Basic Electronics

5/21/2018 Basic Electronics

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e-Notes by Prof.K.R.Sudhindra, BMS College of Engineering, Bangalore

BASIC EEC!R"NICS

Sub#e$t Code% EN-&'()' IA *ar+s% )'

ours er ee+ % /0 E1a* ours /2!otal rs% ') E1a* Mar+s% &//

CAP!ER &

C"N34C!I"N IN SEMIC"N34C!"RS

Electrons and holes in an intrinsic semiconductors, conductivity of a semiconductor,carrier concentrations in an intrinsic semiconductor, donor and acceptor impurities,

charge densities in a semiconductor, Fermi level in a semiconductor having impurities,diffusion, carrier life time, Hall effect. 05 Hrs.

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.

Ato*i$ Stru$ture

Atom is the basic building bloc of all the elements. !t consists of the centralnucleus of positive charge around which small negatively charged particles calledelectrons revolve in different paths or orbits.

An Electrostatic force of attraction between electrons and the nucleus holds upelectrons in different orbits.

Electrostatic force.

"#entrifugal force.

Figure$.$. Atomic structure

$


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%ucleus is the central part of an atom and contains protons and neutrons. A protonis positively charged particle, while the neutron has the same mass as the proton,but has no charge. &herefore ,nucleus of an atom is positively charged.

ato*i$ eight 5 no. of rotons 6 no. of neutrons

An electron is a negatively charged particle having negligible mass. &he chargeon an electron is e'ual but opposite to that on a proton. Also the number ofelectrons is e'ual to the number of protons in an atom under ordinary conditions.&herefore an atom is neutral as a whole.

ato*i$ nu*ber 5 no. of rotons or ele$trons in an ato*

&he number of electrons in any orbit is given by (n(where n is the number of theorbit.

For e)ample, ! orbit contains ()$(*( electrons

!! orbit contains ()((* + electrons

!!! orbit contains ()(* $+ electrons and so on

&he last orbit cannot have more than + electrons.

&he last but one orbit cannot have more than $+ electrons.

Positi7e and negati7e ions

-rotons and electrons are e'ual in number hence if an atom loses an electron ithas lost negative charge therefore it becomes positively charged and is referred aspositive ion.

!f an atom gains an electron it becomes negatively charged and is referred to asnegative ion.

8alen$e ele$trons

The electrons in the outermost orbit of an atom are known as valence electrons .

&he outermost orbit can have a ma)imum of + electrons.

&he valence electrons determine the physical and chemical properties of amaterial.

(


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hen the number of valence electrons of an atom is less than /, the material isusually a metal and a conductor. E)amples are sodium, magnesium andaluminium, which have $,( and valence electrons respectively.

hen the number of valence electrons of an atom is more than /, the material is

usually a nonmetal and an insulator. E)amples are nitrogen, sulphur and neon,which have 5,1 and + valence electrons respectively.

hen the number of valence electrons of an atom is / the material has both metaland nonmetal properties and is usually a semiconductor. E)amples are carbon,silicon and germanium.

9ree ele$trons

&he valence electrons of different material possess different energies. &he greater

the energy of a valence electron, the lesser it is bound to the nucleus. !n certain substances, particularly metals, the valence electrons possess so much

energy that they are very loosely attached to the nucleus.

&he loosely attached valence electrons move at random within the material andare called free electrons.

The valence electrons, which are loosely attached to the nucleus, are known as free

electrons.

Energy bands

!n case of a single isolated atom an electron in any orbit has definite energy.

hen atoms are brought together as in solids, an atom is influenced by the forcesfrom other atoms. Hence an electron in any orbit can have a range of energiesrather than single energy. &hese range of energy levels are nown as Energybands.

ithin any material there are two distinct energy bands in which electrons may

e)ist viz 2alence band and conduction band.


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#onduction band

Forbidden gap

2alence band

Energy level

Figure$.( Energy level diagram

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.

Electrons in conduction band are either escaped from their atoms 3free electrons4 or onlywealy held to the nucleus. &hereby by the electrons in conduction band may be easilymoved around within the material by applying relatively small amount of energy. 3eitherby increasing the temperature or by focusing light on the material etc. 4 &his is the reasonwhy the conductivity of the material increases with increase in temperature.

ut much larger amount of energy must be applied in order to e)tract an electron fromthe valence band because electrons in valence band are usually in the normal orbit arounda nucleus. For any given material, the forbidden gap may be large, small or none)istent.

Classifi$ation of *aterials based on Energy band theory

ased on the width of the forbidden gap, materials are broadly classified as conductors,!nsulators and semiconductors.

3a4 #onductor 3b4 !nsulator 3c4 6emiconductor Condu$tors

/

#onductionband

overlap

2alenceband

#onduction band

2alenceband

#onductionband

2alenceband

Forbidden7a E7*$e2

Forbidden 7ap

E7*1e2


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#onductors are those substances, which allow electric current to pass throughthem.

E)ample8 #opper, Al, salt solutions, etc.

!n terms of energy bands, conductors are those substances in which there is no

forbidden gap. 2alence and conduction band overlap as shown in fig 3a4.

For this reason, very large number of electrons are available for conduction evenat e)tremely low temperatures. &hus, conduction is possible even by a very weaelectric field.

Insulators

!nsulators are those substances, which do not allow electric current to passthrough them.E)ample8 9ubber, glass, wood etc.

!n terms of energy bands, insulators are those substances in which the forbiddengap is very large.

&hus valence and conduction band are widely separated as shown in fig 3b4.&herefore insulators do not conduct electricity even with the application of a largeelectric field or by heating or at very high temperatures.

Se*i$ondu$tors

6emiconductors are those substances whose conductivity lies in between that of a

conductor and !nsulator.E)ample8 6ilicon, germanium, #ealenium, 7allium, arsenide etc.

!n terms of energy bands, semiconductors are those substances in which theforbidden gap is narrow.

&hus valence and conduction bands are moderately separated as shown in fig3#4.

!n semiconductors, the valence band is partially filled, the conduction band is alsopartially filled, and the energy gap between conduction band and valence band isnarrow.

&herefore, comparatively smaller electric field is re'uired to push the electronsfrom valence band to conduction band . At low temperatures the valence band iscompletely filled and conduction band is completely empty. &herefore, at verylow temperature a semiconductor actually behaves as an insulator.

Condu$tion in solids

5


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#onduction in any given material occurs when a voltage of suitable magnitude isapplied to it, which causes the charge carriers within the material to move in adesired direction.

&his may be due to electron motion or hole transfer or both.

Ele$tron *otion

Free electrons in the conduction band are moved under the influence of the appliedelectric field. 6ince electrons have negative charge they are repelled by the negativeterminal of the applied voltage and attracted towards the positive terminal.

ole 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.

! ! " "

2 2

3a4 #onductor 3b4 6emiconductor

Flow of electrons Flow of electrons

Flow of current Flow of holes Flow of current

!n a good conductor 3metal4 as shown in fig 3a4 the current flow is due to freeelectrons only.

!n a semiconductor as shown in fig 3b4. &he current flow is due to both holes andelectrons moving in opposite directions.

&he unit of electric current is Ampere 3A4 and since the flow of electric current isconstituted by the movement of electrons in conduction band and holes in valenceband, electrons and holes are referred as charge carriers.

Classifi$ation of se*i$ondu$tors

6emiconductors are classified into two types.

1


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a4 !ntrinsic semiconductors.

b4 E)trinsic semiconductors.

a: Intrinsi$ se*i$ondu$tors

A semiconductor in an extremely pure form is known as Intrinsic

semiconductor.

E)ample8 6ilicon, germanium.

oth silicon and 7ermanium are tetravalent 3having / valence electrons4.

Each atom forms a covalent bond or electron pair bond with the electrons ofneighboring atom. &he structure is shown below.

6ilicon or 7ermanium

2alence electron

#ovalent bond

Figure$.. #rystalline structure of 6ilicon 3or 7ermanium4

At lo te*erature

At low temperature, all the valence electrons are tightly bounded the nucleushence no free electrons are available for conduction.

&he semiconductor therefore behaves as an !nsulator at absolute zerotemperature.

:


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At roo* te*erature

Free electron

2alence electron

Holes

Figure $./. #rystalline structure of 6ilicon 3or 7ermanium4 at room temperature

At room temperature, some of the valence electrons gain enough thermal energyto brea up the covalent bonds.

&his breaing up of covalent bonds sets the electrons free and are available forconduction.

hen an electron escapes from a covalent bond and becomes free electrons avacancy is created in a covalent bond as shown in figure above. 6uch a vacancy is

called Hole. !t carries positive charge and moves under the influence of an electricfield in the direction of the electric field applied.

%umbers of holes are e'ual to the number of electrons since, a hole is nothing butan absence of electrons.

E1trinsi$ Se*i$ondu$tor

hen an impurity is added to an !ntrinsic semiconductor its conductivity changes.

&his process of adding impurity to a semiconductor is called ;oping and the

impure semiconductor is called e)trinsic semiconductor.

;epending on the type of impurity added, e)trinsic semiconductors are furtherclassified as ntype and ptype semiconductor.

+


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n-tye se*i$ondu$tor

free e

Figure $.5 ntype semiconductor

Fermi level

Figure$.1 Energy band diagram for ntype semiconductor

;hen a s*all $urrent of Penta7alent i*urity is added to a ure

se*i$ondu$tor it is $alled as n-tye se*i$ondu$tor.

Addition of -entavalent impurity provides a large number of free electrons in asemiconductor crystal.

&ypical e)ample for pentavalent impurities are Arsenic, Antimony and-hosphorus etc. 6uch impurities which produce ntype semiconductors are nown

as ;onor impurities because they donate or provide free electrons to thesemiconductor crystal.

&o understand the formation of ntype semiconductor, consider a pure siliconcrystal with an impurity say arsenic added to it as shown in figure $.5.

e now that a silicon atom has / valence electrons and Arsenic has 5 valenceelectrons. hen Arsenic is added as impurity to silicon, the / valence electrons ofsilicon mae covalent bond with / valence electrons of Arsenic.

ve4.

-tye se*i$ondu$tor

hole

Figure $.: ptype semiconductor

$0

6

6

6

6

6

7

6 6 6


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Fermi level

Figure $.+ Energy band diagram for ptype semiconductor

;hen a s*all a*ount of tri7alent i*urity is added to a ure se*i$ondu$tor

it is $alled -tye se*i$ondu$tor.

&he addition of trivalent impurity provides large number of holes in thesemiconductor crystals.

E)ample8 7allium, !ndium or oron etc. 6uch impurities which produce ptypesemiconductors are nown as acceptor impurities because the holes created canaccept the electrons in the semi conductor crystal.

&o understand the formation of ptype semiconductor, consider a pure silicon crystal withan impurity say gallium added to it as shown in figure $.:.

e now that silicon atom has / valence electrons and 7allium has electrons.hen 7allium is added as impurity to silicon, the valence electrons of galliummae covalent bonds with valence electrons of silicon.

&he /thvalence electrons of silicon cannot mae a covalent bond with that of7allium because of short of one electron as shown above. &his absence ofelectron is called a hole. &herefore for each gallium atom added one hole iscreated, a small amount of 7allium provides millions of holes.

;ue to thermal energy, still holeelectron pairs are generated but the number of holes are

very large compared to the number of electrons. &herefore, in a ptype semiconductorholes are ma=ority carriers and electrons are minority carriers. 6ince the currentconduction is predominantly by hole3 " charges4 it is called as ptype semiconductor3 pmeans "ve4

$$

#onduction band

Forbidden gap

2alence band


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3rift and 3iffusion $urrent

&he flow of current through a semiconductor material is normally referred to as one ofthe two types.

3rift $urrent

!f an electron is sub=ected to an electric field in free space it will accelerate in astraight 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 freeelectrons under the influence of electric field will move towards the "ve terminalof the applied voltage but will continuously collide with atoms all the ways asshown in figure $.


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heavy concentration of less concentrationelectrons electrons

;iffusion current

Even distribution

%et diffusion current is zero ith no applied voltage if the number of charge carriers 3either holes or

electrons4 in one region of a semiconductor is less compared to the rest of theregion then there e)ist a concentration gradient.

6ince the charge carriers are either all electrons or all holes they sine polarity ofcharge and thus there is a force of repulsion between them.

As a result, the carriers tend to move gradually or diffuse from the region ofhigher concentration to the region of lower concentration. &his process is called

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

&his process continues until all the carriers are evenly distributed through thematerial. Hence when there is no applied voltage, the net diffusion current will bezero.

$


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9er*i-le7el

!ermi level indicates the level of energy in the forbidden gap.

&. 9er*i-le7el for an Intrinsi$ se*i$ondu$tor

Energy levelE#

Ef

Forbidden gapE2

E# #onduction band energy level

E2 2alence band energy level

Ef fermi level

e now that the !ntrinsic semiconductor acts as an insulator at absolute zerotemperature because there are free electrons and holes available but as thetemperature increases electron hole pairs are generated and hence number ofelectrons will be e'ual to number of holes.

&herefore, the possibility of obtaining an electron in the conduction band will bee'ual to the probability of obtaining a hole in the valence band.

!f Ec is the lowest energy level of #onduction band and Ev is the highest energy

level of the valence band then the fermi level Efis e)actly at the center of thesetwo levels as shown above.

$/

#onduction band

Fermilevel

2alence band


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). 9er*i-le7el in a se*i$ondu$tors ha7ing i*urities


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b: 9er*i-le7el for P-tye se*i$ondu$tor

?et an acceptor impurity be added to an !ntrinsic semiconductor then the acceptorenergy level 3Ea4 shown by dotted lines is very close to the valence band shownby dotted lines is very close to the valence band energy level 3Ev4.

&herefore the valence band electrons of the impurity atom can very easily =umpinto the valence band thereby creating holes in the valence band.

E#

EfEA

E2

EA Energy level of acceptor impurity

Figure $.$$ 3b4 Energy level diagram for -type semiconductor

&he acceptor energy level 3EA4 is =ust above the valence band level as shown infigure $.$$ 3b4.

;ue to large number of holes the probability of holes occupying the energy leveltowards the valence band will be more and hence, the fermi level gets shiftedtowards the valence band.

$1

#onduction band

Fermi level

2alence band


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A E99EC!

!f a piece of metal or semiconductor carrying a current ! is placed in a transversemagnetic field then an electric field E is induced in the direction perpendicular to both !and . &his phenomenon is nown as Hall effect.

@3"ve46urface(

" " " " " " " "d

! 2Hw

3"ve4

6urface $

B 3"ve4Hall effect is normally used to determine whether a semiconductor is ntype or ptype.

!o find hether the se*i$ondu$tor is n-tye or -tye

i4 !n the figure. above, !f ! is in the "ve direction and is in the "ve Bdirection, then a force will be e)erted on the charge carriers 3holes andelectrons4 in the >ve @ direction.

ii4 &his force is independent of whether the charge carriers are electrons or holes.

;ue to this force the charge carriers 3 holes and electrons4 will be forceddownward towards surface >$ as shown.

iii4 !f the semiconductor is %type, then electrons will be the charge carriers andthese electrons will accumulate on surface >$ maing that surface >velycharged with respect to surface >(. Hence a potential called Hall voltageappears between the surfaces $ and (.

iv4 6imilarly when surface >$ is positively charged with respect to surface >(,then the semiconductor is of -type. !n this way, by seeing the polarity of Hallvoltage we can determine whether the semiconductor is of -type or %type.

$:


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Ali$ations of all effe$t

Hall effect is used to determine,

carrier concentration, conductivity and mobility.

&he sign of the current carrying charge. #harge density. !t is used as magnetic field meter.

Carrier lifeti*e


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CAP!ER )

SEMIC"N34C!"R 3I"3E

hen a ptype semiconductor material is suitably =oined to ntype semiconductor the

contact surface is called a pn =unction. &he pn =unction is also called as semiconductordiode.p n

3b4

3a4 ;epletion region

9ig. )./ ve acceptorions by filling in the holes. &herefore a net >ve charge is established on pside ofthe =unction.

hen a sufficient number of donor and acceptor ions is uncovered furtherdiffusion is prevented.

&hus a barrier is set up against further movement of charge carriers. &his is calledpotential barrier or =unction barrier 2o. &he potential barrier is of the order of 0.$

to 0.2.

Note%outside this barrier on each side of the =unction, the material is still neutral. Cnlyinside the barrier, there is a "ve charge on nside and >ve charge on pside. &his region iscalled depletion layer.

).& Biasing% #onnecting a pn =unction to an e)ternal d.c. voltage source is called

$uation

&he current in a diode is given by the diode current e'uation! * !03 e 2G2&>$4

here, ! diode current !0 reverse saturation current 2 diode voltage semiconductor constant *$ for 7e, ( for 6i. 2& 2oltage e'uivalent of temperature* &G$$,100 3&emperature & is in elvin4

%ote !f the temperature is given in 0# then it can be converted to elvin by the helpof following relation, 0#"(: *

).0 3iode e>ui7alent $ir$uit

!t is generally profitable to replace a device or system by its e'uivalent circuit. Cnce thedevice is replaced by its e'uivalent circuit, the resulting networ can be solved bytraditional circuit analysis techni'ue.

switch r f !f

2F 2o2F

3i4 3ii 4

9ig.).0 3iode e>ui7alent $ir$uit.


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)0.& Basi$ 3efinitions

&.Knee 7oltage or Cut-in 8oltage.

!t is the forward voltage at which the diode starts conducting.

). Brea+don 7oltage!t is the reverse voltage at which the diode 3pn =unction4 breas down with sudden risein reverse current.

2. Pea+-in7erse 7oltage ve half cycle of a.c. doesnot e)ceed thepeainverse voltage of the diode.

0. Ma1i*u* 9orard $urrent

!t is the Da). instantaneous forward current that a pn =unction can conduct withoutdamaging the =unction. !f the forward current is more than the specified rating then the=unction gets destroyed due to over heating.

'.Ma1i*u* Poer rating

!t is the ma)imum power that can be dissipated at the =unction without damaging it. &hepower dissipated across the =unction is e'ual to the product of =unction current and the

voltage across the =unction.

(/


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).' REC!I9IERS

?Re$tifiers are the $ir$uit hi$h $on7erts a$ to d$@

9ectifiers are grouped into tow categories depending on the period of conductions.

$. Halfwave rectifier(. Full wave rectifier.

).'.& alf-a7e re$tifier

&he circuit diagram of a halfwave rectifier is shown in fig. (.5 below along with the !G-and CG- waveforms.

3i42i

t

2o

I (I t

3ii4

Fig. (.5 Half wave rectifier 3i4 #ircuit diagram 3ii4 waveforms

&he transformer is employed in order to stepdown the supply voltage and also toprevent from shocs.

&he diode is used to rectify the a.c. signal while , the pulsating d.c. is taen acrossthe load resistor 9?.

(5


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;uring the "ve half cycle, the end of the secondary is "ve and end @ is ve .&hus , forward biasing the diode. As the diode is forward biased, the current flowsthrough the load 9?and a voltage is developed across it.

;uring the >ve halfcycle the end @ is "ve and end is >ve thus, reverse biasing

the diode. As the diode is reverse biased there is no flow of current through 9?thereby the output voltage is zero.

).'.) Effi$ien$y of a re$tifier

&he ratio of d.c. power to the applied imp ac power is nown as rectifier efficiency.

Re$tifier effi$ien$y 5powercainput

outputpowercd

...

...

).'.23eri7ation of re$tifier effi$ien$y of alf a7e re$tifier

?et 2 * 2msinJ be the voltage across the secondary winding rf* diode resistance 9?* load resistance

d.$. oer

avI * dcI *

0

.(

$di *

dRr

VmSin

LL +0

(

$

*

dSinRr

Vm

Lf

+0

43(

*43(

(

Lf Rr

Vm

+ *

!m

d.c. power -dc * !(dcK 9?

* LR

(

!m 3$4

(1


27/113

a.$. oer inut

&he a.c. power input is given by -ac* !(rms3 rf " 9?4

diIrms

=

(

0

(

(

$

6'uaring both sides we get

diIrms =(

0

((

(

$

ut i * !m 6inJ

3 current flows through diode only for duration 0 to L 4

/

(

( mrms

II =

(

m

rms

II =

( )Lfmac RrI

P +

=

(

( 3(4

( )LfL

m

m

ac

dc

Rr

R

I

I

P

P

+

== K

(

(

(

*

L

f

R

r+$/01.0

34

&he efficiency is ma)imum if rfis negligible as compared to 9?

!herefore *a1i*u* re$tifier effi$ien$y 5 0/.

(:

=0

(( 43!m(

$dSinIrms


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).'.0 9ull-a7e re$tifier

Fullwave rectifier are of two types

$. #entre tapped fullwave rectifier

(. ridge rectifier).'.0.&Centre taed full Da7e re$tifier

3i42in

2out

I (I

3ii4

Fig. (.1 #entre tapped Full wave rectifier 3i4 #ircuit diagram 3ii4 waveforms

&he circuit diagram of a center tapped full wave rectifier is shown in fig. (.1

above. !t employs two diodes and a center tap transformer. &he a.c. signal to berectified is applied to the primary of the transformer and the d.c. output is taenacross the load 9?.

;uring the "ve halfcycle end is "ve and end @ is >ve this maes diode ; $forward biased and thus a current i$ flows through it and load resistor 9?.;iode;( is reverse biased and the current i(is zero.

(+


29/113

;uring the >ve halfcycle end @ is "2e and end is >2e. %ow diode ;( isforward biased and thus a current i(flows through it and load resistor 9?. ;iode;$is reversed and the current i$* 0.

3isad7antages

6ince, each diode uses only onehalf of the transformer secondary voltage the d.c.output is comparatively small.

!t is difficult to locate the centertap on secondary winding of the transformer. &he diodes used must have high -eainverse voltage.

).'.0.) Bridge re$tifier

3i4

2out

;$; ;(;/ ;$;/

t

3ii4

9ig. ). 9ull a7e bridge a7e re$tifier


30/113

;uring the >ve halfcycle, end is "ve and end A is >ve thus diodes ;( and ;/are forward biased while the diodes ;$and ;are reverse biased. %ow the flow ofcurrent is through diode ;/load 9? 3 ; to #4 and diode ;(. &hus, the waveform issame as in the case of centertapped full wave rectifier.

Ad7antages

&he need for centertaped transformer is eliminated. &he output is twice when compared to centertapped full wave rectifier.

for the same secondary voltage. &he pea inverse voltage is onehalf3$G(4 compared to centertapped full wave

rectifier. #an be used where large amount of power is re'uired.

3isad7antages

!t re'uires four diodes. &he use of two e)tra diodes cause an additional voltage drop thereby reducing the

output voltage.

).'. Effi$ien$y of 9ull-a7e re$tifier

?et 2 * 2msinJ be the voltage across the secondary winding ! * !msinJ be the current flowing in secondary circuit rf* diode resistance 9?* load resistance

d$ oer outut

LdcRIPdc (= 3$4

==

0

.(

$( diII avdc

=

0

.!m

(

$( dSinIav

m

av

II

(=

3(4

L

m

dc RI

P

((

=

34

0


31/113

inut a$ oer

( )Lfrmsac RrIP += (

---------------------------------------- 3/4

diIrms = 0(

($(

6'uaring both sides we get

diIrms =0

(( $

(

(( m

rms

II =

(

m

rms

II = 354

( )Lfmac RrI

P +

=

(

( 314

( )LfL

m

m

ac

dc

Rr

R

I

I

P

P

+

== K

(

( (

(

*

L

f

R

r+$

+$(.0

3:4

&he efficiency will be ma)imum if rf is negligible as compared to 9?.

Ma1i*u* effi$ien$y 5 F&.)

&his is the double the efficiency due to half wave rectifier. &herefore a Fullwaverectifier is twice as effective as a halfwave rectifier.

$

=0

(( 43!m$

dSinIrms


32/113

). Co*arision of Re$tifiers

-articulars Half wave rectifier #entretapped Fullwave rectifier

ridge rectifier

$. %o. of diodes

(. !dc

. 2dc

/.!rms

5.Efficiency

1.-!2

:.9ipple factor

$

!mG L

2mG L

!mG (

/0.1 M

2m

$.($

(

(!mGL

(2mG L

!mGN (

+$.( M

(2m

0./+

/

(!mGL

(2mG L

!mGN (

+$.( M

2m

0./+

Note%

&he relation between turns ratio and voltages of primary and secondary of thetransformer is given by

o %$G %(* 2pG 2s

9D6 value of voltage and Da). value of voltage is related by the e'uation. 2rms* 2m G N( 3 for fullcycle of ac4

!f 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 cutin voltage , 2O* 0.

(


33/113

). Rile fa$tor

&he pulsating output of a rectifier consists of d.c. component and a.c. component 3 alsonown as ripple4. &he a.c. component is undesirable and account for the pulsations in therectifier output. &he effectiveness of a rectifier depends upon the magnitude of a.c.

component in the output 8 the smaller this component, the more effective is the rectifier." The ratio of rms value of a.c. component to the d.c. component in the rectifier output

is known as ripple factor#

r *Idc

Iac

)..& Rile fa$tor for alf-a7e re$tifi$ation

y definition the effective 3ie rms4 value of total load current is given by

C9 (( dcrmsac III =

here !dc * value of dc component

!ac* rms value of ac component

;ivide both 9.H.6 and ?.H.6. by !dc we get(($dcrms

dcdc

ac IIII

I=

r * $

(

dc

rms

I

I 3$4

for halfwave rectification, we have !(

m

rms

II =

m

dc II =

6ubstituting above values in e'uation 3$4 we get,

rile fa$torr 5 &.)&

!t is clear that a.c. component e)ceeds dc component in the output of a halfwaverectifier.

((

dcacrms III +=


34/113

)..) Rile fa$tor for full-a7e re$tifi$ation

For full wave rectification we have !rms *(

mI

!dc *mI(

6ubstituting above values in e'uation 3$4 we get ,

rile fa$tor r 5 /.0F

&his shows that in the output of Fullwave rectifier, the d.c. component is more than the

a.c. component).F 9I!ERS

e now that the output of the rectifier is pulsating d.c. ie the output obtained by therectifier is not pure d.c. but it contains some ac components along with the dc oGp. &heseac components are called as 9ipples, which are undesirable or unwanted. &o minimizethe ripples in the rectifier output filter circuits are used. &hese circuits are normallyconnected between the rectifier and load as shown below.

2i2o pure dc oGp

-ulsating d.c. output

Filter is a circuit which converts pulsating dc output from a rectifier to a steady dc output.!n otherwords, filters are used to reduce the amplitudes of the unwanted ac components in

the rectifier.Note%A capacitor passes ac signal readily but blocs dc.

/

9ectifier Filter


35/113

).F.& !yes of 9ilters

$. #apacitor Filter 3#Filter4(. !nductor Filter. #hoe !nput Filter 3?#filter4/. #apacitor !nput Filter 3Lfilter4

).F.) Caa$itor 9ilter< C-filter:

3i42in a f

et

b d

c

2$

oGp withoutfilter

t

2ooGp

with filter ae

t

9ig.).F Caa$itor filter < C-filter:


36/113

hen the !nput signal rises from o to a the diode is forward biased therefore itstarts conducting since the capacitor acts as a short circuit for ac signal it getscharged up to the pea of the input signal and the dc component flows through theload 9?.

hen the input signal fall from a to b the diode gets reverse biased . &his ismainly because of the voltage across the capacitor obtained during the period o toa is more when comapared to 2i. &herefore there is no conduction of currentthrough the diode.

%ow the charged capacitor acts as a battery and it starts discharging through theload 9?. Dean while the input signal passes through b,c,d section. hen thesignal reaches the point d the diode is still reverse biased since the capacitorvoltage is more than the input voltage.

hen the signal reaches point e, the input voltage can be e)pected to be more

than the capacitor voltage. hen the input signal moves from e to f the capacitorgets charged to its pea value again. &he diode gets reverse biased and thecapacitor starts discharging. &he final output across 9?is shown in Fig. (.+

&he ripple factor for a Halfwave rectifier with #filer is given by

r* $G(Nf#9?

fthe line fre'uency 3 Hz4#capacitance 3 F49? ?oad resistance 3P4

9ipple factor for fullwave rectifier with #filter is given by r * $G / N f #9?

).F.2 Ad7antages of C-9ilter

low cost, small size and good characteristics. !t is preferred for small load currents 3 upto 50 mA4 !t is commonly used in transistor radio, batteries eliminator etc.

1


37/113

).G Hener 3iode

&he reverse voltage characteristics of a semiconductor diode including the breadownregion is shown below.

2 2z 0

!

9ig. ).G Hener diode $hara$teristi$s

Bener diodes are the diodes which are designed to operate in the breadown region. &heyare also called as readown diode or Avalanche diodes.

&he symbol of Bener diode is shown below

- %

Fig. (.$0 6ymbol of Bener diode

&he breadown in the Bener diode at the voltage 2zmay be due to any of thefollowing mechanisms.

&. A7alan$he brea+don

;epletion region charge carriers striing the atoms

3 4 3 4

ve terminal "ve terminal

minority charge carriers

Fig. (.$$ Avalanche breadown in Bener diode e now that when the diode is reverse biased a small reverse saturation current

!0flows across the =unction because of the minority cariers in the depletion region.

:

%-


38/113

&he velocity of the minority charge carriers is directly proportional to the appliedvoltage. Hence when the reverse bias voltage is increased, the velocity of minoritycharge carriers will also increase and conse'uently their energy content will alsoincrease.

hen these high energy charge carriers stries the atom within the depletionregion they cause other charge carriers to brea away from their atoms and =ointhe flow of current across the =unction as shown above. &he additional chargecarriers generated in this way stries other atoms and generate new carriers bymaing them to brea away from their atoms.

&his cumulative process is referred to as avalanche multiplication which results inthe flow of large reverse current and this breadown of the diode is calledavalanche breadown.

).Hener brea+don

e have electric field strength * 9everse voltageG ;epletion region

;epletion region

3 4 3 4

ve terminal "ve terminal

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

9ig.).&) Hener brea+don in Hener diode

From the above relation we see that the reverse voltage is directly proportional to theelectric field hence, a small increase in reverse voltage produces a very high intensityelectric field with ina narrow ;epletion region.

&herefore when the reverse voltage to a diode is increased, under the influence of highintensity electric filed large numbr of electrons within the depletion region brea thecovalent bonds with their atoms as shown above and thus a large reverse current flowsthrough the diode. &his breadown is referred to as Bener breadown.

+

%-


39/113

).G.& Hener 7oltage regulator

&he circuit diagram of Bener voltage regulator is shown below9s

! !B !?

2in 2B 9? 2o

Fig. (.$ Bener voltage regulator

.A zener diode of breadown voltage 2Bis connected in reverse biased condition acrossthe load 9?such that it operates in breadown region. Any fluctuations in the current areabsorbed by the series resistance 9s. &he Bener will maintain a constant voltage 2B3 e'ual to 2o4 across the load unless the input voltage does not fall below the zenerbreadown voltage 2B.

Case


40/113

!f there is a decrease in the load resistance 9?and the input voltage remains constant thenthere is a increase in load current.

6ince 2inis constant the current cannot come from the source. &his addition load current

is driven from the battery 2Band we now that even for a large decrease in current theBener output voltage 2zremains same. Hence the output voltage across the load is alsoconstant..

/0


41/113

CAP!ER 2

!RANSIS!"RS

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

ayers of other types.

&ransistors are classified into two typesQ&. n transistor

pnp transistor is obtained when a ntype layer of silicon is sandwichedbetween two ptype silicon material.

). nn transisitor

npn transistor is obtained when a ptype layer of silicon is sandwiched

between two ntype silicon materials.

Figure.$ below shows the schematic representations of a transistor which ise'uivalent of two diodes connected bac to bac.

RE R# RE R#E # E #

9ig 2.&%Sy*boli$ reresentation

pnp npn9ig 2.)%S$he*ati$ reresentation

&he three portions of transistors are named as emitter, base and collector. &he=unction between emitter and base is called emitterbase =unction while the =unctionbetween the collector and base is called collectorbase =unction.

&he base is thin and tightly doped, the emitter is heavily doped and it is wider whencompared to base, the width of the collector is more when compared to both base andemitter.

/$

n npp pn


42/113

!n order to distinguish the emitter and collector an arrow is included in the emitter.&he direction of the arrow depends on the conventional flow of current when emitterbase =unction is forward biased.

!n a pnp transistor when the emitter =unction is forward biased the flow of current is

from emitter to base hence, the arrow in the emitter of pnp points towards the base.2.& "erating regions of a transistor

A transistor can be operated in three different regions asa4 active regionb4 saturation regionc4 cutoff region

A$ti7e region

E RE R# #

2E 2# 9ig 2.2% n transistor oerated in a$ti7e region

&he transistor is said to be operated in active region when the emitterbase =unction isforward biased and collector >base =unction is reverse biased. &he collector current is saidto have two current components one is due to the forward biasing of E =unction and theother is due to reverse biasing of # =unction. &he collector current component due to thereverse biasing of the collector =unction is called reverse saturation current 3! #C or !#C4and it is very small in magnitude.

Saturation region

E RE R# #

2E 2# 9ig 2.0% n transistor oerated in Saturation region

/(

p pn

p pn


43/113

&ransistor is said to be operated in saturation region when both E =unction and #=unction are forward biased as shown. hen transistor is operated in saturation region !#increases rapidly for a very small change in 2#.

Cut-off region E RE R# #

2E 2# 9ig 2.'% n transistor oerated in Cut-off region

hen both E =unction and # =unction are reverse biased, the transistor is said to beoperated in cutoff region. !n this region, the current in the transistor is very small andthus when a transistor in this region it is assumed to be in off state.

2.) ;or+ing of a transistor


44/113

biased all the holes are immediately attracted by the >ve potential of thesupply 2#. &hereby giving rise to collector current !#.

&hus we see that !E* !" !# 3$4 3y 2?4

6ince the # =unction is reverse biased a small minority carrier current !#Cflows from base to collector.

2.2 Current $o*onents of a transistor

RE R#

!E!#

!

2E2#

9ig 2.%Current $o*onents of a transistor

Fig .: above shows a transistor operated in active region. !t can be noted from the

diagram the battery 2E forward biases the E =unction while the battery 2# reversebiases the # =unction.

As the E =unction is forward biased the holes from emitter region flow towards the basecausing a hole current !-E. At the same time, the electrons from base region flow towardsthe emitter causing an electron current !%E. 6um of these two currents constitute anemitter current !E* !-E"!%E.

&he ratio of hole current !-Eto electron current !%E is directly proportional to the ratio ofthe conductivity of the ptype material to that of ntype material. 6ince, emitter is highlydoped when compared to baseQ the emitter current consists almost entirely of holes.

%ot all the holes, crossing E =unction reach the # =unction because some of the themcombine with the electrons in the ntype base. !f !-# is the hole current at 3Rc4 #=unction. &here will be a recombination current !-E !-# leaving the base as shown infigure .:.

!f emitter is open circuited, no charge carriers are in=ected from emitter into the base andhence emitter current !E *o. Snder this condition # =unction acts a a reverse biased

//

!-E !-#3hole current4 3hole current4

!%E 3ecurrent4 !#C

!-


45/113

diode and therefore the collector current 3 !#* !#C4 will be e'ual to te reverse saturationcurrent. &herefore when E =unction is forward biased and collector base =unction isreverse biased the total collector current !# * !-#"!#C.2.0 !ransistor $onfiguration

e now that, transistor can be used as an amplifier. For an amplifier, two terminals are

re'uired to supply the wea signal and two terminals to collect the amplified signal. &husfour terminals are re'uired but a transistor is said to have only three terminals &herefore,one terminal is used common for both input and output.

&his gives rise to three different combinations.$. #ommon base configuration 3#4(. #ommon emitter configuration 3#E4. #ommon collector configuration 3##4

&. CB $onfiguration

A simple circuit arrangement of # configuration for pnp transistor is shown below.!E !#

2i ! 9? 2out

2E 2# 9ig 2.F%CB $onfiguration

!n this configuration, base is used as common to both input and output. !t can be noted

that the iGp section has an a.c. source 2ialong with the d.c. source 2E. &he purpose ofincluding 2Eis to eep E =unction always forward biased 3because if there is no 2Ethen the E =unction is forward biased only during the "ve halfcycle of the iGp andreverse biased during the >ve half cycle4. !n # configuration, !E>iGp current, !#>oGpcurrent.

Current relations

&.$urrent a*lifi$ation fa$tor


46/113

!#* !E" !#C

6ince a portion of emitter current !E flows through the base ,let remaining emitter currentbe IE.

IC5 IE 6 ICo

Chara$teristi$s

&. Inut $hara$teristi$s

!E 2#*$02 2#*52

8EB 9ig 2.G% Inut $hara$teristi$s

!Gp characteristics is a curve between !Eand emitter base voltage 2Eeeping 2#constant. !Eis taen along ya)is and 2Eis taen along )a)is. From the graph followingpoints can be noted.

$. For small changes of 2Ethere will be a large change in !E. &herefore inputresistance is very small.

(. !Eis almost independent of 2#

. !G- resistance , 9i

* T2E

G T !E

2#

*constant). "utut $hara$teristi$s

!#

Active region!E* mA

!E*( mA6aturation !E * $ mAregion !E* 0

#utoff region 2#9ig 2.&/%"utut $hara$teristi$s

oGp characteristics is the curve between !#and 2#at constant !E. &he collector current !#is taen along ya)is and 2#is taen along )a)is. !t is clear from the graph that the oGpcurrent !#remains almost constant even when the voltage 2#is increased.

/1


47/113

i.e. , a very large change in 2#produces a small change in !#. &herefore, outputresistance is very high.

CGp resistance 9o * T2EG T !# !E* constant

9egion below the curve !E*0 is nown as cutoff region where !#is nearly zero. &heregion to the left of 2#*0 is nown as saturation region and to the right of 2#*0 isnown as active region.

). CE $onfiguration

!#

9? 2out

!2i !E

2E 2#E9ig 2.&&%CE $onfiguration

!n this configuration the input is connected between the base and emitter while the outputis taen between collector and emitter. For this configuration !is input current and !#isthe output current.

&. Current a*lifi$ation fa$tor


48/113

3 !#G !* V 4

+=

$

Also we have

=

===+=+

$

4$3

4$3

3eri7ation of !otal outut $urrent IC

e have #$"!# III +=

+

++=

+

+

=

$

4$3

$

#$"!

#

#$"!#

III

II$

I

!c * #$"$ II 4$3 ++

!ransistor Chara$teristi$s

&. i( $hara$teristi$s

! 2#E*$02 2#E*52

8EB 9ig 2.&&% i( $hara$teristi$s

/+


49/113

!nput characteristics is a curve between E voltage 32E4 and base current 3!4 atconstant 2#E. From the graph following can be noted.

$. &he input characteristic resembles the forward characteristics of a pn =unctiondiode.

(. For small changes of 2Ethere will be a large change in base current !. i.e., inputresistance is very small.. &he base current is almost independent of 2#E./. !nput resistance , 9i* T2EG T ! 2 #E * constant

). "utut $hara$teristi$s

!#3mA4

Active region 0 A

(0 A $0 A ! *0A

#utoff region 2 #E3volts4

9ig 2.&)% "utut $hara$teristi$s

!t is the curve between 2#Eand !#at constant !. From the graph we can see that,$. 2ery large changes of 2#Eproduces a small change in !#i.e output resistance is

very high.(. output resistance 9o* T2#E G T!# W!* constant

9egion between the curve !*0 is called cutoff region where !is nearly zero. 6imilarlythe active region and saturation region is shown on the graph.

2. CC $onfiguration

!E

9? 2out

!2i !#

2# 2#E9ig 2.&2% CC $onfiguration

!n this configuration the input is connected between the base and collector while theoutput is taen between emitter and collector.Here !is the input current and !Eis the output current.

/


50/113

Current relations

$. #urrent amplification factor 3O4

). Relationshi beteen J and

O *$

!

I

I

O *$

#$

I

II +

divide both %umerator and denominator by !

$

$$

#

II+

=

+=$ 3 V * !#G !4

+=

$$

=

$

$

3eri7ation of total outut $urrent IE

e now that !# * #$"! II +

!E* !" !#

!E* ! " U!E" !#C

!E3$U 4 * !" !#C

!E*

+ $$

#$"$ II

50


51/113

!E* O!" O!#C

!E* O 3!" !#C4

5$


52/113

2.' Co*arison beteen CB, CC and CE $onfiguration

#haracteristics # #E ##$. !nput reistance 39i4

(. Cutput resistance 39o4. #urrent amplification

factor

/. &otal output current

5. -hase relationshipbetween input and output

1. Applications

:. #urrent gain

+. 2oltage gain

low

high

+=

$

#$"!# III +=

!nphase

For high fre'uencyapplications

?ess than unity

2ery high

low

high

=

$

!c * #$"$ II 4$3 ++

Cutof phase

For audio fre'uencyapplications

7reater than unity

7rater than unity

high

low

=

$

$

!E* O!" O!#C

inphase

For impedancematching

2ery high

?ess than unity

5(


53/113

2. !ransistor as an a*lifier

!#

9? 2out

!2i X !E

2E 2#

9ig 2.&2% !ransistor as an a*lifier

#onsider a npn transistor in #E configuration as shown above along with its inputcharacteristics.

A transistor raises the strength of a wea input signal and thus acts as an amplifier. &hewea signal to be amplified is applied between emitter and base and the output is taenacross the load resistor 9#connected in the collector circuit.

!n order to use a transistor as an amplifier it should be operated in active region i.e.emitter =unction should be always F and collector =unction should be 9. &herefore inaddition to the a.c. input source 2itwo d.c. voltages 2Eand 2#Eare applied as shown.&his d.c. voltage is called bias voltage.

As the input circuit has low resistance, a small change in te signal voltage 2 icauses alarge change in the base current thereby causing the same change in collector current3because !#* V!4.

&he collector current flowing through a high load resistance 9#produces a large voltageacross it. &hus a wea signal applied at the input circuit appears in the amplified form atthe output. !n this way transistor acts as an amplifier.

E)ample8 ?et 9#* 5P, 2in*$2, !# *$mA then output 2*!#9#*52

5


54/113

2. Bias stabiliLation

&he process of maing operating point independent of temperature changes or variationin transistor parameters is called the stabilization.

e now that for transistor to operate it should be properly biased so that we can have afi)ed operating point. &o avoid any distortions, the Ypoint should be at the center of theload line.

ut in practice this Ypoint may shift to any operating region 3saturation or curoffregion4 maing the transistor unstable. &herefore in order to avoid this, biasing stabilityshould be maintained.

2.F Causes for Bias instability

ias instability occurs mainly due to two reasons.

$. &emperature(. #urrent gain

&. !e*erature

&he temperature at the =unctions of a transistor depends on the amount of current flowingthrough it. ;ue to increase in temperature following parameters of a transistor willchange.

3a:base-e*itter 7oltage


55/113

2.G Cas$ading transistor a*lifiers

hen the amplification provided by a single stage amplifier is not suffiecient for aparticular purpose or when the input and output impedance is not of the correctmagnituded for the re'uired application then two or more amplifiers are connected in

cascade as shown below.

9ig 2.&0%Cas$ading transistor a*lifiers

Here the output of amplifier $ is connected as the input of amplifier (.

E)ample8 &he gain of a single amplifier is not sufficient to amplify a signal from a weasource such as microphone to a level which is suitablefor the operation of another circuitas loud speaer. !n such cases, amplifiers are used.

hen amplifiers are cascaded, individual amplifiers provides re'uired amplification andinput and output provide impedance matching.

3e$ibel


56/113

2.&/ Single stage RC $ouled A*lifier

2##2o

9$ 9# ## 2o t

969( 9E 9?

2i

9ig 2.&'%single stage RC-$ouled a*lifier

Figure above shows a practical circuit of a single stage 9# coupled amplifier.&hedifferent circuit components and their functions are as described below.

a. Inut $aa$itor


57/113

2.&& !o stage RC $ouled a*lifier

2##

9$ 9# ## 9$$ 9#$

##$

969( 9E #E 9($ #E $

9? 2i 9E$

6tage$ 6tage(

9ig 2.&% !o stage RC $ouled A*lifier

Figure above shows the circuit diagram of a two stage 9# coupled amplifier . &hecoupling capacitor ##connects the output of the first stage to the input of the secondstage. 6ince the coupling from one stage to the ne)t stage is achieved by couplingcapacitor along with a shunt resistor the amplifier is called 9# coupled amplifier.. &he

input signal is first applied to the transistor &$and output is taen at the collector of &$.&he signal at the output will be $+00 out of phase when compared to the input. &heoutput is taen across 9#with the help of a coupling capacitor. &his signal is fed as inputto the ne)t stage i.e transistor &(. &he signal is amplified further and the amplified outputis taen across 9c$ of &(. &he phase of the signal is reversed again. &he output isamplified twice and its is amplified replica of the input signal.

2.&) 9re>uen$y resonse in a*lifer

Fre'uency response is the curve between the gain of the amplifier 3A * 2o G 2i 4verses the fre'uency of the input signal. &he fre'uency response of a typical 9#coupled amplifier is shown below.

Fre'uency response has regions.

$. ?ow fre'uency range(. Did fre'uency range. High fre'uency range

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar 5:


58/113

ain in

dB ?F9 DF9 HF9

f& f) fre>uen$y resonse in a*lifer

o fre>uen$y range


59/113

2.&2 3C oad ine and "erating oint sele$tion

!#

9 9#2#E

!

2 2##9ig 2.&%nn transistor in CE $onfiguration

!#3mA4 Ypoint

A

0 A (0 A $0 A

! *0A

2 #E3volts

9ig.2.&F% outut $hara$teristi$s

#onsisder a #E amplifier along with the output characteristics as shown in figure .$+

above. A straight line drawn on the output characteristic of a transistor which gives thevarious zero signal values 3ie. hen no signal applied4 of 2#Eand !#is called ;# loadline.

Constru$tion of 3C load line

Applying 2? to the collector circuit we get,

2##>!#9#>2#E*0$

2#E * 2##>!#9# (

&he above e'uation is the first degree e'uation and can be represented by astraight line. &his straight line is ;# load line.

&o draw the load line we re'uire two end points which can be found as follows.

$. !f !# *0, e'un ( becomes 2#E * 2##

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar 5


60/113

(. if 2#E* 0, e'un ( becomes 2##* !#9# ie. !#* 2## G9#

2.&0 "erating oint


61/113

CAP!ER 0

"PERA!I"NA AMPI9EIR

IN!R"34C!I"N

CpAmp 3operational amplifier4 is basically an amplifier available in the !# form. &he

word Zoperational[ is used because the amplifier can be used to perform a variety of

mathematical operations such as addition, subtraction, integration, differentiation etc.

Figure $ below shows the symbol of an CpAmp.

"2##2$

!nverting input

2( %oninverting input

2EE

9ig.& Sy*bol of "-A*

!t has two inputs and one output. &he input mared ZZ is nown as !nverting input and

the input mared Z"[ is nown as %oninverting input.

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar 1$


62/113

!f a voltage 2iis applied at the inverting input 3 eeping the noninverting input atground4 as shown below.

2i

2Ct

t

2i 2C

9ig.) "-a* in in7erting *ode

&he output voltage 2o* A2iis amplified but is out of phase with respect to the input

signal by $+00.

!f a voltage 2iis fed at the noninverting input 3 eeping the inverting input at

ground4 as shown below.

2o

2Ct

2i

t

9ig.2 "-A* in Non-in7erting *ode

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar 1(


63/113

&he output voltage 2o* A2iis amplified and inphase with the input signal.

!f two different voltages 2$ and 2( are applied to an ideal CpAmp as shown

below.

2$2C

2(

9ig.0 Ideal "-A*

&he output voltage will be 2o* A32$> 2(4

i.e the difference of the tow volatages is amplified. Hence an CpAmp is also called as a

High gain differential amplifier.

%ote8 CpAmp is + pin !# 3 named as A :/$4 with pin details as shown.

CFF6E& %S?? %C #C%%E#&!C%

"2## !%2E9&!%7 !G-

CS&-S& %C% >!%2E9&!%7 !G-

2EE CFF6E& %S??

9ig. ' Pin details of "-A*


$ +

( : A :/$

1

/ 5


64/113

Blo$+ 3iagra* of an "-AMP

An CpAmp consists of four blocs cascaded as shown above

9ig. Blo$+ diagra* of an "-A*

Inut stage% !t consists of a dual input, balanced output differential amplifier. !ts

function is to amplify the difference between the two input signals. !t provides high

differential gain, high input impedance and low output impedance.

Inter*ediate stage% &he overall gain re'uirement of an CpAmp is very high. 6ince the

input stage alone cannot provide such a high gain. !ntermediate stage is used to provide

the re'uired additional voltage gain.

!t consists of another differential amplifier with dual input, and unbalanced 3 single

ended4 output

Buffer and e7el shifting stage

As the CpAmp amplifies ;.# signals also, the small ;.#. 'uiescent voltage level of

previous stages may get amplified and get applied as the input to the ne)t stage causing

distortion the final output.

Hence the level shifting stage is used to bring down the ;.#. level to ground potential,

when no signal is applied at the input terminals. uffer is usually an emitter followerused for impedance matching.

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar 1/


65/113

"utut stage- !t consists of a pushpull complementary amplifier which provides large

A.#. output voltage swing and high current sourcing and sining along with low output

impedance.

Con$et of 8irtual ground

e now that , an ideal CpAmp has perfect balance 3ie output will be zero when inputvoltages are e'ual4.

Hence when output voltage 2o* 0, we can say that both the input voltages are e'ual ie 2$* 2(.

2$ 2o

9i2(

9ig.


66/113

Ali$ations of "-A*

An CpAmp can be used as$. !nverting Amplifer(. %on!verting Amplifer

. 2oltage follower/. Adder 3 6ummer45. !ntegrator1. ;ifferentiator

3efinitions

&. Slerate


67/113

Chara$teristi$s of an Ideal "-A*

An ideal CpAmp has the following characteristics.

$. !nfinite voltage gain 3 ie A2*]4

(. !nfinite input impedance 39i * ]4

. Bero output impedance39o *04

/. !nfinite andwidth 3.. * #

5. !nfinite #ommon mode re=ection ratio 3ie #D99 *]4

1. !nfinite slew rate 3ie 6*]4

:. Bero power supply re=ection ratio 3 -699 *04ie output voltage is zero whenpower supply 2##*0

+. Bero offset voltage3ie when the input voltages are zero, the output voltage willalso be zero4


68/113

y #? we have($ ii =

f

oi

R

V

R

V

=

00

$

f

oi

R

V

R

V=

$

i

f

" VR

RV

=

$

;here$R

Rfis the gain of the amplifier and negative sign indicates that the output is

inverted with respect to the input.

2C2i

t t

9ig. G ;a7efor*s of In7erting A*lifers

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar 1+


69/113

). Non- In7erting A*lifier

A noninverting amplifier is one whose output is amplified and is inphase with theinput.

9f

i(9$

2$ i$ 7*2i2C

2i

9ig.&/ Non In7erting A*lifiers

y #? we have

($ ii =

f

"ii

R

VV

R

V =

$

0

$

0

$

R

R

V

ViV

R

VV

R

V

f

i

f

i"i

=

=

$$ R

R

V

V f

i

"

=

i

f

i

"

R

R

V

V+=$



70/113

i

fV

R

RV

+=

$

0 $

here

+

$

$RRf is the gain of the amplifier and " sign indicates that the output

is inphase with the input.

2. 8oltage folloer

2C

2i 2C2i

t t

9ig. && 8oltage folloer

2oltage follower is one whose output is e'ual to the input.

&he voltage follower configuration shown above is obtained by short circuiting Z9f[ and

open circuiting Z9$[ connected in the usual noninverting amplifier.

&hus all the output is fed bac to the inverting input of the opAmp.

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar :0


71/113

#onsider the e'uation for the output of noninverting amplifer

i

fV

R

RV

+=

$

0 $

hen 9f* 0 short circuiting 9$* ] open circuiting

iV

+=

0$2C

i" VV =

&herefore the output voltage will be e'ual and inphase with the input voltage. &husvoltage follower is nothing but a noninverting amplifier with a voltage gain of unity.

0. In7erting Adder

!nverting adder is one whose output is the inverted sum of the constituent inputs

9$9f

2$ i$

!f9(

2( i( 7*02C

2 9 i

9ig.&). In7erting Adder

y #? we have

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar :$


72/113

($ iiii f ++=

(

(

$

$ 0000

R

V

R

V

R

V

R

V

f

" +

+

=

,

,

(

(

$

$

9

2

R

V

R

V

R

V

f

"++=

++=

(

(

$

$

9

2

R

V

R

VRV f"

!f 9$* 9(* 9*9 then

[ ]($ VVVR

RV

f

" ++=

!f 9f * 9 then

2C* ^ 2$" 2(" 2 _

Hence it can be observed that the output is e'ual to the inverted sum of the inputs.

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar :(


73/113

'. Integrator

#

i(9$

2$ i$ 7*02C

Fig, $ !ntegratorAn integrator is one whose output is the integration of the input.,y #? we have, $($ = ii

i"

"i

"

"

"

"

ii

VR#dt

dV

dt

dV#

R

V

haveweinand&substituin

dt

dV#iei

i#dt

dV

dti#

V

dti#

V

havewesimiaryand

R

V

R

Vi

havewefi&ureabovethe'rom

$

$,(

,..

$

$

$0

(0

(

(

(

(

$

=

=

=

=

=

=

=

=

dtVR#

V i" = $

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar :


74/113

3ifferentiator

A differentiator is one whose output is the differentiation of the input9

i(

2$ i$ 7*02C

y #? we have

R

V

dt

dV#

haveweinandn&substituti

R

V

R

Vi

havewesimiaryand

dt

dV#i

i#dt

dV

dti#

V

havewefi&ureabovethe'rom

ii

"i

""

i

i

i

=

=

=

=

=

=

=

$(

0

(.

$

$

$

(

$

$

$

($

dt

dVR#V" =

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar :/


75/113

Proble*s

&. 9or an in7erting a*lifer Ri5&//K and Rf5//K. ;hat is the outut 7oltage

for an inut of -28

Soln%7iven8 9$*$00P9f*100P

2i*2 2C *`

e have,

VV

VR

RV

"

i

f

"

$+

$0$00

$0100

$

=

=

=

$. esign an inverting amplifer for output voltage of %&'V and an input voltage of &V.

Soln%

7iven8 2i*$ 2

2C* $02e Have,

i

f

" VR

RV

=

$

$$0$

=R

Rf



76/113

havewe(R&Assu

RRorR

Rf

f

=

==

$min

$0$0

$

$

$

= (Rf $0

9f

i(9$

2$ i$2C

2. 9or an in7erting a*lifier R&5&/K and 8i5&8. Cal$ulate i&and 8".Soln%

7iven8 9$* $0P, 9f*$00P 2i *$ 2

e have,

9f

i(

9$

2$ i$2C



77/113

VVR

RV

mAR

Vi

i

f

"

i

$0$$0$0

$0$00

$.0$0$0

$0

$

$

$

=

=

=

=

=

=

0. 3esing an a*lifier ith a gain of 6G and Rf5&) K using an o-A*

Soln%

6ince the gain is positive8#hoose a noninverting amplifier

&hen we have,

i

f

" VR

RV

+=

$

$

7ain is,

=

==

=

=+

(R

RR

R

R

R

R

f

f

f

5.$

+

$0$(

+

+


78/113

. In the figure shon if 8&56&8, 8)5628 and 8256)8 ith R&5R)5R25)K.

3eter*ine the outut 7oltage.

9$9f

2$ i$!f

9(

2( i( 7*02C

2 9 i

Soln%e have ,

[ ]

[ ]

VV

V

VVVR

RV

"

"

f

"

ve and lower

plate "ve as shown in fig. .

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.

&he se'uence of charge and discharge results in alternating motion of electrons or

an oscillating current. &he energy is alternately stored in the lectric field of the

capacitor # and the magnetic field of the inductance coil ? . &his interchange of

energy between ? and # is repeated over and again resulting in the production of

Cscillations.

;a7efor*- !n practical tan circuit there are resistive and radiation losses in the coil and

dielectric losses in the capacitor. ;uring each cycle a small part of the originally imparted

energy is used up to overcome these losses. &he result is that the amplitude of oscillating

current decreases gradually and eventually it become zero. &herefore tan circuit

produces damped oscillations.

9re>uen$y of os$illations- &he e)pression for fre'uency of oscillation is given by,

L#fr

(

$= 3$4

4nda*ed "s$illations fro* !an+ Cir$uit

A tan circuit produces damped oscillations. !n practice we need continuous undampedoscillations for the successful operation of electronics e'uipment. !n order to mae the

oscillations in the tan circuit undamped it is necessary to supply correct amount of

energy to the tan circuit at the proper time intervals to meet the losses.

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar +


84/113

&he following conditions must be fulfilledQ

$. &he amount of energy supplied be such so as to meet the losses in the tan and the

a.c. energy removed from the circuit by the load. For e)ample if losses in ?#

circuit amount ot 5 m and a.c. output being taen is $00 m, then power of

$05m should be continuously supplied to the circuit.

(. &he applied energy should have the same fre'uency as the of the oscillations in

the tan circuit.

. &he applied energy should be in phase with the oscillations set up in the tan

circuit.

Positi7e feedba$+ A*lifier-"s$illator

$. A transistor amplifier with proper "ve feedbac can act as an oscillator.

S

2out2in 2f

t t t

(. &he circuit needs only a 'uic trigger signal to start the oscillations. Cnce the

oscillations have started, no e)ternal signal source is necessary.

. !n order to get continuous undamped output from the circuit, the following

condition must be metQ

mvA2*$

where A2* voltage gain of amplifier without feedbac.

mv* feedbac fraction.

&his relation is also called arhausen criterion

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar +/

Amplifier

Feedbacnetwor


85/113

Essentials of !ransistor "s$illator

Fig. below shows the bloc diagram of an oscillator. !ts essential components are8

$. &an #ircuit8 !t consists of inductance coil3?4 connected in parallel with

capacitor3# 4. &he fre'uency of oscillations in the circuit depends upon the valuesof inductance of the coil and capacitance of the capacitor.

(. &ransistor Amplifier8 &he transistor amplifier receives d.c. power from the battery

and changes it into a.c. power for supplying to the tan circuit. &he oscillations

occurring in the tan circuit are applied to the input the transistor amplifer. &he

output of the transistor can be supplied to the tan circuit to meet the losses.

. Feedbac circuit8 &he feedbac circuit supplies a part of collector energy to the

tan circuit in correct phase to aid the oscillations. ie. provides positive feedbac.

9ig. Blo$+ diagra* of !ransistor "s$illator

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar +5

&ransistorAmplifier

Feedbaccircuit


86/113

!yes of !ransistor "s$illators

$. Hartley Cscillator

(. #olpitts Cscillator

. -hase 6hift Cscillator/. &uned #ollector Cscillator

5. ein ridge Cscillator

1. #rystal Cscillator

&. artley "s$illator

&he circuit diagram of Hartley Cscillator is as shown in figure below. !t usestwo inductors placed across common capacitor # and the center of two inductors ins

tapped. &he tan circuit is made up of ?$, ?(and # and is given by.

#Lf

*(

$= 3(4

where ?&* ?$" ?(" (D

D* Dutual inductance between ?$and ?(

9igure% Cir$uit diagra* of artley "s$illator

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar +1


87/113

hen the circuit is turned C%, the capacitor is charged. hen this capacitor is

fully charged, it discharges through coils ?$ and ?( setting up oscillations of

fre'uency determined by e)pression $. &he output voltage of the amplifier

appears across ?(and feedbac voltage across ?$. &he voltage across ?$is $+00

out of phase with the voltage developed across ?(.

A phase shift of $+00is produced by the transistor and a further phase shift of

$+00is produced by ?$?(voltage divider circuit. !n this way feedbac is properly

phased to produce continuous undamped oscillations.

9eedba$+ fra$tion- !n Hartley oscillator the feedbac voltage is across ?$and outputvoltage is across ?(.

&herefore feedbac fraction(

$

(

$

L

L

+

+

v

vm

L

L

out

f

v === 34

). ColittXs "s$illator

9igure% Cir$uit diagra* of ColittXs "s$illator

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar +:


88/113

&he tan circuit is made up of #$, #( and ?. &he fre'uency of oscillations is

determined by8

($

($

(

$

##

###where

L#f

*

*

+=

=

3/4

hen the circuit is turned C%, the capacitor #$and #( are charged. &he capacitors

discharge through ? setting up oscillations of fre'uency determined bye)pression. $. &he output voltage appears across #( and feedbac voltage is

developed across #$. &he voltage across #$is $+00out of phase with the voltage

developed across #(32out 4. A phase shift of $+00is produced by the transistor and

a further phase shift of $+00 is produced by #$#( voltage divider. !n this way

feedbac is properly phased to produce continuous undamped oscillations.

9eedba$+ fa$tor

Feedbac factor$

(

(

$

#

#

+

+

V

Vm

#

#

out

f

v === 354

3e*erits of "s$illator using !an+ Cir$uit

$. &hey suffer for fre'uency instability and poor waveform

(. &hey cannot be used to generate low fre'uencies, since they become toomuch

buly and e)pensive too.

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar ++


89/113

2. RC Phase Shift "s$illator

9igure% Cir$uit diagra* of RC hase shift "s$illator

!t consists of a conventional single transistor amplifier and a 9# phase shift

circuit. &he 9# phase shift circuit consists of three sections 9$#$, 9(#(, and

9#.At some particular fre'uency f0the phase shift in each 9# section is 100so

that the total phase shift produced by the 9# networ is $+0 0. &he fre'uency of

oscillation is given by

1(

$

R#fo

= 314

hen the circuit is switched C% it produces oscillations of fre'uency determined

by e'uation $. &he output EC of the amplifier is feedbac to 9# feedbacnetwor. &his networ produces a phase shift of $+00 and the transistor gives

another $+00shift. &hereby total phase shift of the output signal when fed bac is

100

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar +


90/113

Merits-

$. &hey do not re'uire any transformer or inductor thereby reduce the cost.

(. &hey are 'uite useful in the low fre'uency range where tan circuit

oscillators cannot be used.

. &hey provide constant output and good fre'uency stability.

3e*erits D

$. !t is difficult to start oscillations.

(. &he circuit re'uires a large number of components.

. &hey cannot generate high fre'uencies and are unstable as variable fre'uency

generators.

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar


91/113

!ransistor Crystal "s$illator

9igure% Cir$uit diagra* of !ransistor $rystal os$illator

Figure shows the transistor crystal oscillator. &he crystal will act as parallel >tuned circuit. At parallel resonance, the impedance of the crystal is ma)imum.

&his means that there is a ma)imum voltage drop across #(. &his in turn willallow the ma)imum energy transfer through the feedbac networ.

&he feedbac is "ve. A phase shift of $+00is produced by the transistor. A furtherphase shift of $+00is produced by the capacitor voltage divider. &his oscillatorwill oscillate only at fp.

here fp* parallel resonant fre'uency ie the fre'uency at which the vibratingcrystal behaves as a parallel resonant circuit.

m

m

*

*p

##

###where

L#f

+=

= ($

3:4



92/113

Ad7antages

$. Higher order of fre'uency stability(. &he Yfactor of the crystal is very high.

3isad7antages

$. #an be used in low power circuits.(. &he fre'uency of oscillations cannot be changed appreciably.



93/113

CAP!ER

C"MM4NICA!I"N SVS!EMS

Radio Broad$asting, !rans*ission and Re$etion

9adio 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.

!t essentially consists of transmitter, transmission of radio waves and radio receiver.

9)g antenna &ransmitting

Antenna

9igure% Blo$+ diagra* of Co**uni$ation syste*

!rans*itter

!t essentially consists of microphone, audio amplifiers, oscillator and modulator.

A microphone is a device which converts sound waves into electrical waves. &he output

of microphone is fed to multistage audio amplifier for raising the strength of wea signal.

&he =ob of amplification is performed by cascaded audio amplifiers. &he amplified output

from the last audio amplifier is fed to the modulator for rendering the process of

modulation.



94/113

&he function of the oscillation is to produce a high fre'uency signal called a carrier wave.

Ssually crystal oscillator is used for the purpose.

&he amplified audio signal and carrier waves are fed to the modulator. Here the audio

signal is superimposed on the carrier wave in suitable manner. &he resultant waves arecalled modulated waves, and the process is called modulation. &he process of modulationpermits the transmission of audio signal at the carrier signal 3fre'uency4. As the carrierfre'uency is very high, therefore the audio signal can be transmitted to large distances.&he radio waves from the transmitter are fed to the transmitting antenna or aerial fromwhere these are radiated into space.

&he transmitting antenna radiates the radio waves in space in all directions. &hese radio

waves travel with the velocity of light )$0+mGsec. &he radio waves are electromagnetic

waves and possess the same general properties.

Re$ei7er-Cn reaching the receiving antenna, the radio waves induce tiny emf in it. &his smallvoltage is fed to the radio receiver. Here the radio waves are first amplified and thensignal is e)tracted from them by the process of demodulation. &he signal is amplified byaudio amplifiers and then fed to the speaer for reproduction into sound waves.

Need for *odulation

&. Pra$ti$al Antenna length-theory shows that in order to transmit a wave effectively

the length of the transmitting antenna should be appro)imately e'ual to the

wavelength of the wave.

metres,-fre)uencyfre)uency

Veocitywaveen&th

43

$0 +

==

As the audio fre'uencies range from (0 Hz to (0hz, if they are transmitted directly into

space, the length of the transmitting antenna re'uired would be e)tremely large. For

e)ample to radiate a fre'uency of (0 Hz directly into space we would need an antenna

length of )$0+G(0)$0\ $5,000 meters. &his is too long to be constructed practically.



95/113

). "erating Range- &he energy of a wave depends upon its fre'uency. &he greater

the fre'uency of the wave, the greater the energy possessed by it. As the audio signal

fre'uencies are small, therefore these cannot be transmitted over large distances if

radiated directly into space.

2. ;ireless $o**uni$ation- 9adio transmission should be carried out without wires.

Modulation-&he process of changing some characteristics 3e)ample amplitude,

fre'uency or phase4 of a carrier wave in accordance with the intensity of the signal is

nown as modulation.

!yes of *odulation-

$. Amplitude modulation

(. Fre'uency modulation

. -hase modulation

&. A*litude *odulation

hen the amplitude of high fre'uency carrier wave is changed in accordance with theintensity of the signal, it is called amplitude modulation.

&he following points are to be noted in amplitude modulation .

$. &he amplitude of the carrier wave changes according to the intensity of the signal.

(. &he amplitude variations of the carrier wave is at the signal fre'uency f6.

. &he fre'uency of the amplitude modulated wave remains the same ie.carrier

fre'uency f#.



96/113

es

t

ec

t

e

t

Figure8 AD waveforms

Modulation fa$tor

&he ratio of change of amplitude of carrier wave to the amplitude of normal carrier waveis called modulation factor.

m*3amplitude change of carrier wave4 G normal carrier wave3unchanged4



97/113

A " %o 6ignal *

carrier m*0GA * 0M

(A" *

carrier signal

m*3(AA4GA *$

Dodulation factor is very important since it determines the strength and 'uality of the

transmitted signal. &he greater the degree of modulation, the stronger and clearer will be

the audio signal. !t should be noted that if the carrier is overmodulated 3ie m$4 distortion

will occur at reception.

Analysis of a*litude *odulated a7e

signal

mEc

E# Ec

#arrier

AD ave



98/113

A carrier wave is represented by ec* Eccoswct3$4

here ec instantaneous voltage of carrier. Ecamplitude of carrier.

!n amplitude modulation, the amplitude E#of the carrier wave is varied inaccordance with intensity of the signal as shown in figure.

6uppose m*modulation inde), then change in carrier amplitude *mEc.Amplitude or Ema) of the signal * mEc.

es *mEccoswst3(4

where mEcis the amplitude of the signal.

es instantaneous voltage of the signal.

&he amplitude of the carrier varies at signal fre'uency fs. &herefore the amplitude of ADwave is given by, Ec"mEccoswst * Ec3$"mcoswst4

&he instantaneous voltage of AD wave is,

e * Amplitude ) coswct

434cos3(

4cos3(

cos

_4cos34^cos3(

cos

_coscos(^(

cos

coscoscos

cos4cos$3

+++=

+++=

+=

+=+=

twwm!

twwm!

tw!e

twwtwwm!

tw!

twtwm!

tw!

twtwm!tw!

twtwm!e

sc

c

sc

c

c#

scsc

c

c#

cs

#

c#

cs#c#

cs#

&he AD wave is e'uivalent ot thesummatoin of theree sinusoidal waves8 aone

having amplitude E#and fre'uency fc, the second having amplitde mEcG( and

fre'uency 3fc" fs4 and the third having amplitude mEcG( and fre'uency fc> fs..

&he AD wave consists three fre'uencies viz, fc, fc"fs. &he first fre'uency is the

carrier fre'uency. &hus the process of modulation doesnot change the original

carrier fre'uency but produces two new fre'uencies fc"fs and fc> fs.which are

called sideband fre'uencies.



99/113

!n amplitude modulation the bandwidth is from fc> fs.to fc"fs ie (fsie twice thesignal fre'uency.

Fre'uency spectrum of an amplitude modulated wave is shown in figure below.

E#

mE#G(

f#f6 f# f#"f6 fre'uency

9igure% 9re>uen$y Se$tru* of AM a7e



100/113

!ransistor AM *odulator

9igure% !ransistor AM *odulator

A circuit which does amplitude modulation is called AD modulator.

Fig. above shows the circuit of a simple AD modulator. !t is essentially a #E

amplifier having a voltage gain of A. &he carrier signal is the input to the

amplifier. &he modulating signal is applied in the emitter resistance circuit.

&he amplifier circuit amplifies the carrier by a factor ZA[ so that the output is Ae c.

6ince the modulating signal is part of the biasing circuit it produces low

fre'uency variations in the circuit. &his in turn causes variations in ZA[. &he

result is that the amplitude of the carrier varies in accordance with the strength of

the signal. &he amplitude modulated output is obtained across 9?.

%otes by .9.6udhindra 7housia #ollege of Engineering 9amnagar $00


101/113

Poer in AM a7e

4

Basic Electronics

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