since it does not find a place in the covalent

bonds.

This free electron is readily available for conduction.

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MAHESH PRASANNA K., AIET, MOODBIDRI

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If a voltage is applied now, conduction readily takes place.

In this case, the conduction is mainly due to the large number of electrons present

in the material. The electrons are the majority charge carriers.

It should be noted that, holes are also present in the n – type semiconductor

material; but, they are few in number. Hence, holes are the minority charge

carriers.

The arsenic atoms added to the pure semiconductor are termed as donor atoms;

since, each arsenic atom donates or contributes one free electron.

This type of semiconductor is called n – type semiconductor for the simple reason

that, negatively charged electrons are majority charge carriers.

p – type Semiconductor:-

If a trivalent impurity like gallium or indium or aluminium is added to pure

germanium or silicon, a p – type semiconductor results.

Consider a trivalent impurity like gallium added to a small mass of pure silicon.

Each atom of gallium so positions itself inside the crystal lattice of silicon that it

shares all of its valence electrons with three neighboring silicon atoms.

Thus, three silicon atoms complete their valence orbits; and the fourth silicon

atom cannot complete its valence orbit; since, there is no fourth valence electron

which the gallium can contribute.

Hence a hole is created in its covalent bond, as shown in the fig. below.

Now, the silicon has become p – type semiconductor.

Holes which are present in large number are mainly responsible for the

conduction, and hence they are the majority charge carriers.

Electrons are also present in the p – type semiconductor material, but they are few

in number, and they are the minority charge carriers.

The gallium atom added to the pure silicon are termed as acceptor atoms, since

they are ready to accept electrons.

NOTE: - In extrinsic semiconductors, conduction of current is due to both electrons &

holes.

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MAHESH PRASANNA K., AIET, MOODBIDRI

10 ELN

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Fermi Level:-

In a metal, there are a large number of electrons, and these move about randomly

with in the material.

They cannot leave the metal surface, since they do not possess the high energy

required to overcome the surface barrier.

Electrons in the outermost orbits of different atoms of the material possess

slightly different energies.

These different energies are referred as energy states or energy levels.

The probability of an energy state being occupied or not is given by the following

relationship, termed as „Fermi-Dirac probability function‟.

f (E) = 1 „

1 + e (E – E

F)/KT

where E represents a quantum of state with energy E (in Electron-volts)

occupied by an electron.

K is Boltzmann constant in eV/0K

T is the absolute temperature in 0K, and

EF is called „Fermi level‟ or „characteristic energy‟, expressed in Electron-volts.

If we put E = EF in the above expression, we get:

f (E) = 1 „ = 1 = 50%

1 + e0 2

Hence, Fermi level represents the energy state with 50% probability of being filled, if

there were no forbidden energy barrier.

Let the temperature be 00K – Two cases arise.

Case (i): Let E > EF

At T = 00K, we have; f (E) = 1 „ = 0

1 + e∞

It implies that at absolute zero, there can be no quantum state of energy greater than EF

which is occupied (or filled).

Case (ii): Let E < EF

At T = 00K, we have; f (E) = 1 „= 1 = 100%

1 + e-∞

This implies that all quantum levels with energies less than EF are filled, at absolute zero.

Hence, Fermi level represents the maximum energy any electron can possess, at

absolute zero.

It is found that, for majority of metals, the Fermi level is less than 10 eV.

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Energy Band Diagrams for Semiconductors

Energy band diagram for Intrinsic semiconductor:-

In an intrinsic or pure semiconductor material, the number of free electrons

generated by thermal agitation is equal to number of holes.

This implies that, the Fermi level lies midway between the top of the valence

band & the bottom of the conduction band, as shown in the fig. below.

Energy band diagram for Extrinsic semiconductor:-

n – type semiconductor:-

In a n – type semiconductor, there are more electrons in the conduction band than

holes in the valence band.

Hence, the Fermi level tends to move up, towards the conduction band; but it will

be below the donor energy level, as shown in the fig. below.

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p – type semiconductor:-

In a p – type semiconductor, there are more holes in the valence band than

electrons in the conduction band.

Hence, the Fermi level tends to move down, towards the valence band; but it will

be above the acceptor energy level, as shown in the fig. below.

NOTE: - Doping an intrinsic semiconductor with trivalent impurity atoms lowers

the Fermi-level, whereas doping with pentavalent impurity atoms raises the Fermi-

level; as shown in the figures.

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Drift and Diffusion current

The flow of current through a semiconductor material is normally referred to as

one of the two types.

Drift current

If an electron is subjected to an electric field in free space it will accelerate in a

straight line form the –ve terminal to the + ve terminal of the applied voltage.

However in the case of conductor or semiconductor at room temperature, a free

electrons under the influence of electric field will move towards the +ve terminal

of the applied voltage but will continuously collide with atoms all the ways as

shown in figure 1.9.

Electron drift due to field

conduction

when electric

field is

present

+

Conduction

Applied voltage when no electric

field is applied

Each time, when the electron strikes an atom, it rebounds in a random direction

but the presence of electric field does not stop the collisions and random motion.

As a result the electrons drift in a direction of the applied electric field.

The current produced in this way is called as Drift current and it is the usual kind

of current flow that occurs in a conductor.

Diffusion current

The directional movement of charge carriers due to their concentration

gradient produces a component of current known as Diffusion current.

The mechanism of transport of charges in a semiconductor when no electric field

is applied called diffusion. It is encountered only in semiconductors and is

normally absent in conductors.

semiconductor

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heavy concentration of less concentration

electrons electrons

Diffusion current

Even distribution

Net diffusion current is zero

With no applied voltage if the number of charge carriers (either holes or electrons) in

one region of a semiconductor is less compared to the rest of the region then there

exist a concentration gradient.

Since the charge carriers are either all electrons or all holes they sine polarity of

charge and thus there is a force of repulsion between them.

As a result, the carriers tend to move gradually or diffuse from the region of higher

concentration to the region of lower concentration. This process is called diffusion

and electric current produced due to this process is called diffusion current.

This process continues until all the carriers are evenly distributed through the

material. Hence when there is no applied voltage, the net diffusion current will be

zero.

HALL EFFECT

If a piece of metal or semiconductor carrying a current I is placed in a transverse

magnetic field B then an electric field E is induced in the direction perpendicular to both I

and B. This phenomenon is known as Hall Effect.

Y(+ve)

Surface-2

+ + + + + + + +

d

I VH

w

X (+ve)

B

Surface -1

Z (+ve)

Hall effect is normally used to determine whether a semi-conductor is n-type or p-type.

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To find whether the semiconductor is n-type or p-type

i) In the figure. above, If I is in the +ve X direction and B is in the +ve Z

direction, then a force will be exerted on the charge carriers (holes and

electrons) in the –ve Y direction.

ii) This force is independent of whether the charge carriers are electrons or holes.

Due to this force the charge carriers (holes and electrons) will be forced

downward towards surface –1 as shown.

iii) If the semiconductor is N-type, then electrons will be the charge carriers and

these electrons will accumulate on surface –1 making that surface –vely

charged with respect to surface –2. Hence a potential called Hall voltage

appears between the surfaces 1 and 2.

iv) Similarly when surface –1 is positively charged with respect to surface –2,

then the semiconductor is of P-type. In this way, by seeing the polarity of Hall

voltage we can determine whether the semiconductor is of P-type or N-type.

Applications of Hall Effect –

Hall effect is used to determine,

carrier concentration, conductivity and mobility.

The sign of the current carrying charge.

Charge density.

It is used as magnetic field meter.

Carrier lifetime (τ)

In a pure semiconductor, we know that number of holes are equal to the number

of electrons. Thermal agitation however, continues to produce new hole electron pairs

while other hole-electron pair disappear as a result of recombination.

On an average, a hole will exist for τp second and an electron will exist for τn

second before recombination. This time is called the carrier lifetime or Mean lifetime.

The average time an electron or hole can exist in the free state is called carrier

lifetime.

Semiconductor Diode

If a p-type semiconductor & n-type semiconductor are suitably joined by kind of

mechanism, the joint or the contact surface so formed is known as p-n junction.

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Some of the properties of p-n junction are:

• In an unbiased p-n junction, there is a barrier potential across the junction.

• A forward-biased p-n junction readily conducts current, offering very little

resistance.

• A reverse-biased p-n junction offers enormous resistance & hence, it does not

conduct current.

• A p-n junction can function as a rectifier.

• A p-n junction can act as an electronic switch.

Un-biased p-n Junction:-

When a p-n junction is formed by any mechanism, a large number of electrons are

present in the n-region & a large number of holes are present in p-region.

Also, donor atoms (in the form of + ions) are present in the n-region & acceptor

atoms (in the form of – ions) are present in the p-region.

Due to the natural attraction, electrons in the n-region diffuse across the junction

into the p-region & holes in the p-region diffuse across the junction into the n-

region.

When electrons diffuse into the p-region, they combine with holes in the region

near the junction; with the result that, a negative charge is established in this

region.

Similarly, holes diffuse into the n-region, they combine with electrons in the

region near the junction; establishing a positive charge in this region.

As more & more electrons from the n-region and holes from the p-region cross

over the junction, more & more potential builds up.

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The diffusion stops when the negative charge on the p-side repels electrons & the

positive charge on the n-side repels holes.

The potential built up across the junction, after diffusion has stopped, is termed as

„barrier potential‟.

The region on either side of the junction (A-B in the fig. below) is termed as

„depletion region‟ or „depletion layer‟.

In practice, the barrier potential is about 0.3 Volt for Germanium & about 0.7

Volt for Silicon.

This barrier potential has to be overcome, by the application of external bias

voltage, to cause conduction through the junction.

Forward biasing of p-n Junction:-

If an external voltage is applied across the p-n junction, such that, it neutralizes

the barrier potential & causes conduction through the junction, the p-n junction is

said to be forward biased.

Consider a p-n junction whose p-side is connected to the positive terminal of a

battery & n-side is connected to negative terminal, as shown in the fig. above.

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If the applied voltage is less than the barrier potential, there would be no

conduction.

If the applied voltage is more than the barrier potential; it neutralizes the barrier

potential; thus, creates a path of low resistance & conduction of current through

the junction results.

Holes in the p-region are repelled by the positive terminal of the voltage source &

hence, they cross over the junction.

Similarly, the electrons in the n-region are repelled by the negative terminal of the

voltage source & hence, they cross over the junction.

This results in conduction.

If the forward voltage is gradually increased, current through the junction rapidly

rises, as shown in the fig. below.

The volt-ampere characteristics of forward biased p-n junction is as shown in the

above fig.

Reverse biasing of p-n Junction:-

If an external voltage is applied across the p-n junction, such that the depletion

layer widens and the barrier potential increases, with the result that no conduction

occurs, the p-n junction is said to be reverse biased.

Consider a p-n junction whose p-side is connected to the negative terminal of a

battery & n-side is connected to positive terminal, as shown in the fig. above.

The holes in the p-region are attracted towards the negative terminal of the

voltage source & hence, they move away from the junction.

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Similarly, the electrons in the p-region are attracted by the positive terminal, and

they move away from the junction.

Since, the applied voltage establishes an electric field in the same direction as the

field due to the barrier potential, the barrier potential will be increased. Hence, no

conduction occurs.

The junction develops a very high resistance.

But, in practice, a small reverse current flows. This is due to the movement of

minority charge carriers through the junction.

As the reverse voltage is gradually increased, a stage would be reached, when a

very large reverse current flows.

This reverse voltage which causes breakdown is termed as breakdown voltage.

The breakdown of the p-n junction at the reverse breakdown voltage is due to avalanche

effect.

When the reverse voltage is increased, the velocity of minority charge carriers

increases & they acquire very large kinetic energy.

As they move through the depletion region, they cause ionization of the atoms in

the region, which creates more charge carriers.

At the breakdown voltage, the number of charge carriers is so large that, a very

high reverse current results.

NOTE: - The p-n junction conducts when it is forward biased & it blocks conduction

when it is reverse biased. It can therefore be used as rectifier for converting alternating

current into direct current.

The breakdown of a p-n junction may occur due to one more effect called zener effect.

When p-n junction is heavily doped the depletion region is narrow.

So, under reverse bias conditions, the electric field across the depletion layer is

very intense.

Such an intense field is enough to pull the electrons out of the valence bands.

Such a creation of free electrons is called zener effect.

These minority carriers constitute very large current & the mechanism is called

zener breakdown.

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Diode equivalent circuit

It is generally profitable to replace a device or system by its equivalent circuit.

Once the device is replaced by its equivalent circuit, the resulting network can be solved

by traditional circuit analysis technique.

switch Rf

If

VF Vo

VF

Diode circuit symbol Diode equivalent circuit

The forward current If flowing through the diode causes a voltage drop in its internal

resistance Rf. Therefore the forward voltage VF applied across the actual diode has to

overcome

1. potential barrier Vo

2. internal drop If Rf

Vf = Vo + If rf

For silicon diode Vo=0.7V whereas for Germanium diode Vo = 0.3 V.

For ideal diode Rf =0.

V – I Characteristics of p-n Junction Diode

R A

Diode

V V

IF(mA)

Break over

Voltage

VR Knee voltage VF

IR(μA)

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The V-I characteristics of a semiconductor diode can be obtained with the help of

the circuit shown in above fig.

The supply voltage V is a regulated power supply, the diode is forward biased in the

circuit shown. The resistor R is a current limiting resistor. The voltage across the

diode is measured with the help of voltmeter and the current is recorded using an

ammeter.

By varying the supply voltage different sets of voltage and currents are obtained.

By plotting these values on a graph, the forward characteristics can be obtained.

It can be noted from the graph the current remains zero till the diode voltage

attains the barrier potential.

For silicon diode, the barrier potential is 0.7 V and for Germanium diode, it is 0.3

V. The barrier potential is also called as knee voltage or cur-in voltage.

The reverse characteristics can be obtained by reverse biasing the diode. It can be

noted that at a particular reverse voltage, the reverse current increases rapidly.

This voltage is called breakdown voltage.

Basic Definitions

1. Knee voltage or Cut-in Voltage.

It is the forward voltage at which the diode starts conducting.

2. Breakdown voltage

It is the reverse voltage at which the diode (p-n junction) breaks down with

sudden rise in reverse current.

3. Peak-inverse voltage (PIV)

It is the maximum reverse voltage that can be applied to a p-n junction without

causing damage to the junction.

If the reverse voltage across the junction exceeds its peak-inverse voltage, then

the junction exceeds its Peak-inverse voltage, then the junction gets destroyed because of

excessive heat. In rectification, one thing to be kept in mind is that care should be taken

that reverse voltage across the diode during –ve half cycle of a.c. doesnot exceed the

peak-inverse voltage of the diode.

4. Maximum Forward current

It is the maximum instantaneous forward current that a p-n junction can conduct

without damaging the junction. If the forward current is more than the specified rating

then the junction gets destroyed due to overheating.

5. Maximum Power rating

It is the maximum power that can be dissipated at the junction without damaging

it. The power dissipated across the junction is equal to the product of junction current and

the voltage across the junction.

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Diode Current Equation:-

It can be shown that, the general characteristics of an p-n junction are defied by the

following diode current equation:

ID = IS [e (KV

D / T

K) – 1]

ID = Current through the p-n junction,

IS = Reverse saturation current,

VD = External bias voltage,

TK = Temperature in 0K, given as TK =

0C + 273, and

K = 11,600/η, where η = 1 for Germanium &

η = 2 for Silicon.

For VD = 0, i.e., for an un-biased p-n junction: ID = IS [e0 – 1] = 0.

Effect of Temperature on Diode

RECTIFIERS

“Rectifiers are the circuit which converts ac to dc”

Rectifiers are grouped into two categories depending on the period of conductions.

1. Half-wave rectifier

2. Full- wave rectifier.

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Half-wave rectifier

The circuit diagram of a half-wave rectifier is shown in the following fig. along

with the I/P and O/P waveforms.

Vi

t

Vo

π 2π t

Half wave rectifier: Circuit diagram and waveforms

The transformer is employed in order to step-down the supply voltage and also to

prevent from shocks.

The diode is used to rectify the a.c. signal while , the pulsating d.c. is taken across

the load resistor RL.

During the +ve half cycle, the end X of the secondary is +ve and end Y is -ve .

Thus , forward biasing the diode. As the diode is forward biased, the current flows

through the load RL and a voltage is developed across it.

During the –ve half-cycle the end Y is +ve and end X is –ve thus, reverse biasing

the diode. As the diode is reverse biased there is no flow of current through RL

thereby the output voltage is zero.

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Efficiency of a rectifier

The ratio of dc output power to the applied ac power is known as rectifier

efficiency.

Rectifier efficiency η = powercainput

outputpowercd

...

...

Derivation of rectifier efficiency of Half wave rectifier

Let V = Vmsinθ be the voltage across the secondary winding, Vm is the peak

value

rf = diode resistance

RL = load resistance

To find dc output power

avI = dcI = 0

.2

1di = d

Rr

VmSin

LL0

2

1

= dSinRr

Vm

Lf0

)(2

=)(2

2

Lf Rr

Vm =

Im

Hence, dc output power Pdc = I2

dc * RL

= LR

2

Im ------------------------------ (1)

To find ac power input

The ac power input is given by Pac = I2

rms ( rf + RL)

diI rms

2

0

2

2

1

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Squaring both sides we get

diI rms

2

0

22

2

1 But i = Im Sinθ Hence,

(current flows through diode only for duration 0 to Π )

4

22 mrms

II

or 2

m

rms

II

Lf

m

ac RrI

P

2

2 --------------------------------------------(2)

Lf

L

m

m

ac

dc

Rr

R

I

I

P

P*

2

2

2

η =

L

f

R

r1

406.0 -------------------------------------------------(3)

The efficiency is maximum if rf is negligible as compared to RL

Therefore maximum rectifier efficiency = 40.6 %

Full-wave rectifier

Full-wave rectifiers are of two types –

1. Centre tapped full-wave rectifier

2. Bridge rectifier

0

22 )(Im2

1dSinI rms

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Centre tapped full –wave rectifier

Vin

Vout

π 2π

Centre tapped Full wave rectifier Circuit diagram and waveforms

The circuit diagram of a center tapped full wave rectifier is shown in fig. 2.6

above. It employs two diodes and a center tap transformer. The a.c. signal to be

rectified is applied to the primary of the transformer and the d.c. output is taken

across the load RL.

During the +ve half-cycle end X is +ve and end Y is –ve this makes diode D1

forward biased and thus a current i1 flows through it and load resistor RL.Diode D2

is reverse biased and the current i2 is zero.

During the –ve half-cycle end Y is +Ve and end X is –Ve. Now diode D2 is

forward biased and thus a current i2 flows through it and load resistor RL. Diode

D1 is reversed and the current i1 = 0.

Disadvantages

Since, each diode uses only one-half of the transformer secondary voltage the d.c.

output is comparatively small.

It is difficult to locate the center-tap on secondary winding of the transformer.

The diodes used must have high Peak-inverse voltage.

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Bridge rectifier

Vout

D1D3 D2D4 D1D4

t

Full wave bridge wave rectifier Circuit diagram and waveforms

The circuit diagram of a bridge rectifer is shown above. It uses four diodes and a

transformer.

During the +ve half-cycle, end A is +ve and end B is –ve thus diodes D1 and D3

are forward bias while diodes D2 and D4 are reverse biased thus a current flows

through diode D1, load RL ( C to D) and diode D3.

During the –ve half-cycle, end B is +ve and end A is –ve thus diodes D2 and D4

are forward biased while the diodes D1 and D3 are reverse biased. Now the flow

of current is through diode D4 load RL ( D to C) and diode D2. Thus, the

waveform is same as in the case of center-tapped full wave rectifier.

Advantages

The need for center-taped transformer is eliminated.

The output is twice when compared to center-tapped full wave rectifier.

for the same secondary voltage.

The peak inverse voltage is one-half(1/2) compared to center-tapped full wave

rectifier.

Can be used where large amount of power is required.

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Disadvantages

It requires four diodes.

The use of two extra diodes cause an additional voltage drop thereby reducing the

output voltage.

Efficiency of Full-wave rectifier

Let V = Vmsinθ be the voltage across the secondary winding, Vm = peak value

I = Imsinθ be the current flowing in secondary circuit

rf = diode resistance

RL = load resistance

To find dc power output

Ldc RIPdc 2 -----------------------------(1)

0

.2

12 diII avdc

0

.Im2

12 dSinI av

m

av

II

2 -------------------------------------------------------- (2)

L

m

dc RI

P

22

------------------------------------------ (3)

To find input ac power

Lfrmsac RrIP 2---------------------------------------- (4)

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diI rms

0

2

2

12

Squaring both sides we get

diI rms

0

22 1

2

2

2 m

rms

II or

2

m

rms

II -------------------------------------------- (5)

Lf

m

ac RrI

P

2

2 --------------------------------------------(6)

Lf

L

m

m

ac

dc

Rr

R

I

I

P

P*

2

22

2

η =

L

f

R

r1

812.0 -------------------------------------------------(7)

The efficiency will be maximum if rf is negligible as compared to RL.

Maximum efficiency = 81.2 %

This is the double the efficiency due to half wave rectifier. Therefore a Full-wave

rectifier is twice as effective as a half-wave rectifier.

0

22 )(Im1

dSinI rms

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Comparison of Rectifiers

Particulars Half wave rectifier Centre-tapped Full

wave rectifier

Bridge rectifier

1. No. of diodes

2. Idc

3. Vdc

4.Irms

5.Efficiency

6.PIV

7.Ripple factor

1

Im / Π

Vm / Π

Im / 2

40.6 %

Vm

1.21

2

2Im /Π

2Vm / Π

Im /√ 2

81.2 %

2Vm

0.48

4

2Im /Π

2Vm / Π

Im /√ 2

81.2 %

Vm

0.48

NOTE:-

The relation between turns ratio and voltages of primary and secondary of the

transformer is given by

o N1 / N2 = Vp / Vs

RMS value of voltage and Max. value of voltage is related by the equation.

Vrms = Vm / √2 ( for full-cycle of ac)

If the type of diode is not specified then assume the diode to be of silicon type.

For an ideal diode, forward resistance rf = 0 and cut-in voltage , Vγ = 0.

Ripple factor

The pulsating output of a rectifier consists of d.c. component and a.c. component

(also known as ripple). The a.c. component is undesirable and account for the pulsations

in the rectifier output. The effectiveness of a rectifier depends upon the magnitude of a.c.

component in the output : the smaller this component, the more effective is the rectifier.

“ The ratio of rms (or effective) value of a.c. component to the d.c. component in the

rectifier output is known as ripple factor”

r = componentcd

componentscvalueofaeffectiverms

..

..)(=

Idc

Iac

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Ripple factor for Half-wave rectification

By definition the effective (ie rms) value of total load current is given by

OR 22

dcrmsac III

Where Idc = value of dc component

Iac = rms value of ac component

Divide both R.H.S and L.H.S. by Idc we get

221dcrms

dcdc

ac IIII

I OR r = 1

2

dc

rms

I

I ------------------------------------(1)

We have, for half-wave rectification, 2

m

rms

II

mdc

II

Substituting above values in equation (1) & simplifying, we get,

Ripple factor, r = 1.21

It is clear that a.c. component exceeds dc component in the output of a half-wave

rectifier.

Ripple factor for full-wave rectification

We have, for full wave rectification, Irms = 2

mI

Idc = mI2

Substituting above values in equation (1) & simplifying, we get,

Ripple factor, r = 0.48

This shows that in the output of Full-wave rectifier, the d.c. component is more than the

a.c. component

FILTERS

We know that the output of the rectifier is pulsating d.c. ie the output obtained by

the rectifier is not pure d.c. but it contains some ac components along with the dc o/p.

These ac components are called as Ripples, which are undesirable or unwanted. To

minimize the ripples in the rectifier output filter circuits are used. These circuits are

normally connected between the rectifier and load as shown below.

22

dcacrms III

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Vi

Vo pure dc o/p

Pulsating d.c. output

Filter is a circuit which converts pulsating dc output from a rectifier to a

steady dc output. In other words, filters are used to reduce the amplitudes of the

unwanted ac components in the rectifier.

“A filter is an electronic circuit which is interposed between the rectifier

and load and its main function is to filter out the ripples (or pulsations) from the

rectifier output”

Note: A capacitor passes ac signal readily but blocks dc.

Types of Filters

1. Capacitor Filter (C-Filter)

2. Inductor Filter

3. Choke Input Filter (LC-filter)

4. Capacitor Input Filter (Π-filter)

Capacitor Filter (C-filter)

Rectifier

Filter

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Vin

a f

e

t

b d

c V1

o/p without

filter

t

Vo

o/p

with filter a

e

t

Capacitor filter (C-filter) – Circuit diagram & waveforms

When the Input signal rises from o to a the diode is forward biased therefore it

starts conducting since the capacitor acts as a short circuit for ac signal it gets

charged up to the peak of the input signal and the dc component flows through the

load RL.

When the input signal fall from a to b the diode gets reverse biased. This is

mainly because of the voltage across the capacitor obtained during the period o to

a is more when compared to Vi. Therefore there is no conduction of current

through the diode.

Now the charged capacitor acts as a battery and it starts discharging through the

load RL. Mean while the input signal passes through b, c, d section. When the

signal reaches the point d the diode is still reverse biased since the capacitor

voltage is more than the input voltage.

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When the signal reaches point e, the input voltage can be expected to be more

than the capacitor voltage. When the input signal moves from e to f the capacitor

gets charged to its peak value again. The diode gets reverse biased and the

capacitor starts discharging. The final output across RL is shown in Fig.

The ripple factor for a Half-wave rectifier with C-filer is given by

r= 1/2√3fCRL

where, f-----the line frequency ( Hz)

C-----capacitance ( F)

RL------- Load resistance (Ω)

The ripple factor for full-wave rectifier with C-filter is given by r = 1/ 4 √3 f C RL

Advantages of C-Filter

Low cost, small size and good characteristics.

It is preferred for small load currents ( upto 50 mA)

It is commonly used in transistor radio, batteries eliminator etc.

Zener Diode

The reverse voltage characteristics of a semiconductor diode including the

breakdown region is shown below.

V Vz 0

I

Zener diode characteristics

Zener diodes are the diodes which are designed to operate in the breakdown

region. They are also called as Breakdown diode or Avalanche diodes.

The symbol of Zener diode is shown below

P N

Symbol of Zener diode

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The breakdown in the Zener diode at the voltage Vz may be due to any of the

following mechanisms.

1. Avalanche breakdown

Depletion region charge carriers striking the atoms

( ) ( )

-ve terminal +ve terminal

minority charge carriers

We know that when the diode is reverse biased a small reverse saturation current I0 flows

across the junction because of the minority carriers in the depletion region.

The velocity of the minority charge carriers is directly proportional to the applied voltage.

Hence when the reverse bias voltage is increased, the velocity of minority charge carriers

will also increase and consequently their energy content will also increase.

When these high energy charge carriers strikes the atom within the depletion region they

cause other charge carriers to break away from their atoms and join the flow of current

across the junction as shown above. The additional charge carriers generated in this way

strikes other atoms and generate new carriers by making them to break away from their

atoms.

This cumulative process is referred to as avalanche multiplication which results in the

flow of large reverse current and this breakdown of the diode is called avalanche

breakdown.

2. Zener breakdown

We have electric field strength = Reverse voltage/ Depletion region

Depletion region

( ) ( )

-ve terminal +ve terminal

electrons pulled out of their covalent bonds because of high intensity electric field

N

P

N

P

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From the above relation we see that the reverse voltage is directly proportional to the

electric field hence, a small increase in reverse voltage produces a very high intensity electric

field within a narrow Depletion region.

Therefore when the reverse voltage to a diode is increased, under the influence of high

intensity electric filed large number of electrons within the depletion region break the covalent

bonds with their atoms as shown above and thus a large reverse current flows through the diode.

This breakdown is referred to as Zener breakdown.

Zener voltage regulator

The circuit diagram of Zener voltage regulator is shown below

Rs

I IZ IL

Vin VZ RL Vo

Zener voltage regulator

. A zener diode of breakdown voltage VZ is connected in reverse biased condition

across the load RL such that it operates in breakdown region. Any fluctuations in the

current are absorbed by the series resistance Rs. The Zener will maintain a constant

voltage VZ (equal to Vo) across the load unless the input voltage does not fall below the

zener breakdown voltage VZ.

Case (i): When input voltage Vin varies and RL is constant

Rs

I IZ IL

Vin VZ RL Vo

If the input voltage increases, the Zener diode which is in the breakdown region is

equivalent to a battery VZ as shown in figure. The output voltage remains constant at VZ

(equal to Vo) and the excess voltage is dropped across the series resistance RS. We know

that for a zener diode under breakdown region large change in current produces very

small change in voltage, thereby the output voltage remains constant.

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Case (ii): When Vin is constant and RL varies.

Rs

I IZ IL

Vin VZ RL Vo

If there is a decrease in the load resistance RL and the input voltage remains

constant then there is a increase in load current.

Since Vin is constant the current cannot come from the source. This addition load

current is driven from the battery VZ and we know that even for a large decrease in

current the Zener output voltage Vz remains same. Hence the output voltage across the

load is also constant..

Junction Diode as Electronic Switch

A diode readily conducts when forward biased, and blocks conduction when

reverse biased. This is similar to the ON and OFF action of a mechanical switch. Thus,

the diode is ON under forward bias and OFF under reverse bias condition.

If a diode is FB and RB alternately due to the application of an alternating e.m.f.,

it moves from ON state to OFF state, and again from OFF state to ON state; and this

sequence repeats.

But the response of a practical diode to reversal of polarity in not instantaneous;

there is a certain time-lag.

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Most diodes switch rapidly into the FB condition. But, the switch-off time is

somewhat longer. This is due to the diffusion capacitance of the junction diode.

The rate of change of injected charge with applied voltage is called diffusion

capacitance.

i.e., CD = dV

dQ

The forward recovery time does not pose any problem. It is the reverse recovery

time that needs to be considered carefully.

Consider the diode in ON or conducting state. If RB is applied, say at t = t1, the

current simply reverses instantaneously as shown; and it is only after a definite lapse of

time that it reaches the reverse saturation level Is.

When FB is applied, electrons from N-side diffuse into P-region and holes from

the P-region diffuse into the N-region. In these new regions they constitute minority

charge carriers. Now, if a RB is applied abruptly, when conduction is ON, since this

applied bias acts as FB for the large number of minority charge carriers; the current

simply reverses. Certain time ts is required or these minority charge carriers to move into

opposite material and become majority charge carriers. Hence during this time interval ts

called storage time, the reverse current says at the same level, as shown. After this

transition has passed off, the reverse current decreases exponentially over a time t t and

eventually becomes equal to Is, the reverse saturation current.

The time interval tt is termed as transition time. The reverse recovery time, trr is

given as the sum of storage time and transition time. i.e., trr = ts + tt.

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UNIT – 2

TRANSISTORS

A transistor is a sandwich of one type of semiconductor (P-type or n-type)

between two layers of other types.

Transistors are classified into two types;

1. pnp transistor

pnp transistor is obtained when a n-type layer of silicon is sandwiched

between two p-type silicon material.

2. npn transisitor

npn transistor is obtained when a p-type layer of silicon is sandwiched

between two n-type silicon materials.

Figure below shows the schematic representations of a transistor which is equivalent

of two diodes connected back to back.

JE JC JE JC

E C E C

B B

Symbolic representation

pnp npn

Schematic representation

The three portions of transistors are named as emitter, base and collector. The

junction between emitter and base is called emitter-base junction while the junction

between the collector and base is called collector-base junction.

The base is thin and tightly doped, the emitter is heavily doped and it is wider when

compared to base, the width of the collector is more when compared to both base and

emitter.

n n

p

p p

n

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In order to distinguish the emitter and collector an arrow is included in the emitter.

The direction of the arrow depends on the conventional flow of current when emitter

base junction is forward biased.

In a pnp transistor when the emitter junction is forward biased the flow of current is

from emitter to base hence, the arrow in the emitter of pnp points towards the base.

Operating regions of a transistor

A transistor can be operated in three different regions as

a) active region

b) saturation region

c) cut-off region

Active region

E JE B JC C

VEB VCB

pnp transistor operated in active region

The transistor is said to be operated in active region when the emitter-base

junction is forward biased and collector –base junction is reverse biased. The collector

current is said to have two current components one is due to the forward biasing of EB

junction and the other is due to reverse biasing of CB junction. The collector current

component due to the reverse biasing of the collector junction is called reverse saturation

current (ICO or ICBO) and it is very small in magnitude.

Saturation region

E JE B JC C

VEB VCB

pnp transistor operated in Saturation region

p p

n

p p

n

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Transistor is said to be operated in saturation region when both EB junction and

CB junction are forward biased as shown. When transistor is operated in saturation region

IC increases rapidly for a very small change in VC.

Cut-off region

E JE B JC C

VEB VCB

pnp transistor operated in Cut-off region

When both EB junction and CB junction are reverse biased, the transistor is said

to be operated in cut-off region. In this region, the current in the transistor is very small

and thus when a transistor in this region it is assumed to be in off state.

Transistor Configurations:-

A transistor is a semiconductor device having three terminals – emitter, base &

collector.

If it is to be incorporated in a circuit, four terminals would be required – two input

terminals & two output terminals.

Therefore, it is common practice to make one of the terminals common to both

input & output circuits.

Hence, there are three different types of modes of transistor operations:

1. Common Base connection (CB),

2. Common Emitter connection (CE), and

3. Common Collector (CC) connection.

These configurations, in relation to n-p-n transistor, are symbolically shown in the

fig. below.

p p

n

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Working of a p-n-p transistor

E JE B JC C

IE IC

IB

VEB VCB

Consider a pnp transistor operated in active region as shown in above figure.

Since the EB junction is forward biased large number of holes present in the

emitter as majority carriers are repelled by the +ve potential of the supply

voltage VEB and they move towards the base region causing emitter current IE.

Since the base is thin and lightly doped very few of the holes coming from the

emitter recombine with the electrons causing base current IB and all the

remaining holes move towards the collector. Since the CB junction is reverse

biased all the holes are immediately attracted by the –ve potential of the

supply VCB. Thereby giving rise to collector current IC.

Thus we see that, IE = IB + IC -----------------(1) (By KVL)

Since the CB junction is reverse biased a small minority carrier current ICO

flows from base to collector.

IE IC

p p ICBO

ICO

n

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Working of a n-p-n transistor

E JE B JC C

IE IC

IB

VEB VCB

Consider a npn transistor operated in active region as shown in above figure.

Since the EB junction is forward biased large number of holes present in the

emitter as majority carriers are repelled by the –ve potential of the supply

voltage VEB and they move towards the base region causing emitter current IE.

Since the base is thin and lightly doped very few of the holes coming from the

emitter recombine with the electrons causing base current IB and all the

remaining holes move towards the collector. Since the CB junction is reverse

biased all the holes are immediately attracted by the +ve potential of the

supply VCB. Thereby giving rise to collector current IC.

Thus we see that, IE = IB + IC -----------------(1) (By KVL)

Since the CB junction is reverse biased a small minority carrier current ICO

flows from base to collector.

IE IC

n p n ICBO

ICO

n

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Current components of a transistor JE JC

IE

IC

IB

VEB VCB

Figure above shows a transistor operated in active region. It can be noted from the

diagram the battery VEB forward biases the EB junction while the battery VCB reverse

biases the CB junction.

As the EB junction is forward biased the holes from emitter region flow towards

the base causing a hole current IPE. At the same time, the electrons from base region flow

towards the emitter causing an electron current INE. Sum of these two currents constitute

an emitter current

IE = IPE +INE.

The ratio of hole current IPE to electron current INE is directly proportional to the

ratio of the conductivity of the p-type material to that of n-type material. Since, emitter is

highly doped when compared to base; the emitter current consists almost entirely of

holes.

Not all the holes, crossing EB junction reach the CB junction because some of the

them combine with the electrons in the n-type base. If IPC is the hole current at (Jc) CB

junction. There will be a recombination current IPE - IPC leaving the base as shown in

figure.

If emitter is open circuited, no charge carriers are injected from emitter into the

base and hence emitter current IE =o. Under this condition CB junction acts a a reverse

biased diode and therefore the collector current (IC = ICO) will be equal to te reverse

saturation current. Therefore when EB junction is forward biased and collector base

junction is reverse biased the total collector current IC = IPC +ICO.

Transistor Static Characteristics:-

A transistor can be visualized as a two port network; – an input port & an output

port.

It is possible to vary the output current by varying the output voltage, at constant

input current. This provides the date required to plot the static output

characteristics of a transistor.

IPE IPC

(hole current) (hole current)

INE (e- current) ICO

IPB

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Similarly, it is possible to vary the input current by varying the input voltage, at

constant output voltage. This provides the data required to plot the static input

characteristics of a transistor.

Transistor in CB configuration

A simple circuit arrangement of CB configuration for pnp transistor is shown

below.

IE IC

Vi IB RL Vout

VEB VCB

In this configuration, base is used as common to both input and output. It can be

noted that the i/p section has an a.c. source Vi along with the d.c. source VEB. The

purpose of including VEB is to keep EB junction always forward biased (because if there

is no VEB then the EB junction is forward biased only during the +ve half-cycle of the i/p

and reverse biased during the –ve half cycle). In CB configuration, IE –i/p current, IC –o/p

current.

Current relations

Current amplification factor (α)

It is defined as the ratio of d.c. collector current to d.c. emitter current

α = E

O

I

I

Total o/p current

We know that CB junction is reverse biased and because of minority charge carriers a

small reverse saturation current ICO flows from base to collector.

IC = IE + ICO

Since a portion of emitter current IE flows through the base ,let remaining emitter current

be αIE.

IC = αIE + ICo

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Static Input characteristics

IE

VCB=10V VCB=5V

VEB

I/p characteristics is a curve between IE and emitter base voltage VEB keeping VCB

constant. IE is taken along y-axis and VEB is taken along x-axis. From the graph following

points can be noted.

1. For small changes of VEB there will be a large change in IE. Therefore input

resistance is very small.

2. IE is almost independent of VCB

3. I/P resistance , Ri = ΔVEB / Δ IE VCB =constant

Static Output characteristics

IC

Active region

IE=3 mA

IE =2 mA

Saturation IE = 1 mA

region IE = 0

Cut-off region VCB

O/p characteristics is the curve between IC and VCB at constant IE. The collector

current IC is taken along y-axis and VCB is taken along x-axis. It is clear from the graph

that the o/p current IC remains almost constant even when the voltage VCB is increased.

i.e. , a very large change in VCB produces a small change in IC. Therefore, output

resistance is very high.

O/p resistance Ro = ΔVEB / Δ IC IE = constant

Region below the curve IE =0 is known as cut-off region where IC is nearly zero.

The region to the left of VCB =0 is known as saturation region and to the right of VCB =0

is known as active region.

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Transistor in CE configuration

IC

RL Vout

IB

Vi IE

VEB VCE

In this configuration the input is connected between the base and emitter while the

output is taken between collector and emitter. For this configuration IB is input current

and IC is the output current.

Current amplification factor (β)

It is the ratio of d.c. collector current to d.c. base current.

i.e., β = IC / IB

Relationship between α and β

We know that α = E

C

I

I

α = CB

C

II

I

Divide both numerator and denominator of RHS by IC, we get;

1

1

C

B

I

I OR

(IC/IB=β)

1

11

1

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Also we have;

1

)1(

)1(

Derivation of Total output current IC

We have CBOEC III

1

)1(

1

CBOEC

CBOEC

III

IIB

I

Ic = CBOB II )1(

Static Input Characteristics

IB

VCE=10V VCE=5V

VEB

Input characteristics is a curve between EB voltage (VEB ) and base current (IB ) at

constant VCE. From the graph following can be noted.

1. The input characteristic resembles the forward characteristics of a p-n junction

diode.

2. For small changes of VEB there will be a large change in base current IB. i.e., input

resistance is very small.

3. The base current is almost independent of VCE.

4. Input resistance , Ri = ΔVEB / Δ IB V CE = constant

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Output characteristics

IC

(mA)

Active region

30 μA

20 μA

10 μA

IB =0μA

Cut-off region

VCE(volts)

It is the curve between VCE and IC at constant IB. From the graph we can see that,

1. Very large changes of VCE produces a small change in IC i.e output resistance is

very high.

2. output resistance Ro = ΔVCE / ΔIC │IB = constant

Region between the curve IB =0 is called cut-off region where IB is nearly zero. Similarly

the active region and saturation region is shown on the graph.

Transistor in CC configuration

IE

RL Vout

IB

Vi IC

VCB VCE

In this configuration the input is connected between the base and collector while

the output is taken between emitter and collector.

Here IB is the input current and IE is the output current.

Current relations

Current amplification factor (γ)

γ = B

E

I

I

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Relationship between α β and γ

γ = B

E

I

I

γ = B

CB

I

II

Divide both Numerator and denominator by IB

1

1B

C

II

1 (β = IC / IB)

11

1

1

Derivation of total output current IE

We know that IC = CBOE II and IE = IB + IC Hence,

IE = IB + αIE + ICBO or IE(1-α ) = IB + ICBO

IE = 11

CBOB II or IE = γIB + γICBO Hence,

IE = γ (IB + ICBO)

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Comparison between CB, CC and CE configuration

Characteristics CB CE CC

1. Input reistance (Ri)

2. Output resistance (Ro)

3. Current amplification

factor

4. Total output current

5. Phase relationship

between input and output

6. Applications

7. Current gain

8. Voltage gain

9. Practical application

low

high

1

CBOEC III

In-phase

For high frequency

applications

Less than unity

Very high

For high frequency

application

low

high

1

Ic = CBOB II )1(

Out-of phase

For audio frequency

applications

Greater than unity

Greater than unity

For audio frequency

application

high

low

1

1

IE = γIB + γICBO

in-phase

For impedance

matching

Very high

Less than unity

For impedance

matching

Early Effect:-

Consider a p-n-p transistor operating in the CE mode.

EB junction is FB; hence, large number of holes present in the emitter region (p-

material) diffuse into the base region (n-material).

Since the base is lightly doped, there are few electrons in this region. About 3% of

the holes reaching the base combine with these electrons, forming the base

current.

CB junction is RB; hence, remaining 97% of holes are collected by the collector

& this results in collector current. If the RB of the CB junction is increased, the

depletion region across the junction widens.

Since the base is very lightly doped, the depletion layer penetrates deeply into the

base.

This effectively reduces the width of the base, and more holes from the emitter

region reach the collector, without recombination with the electrons in the base

region. This results in an increase of collector current.

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The dependence of the effective width of the base on the CB voltage is termed as

early effect.

Due to this reduction of effective width of the base, there can be slight increase of

collector current, even if the FB voltage of the EB junction is slightly reduced.

Transistor as an amplifier

iC

RL Vout

iB

Vi ` iE

VBB VCC

Consider a npn transistor in CE configuration as shown above along with its input

characteristics.

A transistor raises the strength of a weak input signal and thus acts as an

amplifier. The weak signal to be amplified is applied between emitter and base and the

output is taken across the load resistor RC connected in the collector circuit.

In order to use a transistor as an amplifier it should be operated in active region

i.e. emitter junction should be always FB and collector junction should be RB. Therefore

in addition to the a.c. input source Vi two d.c. voltages VBB and VCC are applied as

shown. This d.c. voltage is called bias voltage. The following waveforms are drawn

according to the circuit diagram.

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Let VBB supplies direct current IB.

Vi causes the alternating current ib.

Total base current, iB = IB + ib, which is alternating in nature as shown.

As the input circuit has low resistance, a small change in the signal voltage Vi

causes a large change in the base current thereby causing the same change in collector

current (because iC = βiB).

The collector current flowing through a high load resistance RC produces a large

voltage across it. Thus a weak signal applied at the input circuit appears in the amplified

form at the output. In this way transistor acts as an amplifier.

Example: Let RC = 5KΩ, Vin =1V, IC =1mA then output V=ICRC =5V

The phase displacement of 1800 between input and output signals can be explained as

follows:

Consider the positive half cycle of input. Since base current iB = IB + ib, the FB on

base-emitter junction increases. This increases the base current. Since, the collector

current is β times the base current, collector current will also increase. This increases the

voltage drop across RC. Since VC = VCC – ICRC, the increase in IC results in a drop in

collector voltage VC, as VCC is constant. Thus, as Vi increases in a positive direction, V0

goes in a negative direction. Similar explanation hold good for negative half cycle of

input signal.

Transistors can work in CB, CE, and CC modes.

In CE mode of operation, a transistor has both current gain & voltage gain.

Hence, it has power gain also.

Therefore, junction transistors are used as both voltage amplifiers & power

amplifiers.

DC Load Line and Operating point selection

IC

RB RC VCE

IB

VBB VCC

NPN transistor in CE configuration

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IC

(mA) Q-point

A

30 μA

20 μA

10 μA

IB =0μA

B V CE(volts

Output characteristics

Consisder a CE amplifier along with the output characteristics as shown in figure

above. A straight line drawn on the output characteristic of a transistor which gives the

various zero signal values (ie. When no signal applied) of VCE and IC is called DC load

line.

Construction of DC load line

Applying KVL to the collector circuit we get,

VCC –ICRC –VCE =0-------------------1

VCE = VCC –ICRC ----------------------2

The above equation is the first degree equation and can be represented by a

straight line. This straight line is DC load line.

To draw the load line we require two end points which can be found as follows.

1. If IC =0, equn 2 becomes VCE = VCC

2. if VCE = 0, equn 2 becomes VCC = ICRC ie. IC = VCC /RC

Operating point (Q- Point)

A point on the d.c. load line which represent the zero signal values of VCE and IC

in a transistor is called as operating point or silent point or quiescent point or Q-point.

The Q-point is selected where the DC load line intersects the curve of output

characteristics for particular value of zero signal current.

i.e. Q-point = (VCE ,IC)

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UNIT – 3

BIASING METHODS

Biasing of Transistor:-

The most important application of a transistor is its use in electronic circuits as

amplifier.

Amplification is the process of strengthening of a weak signal (i.e., increasing its

amplitude), without distortion of its wave shape.

For faithful amplification, it is essential that the –

The emitter-base junction remains forward-biased, and

The collector-base junction remains reverse-biased, throughout the signal

period.

Hence, if the junctions are not properly biased, there would be distortion of the

output voltage.

Thus, one must ensure that, emitter diode is forward biased, and the collector

diode is reverse biased. This is termed as transistor biasing.

There are three basic requirements to be satisfied, in order to achieve faithful

amplification in a transistor amplifier. These are:

It should have proper zero signal collector current.

It should have minimum proper emitter-base voltage at any instant, and

It should have minimum proper collector-emitter voltage at any instant.

Biasing can be brought about by associating a circuit called biasing circuit with the

transistor. The following are the effective and widely used methods of biasing:

1. Fixed current bias method or Base resistor method

2. Collector-to-base bias method

3. Self bias method or Voltage divider bias method

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In this method, a base resistor RB is connected as shown, between the base

terminal and battery VCC, such that the base-emitter junction gets forward biased. RB and

VCC are so chosen that the required base current IB flows, even in the absence of signal.

Applying KVL to the base circuit, we get;

VCC – IBRB – VBE = 0 Solve for IB.

Applying KVL to the collector circuit, we get;

VCC – ICRC – VCE = 0 Solve for IC.

Advantages of Base Bias:-

This is a simple circuit which uses very few components.

The operating point can be anywhere in the active region, by simply changing the

value of RB.

Disadvantages of Base Bias:-

The thermal stability is not provided by this circuit. i.e., circuit does not provide

any check on the collector current, which increases with raise in temperature.

Thus, the operating point is not maintained.

Since, IC = βIB + ICEO - the collector current, IC depends on the value of β,

which changes unit to unit; and shifts the operating point.

Since, IC = βIB - the change in IC changes the operating point; hence, fixed

bias circuit is unsatisfactory if the transistor is replaced by another of same type.

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This is an improvement over the previous method. The base resistor RB is

connected to the collector terminal C, and the load resistor/collector resistor RC is

connected to the VCC as shown in the above figure.

This circuit causes a base current IB required under zero-signal condition, to flow

through RB and this current is controlled by collector-base voltage, rather than VCC.

Applying KVL to the base circuit, we get;

VCC – (IB + IC) RC – IBRB – VBE = 0

Or VCC – (IB + βIB) RC – IBRB – VBE = 0 Solve for IB.

Applying KVL to the collector circuit, we get;

VCC – (IB + IC) RC – VCE = 0 Solve for IC.

Advantages of Collector-to-Base Bias:-

In this method, increase in value of β increases IC. As a result drop across RC also

increases.

But, supply VCC is constant. Due to increase in ICRC, VCE decreases.

Hence, IB also reduces.

As IC depends on IB, decrease in IC reduces the original increase in IC .

The result is that, the original circuit tends to maintain a stable value of collector

current, which in tern keeps the Q-point fixed.

In this circuit, a part of output is fed back to the input; and increase in collector

current decreases the base current.

Thus, negative feedback exists in this circuit. So, this circuit is also called as

Voltage Feedback Bias Circuit.

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In this method of biasing, two suitably chosen resistors R1 and R2 are connected

across the bias battery VCC so that they form a potential divider. The voltage drop VB

across R2 remains fairly constant and provides the necessary fixed bias for the emitter-

base junction. Current IB flows into the base and the emitter diode is always forward-

biased.

This is the most widely used of all biasing methods.

The signal (weak a.c. voltage) to be amplified is applied across the input

terminals, and the amplified output is taken across the load resistor RL.

Advantages of Voltage Divider Bias:-

In this method, if the collector current increases, due to change in temperature or

change in value of β, the emitter current IE also increases.

Hence, the voltage drop across RE increases; which reduces the VBE.

Due to reduction in VBE, base current IB reduces; which in turn reduces the

collector current IC.

This reduction in collector current IC, compensates for the original change in IC.

Therefore, we can say that, negative feedback exists in the emitter bias circuit.

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Bias stabilization

The process of making operating point independent of temperature changes or

variation in transistor parameters is called the stabilization.

We know that for transistor to operate it should be properly biased so that we can

have a fixed operating point. To avoid any distortions, the Q-point should be at the center

of the load line.

But in practice this Q-point may shift to any operating region (saturation or cur-

off region) making the transistor unstable. Therefore in order to avoid this, biasing

stability should be maintained.

Causes for Bias instability

Bias instability occurs mainly due to two reasons.

1. Temperature

2. Current gain

1. Temperature (T)

The temperature at the junctions of a transistor depends on the amount of current

flowing through it. Due to increase in temperature following parameters of a transistor

will change.

(a) Base-emitter voltage (VBE)

VBE increases at a rate of 2.4mV/0C. With increase in temperature the base

current IB will increase and since IC= βIB, IC is also increased hence, changing the

Q-point.

(b) Reverse saturation current ( ICBO/ICEO)

We know that IC = βIB + (1+β) ICBO where ICBO is the reverse saturation

current. As the temperature increases ICBO increases thereby increase in IC and

hence changing the Q-point.

2. Current gain (β)

In the process of manufacturing the transistors different transistors of same type

may have different parameters (i.e. if we take two transistor units of same type and use

them in the circuit there is a change in the β value in actual practice). The biasing circuit

will be designed according to the required β value but due to the change in β from unit to

unit the operating point may shift.

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Thermal Stability of Biasing Circuits:-

Biasing circuit should be designed to fix the operating point or Q-point at the

centre of the active region.

So, while designing the biasing circuit, care should be taken so that, the Q-point

will not shift into an undesirable region.

Here, two important factors are to be considered –

I. Temperature:

ICEO

The flow of current in the circuit produces heat at the junction. This

heat increases the temperature at the junction.

We know that; IC = βIB + ICEO .

The increase in temperature increases ICEO, which in turn increases IC.

This increase in IC further raises the temperature, and the same cycle

repeats.

The increase in the collector current IC, increases the power dissipated

at the collector junction; PD = VCIC .

This in turn increases the temperature at the junction; and hence,

increases the collector current.

The excess heat generated at the collector junction even burn and

destroy the transistor. This situation is called Thermal runway of the

transistor.

So, power dissipation should not cross the maximum power rating of a

transistor.

VBE

As the temperature increases, VBE also changes.

As IB depends on VBE , the collector current IC also changes.

Change in collector current changes the Q-point.

βdC

The value of β is also temperature dependent.

As β varies IC also varies; since IC = βIB.

The change in collector current changes the operating point.

Thus, thermal stability must be taken care while designing the biasing

circuits.

II. Transistor current gain β :

The value of β changes from transistor to transistor.

The biasing circuit is designed according to the required β value.

But, due to change in β from unit to unit, the operating point may shift.

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UNIT – 5

AMPLIFIERS & OSCILLATORS

Transistor as an amplifier

iC

RL Vout

iB

Vi ` iE

VBB VCC

Consider a npn transistor in CE configuration as shown above along with its input

characteristics.

A transistor raises the strength of a weak input signal and thus acts as an

amplifier. The weak signal to be amplified is applied between emitter and base and the

output is taken across the load resistor RC connected in the collector circuit.

In order to use a transistor as an amplifier it should be operated in active region

i.e. emitter junction should be always FB and collector junction should be RB. Therefore

in addition to the a.c. input source Vi two d.c. voltages VBB and VCC are applied as

shown. This d.c. voltage is called bias voltage. The following waveforms are drawn

according to the circuit diagram.

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Let VBB supplies direct current IB.

Vi causes the alternating current ib.

Total base current, iB = IB + ib, which is alternating in nature as shown.

As the input circuit has low resistance, a small change in the signal voltage Vi

causes a large change in the base current thereby causing the same change in collector

current (because iC = βiB).

The collector current flowing through a high load resistance RC produces a large

voltage across it. Thus a weak signal applied at the input circuit appears in the amplified

form at the output. In this way transistor acts as an amplifier.

Example: Let RC = 5KΩ, Vin =1V, IC =1mA then output V=ICRC =5V

The phase displacement of 1800 between input and output signals can be explained as

follows:

Consider the positive half cycle of input. Since base current iB = IB + ib, the FB on

base-emitter junction increases. This increases the base current. Since, the collector

current is β times the base current, collector current will also increase. This increases the

voltage drop across RC. Since VC = VCC – ICRC, the increase in IC results in a drop in

collector voltage VC, as VCC is constant. Thus, as Vi increases in a positive direction, V0

goes in a negative direction. Similar explanation hold good for negative half cycle of

input signal.

Transistors can work in CB, CE, and CC modes.

In CE mode of operation, a transistor has both current gain & voltage gain.

Hence, it has power gain also.

Therefore, junction transistors are used as both voltage amplifiers & power

amplifiers.

DC Load Line:- In order to illustrate the phase reversal of the amplified output voltage, a

knowledge of load line is essential.

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Consider the circuit set-up shown in the figure above. In the absence of signal, we

have, IE = IB + IC and VCE = VCC – ICRC or IC = (– 1/RC) VCE + VCC / RC

Putting IC = 0; we get, VCE = VCC

Putting VCE = 0; we get, IC = VCC / RC

Hence, we get two points; A (VCC, 0) and B (0, VCC/RC)

The output characteristics of the CE transistor are plotted for different levels of

input current IB. The points A and B are plotted. Line AB is drawn. This line is termed as

DC Load line. Point A is called cut-off point and the point B is called saturation point.

The point of intersection of the load line with an output characteristic is termed as

operating point or quiescent point (point Q in the figure).

Let point Q denote the operating point, corresponding to IB = IB2. If a signal is

applied across the input terminals, there would be variations of base current and

corresponding variations of collector current.

From the graph it is seen that, when the input voltage is passing through a positive

half-cycle, the output voltage is passing through a negative half-cycle. Thus, there is a

phase displacement of 1800 between the input signal and the output voltage.

Decibel (dB)

Many a times it is convenient to represent the gain of an amplifier on a log scale

instead of a linear scale. The unit of this log scale is called decibel.

By definition, we have;

Power gain =log10 (Pout /Pin) bel

Since, 1 bel= 10 decibels, we have;

Power gain in dB =10 log10 (Pout /Pin) dB

Since, power is α current2 or voltage

2, we have;

Voltage gain = 10 log10 (Vout /Vin)2 dB = 20 log10 (Vout /Vin) dB

Current gain = 10 log10 (Iout /Iin)2 dB = 20 log10 (Iout / Iin ) dB

Note: - For a multistage amplifier if AV1, AV2, and AV3 are the voltage gains of amplifier

1, 2, and 3 respectively then the overall voltage gain AV = AV1 x AV2 x AV3.

If it is expressed in dB the AV (dB) = AV1 (dB) + AV2 (dB) +AV3 (dB)

Similarly for four or more stages.

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Frequency Response Curve

The voltage of an amplifier depends upon the signal frequency. Thus if the signal

frequency is gradually increased, it is found that the gain also increases until it attains a

maximum value at a frequency termed as resonant frequency. If the frequency is further

increased, the gain starts decreasing. The graph of voltage gain plotted against signal

frequency is shown in the following figure. This curve is known as frequency response

curve or frequency response characteristic.

The frequency response of an amplifier is a very important parameter, since the

performance of the amplifier can be judged on the basis of its frequency response.

Band-width

Band-width may be defined as the range of frequencies in which the gain is

either equal to or greater than 70.7% of the maximum frequency.

From the frequency response curve shown in the figure above, we have;

Band-width = (f2 – f1) Hz.

It can be seen from the graph that – at f1 and f2, the voltage gain = 0.707 Gm; i.e.,

there is a decrease of gain from maximum value to 70.7% of maximum value.

We have; Gain = 20 log10 (Vout/Vin) = 20 log10 (Av)

Hence; Drop in gain = 20 log10 (Gm/0.707Gm)

= 20 log10 (1.4142) dB = 3 db.

On this basis, band-width may be defined as the range of frequencies at the limit

of which, the voltage gain falls by 3dB.

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Classification of transistor amplifier:-

I. Classification based upon signal frequency –

1) Audio amplifier – can amplify signals of frequency 15Hz to 20KHz.

2) Video amplifier – can amplify signals of frequency far above 20KHz to few MHz.

3) Radio frequency amplifier – can amplify signals of frequency ranging from a few

KHz to several MHz.

II. Classification based on the electrical quantity amplified –

1) Voltage amplifier – voltage level of the input signal is amplified.

2) Power amplifier – the output power is larger than the input power.

III. Classification based upon mode of operation –

1) Class A amplifier – collector current flows throughout the input signal cycle.

2) Class B amplifier – collector current flows only during the +ve half cycles of the

input signal.

3) Class C amplifier – collector current flows for less than half of the period of the

input signal.

4) Class AB amplifier – collector current flows for more than half of the input signal

period, but not throughout the full cycle.

IV. Classification based on coupling elements –

* R-C coupled amplifier,

* Transformer-coupled amplifier, and

* Direct-coupled amplifier.

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An audio amplifier can amplify signals of frequencies lying between 20 Hz and

20 KHz. This amplification is usually achieved in several stages. In the earlier stages, the

voltage level of the weak input signal is progressively raised and in the final stage, there

is considerable increase of power so that the last stage can drive a load like loud-speaker.

During each stage, the input signal gets strengthened and the output of each stage forms

the input to the next stage.

Power amplification is achieved in two stages as shown: the driver stage and the

output stage. In order to ensure maximum power transfer from output stage to the

speaker, impedance matching can be used.

Single stage RC coupled Amplifier

VCC

Vo

R1 RC CC

Vo t

RS

R2 RE RL

Vi

Figure above shows a practical circuit of a single stage RC coupled amplifier. The

different circuit components and their functions are as described below.

a. Input capacitor(Cin)- This capacitor is used to couple the input signal to the base

of the transistor if it is not used, the signal source resistance RS gets in parallel

with R2 thus changing the bias. The capacitor Cin blocks any d.c. component

present in the signal and passes only a.c. signal for amplification.

b. Biasing circuit –The resistances R1, R2 and RE forms the biasing and stabilization

circuit for the CE amplifier. It sets the proper operating point for the amplifier.

c. Emitter bypass capacitor (CE)-This capacitor is connected in parallel with the

emitter resistance RE to provide low reactance path to the amplified a.c. signal. If

it is not used, the amplified a.c. signal passing through RE will cause voltage drop

across it thereby reducing the output voltage of the amplifier.

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d. Coupling capacitor(Cc)- This capacitor couples the output of the amplifier to the

load or to the next stage of the amplifier. If it is not used, the biasing conditions of

the next stage will change due to the parallel effect of collector resistor RC.

i.e. RC will come in parallel with the resistance R1 of the biasing network of the

next stage thus changing the biasing conditions of the next stage amplifier.

Two stage RC coupled amplifier

VCC

R1 RC CC R11

RC1

CC1

RS

R2 RE CE R21

CE 1

RL

Vi RE1

Stage-1 Stage-2

Figure above shows the circuit diagram of a two stage RC coupled amplifier. The

coupling capacitor CC connects the output of the first stage to the input of the second

stage. Since the coupling from one stage to the next stage is achieved by coupling

capacitor along with a shunt resistor the amplifier is called RC coupled amplifier.. The

input signal is first applied to the transistor T1 and output is taken at the collector of T1.

The signal at the output will be 1800

out of phase when compared to the input. The output

is taken across RC with the help of a coupling capacitor. This signal is fed as input to the

next stage i.e transistor T2. The signal is amplified further and the amplified output is

taken across Rc1 of T2. The phase of the signal is reversed again. The output is amplified

twice and its is amplified replica of the input signal.

Frequency response curve of RC coupled amplifiers

Frequency response is the curve between the gain of the amplifier (A = Vo / Vi )

verses the frequency of the input signal. The frequency response of a typical RC-

coupled amplifier is shown below.

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Frequency response has 3 regions.

1 Low frequency range

2 Mid frequency range

3 High frequency range

Gain in

dB LFR MFR HFR

f1 f2 freq (Hz)

Low frequency range (< 50 Hz)

We have;

Xc =fC2

1 where XC ----- reactance of capacitor and

f----- frequency

Since frequency is inversely proportional to the reactance, the reactance of the

coupling capacitor CC will be quite high at low frequencies.

Hence very small amount of signal will pass through one stage to the next stage.

Moreover CE cannot shunt the emitter resistance RE effectively because of its large

reactance at low frequency. These two factors causes the fall of voltage gain at low

frequencies.

Mid frequency range (50Hz –20KHz)

In this range of frequencies, voltage gain of the amplifier is constant. The effect of

coupling capacitor in this range is as such to maintain a uniform voltage gain.

High frequency range (> 20 KHz)

In this range of frequency, the reactance of the coupling capacitor CC is very

small and it behaves as a short circuit. This increases the loading effect of next stage ( RC

will comes in parallel with R1) and reduces the voltage gain. This reduces the current

amplification there by the voltage drops at high frequencies.

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Advantages of RC coupled amplifier

1. Low cost-Because only resistors and capacitors are used for biasing and coupling

which are cheap.

2. Compact-Because modern resistor and capacitors are small and light

3. Good frequency response- The gain is constant over the audio frequency range

and hence suitable for audio frequency amplification.

4. The gain of RC coupled amplifier is constant in audio frequency range, and

hence, offers high fidelity.

Demerits of RC coupled amplifier:-

1. Due to the loading effect, the gain of RC coupled amplifier is low.

2. RC coupled amplifier tend to become noisy with age.

3. Impedance matching is not possible, since, the output impedance of RC coupled

amplifier is high.

Feed-back Concept

In a practical amplifier, although voltage gain and power gain are achieved, there

are certain draw-backs which must overcome with view of enhancing the usefulness of

the device. Thus an amplifier suffers from –

i. Distortion

ii. Noise and

iii. Gain instability

In an ideal amplifier, the amplified output voltage wave is an exact replica of the

input signal. But this is not so in practice. There is some amount of distortion – i.e.,

distortion in amplitude or frequency or phase. Distortion of any sort is not desirable and it

must be minimized, if it cannot be totally eliminated.

The percentage of distortion can be minimized, the noise level can be reduced and

the gain can be made independent of transistor parameters by the mechanism of negative

feed-back.

Feed-back is the process of supplying a part of the output back to the input. In a

transistor amplifier, a fraction of output voltage may be fed back to the input

terminals.

There are two possibilities –

If the feed-back voltage is so applied to the input port that, it aids the input signal

(i.e., it is in phase with input signal), the process is called positive feed-back.

If the feed-back voltage is so applied to the input port such as to oppose the input

signal (i.e., it is out of phase with input signal), the process is called negative feed-

back.

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Positive feed-back increases the gain of the amplifier, but it increases distortion

and gain instability. Negative feed-back, on the other hand, decreases the gain, which is

not desirable, but it reduces distortion and noise level, and brings about gain stability

and operation stability. In view of these advantages, negative feed-back is employed in

practical amplifiers.

Consider an amplifier with negative feed-back as shown in the following figure.

Vin Vout

VS

Vf

From the fig., the gain of the amplifier (open-loop gain) is given by; in

outV

V

VA

And, the feed-back factor is given by; out

f

V

Vk

Also, the closed-loop gain is given by; S

outf

V

VA

Since, for negative feed-back, Vin = VS – Vf ; VS = Vin + Vf.

Therefore; Af fin

out

VV

V

outin

out

kVV

V (Because, Vf = kVout)

inout

inout

VkV

VV

/1

/ (By dividing both Nr and Dr by Vin)

Or, V

Vf

kA

AA

1 (For negative feed-back circuits)

This is the general feed-back equation.

Amp AV +

Feed-back Network

k

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Advantages of Negative Feed-back

1. Gain stability

2. Gain control

3. Constant gain for al frequencies

4. Increase in bandwidth

NOTE: - The gain-bandwidth product is always a constant.

Gain * Bandwidth = Constant

i.e., if bandwidth increases, the gain decreases.

Normally for an amplifier, the higher input impedance and lower output impedance are

desired. These are achieved with a voltage-series feed-back. Hence voltage-series feed-

back shown in the following figure is normally used in amplifiers.

THEORY OF SINUSOIDAL OSCILLATORS

Oscillator is an electronic device which generates electrical oscillations of

desired frequency and wave form.

The frequency required for the practical purpose ranges from a few Hz to several

MHz.

The wave form is either sinusoidal or non-sinusoidal.

A transistor amplifier with positive feedback generates sinusoidal oscillations & it

is termed as sinusoidal oscillator.

An oscillator which produces non-sinusoidal oscillations is usually called

relaxation oscillator.

E.g..:- Multivibrators which generates square-wave oscillations, Voltage sweep

generators, Current sweep generators, etc.

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An oscillator is a device which mainly converts AC energy of the required

frequency.

Oscillators find extensive use in various electronic equipments like radio, radar,

television, etc.

Types of Sinusoidal Oscillations

1. Damped Oscillations

2. Un-damped Oscillations

1. Damped Oscillations-The electrical oscillations whose amplitude goes on

decreasing with time are called damped oscillations.

2. Un-damped Oscillations- The electrical oscillations whose amplitude remains

constant with time are called un-damped oscillations.

e e

t t

Damped oscillations Un-damped oscillations

Positive feedback Amplifier-Oscillator

1. A transistor amplifier with proper +ve feedback can act as an oscillator.

S

Vout

Vin Vf

t t t

V

Vf

kA

AAgainClosedloop

1,

2. The circuit needs only a quick trigger signal to start the oscillations. Once the

oscillations have started, no external signal source is necessary.

Amplifier

Av

Feedback

network, k

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3. In order to get continuous un-damped output from the circuit, the following

condition must be met;

kAV =1

where AV = voltage gain of amplifier without feedback.

k = feedback fraction.

This relation is also called Barkhausen criterion.

The Barkhausen Criterion states that:

1.] The total phase shift around a loop, as the signal proceeds from input

through the amplifier, feedback network back to input again, completing a loop, is

precisely 00 or 360

0 .

2.] The magnitude of the product of the open loop gain of the amplifier (AV)

and the magnitude of the feedback factor (k) is unity. i.e., | k AV | = 1.

If Barkhausen criterion is satisfied at only one frequency, sinusoidal oscillations

result.

If the above requirement is met over a band of frequencies, then non-sinusoidal

oscillations would result.

Essentials of Transistor Oscillator

Fig. below shows the block diagram of an oscillator. It‟s essential components are:

1. Tank Circuit: It consists of inductance coil(L) connected in parallel with

capacitor(C ). The frequency of oscillations in the circuit depends upon the values

of inductance of the coil and capacitance of the capacitor.

2. Transistor Amplifier: The transistor amplifier receives d.c. power from the

battery and changes it into a.c. power for supplying to the tank circuit. The

oscillations occurring in the tank circuit are applied to the input the transistor

amplifier. The output of the transistor can be supplied to the tank circuit to meet

the losses.

3. Feedback circuit: The feedback circuit supplies a part of collector energy to the

tank circuit in correct phase to aid the oscillations. ie., provides positive feedback.

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L C

Block diagram of Transistor Oscillator

Oscillatory circuit

A transistor amplifier with positive feed-back can function as an oscillator.

It can generate un-damped oscillations, provided the following conditions are

satisfied.

1) There must be an oscillatory circuit, which generates electrical oscillations.

2) There must be an amplifier, which supplies energy for the loss of power occurring

in the oscillatory circuit.

3) There must be a feed-back circuit, in order to supply energy to the oscillatory

circuit in correct phase & magnitude.

The simplest form of oscillatory circuit is a tank circuit. A circuit, which produces

electrical oscillations of any desired frequency, is known as an oscillatory circuit

or tank circuit.

A tank circuit consists of an inductor L & a capacitor C, connected in parallel.

A simple oscillatory circuit consists of a capacitor C and inductance coil L in

parallel as shown in figure below. This electrical system can produce electrical

oscillations of frequency determined by the values of L and C.

S

++ ++

L C _ _ _ _

Fig. 1 Fig. 2

Transistor

Amplifier

Feedback

circuit

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_ _

+ +

Fig. 3 Fig. 4

Circuit operations- Assume capacitor is charged from a d. c. source with a polarity as

shown in figure 1.

When switch S is closed as shown in fig.ii, the capacitor will discharge through

inductance and the electron flow will be in the direction indicated by the arrow.

This current flow sets up magnetic field around the coil. Due to the inductive

effect, the current builds up slowly towards a maximum value. The circuit current

will be maximum when the capacitor is fully discharged. Hence the electrostatic

energy across the capacitor is completely converted into magnetic field energy

around the coil.

Once the capacitor is discharged, the magnetic field will begin to collapse and

produce a counter emf. According to Lenz‟s law the counter emf will keep the

current flowing in the same direction. The result is that the capacitor is now

charged with opposite polarity making upper plate of capacitor –ve and lower

plate +ve as shown in fig. 3.

After the collapsing field has recharged the capacitor, the capacitor now begins to

discharge and current now flows in the opposite direction as shown in fig. iv.

The sequence of charge and discharge results in alternating motion of electrons or

an oscillating current. The energy is alternately stored in the lectric field of the

capacitor C and the magnetic field of the inductance coil L . This interchange of

energy between L and C is repeated over and again resulting in the production of

Oscillations.

Waveform- In practical tank circuit there are resistive and radiation losses in the coil and

dielectric losses in the capacitor. During each cycle a small part of the originally imparted

energy is used up to overcome these losses. The result is that the amplitude of oscillating

current decreases gradually and eventually it become zero. Therefore tank circuit

produces damped oscillations.

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Frequency of oscillations- The expression for frequency of oscillation is given by,

LCf r

2

1-------------------------------(1)

Un-damped Oscillations from Tank Circuit

A tank circuit produces damped oscillations. In practice we need continuous un

damped oscillations for the successful operation of electronics equipment. In order to

make the oscillations in the tank circuit un-damped it is necessary to supply correct

amount of energy to the tank circuit at the proper time intervals to meet the losses.

The following conditions must be fulfilled;

1. The amount of energy supplied be such so as to meet the losses in the tank and the

a.c. energy removed from the circuit by the load. For example if losses in LC

circuit amount ot 5 mW and a.c. output being taken is 100 mW, then power of

105mW should be continuously supplied to the circuit.

2. The applied energy should have the same frequency as the of the oscillations in

the tank circuit.

3. The applied energy should be in phase with the oscillations set up in the tank

circuit.

Classification of Oscillators:-

Oscillators may be classified as –

i. Sinusoidal oscillators, and

ii. Non-sinusoidal oscillators.

Based upon the frequency of oscillations generated –

i. Audio frequency oscillators,

ii. Radio frequency oscillations,

iii. Ultra high frequency oscillators, and

iv. Microwave oscillations.

Oscillators may be categorized as –

i. Feed-back oscillators, and

ii. Negative resistance oscillators.

Sinusoidal oscillators which employ different circuit elements for the feed-back

network may be classified into two groups as –

i. R – C oscillators – Low frequency: 20 Hz to 20 KHz, and

ii. L – C tuned oscillators – High frequency more than 300 KHz.

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Based upon how energy is supplied to the tank circuit to overcome the losses,

practical transistor oscillators are classified as –

i. Tuned collector oscillator,

ii. Wein – bridge oscillator,

iii. Phase shift oscillator,

iv. Hartley oscillator, and

v. Colpitt‟s oscillator.

In addition to these, we have crystal oscillators, whose operation is based on

piezo-electric effect.

Tuned Collector Oscillator

The circuit shown above, comprising of inductor L1, and capacitor C1 is

incorporated in the collector circuit of the CE transistor. Another inductive coil L2 is

magnetically coupled to L1. L1 and L2 may be the primary and secondary windings of a

transformer.

Resistors R1, R2 and RE form the biasing and stabilization circuit. CE is the emitter

bypass capacitor and the capacitor C provides a path of low reactance to the electrical

oscillations generated by the tank circuit.

The frequency of oscillations is given by; 112

1

CLfR

When the switch S is closed, the capacitor C1 gets charged by the collector

current. After it is fully charged, it begins to discharge through L1. Hence electrical

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oscillations of frequency fR is generated. Since L2 is inductively coupled to L1, an

alternating e.m.f. of the same frequency fR is induced in it. Its magnitude is decided by

number of turns in the coils L1 and L2, and the mutual inductance between them. This

e.m.f. is applied to the input terminals of the transistor.

The transistor amplifier causes a phase shift of 1800 between input and output

voltages. The output voltage when applied across the primary winding L1, undergoes

further phase shift of 1800 in the secondary winding L2. Thus a total phase shift of 3600 or

00 is achieved; hence, a positive feedback results.

Hartley Oscillator

The circuit diagram of Hartley Oscillator is as shown in figure below. It uses two

inductors placed across common capacitor C and the center of two inductors ins tapped.

The tank circuit is made up of L1 , L2 and C and is given by.

CLf

T2

1

where LT = L1 + L2 + 2M

M = Mutual inductance between L1 and L2

When the circuit is turned ON, the capacitor is charged. When this capacitor is

fully charged, it discharges through coils L1 and L2 setting up oscillations of

frequency determined by expression 1. The output voltage of the amplifier

appears across L2 and feedback voltage across L1. The voltage across L1 is 1800

out of phase with the voltage developed across L2.

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A phase shift of 1800 is produced by the transistor and a further phase shift of

1800 is produced by L1-L2 voltage divider circuit. In this way feedback is properly

phased to produce continuous un-damped oscillations.

Feedback fraction- In Hartley oscillator the feedback voltage is across L1 and output

voltage is across L2.

Therefore feedback fraction 2

1

2

1

L

L

X

X

v

vm

L

L

out

f

v

Colpitt’s Oscillator

The tank circuit is made up of C1, C2 and L. The frequency of oscillations is

determined by:

21

21

2

1

CC

CCCwhere

LCf

T

T

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When the circuit is turned ON, the capacitor C1 and C2 are charged. The

capacitors discharge through L setting up oscillations of frequency determined by

expression. 1. The output voltage appears across C2 and feedback voltage is

developed across C1. The voltage across C1 is 1800 out of phase with the voltage

developed across C2 (Vout ). A phase shift of 1800 is produced by the transistor and

a further phase shift of 1800 is produced by C1-C2 voltage divider. In this way

feedback is properly phased to produce continuous un-damped oscillations.

Feedback factor

Feedback factor 1

2

2

1

C

C

X

X

V

Vm

C

C

out

f

v

Demerits of Oscillator using Tank Circuit

1. They suffer for frequency instability and poor waveform.

2. They cannot be used to generate low frequencies, since they become too-much

bulky and expensive too.

RC Phase Shift Oscillator

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It consists of a conventional single transistor amplifier and a RC phase shift

circuit. The RC phase shift circuit consists of three sections R1C1, R2C2, and

R3C3.At some particular frequency f0 the phase shift in each RC section is 600 so

that the total phase shift produced by the RC network is 1800. The frequency of

oscillation is given by,

62

1

RCfo

When the circuit is switched ON it produces oscillations of frequency determined

by above equation. The output VO of the amplifier is feedback to RC feedback

network.

Let the voltage drop across R1or R2 or R3 = VR = IR.

Let the voltage drop across C1 or C2 or C3 = VC = IXC.

The total voltage will be the phaser sum given by; V = VR + VC.

It is seen from the phaser diagram that voltage VR leads V by an angle θ where,

tan θ = fCRR

X C

2

1 Since,

fCX C

2

1

Therefore, θ = tan –1

fCR2

1

By choosing proper values of R and C, θ can be made equal to 600.Thus, since

there are three legs of the ladder network, it produces a total phase shift of 1800

and the transistor gives another 1800 shift. Thereby total phase shift of the output

signal when fed back is 3600.

Merits-

1. They do not require any transformer or inductor thereby reduce the cost.

2. They are quite useful in the low frequency range where tank circuit

oscillators cannot be used.

3. They provide constant output and good frequency stability.

4. The circuit is simple to design.

5. Can produce output over audio frequency range.

6. Produces sinusoidal output waveform.

7. It is a fixed frequency oscillator.

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Demerits –

1. It is difficult to start oscillations.

2. The circuit requires a large number of components.

3. The frequency stability is poor, due to the effect of temperature, aging, etc.

4. They cannot generate high frequencies and are unstable as variable frequency

generators.

5. The frequency of oscillation can be changes by changing the value of R and C.

But the values of R and C of all the three sections must be changed

simultaneously to satisfy the oscillating conditions. This is practically impossible.

Hence, phase shift oscillator is considered as a fixed frequency oscillator.

Transistor Crystal Oscillator

Certain crystals like Quartz, Rochelle salt, and Tourmaline exhibit a particular

property.

When an alternating voltage at a certain frequency is applied across such a crystal,

it is observed that the crystal vibrates at the same frequency.

The converse is also found to be true.

That is; when the crystal is made to vibrate mechanically by subjecting it to

compression or mechanical strain, it is found to be generate an alternating voltage.

This phenomenon is termed as piezo-electric effect.

When a crystal is to be used in an electronic circuit, it is usual to mount a thin

wafer of the crystal in between two metal plates, as shown in the fig.

Since the crystal is dielectric material, and it is placed between metal plates, the

combination acts as a capacitor.

Cm denotes the capacitance, which is termed as mounting capacitance.

When an alternating voltage is applied, the crystal vibrates at the frequency of the

applied voltage.

The crystal can be visualized as a tuned circuit, having resistance, inductance and

capacitance forming a series circuit.

The equivalent circuit of the crystal when it is vibrating is shown in the fig.

below.

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The series resonant frequency is given by –

LC

fR2

1

The parallel resonant frequency is given by –

eq

RLC

f2

1 where,

'

'

CC

CCCeq

The quality factor of the crystal is defined as the ratio of inductive/capacitive reactance

to the resistance of the crystal.

i.e., R

Lf

R

wL

R

CLQ R2/

Crystal Oscillator Circuit Diagram –

Figure shows the transistor crystal oscillator. The crystal will act as parallel –

tuned circuit. At parallel resonance, the impedance of the crystal is maximum.

This means that there is a maximum voltage drop across C2. This in turn will

allow the maximum energy transfer through the feedback network.

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The feedback is +ve. A phase shift of 1800 is produced by the transistor. A further

phase shift of 1800 is produced by the capacitor voltage divider. This oscillator

will oscillate only at fp.

Where fp = parallel resonant frequency ie the frequency at which the vibrating

crystal behaves as a parallel resonant circuit.

m

m

T

T

p

CC

CCCwhere

LCf

2

1

Advantages

1. Higher order of frequency stability

2. The Q-factor of the crystal is very high.

Disadvantages

1. Can be used in low power circuits.

2. The frequency of oscillations cannot be changed appreciably.

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UNIT – 6

INTRODUCTION TO OPERATIONAL AMPLIFIERS

INTRODUCTION

An integrated circuit (IC) consists of a single crystal chip of silicon of very small

dimensions, containing both active & passive elements.

A number of processes are involved in the manufacture of IC‟s. these include –

• Preparing of wafer,

• Epitaxial growth,

• Diffusion of impurities,

• Ion implantation,

• Oxide growth,

• Photolithography,

• Chemical etching,

• Metallization, etc.

An IC does not have discrete components; instead, all components (both active &

passive) are an integral part of it.

Whole circuit comprising of various elements are properly interconnected, and

created on a single chip, and no external wiring is required.

A number of terminals are brought out of the chip, and external connections are

made from these terminals.

The main advantages of IC’s are –

o Increased reliability,

o Smaller size,

o Increased speed of operation, and

o Reduced cost due to mass production techniques.

Classification of IC’s –

1. Linear IC’s (Analog IC’s) – outputs are proportional to inputs. Both inputs &

outputs can take continues values.

• Applications:- amplifiers, voltage regulators, operational

amplifiers, etc.

2. Digital IC’s – input & output can take only two vales: 0 or 1 (or low level & high

level).

• Applications:- micro-processors, logic gates, memory

chips, counters, clock chips, flip-flops, etc.

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Types of IC’s –

i. Monolithic circuits,

ii. Thin-film circuits,

iii. Thick-film circuits, and

iv. Hybrid circuits.

Based on the number of circuits contained in an IC package, integrated circuits

are classified as –

I. Small Scale Integration (SSI) – contains less than 30 circuits.

II. Medium Scale Integration (MSI) – contains about 30 to 100 circuits.

III. Large Scale Integration (LSI) – contains about 100 to 1,00,000 circuits.

IV. Very Large Scale Integration (VLSI) – a single IC contains more than 1,00,000

circuits.

To study about an operational amplifier, knowledge of differential amplifier is essential.

A differential amplifier mainly comprises of two transistors with their emitters

connected as shown in the figure below.

An operational amplifier is basically a differential amplifier with two inputs –

non-inverting & inverting, and one output.

It is used in analog computers to do several mathematical operations like

summing, differentiation, integration, inversion, etc. it is also used in control

systems.

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An operational amplifier is basically a very high gain, direct-coupled amplifier

with high input impedance & low output impedance.

The figure below shows the symbol of an Op-Amp.

If a voltage Vi is applied at the inverting input (keeping the non-inverting input at

ground) as shown below.

Vi

VO

t

t

Vi VO

Op-amp in inverting mode

The output voltage Vo= -AVi is amplified but is out of phase with respect to the input

signal by 1800.

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If a voltage Vi is fed at the non-inverting input (keeping the inverting input at

ground) as shown below.

Vo

VO

t

Vi

t

Op-Amp in Non-inverting mode

The output voltage Vo= AVi is amplified and in-phase with the input signal.

Note: - Op-Amp is 8 pin IC (named as μA 741) with pin details as shown.

OFFSET NULL NO CONNECTION

INVERTING I/P +VCC

NON-INVERTING I/P OUTPUT

-VEE OFFSET NULL

Pin details of Op-Amp

Block Diagram of an Op-AMP

1 8

2 7

μA 741

3 6

4 5

3 6

4 5

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An Op-Amp consists of four blocks cascaded as shown above.

Input stage: It consists of a dual input, balanced output differential amplifier. Its

function is to amplify the difference between the two input signals. It provides high

differential gain, high input impedance and low output impedance.

Intermediate stage: The overall gain requirement of an Op-Amp is very high. Since the

input stage alone cannot provide such a high gain. Intermediate stage is used to provide

the required additional voltage gain.

It consists of another differential amplifier with dual input, and unbalanced (single

ended) output

Buffer and Level shifting stage: As the Op-Amp amplifies D.C signals also, the small

D.C. quiescent voltage level of previous stages may get amplified and get applied as the

input to the next stage causing distortion the final output.

Hence the level shifting stage is used to bring down the D.C. level to ground

potential, when no signal is applied at the input terminals. Buffer is usually an emitter

follower used for impedance matching.

Output stage: It consists of a push-pull complementary amplifier which provides large

A.C. output voltage swing and high current sourcing and sinking along with low output

impedance.

Characteristics of an ideal OP-AMP:-

The voltage gain is infinity.

The input impedance is infinity.

The output impedance is zero.

The band-width is infinity.

When equal voltages are applied at the two input terminals, the output is zero.

There is no change in the characteristic feature, with changes of temperature.

Practical OP-AMPs have –

Large voltage gain.

Very high input impedance.

Very low output impedance.

A small output voltage appears even when equal voltages

are applied at the two input terminals.

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Some Important Definitions:-

Common mode rejection ratio [CMRR] is defined as the ratio of differential gain

of the amplifier to the common-mode gain.

CMMR = |Ad/AC| or CMMR = 20 log |Ad/AC|

Slew rate [SR] is defined as the maximum rate at which an OP-AMP output can

change, and it is expressed in terms of Volts/µsec. SR = ΔV0/Δt

The SR represents the ability of an amplifier to handle the varying

signals like large step-input signals.

Input bias current [IIB]

A difference amplifier mainly consists of two identical transistors which are

direct-coupled. During normal operation, transistors are biased properly using d.c.

voltage sources. Ideally, in an OPAMP, there should be equal d.c. bias currents at

both non-inverting and inverting input terminals. But in practice, the transistors may

not match perfectly, with the result that the d.c. bias currents at + and – inputs are not

exactly equal.

The bias currents are denoted as IIB+ and IIB

–. The average of these two

currents is termed as input bias current given as;

IIB = ½ (IIB+ + IIB

-)

The DC voltage which makes the output voltage zero, when the other terminal is

grounded is called input offset voltage.

The voltage existing at the output, when the inputs are zero, is called output offset

voltage. It is usually caused by input bias current & the input offset voltage.

Power supply rejection ratio [PSRR] is defined as the ratio of the change in the

input offset voltage due to the change in supply voltage producing it, keeping the

other power supply voltage constant. It is also called power supply sensitivity.

PSSR = ΔVios /ΔVCC | VEE constant.

Open Loop Voltage Gain (AV) is the ration of output voltage to input voltage in

the absence of feedback. It‟s typical value is AV = 2x105

Input Impedance (Ri) is defined as “The impedance seen by the input (source)

applied to one input terminal when the other input terminal is connected to

ground”. Ri ≈ 2MΩ

Output Impedance (RO) is defined as “ The impedance given by the output (load)

for a particular applied input”. Ro ≈ 75Ω

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Typical values of several parameters for A741 OP-AMP –

• Input offset voltage 1 mV.

• Input bias current 80 nA.

• Slew rate at unity gain 0.5 V/µsec.

• CMRR 90 dB.

Virtual ground concept

We know that, an ideal Op-Amp has perfect balance (i.e., output will be zero

when input voltages are equal).

Hence when output voltage Vo = 0, we can say that both the input voltages are

equal ie V1 = V2.

V1

Ri VO

V2

Since the input impedances of an ideal Op-Amp is infinite (Ri = ∞). There is no

current flow between the two terminals.

Hence when one terminal (say V2) is connected to ground (i.e., V2 = 0) as shown.

VCC

V1 =V2 =0

Ri VO

V2=0

VEE

Then because of virtual ground V1 will also be zero.

Example: Assuming that the input impedance of an OPAMP to be very large and output

impedance to be very small, the equivalent circuit of the OPAMP is shown in the

following figure.

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From the figure; VO = – A Vi where, A is the gain of OPAMP (open loop gain)

Also, overall gain = VO/V1 where, overall gain is the gain of the circuit (closed loop gain)

In practice, the voltage gain A is very high, say A = 10,000

Let VO = – 5V. ( – sign is used since output is out of phase with the input)

Therefore, Vi = – VO/A = 5/10,000 = 0.5 mV.

Now, let the overall gain VO/V1 be unity. (This can be set by adjusting the values of

resistances R1 and Rf).

Then, V1 = VO = 5V.

Now, it can be seen that the voltage Vi is quite small, as compared to all other

voltages. If the assumption is made that Vi = 0, it means that there is a short-circuit at the

input terminals of the OPAMP. But, since the input impedance is infinite, there can be no

flow of current through the short. Hence, the short is not true short-circuit, but it is only a

virtual short-circuit or virtual ground. As a result, the current I flowing through R1

also flows through Rf.

Applications of Op-Amp – An OPAMP can be used as

1. Inverting Amplifier

2. Non-Inverting Amplifier

3. Voltage follower

4. Adder ( Summer)

5. Integrator

6. Differentiator

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Inverting Amplifier

An inverting amplifier is one whose output is amplified and is out of phase

by 1800 with respect to the input.

Rf

i2

R1

V1 i1 G

VO

The point “G” is called virtual ground and is equal to zero.

By KCL we have

21 ii

f

oi

R

V

R

V 00

1

f

oi

R

V

R

V

1

i

f

O VR

RV

1

Where 1R

R f is the gain of the amplifier and negative sign indicates that the output is

inverted with respect to the input.

VO

Vi

t t

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Non- Inverting Amplifier

A non-inverting amplifier is one whose output is amplified and is in-phase with

the input.

Rf

i2

R1

V1 i1 G=Vi

VO

Vi

By KCL we have

21 ii

f

Oii

R

VV

R

V

1

0

1

0

1

R

R

V

ViV

R

VV

R

V

f

i

f

iOi

1

1R

R

V

V f

i

O

i

f

i

O

R

R

V

V1

i

fV

R

RV

1

0 1

Where 1

1R

R fis the gain of the amplifier and + sign indicates that the output is

in-phase with the input.

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Voltage follower

A voltage follower is one whose output is equal to the input.

VO

Vi VO

Vi

t t

The voltage follower configuration shown above is obtained by short circuiting

“Rf” and open circuiting “R1” connected in the usual non-inverting amplifier.

Thus all the output is fed back to the inverting input of the op-Amp.

Consider the equation for the output of non-inverting amplifier

i

fV

R

RV

1

0 1

When Rf = 0 short circuiting

R1= ∞ open circuiting

iV0

1V O

iO VV

Therefore the output voltage will be equal and in-phase with the input voltage.

Thus voltage follower is nothing but a non-inverting amplifier with a voltage gain of

unity.

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Inverting Adder

Inverting adder is one whose output is the inverted sum of the constituent inputs.

R1

Rf

V1 i1

If

R2

V2 i2 G=0

VO

V3 R3 i3

By KCL we have, 321 iiii f

i.e., 3

3

2

2

1

1 0000

R

V

R

V

R

V

R

V

f

O

or, 3

3

2

2

1

1

R

V

R

V

R

V

R

V

f

O

Hence, 3

3

2

2

1

1

R

V

R

V

R

VRV fO

If R1 = R2 = R3 =R, then ;

321 VVVR

RV

f

O

If Rf = R, then ;

VO = – [V1 + V2 + V3]

Hence it can be observed that the output is equal to the inverted sum of the inputs.

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Integrator

An integrator is one whose output is the integration of the input.

C

i2

R1

V1 i1 G=0

VO

By KCL we have,

121 ii

iO

Oi

O

O

O

O

ii

VRCdt

dV

dt

dVC

R

V

haveweinandgSubstituin

dt

dVCiei

iCdt

dV

dtiC

V

dtiC

V

havewesimilarlyand

R

V

R

Vi

havewefigureabovetheFrom

1

132

3..

1

1

10

20

2

2

2

2

1

dtVRC

V iO

1

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Differentiator

A differentiator is one whose output is the differentiation of the input.

R

i2

V1 i1 G=0

VO

By KCL we have,

R

V

dt

dVC

haveweinandngsubstituti

R

V

R

Vi

havewesimilarlyand

dt

dVCi

iCdt

dV

dtiC

V

havewefigureabovetheFrom

ii

Oi

OO

i

i

i

132

30

2.

1

1

1

2

1

1

1

21

dt

dVRCVO

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Non-inverting Adder

Non-inverting adder is one whose output is the sum of the constituent inputs.

Let the voltage at the inverting input terminal be Vm. Because of virtual ground at

the input terminals, the voltage at G is also Vm.

Applying KCL at the node G, we have;

The OPAMP along with the resistors R and Rf acts as a non-inverting amplifier.

Hence, closed loop voltage gain =

Therefore,

Substituting for Vm, the expression (1) becomes;

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Let R1 = R2 = R3 = R = Rf/2; then we have, VO = V1 + V2 + V3, on simplification.

The output is the sum of the input voltages, without change of sign. Hence, the

name non-inverting adder.

Subtractor

An OPAMP can function as a subtractor, giving an output voltage which is the

difference of two input voltages. The circuit is mainly a basic differential amplifier in

which all resistors are of equal magnitude.

The output of the amplifier can be computed on the basis of the principle of

superposition. – The waited sum of the output is equal to the sum of the outputs when

separate inputs are considered.

Consider the following subtractor figure.

V1 and V2 are the input voltages at the non-inverting and inverting terminals

respectively. R is the resistor in the feedback path.

Case(1): V01 denote the output with V1 applied and V2 set equal to zero, as shown below.

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Let the potential at node M be VM. We have;

The circuit is non-inverting amplifier with an input V1/2 at the non-inverting input

terminal, and the inverting terminal is grounded through resistance R. Hence, the outpot

may be directly obtained as;

Case(2): Let V02 denote the output with V2 applied and V1 set equal to zero, as shown

below.

This circuit is basically an inverting amplifier whose output is given as;

Hence, when both inputs V1 and V2 are applied, we have the output given by the

principle of superposition as;

V0 = V1 – V2

Hence, the circuit is subtractor.

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Problems

1. For an inverting amplifier Ri=100KΩ and Rf=600KΩ. What is the output

voltage for an input of -3V?

Soln:

Given: R1=100KΩ

Rf=600KΩ

Vi=-3V

VO =?

We have,

VV

VR

RV

O

i

f

O

18

310100

106003

3

1

2. Design an inverting amplifier for output voltage of -10V and an input voltage

of 1V.

Soln:

Given: Vi =1 V

VO= -10V

We Have,

i

f

O VR

RV

1

i.e., 1101R

R f

haveweKRgAssu

RRorR

Rf

f

1min

1010

1

1

1

KR f 10 Rf

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Rf

i2

R1

V1 i1

VO

3. For an inverting amplifier R1=10KΩ and Vi =1V. Calculate i1and VO.

Soln:

Given: R1 = 10KΩ, Rf=100KΩ Vi =1 V

We have,

Rf

i2

R1

V1 i1

VO

VVR

RV

mAR

Vi

i

f

O

i

1011010

10100

1.01010

10

3

3

1

31

1

4. Design an amplifier with a gain of +9 and Rf =12 KΩ using an op-Amp

Soln:

Since the gain is positive:

Choose a non-inverting amplifier

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Then we have,

i

f

O VR

RV

1

1

Gain is,

KR

RR

R

R

R

R

f

f

f

5.1

8

1012

8

8

91

1

3

1

1

1

5. In the figure shown if V1=+1V, V2=+3V and V3=+2V with R1=R2=R3=2KΩ.

Determine the output voltage.

R1

Rf

V1 i1

If

R2

V2 i2 G=0

VO

V3 R3 i3

Soln: We have ,

VV

V

VVVR

RV

O

O

f

O

9

231102

1033

3

321

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6. Design an Adder using Op-Amp to give the output voltage VO= -

[2V1+3V2+5V3]

Soln:

Given 1532 321 VVVVO

We Have,

23

3

2

2

1

1

3

3

2

2

1

1

VR

RV

R

RV

R

RV

R

V

R

V

R

VRV

fff

O

fO

Equating eqn 1 and 2 we get,

5;3;2321 R

R

R

R

R

R fff

Assuming Rf =100KΩ, We get,

KRR

R

KRR

KRR

R

f

f

f

205

33.333

R

502

33

22

11

Note: If design is asked after finding the values of Rf and R1 circuit diagram should be

written.

7. Design a summing amplifier to add three input voltages. The output of the

amplifier should be twice the negative sum of the inputs.

Soln:

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KRthenKRLet

RRR

R

getweEquating

VVVR

RVhavewe

VVVV

f

f

f

f

O

O

2010

22

,

2

321

321

8. A 5 mV peak voltage, 1 KHz signal is applied to the input of an Op-Amp

integrator for which R=100KΩ and C=1μF. Find the output voltage.

Soln: Given: R=100KΩ, C=1μF, Vm =5mV, F=1KHz, V0 =?

We have Vi = Vm Sinwt = VmSin2пft or Vi=5sin200пt mV

For an integrator,

mVtV

solvingon

dtVRC

V

O

iO

200cos40

1

,

1

9. The input to a differentiator is a sindusoidal voltage of peak value 5mV and

frequency 2KHz . Find the output if R = 100KΩ and C=1μF.

Soln: Given:

mVtVsolvingon

dt

dVRCVatordifferentifor

mVtV

O

i

O

i

4000cos2000

400sin5

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CATHODE RAY OSCILLOSCOPE

Cathode Ray Tube

The cathode ray tube (CRT) is the heart of CRO. The CRT generates the electron

beam, deflects the beam & also has a screen where beam becomes visible as a

spot. The main parts of CRT are –

– Electron gun

– Deflection system

– Fluorescent screen

– Glass tube or envelope

– Base

1.] Electron Gun:-

● Provides a sharply focused electron beam directed towards the fluorescent

screen. The thermally heated cathode emits the electrons.

● The control grid controls the number of electrons in the beam, going to the

screen.

● The movement of electrons determines the intensity, or brightness of the

light on the screen.

● Since the electron beam consists of many electrons, the beam tends to diverge.

The focusing anode is used to compensate this.

MP, ECE, AIET 3

Cathode

Grid

Focusinganode

Acceleratinganode

HorizontalDeflection

PlatesScreen

VerticalDeflection

PlatesElectron gun

Aquadag

Cathode Ray Oscilloscope

Base

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2.] Deflection System:-

● The deflection system consists of two pairs of parallel plates, called as vertical

& horizontal deflection plates.

● One of the plate is connected to zero volts & the other plate is applied with an

external deflection voltage.

● As the electron beam passes though these plates, it gets deflected vertically &

horizontally, as per the voltage applied.

3.] Fluorescent Screen:-

● The screen is coated with a fluorescent material called phosphor (like

Willemite), which emits light when bombarded by electrons.

● The light produced by the screen does not disappear immediately when the

bombardment by the electrons ceases.

● The time period for which the trace remains on the screen after the signal

becomes zero is known as “persistence” or “fluorescence”. It depends upon the type of

phosphor used.

4.] Glass Tube:-

● All the components of a CRT are enclosed in an evacuated glass tube called

envelope. This allows the emitted electrons to move about freely from one end of the tube

to the other end.

5.] Base:-

● The base is provided to the CRT through which the connections are made to

the various parts.

Block Diagram of CRO

MP, ECE, AIET 7

Block Diagram of CRO

Screen

Delayline

Verticalamplifie

r

Horizontalamplifier

Time baseGenerator

TriggerCircuit High Voltage

Low Voltage

PowerSupply

+ve To all circuits

-ve

InputSignal

CRT

V

V

H

HElectrongun

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The various blocks of block diagram of CRO are –

1.] CRT:-

● The CRT generates the electron beam, deflects the beam & also

has a screen where beam becomes visible as a spot.

2.] Vertical Amplifier:-

● The input signals are generally not strong to provide measurable

deflection on the screen. Hence, the vertical amplifier stage is used to

amplify the input signals.

3.] Delay Line:-

● The delay line is used to delay the signal for some time;

otherwise, a part of the signal gets lost.

4.] Trigger Circuit:-

● To synchronize horizontal deflection with vertical deflection, a

synchronizing or triggering circuit is used.

5.] Time Base Generator:-

● It is used to generate the saw-tooth voltage, which is required to

deflect the beam in horizontal section. This voltage deflects the spot at a

constant time dependent rate; thus, the X-axis on the screen represents

the time.

6.] Horizontal Amplifier:-

● The saw-tooth voltage produced by the time base generator may

not be of sufficient strength. Hence, it is amplified by using

horizontal amplifiers.

7.] Power Supply:-

● The power supply block provides the voltage required by CRT to

generate & accelerate the electron beam and the voltages required

by the other circuits of the oscilloscope like horizontal amplifier,

vertical amplifier, etc.

ABSTRACT:

The cathode ray oscilloscope [CRO] is an electronic device, which is capable of giving a

visual indication of a signal waveform. It is widely used for trouble shooting radio and

television receivers as well as laboratory work involving research and design. In addition

the oscilloscope can also be used for measuring voltage, frequency and phase shift.

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Cathode Ray Tube

A cathode ray tube is the heart of the oscilloscope. It is a vacuum tube of special

geometrical shape and converts an electrical signal into visual one. A cathode ray tube

makes available plenty of electrons. These electrons are accelerated to high velocity and

are brought to focus on a fluorescent screen. The electron beam produces a spot of light

wherever it strikes. The electron beam is deflected on its journey in response to the

electrical signal under study. The result is that electrical signal waveform is displayed

visually.

Electron Gun Assembly- The arrangement of electrodes which produce a focused

beam of electrons is called the electron gun. It essentially consists of an indirectly

heated cathode, control grid, a focusing anode, and an accelerating anode. The control

grid is held at negative potential with respect to cathode whereas the two anodes are

maintained at high potential with respect to cathode.

The cathode consists of a nickel cylinder coated with oxide coating and provides

plenty of electrons. The focusing anode focuses the electron beam into a sharp pin –

point by controlling the positive potential on it. The positive potential ( about 10,000

V) on the accelerating anode is much higher than on the focusing anode. Therefore

this anode accelerates the narrow beam to a high velocity.

Deflection plate assembly-

1. Vertical deflection plates

2. Horizontal deflection plates

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The vertical deflection plates are mounted horizontally in the tube. By applying

proper potential to these plates, the electron beam can be made to move up and down

vertically on the fluorescent screen. An appropriate potential on horizontal plates can

cause the electron beam to move right and left horizontally on the screen.

Screen-The screen is the inside face of the tube and is coated with some fluorescent

material such as Zinc Orthosilicate, Zinc oxide etc. When high velocity electron beam

strikes the screen, a spot of light is produced at the point of impact.

Action of CRT

O1

+ + + + + +

_ _ _ _ _ _ O

O2

When the cathode is heated, it emits plenty of electrons. The control grid

influences the amount of current flow. As the electron beam leaves the control

grid, it comes under the influence of focusing and accelerating anode. As the two

anodes are maintained at high potential, therefore they produce a field which acts

as an electrostatic lens to converge the electron beam at a point on the screen.

As the electron beam leaves the accelerating anode, it comes under the influence

of vertical and horizontal deflection plates. If no voltage is applied to the

deflection plates, the electron will produce spot of light at the center (point O ) of

the screen. If the voltage is applied to vertical plates only, the electron beam and

hence the spot of light will be deflected upwards (point O1 ). The spot of light will

be deflected downwards (O2) of the potential on the plate is reversed. Similarly

the spot of light can be moved horizontally by applying voltage across the

horizontal plates.

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Signal Pattern on Screen

CRO Screen

2

1 3 5

t

4

_ +

Saw-tooth wave applied across horizontal plate

If the signal voltage is applied to the vertical plates and saw tooth wave to the

horizontal plates, we get the exact pattern of the signal as shown in figure.

When the signal is at instant 1, its amplitude is zero. But at this instant, maximum

voltage is applied to the horizontal plates. The result is that the beam is at the

extreme left on the screen as shown. When the signal is at instant 2, its amplitude

is maximum. However the –ve voltage on he horizontal plate is decreased.

Therefore the beam is deflected upwards by the signal and towards the right by

the saw tooth wave. The result is that the beam now strikes the screen at point 2.

On similar reasoning, the beam strikes the screen at points 3,4 and 5. Therefore

exact signal pattern appears on the screen.

Various controls on CRO

In order to facilitate the proper functioning of CRO, various controls are provided

on the front panel of the CRO.

1. Intensity Control-The knob of intensity control regulates the bias on the control

grid and affects the electron beam intensity.If the negative bias on the grid is

increased, the intensity of electron beam is decreased, thus reducing the

brightness of the spot.

2

1 3 5

4

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2. Focus Control- It regulates the positive potential on the focusing anode. If the

positive potential on this anode is increased, the electron beam becomes quite

narrow and the spot on the screen is a pin-point.

3. Vertical position control- The knob of vertical position control regulates the

amplitude of d.c. potential which is applied to the vertical deflection plates in

addition to the signal. By adjusting this control, the image can be moved up or

down as required.

Applications of CRO

1. Examination of waveforms

2. Voltage measurements

3. Frequency measurements

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UNIT – 7

COMMUNICATION SYSTEMS

Radio Broadcasting, Transmission and Reception

Radio communication means the radiation of radio waves by the transmitting station, the

propagation of these waves through space and their reception by the radio receiver.

Fig. below shows the general principle of radio broadcasting, transmission and reception.

It essentially consists of transmitter, transmission of radio waves and radio receiver.

Receiving

Transmitting Antenna

Antenna

Block diagram of Radio Communication system

Transmitter-

It essentially consists of microphone, audio amplifiers, oscillator and modulator.

A microphone is a device which converts sound waves into electrical waves. The

output of microphone is fed to multistage audio amplifier for raising the strength of weak

signal.

The job of amplification is performed by cascaded audio amplifiers. The

amplified output from the last audio amplifier is fed to the modulator for rendering the

process of modulation.

The function of the oscillation is to produce a high frequency signal called a

carrier wave. Usually crystal oscillator is used for the purpose.

Audio

Amplifiers

Oscillator Modulator Radio

receiver

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The amplified audio signal and carrier waves are fed to the modulator. Here the

audio signal is superimposed on the carrier wave in suitable manner. The resultant waves

are called modulated waves, and the process is called modulation. The process of

modulation permits the transmission of audio signal at the carrier signal (frequency). As

the carrier frequency is very high, therefore the audio signal can be transmitted to large

distances. The radio waves from the transmitter are fed to the transmitting antenna or

aerial from where these are radiated into space.

The transmitting antenna radiates the radio waves in space in all directions. These

radio waves travel with the velocity of light 3x108m/sec. The radio waves are

electromagnetic waves and possess the same general properties.

Receiver-

On reaching the receiving antenna, the radio waves induce tiny emf in it. This

small voltage is fed to the radio receiver. Here the radio waves are first amplified and

then signal is extracted from them by the process of demodulation. The signal is

amplified by audio amplifiers and then fed to the speaker for reproduction into sound

waves.

Need for modulation

1. Practical Antenna length- Theory shows that in order to transmit a wave effectively

the length of the transmitting antenna should be approximately equal to the

wavelength of the wave.

metresHzfrequencyfrequency

Velocitywavelength

)(

103 8

As the audio frequencies range from 20 Hz to 20 KHz, if they are transmitted

directly into space, the length of the transmitting antenna required would be extremely

large. For example to radiate a frequency of 20 KHz directly into space we would need

an antenna length of 3x108 /20x10

3 ≈ 15,000 meters. This is too long to be constructed

practically. However, if the audio-signals are transmitted, after modulation using carrier

waves, at 1MHz frequency, the height of the antennas needed is only 75 meters.

2. Operating Range- The energy of a wave depends upon its frequency. The greater

the frequency of the wave, the greater the energy possessed by it. As the audio signal

frequencies are small, therefore these cannot be transmitted over large distances if

radiated directly into space.

3. Wireless communication- Radio transmission should be carried out without wires.

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4. Audio signals (20 Hz to 20 KHz) cannot be directly transmitted over long distances,

because it would require enormous power.

5. If several audio frequencies within 20 KHz, are transmitted directly without

modulation, then they get mixed-up, and proper reception of the signals at the

receiving end would pose problem. On the other hand, if these signals are transmitted

at radio frequency with due modulation, separation of the signals at the receiving end

is quite easy, with the use of proper tuned circuits.

The process of controlling the radio-frequency carrier wave by altering its

amplitude or frequency in accordance with the strength of an audio signal is called as

modulation.

Types:-

1. In amplitude modulation, the amplitude of the carrier wave is altered in

accordance with the strength of the modulating signal.

2. In frequency modulation, the frequency of the carrier wave is altered in

accordance with the strength of the modulating signal.

The process of getting back the modulating signal from the modulated carrier

wave is termed as demodulation or detection.

Amplitude modulation

When the amplitude of high frequency carrier wave is changed in accordance with

the intensity of the signal, it is called amplitude modulation.

The following points are to be noted in amplitude modulation.

1. The amplitude of the carrier wave changes according to the intensity of the signal.

2. The amplitude variation of the carrier wave is at the signal frequency fS.

3. The frequency of the amplitude modulated wave remains the same i.e., carrier

frequency fC.

Modulation factor

The ratio of change of amplitude of carrier wave to the amplitude of normal

carrier wave is called modulation factor.

M = {amplitude change of carrier wave} / {normal carrier wave (unchanged)}

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vm

t

vc

t

vAM

t

AM waveforms

A

+ No signal =

Carrier m = 0/A = 0%

2A

+ =

Carrier Signal

m = (2A-A)/A = 1

Illustration of modulation factor (modulation index)

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Modulation factor is very important since it determines the strength and quality of

the transmitted signal. The greater the degree of modulation, the stronger and clearer will

be the audio signal. It should be noted that if the carrier is over modulated (i.e., m > 1)

distortion will occur at reception.

Analysis of amplitude modulated wave

Signal

mVc

Vc Vc

Carrier

AM Wave

A carrier wave is represented by, vc = Vc sin wct

A modulating signal is represented by, vm = Vm sin wmt

Note that Vm < Vc

Where vc ------instantaneous voltage of carrier,

Vm____ instantaneous voltage of signal,

Vc____ amplitude of carrier,

Vm____ amplitude of modulating signal.

In amplitude modulation, the amplitude Vc of the carrier wave is varied in

accordance with the instantaneous amplitude vm (= Vm sin wmt) of the modulating signal.

Let A be the amplitude of the modulated wave. It is evident that A = VC + vm, since VC is

made proportional to instantaneous value vm of the modulating signal. Hence,

A = Vc + Vm sin wmt (since, vm = Vm sin wmt)

=

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A =

Now, the instantaneous value of the amplitude modulated wave vAM is given by;

vAM

)1()cos(2

)cos(2

cos

])cos()[cos(2

sin

]sinsin2[2

sin

sinsinsin

sin)sin1(

sin

twwmV

twwmV

twV

twwtwwmV

twV

twtwmV

twV

twtwmVtwV

twtwmV

twA

mcc

mcc

cC

mcmcc

cC

cmC

cC

cmCcC

cmC

c

The AM wave is equivalent to the summation of three sinusoidal waves: one

having amplitude Vc and frequency fc, the second having amplitude mVc/2 and

frequency (fc + fm) and the third having amplitude mVc/2 and frequency fc – fm.

The AM wave consists three frequencies viz, fc, fc + fm and fc – fm. The first

frequency is the carrier frequency. Thus the process of modulation does not

change the original carrier frequency but produces two new frequencies fc + fs

and fc – fs. which are called side-band frequencies.

In amplitude modulation the bandwidth is from fc – fm. to fc + fm i.e., 2fm i.e.,

twice the signal frequency.

Frequency spectrum of an amplitude modulated wave is shown in figure below.

` Vc

mVc/2

fC-fm fC fC+fm frequency

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Transistor AM modulator

A circuit which does amplitude modulation is called AM modulator.

Fig. above shows the circuit of a simple AM modulator. It is essentially a CE

amplifier having a voltage gain of A. The carrier signal is the input to the amplifier. The

modulating signal is applied in the emitter resistance circuit.

The amplifier circuit amplifies the carrier by a factor “A” so that the output is Aec.

Since the modulating signal is part of the biasing circuit it produces low-frequency

variations in the circuit. This in turn causes variations in “A”. The result is that the

amplitude of the carrier varies in accordance with the strength of the signal. The

amplitude modulated output is obtained across RL.

Power in AM wave

We have, from equation (1);

twwmV

twwmV

twVv mcc

mcc

cCAM )cos(2

)cos(2

cos

Hence,

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)5(12,mod,

)4(2

1,

2

2

2

21

242

)3(4

2222

)2(2

2

2

22

22222

22

22

2

2

C

T

CT

C

CCC

SCT

C

CC

S

C

c

C

P

Pmexulationindor

mPPor

m

R

V

m

R

V

R

Vm

R

V

PPPwaveAMofpowerTotal

R

Vm

R

mV

R

mV

PsidebandsofpowerTotal

R

V

R

V

PpowerCarrier

Transmission Efficiency of AM

)6(%100*22)4(

)3(2

2

2

2

m

m

m

m

equn

equn

P

P

T

S

Vc + Vm sin wmt

Equation for Modulation Index Vm

Vmin Vmax

Vc Vm

- (Vc + Vm sin wmt)

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From the figure shown above;

We know that; the modulation index

Limitations of Amplitude Modulation

1. Noisy Reception- In an AM wave, the signal is in the amplitude variations of the

carrier. Practically all the natural and man-made noises consist of electrical

amplitude disturbances. As a radio receiver cannot distinguish between amplitude

variations that represent noise and those that contain the desired signal. Therefore

reception is very noisy.

2. Low efficiency- In AM useful power is in the sidebands as they contain the

signal. An AM wave has low sideband power.

For example even if modulation is 100 % ie m=1.

33.012

1

2 2

2

m

m

p

P

T

S or, PS=33% of PT

Sideband power is only one-third of the total power of AM wave. Hence

efficiency of this type of modulation is low.

3. Lack of audio quality- In order to attain high fidelity reception, all audio

frequencies up to 15 KHz must be reproduced. This necessitates a bandwidth of

30 KHz since both sidebands must be reproduced (2fs). But AM broadcasting

stations are assigned with bandwidth of only 10 KHz to minimize the interference

from adjacent broadcasting stations. This means that the highest modulating

frequency can be 5 KHz which is hardly sufficient to reproduce the music

properly.

Frequency modulation

“When the frequency of carrier wave is altered in accordance with the intensity of

the signal, it is called frequency modulation”.

Here the amplitude of the modulated wave remains the same. i.e., carrier wave

amplitude.

The frequency variations of carrier wave depend upon the instantaneous

amplitude of the signal.

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When the signal approaches positive peaks as the B and F, the carrier frequency is

increased to maximum and during negative peak; the carrier frequency is reduced

to minimum as shown by widely spaced cycles.

Signal

b f

a c e g t

d

t

Carrier

t

FM wave

Let the un-modulated carrier wave be represented as; vc = Vc cos (wct + φ)

Where, Vc = amplitude of the carrier wave,

wc = angular frequency of the carrier, and

Φ = the phase of the carrier wave.

When a sinusoidal modulating wave is superimposed on the carrier wave, the

frequency of the carrier wave changes – the change in the carrier frequency is termed as

frequency deviation, and is denoted as Δf.

Let vm = Vm cos wmt be represent the modulating signal.

The instantaneous angular frequency of the modulated wave is given as;

w = wc + wcKvm where, K is a constant.

= wc + wcKVm cos wmt

= wc[1 + KVm cos wmt]

Putting w = 2∏f and wc = 2∏fc, the above expression becomes;

f = fc[1 + KVm cos wmt]

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Also, for maximum frequency deviation, i.e., when cos wmt = 1; we have,

f = fc[1 KVm]

From the above expression, we get the frequency deviation Δf = KVmfc

The modulated signal may be represented in the general form as, vFM = Vc cos θ

where, Vc = amplitude of the carrier wave,

θ = an angle to be evaluated.

It is obvious that θ depends upon the angular frequencies wc and wm.

i.e., θ = F(wc, wm)

Hence, we have; θ =

=

=

=

Or, θ =

Putting KVmfc = Δf; θ =

Where the ratio is termed as modulation index, and is denoted as mf. It is defined as

the ratio of maximum frequency deviation to the frequency of the modulating signal.

Therefore; θ =

Now, the modulated carrier wave is represented as; vFM = Vc cos θ

Or, vFM = Vc cos

Band-width (BW)

In AM, there are only two side-bands. But, in FM, we can show that, there are

infinite side-bands in the frequency spectrum of FM carrier-wave.

The band-width of a FM wave is usually obtained on the basis of Carsen’s Rule;

given as – (BW)FM = 2(Δf + fm) = 2(mf + 1)fm

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Advantages of Frequency Modulation:-

1. The amplitude of FM wave is constant; and hence, the transmitted power is also

constant; independent of modulation depth. Hence, low level modulation can be

employed in FM transmitters.

2. All the transmitted power is useful in FM.

3. FM reception is more immune to noise as amplitude limiter circuits are used.

4. The signal-to-noise ratio can be increased in FM by increasing the frequency

deviation.

5. Adjacent channel interference is less in FM, as guard band is provided in

commercial FM transmitters.

6. Since FM operates in VHF & UHF range, the propagation is line of sight

propagation by space wave. Also, the noise effect will be less in this range.

Disadvantages of Frequency Modulation:-

1. A much larger bandwidth is required for the FM transmission.

2. The service area of FM transmitter is much less as FM reception is limited to

radio horizon up to line of sight.

3. FM transmitter & receiver are complicated.

Comparison – AM & FM

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Demodulation

The process of recovering the audio signal from the modulated wave is known as

demodulation or detection.

At the broadcasting station, modulation is done to transmit the audio signal over

larger distances. When the modulated wave is picked up the receiver, it is necessary

to recover the audio signal from it. This process is accomplished in the radio receiver

and is called demodulation.

AM diode detector

The fig. below shows a simple diode detector employing a diode and a filter

circuit. A detector circuit performs the following two functions.

1. It rectifies the modulated wave.

2. It separates the audio signal from the carrier.

Speaker

Audio output

Rectified

Wave

AM Wave

The modulated wave of desired frequency is selected by the parallel tuned circuit

L1C1 and is applied to the diode. During positive half cycles of the modulated

wave the diode conducts, while during negative half cycles it does not. The result

is the output of diode consists of positive half cycle of modulated wave as shown

in figure.

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The rectified output consists of r.f. component and the audio signal which cannot

be fed to the speaker for sound reproduction. The r.f. component is filtered by the

capacitor „C‟ shunted across the speaker. The value of „C‟ is large enough to

present low reactance to the r.f. component. fc+fs Therefore signal is passed to the

speaker.

AM Radio Receiver In order to reproduce the AM wave into sound waves, every radio receiver must

perform the following functions.

1. The receiving aerial must intercept a portion of the passing radio waves.

2. The radio receiver must select the desired radio from a number of radio

waves intercepted by the receiving aerial. For this purpose tuned parallel

LC circuits must be used. These circuits will select only that radio

frequency which is resonant with them.

3. The selected radio wave must be amplified by the tuned frequency

amplifiers.

4. The audio signal must be recovered from the amplified radio wave.

5. The audio signal must be amplified by suitable number of audio-

amplifiers.

6. The amplified audio signal should be fed to the speaker for sound

reproduction.

Types of AM radio receivers

1. Straight Radio receiver

2. Superhetrodyne radio receiver

Straight Radio Receiver

Receiving antenna

RF amplifier

The Receiving antenna is receiving radio waves from different broadcasting

stations. The desired radio wave is selected by the tuned RF amplifier which

Detector

AF

amplifier

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C3 L3

c2

employs tuned parallel circuit. The selected radio wave is amplified by the rf

amplifier.

The amplified radio wave is fed to the detector circuit. This circuit extracts the

audio signal from the radio wave. The output of the detector is the audio signal

which is amplified by one or more stages of audio-amplifications. The amplified

audio signal is fed the speaker for sound reproduction.

Limitations-

1. In straight radio receivers, tuned circuits are used. As it is necessary to change the

value of variable capacitors (gang capacitors) for tuning to the desired station,

there is a considerable variation of Q between the closed and open positions of the

variable capacitors. This changes the sensitivity and selectivity of the radio

receivers.

2. There is too much interference of adjacent stations.

Superhetrodyne Receiver

Here the selected radio frequency is converted to a fixed lower value called

intermediate frequency (IF). This is achieved by special electronic circuit called mixer

circuit. The production of fixed intermediate frequency (455 KHz) is an important feature

of superhetrodyne circuit. At this fixed intermediate frequency, the amplifier circuit

operates with maximum stability, selectivity and sensitivity.

The block diagram of superhetrodyne receiver is a shown in figure below.

Receiving antenna

RF amplifier Mixer 455KHz

C1 L1 L2

Speaker

Ganged to a common shaft Local Oscillator

Detector

AF amplifier

IF

Amplifier

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Working: The following are the stages involved in the working of superheterodyne

receiver.

1. RF amplifier stage- The RF amplifier stage uses a tuned parallel circuit L1C1

with a variable capacitor C1. The radio waves from various broadcasting stations

are intercepted by the receiving aerial and are coupled to this stage. This stage

selects the desired radio wave and raises the strength of the wave to the desired

level.

2. Mixer stage- The amplified output of RF amplifier is fed to the mixer stage

where it is combined with the output of a local oscillator. The two frequencies

beat together and produce an intermediate frequency (IF).

IF= Oscillator frequency –radio frequency

The IF is always 455 KHz regardless of the frequency to which the

receiver is tuned. The reason why the mixer will always produce 455 KHz

frequency above the radio frequency is that oscillator always produces a

frequency 455 KHz above the selected frequency. In practice, capacitance of C3 is

designed to tune the oscillator to a frequency higher than radio frequency by 455

KHz.

3. IF amplifier stage- The output of mixer is always 455 KHz and is fed to fixed

tuned IF amplifiers. These amplifiers are tuned to one frequency (i.e., 455 KHz).

4. Detector stage- The output from the last IF amplifier stage is coupled to the input

of the detector stage. Here the audio signal is extracted from the IF output.

Usually diode detector circuit is used because of its low distortion and excellent

audio fidelity.

5. AF amplifier stage- The audio signal output of detector stage is fed to a

multistage audio amplifier. Here the signal is amplified until it is sufficiently

strong to drive the speaker. The speaker converts the audio signal into sound

waves corresponding to the original sound at the broadcasting station.

Advantages of Superhetrodyne Circuit –

1. High RF amplification

2. Improved selectivity-losses in the tuned circuits are lower at intermediate

frequency. Therefore the quality factor Q of the tuned circuits is increased. This

makes amplifier circuits to operate with maximum selectivity.

3. Lower cost.

Radio Telegraphy & Radio Telephony

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NUMBER SYSTEMS

Comparison between Analog (Linear) IC’s & Digital IC’s:-

Analog IC / Electronics Digital IC / Electronics

1. The inputs & outputs can assume

continuously varying values. 1. The input & output voltages

can take only two possible

values – a low voltage & a

high voltage. 2. Outputs are proportional to

inputs. 2. Output need not be

proportional to input. 3. With reference to a load line,

adjacent points on the load line

are used, so that the output

voltage is continuous.

3. Only two non-adjacent

points on the load line,

preferably cut-off &

saturation, are chosen for

operation. 4. Continuous variation of output

results if the input is sinusoidal. 5. Digital operation results if

the input is a square wave.

Binary Number System:-

Digital operation implies two stages – a high state or a low state of

voltage.

A high voltage is represented by a 1, and a low voltage is represented

by a 0.

Since only two states – a high voltage level & a low voltage level – are

involved, the operation is binary in nature.

The digits 0 & 1 are accordingly termed as binary digits or bits.

Some examples of binary operation are –

– A mechanical switch: Either closed or open,

– A lamp: Either lighted or dark,

– A statement: Either true or false,

– Punched cards: A hole in the card stands for 1 & absence of hole

represents 0,

– The presence or absence of a signal.

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Decimal, Binary, Octal & Hexadecimal Numbers:-

Binary to Decimal Conversion:-

In decimal system, any number can be expressed as a string of ten

digits – 0 through 9. Since, the system uses 10 digits, it is called base

10 system.

Similarly a binary system uses only two symbols or bits – 0 & 1.

hence, it is termed as base 2 system.

In any system, each digit has two values – its intrinsic value & the

positional value.

For the base 10 system, the positional values of the digits are – … 104

,

103

, 102

, 101

, 100

, 10-1

, 10-2

, 10-3

, 10-4

…

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For the base 2 system, the positional values of the digits are – … 24

,

23

, 22

, 21

, 20

, 2-1

, 2-2

, 2-3

, 2-4

…

Each digit is multiplied with its positional value & the sum of all the

products gives the equivalent decimal number.

Examples:-

1. Consider the decimal number 34568. we have;

34568 = 3 * 104

+ 4 * 103

+ 5 * 102

+ 6 * 101

+ 8 * 100

.

= 30,000 + 4,000 + 500 + 60 + 8.

2. Consider the binary number 1110. We have;

1110 = 1 * 23

+ 1 * 22

+ 1 * 21

+ 0 * 20

.

Problems:-

1) Show that 110112 = 27

10.

110112 = 1 * 2

4

+ 1 * 23

+ 0 * 22

+ 1 * 21

+ 1 * 20

.

= 16 + 8 + 0 + 2 + 1 = 27.

2) Convert binary 0.11011 to equivalent decimal fraction.

0.110112 = 1 * 2

-1

+ 1 * 2-2

+ 0 * 2-3

+ 1 * 2-4

+ 1 * 2-5

.

= ½ + ¼ + 0 + 1/16 + 1/32 = 27/32 = 0.84375.

3) Convert binary 111011.1011 to equivalent decimal fraction.

111011.10112 = (1 * 2

5

+ 1 * 24

+ 1 * 23

+ 0 * 22

+ 1 * 21

+ 1 * 20

)

+

(1 * 2-1

+ 0 * 2-2

+ 1 * 2-3

+ 1 * 2-4

).

= (32 + 16 + 8 + 0 + 2 + 1) + (1/2 + 0 + 1/8 + 1/16).

= (59) + (11/16) = 59.687510

.

Homework:-

1. Show that 1011012 = 45

10.

2. Show that 11102 = 14

10.

3. Convert binary 0.1011 to equivalent decimal fraction.

4. Convert 10000011.1001 into equivalent decimal number.

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Decimal to Binary Conversion:-

(A) Integers:-

The given decimal number (i.e., integer) is repeatedly divided by 2,

which is the base number of binary system, until the remainder

becomes 0 or 1.

The string of remainders from bottom to the top is the binary

equivalent of the decimal integer.

(B) Fractions:-

In order to convert a decimal fraction to its binary equivalent, the

fraction is repeatedly multiplied by 2 (i.e., divided by 2-1

) until the

fraction becomes zero, or to the desired number of places after the

binary point.

The string of integers obtained from top to the bottom gives the

equivalent fraction in the binary number system.

Problems:-

1) Convert decimal 127 into equivalent binary number.

2) Find the binary equivalent of the decimal fraction 0.375.

3) Convert 131.562510

into equivalent binary number.

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Homework:-

1. Convert decimal 64 into equivalent binary number.

2. Convert decimal 43 into equivalent binary number.

3. Convert the decimal fraction 0.4375 to its binary equivalent.

4. Convert 5.410

to equivalent binary number.

5. If 409610

= X2, find X.

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Binary Addition:-

Basic rules of binary addition are –

a) 0 + 0 = 0

b) 0 + 1 = 1

c) 1 + 0 = 1

d) 1 + 1 = 10 (0, with carry 1)

Problems:-

1) Add 1011 and 1110. 2) Add 101.10 and 111.11.

Homework:-

1. Add (a) 101101 and 110101, & (b) 101.10 and 111.11.

2. Convert decimal numbers 25 and 37 to equivalent binary numbers,

add them, and express the sum in binary form. Check your answer.

Binary Subtraction:-

I method:- The sign of the number to be subtracted (i.e., subtrahend) is

changed & it is added to the other number (i.e., minuend).

Example:-

1) 11 1011 2) 6.625 110.101

- 07 - 0111 - 4.875 - 100.111

04 0100 1.750 001.110

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II method [2’s Compliment]:-

Case 1: If the subtrahend is less than the minuend: The 2’s

compliment of the subtrahend is added to the minuend, and the carry

at the leftmost position (i.e., the most significant digit) is disregarded.

Example:-

{ 10’s Complement }

1) 49 – 24: 49

+ 76 Replace 24 by 10’s compliment (99 – 24 + 1).

1 25 Ignore the carry.

{ 2’s Complement }

10102 – 111

2: 1000 1’s compliment of 111.

+ 1 Add 1 to get 2’s compliment.

1001

+ 1010 Add the minuend.

1 0011 Ignore the carry.

II method [2’s Compliment]:-

Case 2: If the subtrahend is greater than the minuend: The 2’s

compliment of the subtrahend is added to the minuend. Now, take 2’s

compliment of the sum obtained & assign the – sign.

Example:-

{ 10’s Complement }

1) 24 – 49: 24

+ 51 Replace 49 by 10’s compliment (99 – 49 + 1).

0 75 Take 10’s compliment & assign – sign.

- 25 Answer

{ 2’s Complement }

1112 – 1010

2: 0101 1’s compliment of subtrahend.

+ 1 Add 1 to get 2’s compliment.

0110

+ 0111 Add the minuend.

1101 Take 2’s compliment & assign – sign.

- 0011 Answer.

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III method [1’s Compliment]:-

Case 1:If the subtrahend is less than the minuend: The 1’s

compliment of the subtrahend is added to the minuend, and the carry

1 at the extreme left is carried to the extreme right, and it is added.

Example:-

{ 9’s Compliment }

1) 49 – 24: 49

+ 75 Replace 24 by 9’s compliment: (99 – 24).

1 24

+ 1 Add the carry here.

25

{ 1’s Compliment }

10102 – 111

2: 1000 1’s compliment of 111.

+ 1010 Add the minuend.

1 0010

+ 1 Add the carry here.

0011

III method [1’s Compliment]:-

Case 2:If the subtrahend is greater than the minuend: The 1’s

compliment of the subtrahend is added to the minuend, and then get

the 1’s compliment of sum, and assign a – sign to it.

Example:-

{ 9’s Compliment }

1) 24 – 49: 24

+ 50 Replace 49 by 9’s compliment: (99 – 49).

0 74 Compliment the sum & assign - sign, since the

- 25 subtrahend is greater.

{ 1’s Compliment }

1112 – 1010

2: 0101 1’s compliment of 1010.

+ 0111 Add the minuend.

0 1100 Compliment the sum & assign – sign, since

- 0011 the subtrahend is greater.

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Problems:-

1) Subtract 100100 from 1010010 in 2’s & 1’s compliment methods.

1011011 1’s compliment of 100100.

+ 1 Add 1 to get the 2’s compliment.

1011100

+1010010 Add the minuend.

1 0101110 Ignore the carry. { 2’s compliment method }

1011011 1’s compliment of 100100.

+ 1010010 Add the minuend.

1 0101101

+ 1 Add the carry generated.

0101110 { 1’s compliment method }

2) Subtract 1010010 from 100100 in 2’s & 1’s compliment methods.

0101101 1’s compliment of 1010010 (subtrahend).

+ 1 Add 1 to get the 2’s compliment.

0101110

+ 0100100 Add the minuend.

1010010 Take 2’s compliment & assign – sign.

- 0101110 Answer. {2’s compliment}

0101101 1’s compliment of 1010010 (subtrahend).

+ 0100100 Add the minuend.

1010001 Take 1’s compliment & assign - sign,

- 0101110 since subtrahend is greater. {1’s compliment}

Note:-

1’s Compliment Method:-

o If Carry is 1: Add it to LSB.

o If Carry is 0: Take 1’s compliment & Assign –ve sign.

2’s Compliment Method:-

o If Carry is 1: Ignore that.

o If Carry is 0: Take 2’s compliment & Assign –ve sign.

To Convert from Base N to Decimal:-

o Multiply Position Value.

To Convert from Decimal to Base N:-

o Divide by N.

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Homework:-

1. Subtract 36 from 82, putting both numbers in binary form & check the

result.

2. Subtract 246 from 213 by expressing them in binary form & check

your answer.

3. Find, using 2’s compliment method, 11112 – 101

2.

Octal to Decimal Conversion:-

In decimal system, any number can be expressed as a string of ten

digits – 0 through 9. Since, the system uses 10 digits, it is called base

10 system.

Similarly a octal system uses only eight symbols – 0 through 7.

In any system, each digit has two values – its intrinsic value & the

positional value.

For the base 10 system, the positional values of the digits are – … 104,

103, 102, 101, 100, 10-1, 10-2, 10-3, 10-4 …

For the octal system, the positional values of the digits are – … 84, 83,

82, 81, 80, 8-1, 8-2, 8-3, 8-4 …

Each digit is multiplied with its positional value & the sum of all the

products gives the equivalent decimal number.

Problems:-

1) Convert 158 to decimal.

158 = 1 * 81 + 5 * 80.

= 8 + 5 = 1310.

2) Convert octal 15.6 into equivalent decimal number.

15.68 = (1 * 81 + 5 * 80 ) + (6 * 8-1 ).

= (8 + 5) 0 + 6/8 = 13.7510.

Homework:-

1. Find the decimal equivalent of octal number 72.

2. Convert the octal number 2376 into decimal number.

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Decimal to Octal Conversion:-

(A) Integers:-

The given decimal number (i.e., integer) is repeatedly divided by 8,

which is the base number of octal system, until the remainder

becomes 0 through 7.

The string of remainders from bottom to the top is the binary

equivalent of the decimal integer.

(B) Fractions:-

In order to convert a decimal fraction to its octal equivalent, the

fraction is repeatedly multiplied by 8 (i.e., divided by 8-1), until the

fraction becomes zero, or to the desired number of places after the

binary point.

The string of integers obtained from top to the bottom gives the

equivalent fraction in the octal number system.

Problems:-

1) Convert 47 to equivalent octal number.

2) Convert the decimal number 4429.625 into equivalent octal number.

Homework:-

1. Convert 93210 to equivalent octal number.

2. If 33210 = X8, find X.

3. If 632.9710 = X8, what is the value of X.

Hexadecimal to Decimal Conversion:-

In decimal system, any number can be expressed as a string of ten

digits – 0 through 9. Since, the system uses 10 digits, it is called base

10 system.

Similarly a hexadecimal system uses sixteen symbols – 0 through 9 &

A through F.

In any system, each digit has two values – its intrinsic value & the

positional value.

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For the base 10 system, the positional values of the digits are – … 104,

103, 102, 101, 100, 10-1, 10-2, 10-3, 10-4 …

For the hexadecimal system, the positional values of the digits are – …

164, 163, 162, 161, 160, 16-1, 16-2, 16-3, 16-4 …

Each digit is multiplied with its positional value & the sum of all the

products gives the equivalent decimal number.

Problems:-

1) Convert 1516 to decimal.

1516 = 1 * 161 + 5 * 160.

= 16 + 5 = 2110.

2) Convert B816 to decimal.

B816 = B * 161 + 8 * 160.

= 11 * 16 + 8 = 18410.

3) Convert 9.1A16 into equivalent decimal number.

9.1AH = (9 * 160 ) + (1 * 16-1 + A * 16-2).

= (9) + (1/16 + 10/256) = 9.101562510.

Homework:-

1. Find the decimal equivalent of AB416 & B6C7H.

2. Convert into decimal equivalent of FACEH & CAD.BFH.

Decimal to Hexadecimal Conversion:-

(A) Integers:-

The given decimal number (i.e., integer) is repeatedly divided by 16,

which is the base number of hexadecimal system, until the remainder

becomes 0 through 15.

The string of remainders from bottom to the top is the binary

equivalent of the decimal integer.

(B) Fractions:-

In order to convert a decimal fraction to its hexadecimal equivalent,

the fraction is repeatedly multiplied by 16 (i.e., divided by 16-1), until

the fraction becomes zero, or to the desired number of places after the

binary point.

The string of integers obtained from top to the bottom gives the

equivalent fraction in the octal number system.

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Problems:-

1) Convert 24 to equivalent hexadecimal number.

Homework:-

1. Convert 51310 to equivalent hexadecimal number.

2. If 134310 = XH, find X.

3. Convert 2604.10546875 to hexadecimal number.

Addition & Subtraction in Octal Arithmetic:-

Addition:-

The sum of two octal digits is the same as their decimal sum, provided

the decimal sum is less than 8.

If the decimal sum is 8 or greater, subtract 8 to obtain the octal digit.

A carry of 1 is produced when the decimal sum is corrected this way.

Problems:-

1) Add (a) 6478 and 5668, & (b) 27.348, and 11.768.

6 4 7 2 7 . 3 4

+ 5 6 6 + 1 1 . 7 6

1 4 3 58 4 1 . 3 28

Subtraction:-

1. Using 8’s Compliment:-

Find the 8’s compliment of the subtrahend & then add it to the

minuend.

If the carry is produced, discard it & the answer is positive.

If the carry is 0, find the 8’s compliment of the sum & assign a – sign.

2. Using 7’s Compliment:-

Find the 7’s compliment of subtrahend & then add it to the minuend.

If carry is produced in the addition, add carry in the LSB of the sum.

If the carry is zero, find the 7’s compliment of the sum & assign –

sign.

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Problems:-

1) 3648 – 1268 .

6 5 2 8’s compliment of 126 (777-126+1).

+ 3 6 4 Add to the minuend.

1 2 3 68 Discard the carry.

2) 34.228 – 417.548.

3 6 0 . 2 4 8’s compliment of 417.54 (777-417.54+1).

+ 0 3 4 . 2 2 Add to the minuend.

4 1 4 . 4 68 No carry – hence, take 8’s compliment.

7 7 7 . 0 0

- 4 1 4 . 4 6

3 6 2 . 3 2 + 1 = - 3 6 3 . 3 2

Problems:-

1) 3648 – 1268 .

6 5 1 7’s compliment of 126 (777-126).

+ 3 6 4 Add to the minuend.

1 2 3 58 Add the carry generated.

+ 1

2 3 6

2) 34.228 – 417.548.

3 5 7 . 2 4 7’s compliment of 417.54 (777-417.54).

+ 0 3 4 . 2 2 Add to the minuend.

4 1 3 . 4 68 No carry – hence, take 7’s compliment.

- 3 6 3 . 3 28 (777.00 - 413.46 = 363.32)

Homework:-

1. Add (a) 1678 and 3258, & (b) 3418, 1258, 4728, and 5778.

2. (a) 2418 – 6538, & (b) 250.458 – 415.158.

Addition & Subtraction in Hexadecimal Arithmetic:-

Addition:-

The sum of two hexadecimal digits is the same as their decimal sum,

provided the decimal sum is less than 16.

If the decimal sum is 16 or greater, subtract 16 to obtain the octal

digit.

A carry of 1 is produced when the decimal sum is corrected this way.

BASIC ELECTRONICS

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Problems:-

1) Add (a) 3F7H and 5B4H.

3 F 7

+ 5 B 4

9 A BH

Subtraction:-

1. Using 16’s Compliment:-

Find the 16’s compliment of the subtrahend & then add it to the

minuend.

If the carry is produced, discard it & the answer is positive.

If the carry is 0, find the 16’s compliment of the sum & assign a –

sign.

2. Using 15’s Compliment:-

Find the 15’s compliment of subtrahend & then add it to the minuend.

If carry is produced in the addition, add carry in the LSB of the sum.

If the carry is zero, find the 15’s compliment of the sum & assign –

sign.

Problems:-

1) C9B4H – AC4FH.

5 3 B 1 16’s compliment of AC4F (FFFF-AC4F+1).

+ C 9 B 4 Add to the minuend.

1 D 6 5H Discard the carry if any.

2) B4D.A2H – 7C9.EDH.

8 3 6 . 1 3 16’scompliment of 7C9.ED (FFFF-7C9.ED+1).

+ B 4 D. A 2 Add to the minuend.

1 3 8 3 . B 5H Ignore carry.

Problems:-

1) C9B4H – AC4FH.

5 3 B 0 15’s compliment of AC4F (FFFF-AC4F).

+ C 9 B 4 Add to the minuend.

1 1 D 6 4H Add the carry to LSB.

1 D 6 5H

2) B4D.A2H – 7C9.EDH.

8 3 5 . 1 3 15’scompliment of 7C9.ED (FFFF-7C9.ED).

+ B 4 D. A 2 Add to the minuend.

1 3 8 2 . B 5H Add the carry to the LSB.

3 8 3. B 5H

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Homework:-

1. 4D7H – B9EAH.

Conversion from Binary to Octal:-

The binary digits are arranged in groups of three (on both sides of

decimal point).

Each group of binary digit is replaced by its octal equivalent.

Problem:-

1) Convert 111101101110.1001111012 to octal.

111 101 101 110 . 100 111 1012 = 7 5 5 6 . 4 7 58.

Conversion from Octal to Binary:-

Each of octal digits is replaced by the equivalent group of three binary

digits.

Problem:-

1) Convert 3146.528 to binary.

3 1 4 6 . 5 2 8 = 011 001 100 110 . 101 0102.

Conversion from Binary to Hexadecimal:-

The binary digits are arranged in groups of four (on both sides of

decimal point).

Each group of binary digit is replaced by its hexadecimal equivalent.

Problem:-

1) Convert 111.112 to hexadecimal.

0111 . 11002 = 7 . C16

Conversion from Hexadecimal to Binary:-

Each hexadecimal digits are replaced by the equivalent group of four

binary digits.

Problem:-

1) Convert AA.1A16 to binary.

A A . 1 AH = 1010 1010 . 0001 10102

Binary Coded Decimal (BCD): - Self Study.

By – Mahesh Prasanna K.,

Dept. of E & C,

AIET, Moodbidri.