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Analog Electronics Circuits Lab Manual III Semester ECE

ANALOGELECTONICS LAB

MANUAL

SUBJECT CODE:

10ESL37

STAFF INCHARGE:

Prof. Akashkumar. BhagatProf. Mahesh. MahendrakarProf.Pooja. ShanbhagProf.Saritha. MProf. Rajadheep. DharProf. Vinay Sangolli

DEPARME! "# E$E%R"!&%S A!D %"MM'!&%A&"!S(A&! %"$$E)E "# E!)&!EER&!)* BE$)A'M




Prepared By+ Prof. Vinay Sangolli

CONTENTS

SI.NO EXPERIMENTS PAGE NO

1 RC-Coupled Amplifer

2 Darlington Emitter Follower

3 Voltage Series Feedback Amplifer

4 R.C. Pase Si!t "scillator

5 #artle$ and Colpitts "scillator

6 Cr$stal "scillator

7 Diode Cliping Circuits

Diode Clamping Circuits

! Class %&' Pus-Pull Amplifer

10 Rectifer Circuits

11 Verifcation o! (e)inin's (eorem and

*a+imum Power (rans!er teorem

12 Series , Parallel Resonance Circuits





E,PER&ME! !" -

R% %"'P$ED AMP$&#&ER

A&M: To design and wire a RC Coupled single stage BJT amplifier and determination of te gainfre!uenc" response# input and output impedances$

%"MP"!E! RE'&RED:

Sno %omponent uantity Spe/ifi/ation% Transistor % BJT SL%&&' Capacitor ( %)*# &$+,)*( Resistor - %&./ 0'1# %./# 2+$(./ 3 4RB+ Signal 5enerator % 3- CR6 % 37 4ecade Resistance Bo8 % 3

, Connecting wires 3 3

%&R%'& D&A)RAM+





&!P' &MPEDE!%E+

"'P' &MPEDE!%E

DES&)!+

Spe/ifi/ations+

6perating points 0IC9%mA# CE9 -1

Stabilit"9%&

oltage 5ain A;9%&

*L9-&&<=

De0i/e +

SL%&& 0Si npn transistor1

BE9 &$,

>9%&&

%?oe 9 %&&./

o find R E +





IE9IC@ IC ? > 9 %mA@%?%&& mA9 %$&%mA

R E 9 E ? IE 9% ? %$&% mA 9&/

Let R E 9 %.Ω 0 standard1

o find R % +

A; 9 Rc ? R E

%&9Rc ? % .Ω

Rc9 %& .Ω

o #ind V%%

*rom te collector loop writing L we get CC 9 ICR C @ CE @ E

∴ cc9 0%mA%& .Ω 1@-@0%mA% .Ω 1

9 %7

o find R b- and R b1+

S2 R b' 3 R E

%&2 R b'?%$&%mA

R b'9%& .Ω

Driting te base loop L we get B 9 E @ BE 9 % @ &$,B 9 %$,

ow B 9 0R b' ? R b% @R b'1 cc

%$, 9 0%& .Ω? R b%@ %& .Ω 1%7

9 2+$( .Ω

o #ind &nput &mpedan/e+

Fin 2 R b% GG R b' 44 ie

ie9 > r e

r e9 '7m? IE





Fin9 %$- .Ω

o #ind "utput &mpedan/e+

Fo 2 R c GG 0%? oe1 9 %& .ΩGG %&& .Ω 9%& .Ω

o find %%i * %%o +

Hcci9 &$%Fin

*L9-&&<=

CCi 9%$7()* 9 %)* 0Standard1

Hcco9 &$%Fo

*L9-&&<=

CCo 9&$(-)* 9 &$+,)* 0Standard1

Hce9 &$%R E

*L9-&&<=

CCe 9($%2) 04CB1

oltage 5ain A;9 3feRc ? ie

9 3+&&

"BSERVA&"!+

Table % : 4C Conditions Input and 6utput Impedence

arameter CE cc Ic

Teoretical

ractical

Table ' : Input and 6utput Impedence

arameter Fi Fo

Teoretical

ractical





Table (: *re!uenc" response: in 9 olts

heory+

Te common emitter RC coupled amplifier is one of te simplest and elementar" transistor amplifier tat can be made$ 4ont e8pect muc boom from tis little circuit# te main purpose of


* = o0p3p1 A 9 o0p3p1 ? in0p3p1$ A in db 9 '&log A

%&&

'&&

(&&

-&&

%.

(.

-.

%-.

'&.

'-.

(&.

-&.

%&&.

'&&.

(&&.

+&&.

-&&.

,&&.

&&.%M




tis circuit is pre3amplification i$e to ma.e wea. signals strong enoug for furter processing or amplification$ If designed properl"# tis amplifier can pro;ide e8cellent signal caracteristics$

Input Capacitor Cin is te input 4C decoupling capacitor wic bloc.s an" 4C component if present in te input signal from reacing te K% base$ If an" e8ternal 4C ;oltage reaces te

base of K%# it will alter te biasing conditions and affects te performance of te amplifier$

Te resistors R% and R' are te biasing resistors$ Tis networ. pro;ides te transistor K%s basewit te necessar" bias ;oltage to dri;e it into te acti;e region$ Te region of operation werete transistor is completel" switced of is called cut3off region and te region of operation werete transistor is completel" switced 6 0li.e a closed switc1 is called saturation region$ Teregion in between cut3off and saturation is called acti;e region$ *or a transistor amplifier tofunction properl"# it sould operate in te acti;e region$ Let us consider tis simple situationwere tere is no biasing for te transistor$ As we all .now# a silicon transistor re!uires &$, ;oltsfor switc 6 and surel" tis &$, will be ta.en from te input audio signal b" te transistor$ Soall parts of tere input wa;e form wit amplitude &$, will be absent in te output wa;eform$

In te oter and if te transistor is gi;en wit a ea;" bias at te base #it will enter intosaturation 0full" 61 and bea;es li.e a closed switc so tat an" furter cange in te basecurrent due to te input audio signal will not cause an" cange in te output$ Te ;oltage acrosscollector and emitter will be &$' at tis condition 0ce sat 9 &$'1$ Tat is w" proper biasingis re!uired for te proper operation of a transistor amplifier$

Pro/edure+

a$ o plot #re5uen/y response : %$ Connections are made as per te gi;en circuit diagram$

'$ Switc on te 4$C$ power suppl" and cec. te 4$C$ conditions witout an" inputsignal and record in table %$($ Select sine wa;e input and set te input signal fre!uenc" at %&.<= constant# and

obser;e te input wa;e and output wa;e on te CR6 and adNust te input amplitudesuc tat te output is undistorted wa;eform$ Calculate midband gain using A 9o0p3p1 ? in0p3p1$

+$ eeping te input amplitude constant# ;ar" te fre!uenc" from -&= to %M= and-$ note down te corresponding output ;oltage0p3p1 in te table '$7$ Calculate gain in db and plot te fre!uenc" response cur;e and find te bandwidt$

b. o find input impedan/e 6in+%$Connect 4RB in series wit te input signal and set it to FER6$

'$ Set te signal fre!uenc" to %&.<= and measure te output ;oltage0p3p1$

($ ar" te 4RB from & to ma8imum suc tat te output signal ;oltage reduces to alfits ;alue$ Tis ;alue of 4RB at wic te output signal reduces to alf its ;alue if teinput impedance Fin$





/. o find output impedan/e 6o+%$Connect 4RB in parallel wit te output and set it to MAHIMOM$

'$ Set te signal fre!uenc" to %&.<= and measure te output ;oltage0p3p1$

($ ar" te 4RB from ma8imum to minimum suc tat te output signal ;oltage reducesto alf its ;alue$ Tis ;alue of 4RB at wic te output signal reduces to alf its ;alue ifte output impedance Fo$

E,PE%ED)RAP7+

0Frequency

A

v

i

n

d

b

f1 f2

B W

3db

Bandwidt 9 f ' 3 f %

RES'$+

E,PER&ME! !" 1





DAR$&!)"! EM&ER #"$$"8ER

A&M: To design a 4arlington Emitter *ollower and determine te gain and input# outputimpedance$

%"MP"!E!S RE'&RED:

S6

Components ame Kuantit" Specification

% Transistor ' SL%&&' Resistors ( -&.P # ((.P # %.P( Capacitors % &$%-)* # '-2)*+ Connecting wires 3 3

- *unction 5enerator % 37 Bread Board % 3, 4c Ammeter % mA2 4c oltmeter % 3 4RB % 3

%&R%'& D&A)RAM+

DES&)!+

IC 9 (mA # CE 9 2 # Fin 9 '&.P # ie 9 %$' .P

%$ Assume > 0 for bot transistors1 9 %&&





E 9 ( #R E 9V

E

I E but IE 9IC

R E 9 % .P

'$ CC9 CE @E9 %%

($ B9 E @BE9 +$+ but B9 CC [ R2

R1+ R

2 ] 9 %$- Q R '

+$ Fin 9 R % R ' >' R ELet 9 R % R ' 9 R Fin 9 R >' R ER 9 '&$&+.P

-$ Input Coupling CapacitanceHCci 9 &$% Fin at lowest fre!uenc" of interestf L 9 -&& <= Cci 9 &$%- )* &$%)*

7$ 6utput ImpedanceFo 9 URs RB@ ie V? fe W R E

9 %'$(+P,$ 6utput Coupling Capacitance Cco

HCo 9 &$% Fo at lowest fre!uenc" of interestf L 9 -&& <= Cco9 '-2)* %&&)* parallel to %&&)*

"BSERVA&"!S+

-. D% %onditions:

cc Ic CE

Teoreticalractical

1. &nput and output &mpedan/e +

Fin FoTeoreticalractical





9. #re5uen/y Response+ *or ;i933333333W

#re5uen/y

:7;<

Vo

:0olts<

)ain in dB

2 1= log :0o30i<

7E"R> +

A ;er" popular connection of two BJTs for operation as one super beta transistor is te4arlington connection$ Te main feature of 4arlington connection is tat te compositetransistor acts# as a single unit wit a current gain is e!ual to product of indi;idual current gains$

i$e$ >4 9 >%8>' if >%9 >' 9 > Ten >4 9 >'

To ma.e te two transistors 4arlington pair# te emitter terminal of te first transistor isconnected to te base of te second transistor and te collector terminals of te two transistorsare connected togeter$ Te result is tat emitter current of te first transistor is te base currentof te second transistor$ Te biasing anal"sis is similar to tat for single transistor e8cept tat twoBE drops are to be considered$ In a circuit if te output ;oltage is appro8imatel" e!ual to teinput ;oltage# suc a circuit is .nown as emitter follower$ In te transistor emitter follower circuit te output is ta.en from te emitter terminal$ Te ;oltage gain is appro8imatel" e!ual tounit" and output ;oltage is in pase wit te input ;oltage$ Te emitter follower configuration isfre!uentl" used for impedance matcing and to increase te current gain$ Sometimes te current

gain and input impedance of emitter followers are insufficient to meet te re!uirement$ In order to increase te o;erall ;alues of circuit current gain 0Ai1 and input impedance# two transistors areconnected in series in emitter follower configuration to obtain 4arlington connection$

PR"%ED'RE +

%$ Connections are made as per te circuit diagram$'$ Te suppl" ;oltages of %% and te input pea. to pea. ;oltage .ept at &$% at %<=

fre!uenc" from signal generator is applied to te circuit($ Te fre!uenc" of te input is ;aried from -&<= to 'M<= and output ;oltage o is noted

down$+$ 5ain is calculated in dB and a grap of fre!uenc" ;ersus gain in dB is plotted on semilog

seet$?. 4C Conditions and input impedance and output impedance are also noted down$





E,PE%ED )RAP7+

RES'$+

E,PER&ME! !" 9





V"$A)E SER&ES #EEDBA%@ AMP$&#&ER

A&M:

4esign of a *ET oltage series feedbac. amplifier and determine te gain# fre!uenc"response# input and output impedances wit and witout feedbac.$

%"MP"!E!S RE'&RED:

S6


% *ET % B*D%&' Resistors ( %$-.P # %.P# 'MP# %7-P#

72P( Capacitors ' &$&%)* # &$+,)*

+ Connecting wires 3 3- Bread Board % 37 4C power suppl" % 3

%&R%'& D&A)RAM+





DES&)!+





7E"R> +

Amplifiers use current or ;oltage as input and output# so four t"pes of amplifier are possible$ Tefour basic t"pes of amplifiers are as follows:

oltage amplifier X Tis is te most common t"pe of amplifier$ An input ;oltage is amplified to alarger output ;oltage$ Te amplifierYs input impedance is ig and teoutput impedance is low$Current amplifier X Tis amplifier canges an input current to a larger output current$ TeamplifierYs input impedance is low and te output impedance is ig$Transconductance amplifier X Tis amplifier responds to a canging input ;oltage b" deli;eringrelated canging output current$ Transresistance amplifier X Tis amplifier responds to acanging input current b" deli;ering a related canging output ;oltage$ 6ter names for tede;ice are transimpedance amplifier and current3to3;oltage con;erter $An" of tese four coices ma" be te open3loop amplifier used to construct te feedbac.

amplifier$ Te obNecti;e for te feedbac. amplifier also ma" be an" one of te four t"pes of amplifier# not necessaril" te same t"pe as te open3loop amplifier$ *or e8ample# an op amp0;oltage amplifier1 can be arranged to ma.e a current amplifier instead$ Te con;ersion from onet"pe to anoter is implemented using different feedbac. connections# usuall" referred to as seriesor sunt 0parallel1 connections$

PR"%'D'RE +

Rig up te circuit as sown in te circuit diagram$

%$ Cec. K conditions i$e#$ measure 4S and 5S$

'$ Set i 9 % or ' at %& <= on Audio signal 5enerator and measure gain A witoutfeedbac.$($ 4isconnect sort of 5reen and Blac. terminal of signal generator to a;oid grounding

problem or isolate ground of signal generator$+$ Measure 6 wit feedbac. Z find Af te gain wit feedbac.$ ote Af is less tan A-$ To plot fre! response# note output ;oltage wit and witout feedbac. from %&&<= to

%&M<=





E,PE%ED )RAP7+

RES'$+

*ET oltage series feedbac. amplifier is designed and te fre!uenc" response# input and outputimpedence# gain wit and witout feedbac. are determined $

E,PER&ME! !"


http://en.wikipedia.org/wiki/Output_impedance

http://en.wikipedia.org/wiki/Current-to-voltage_converter

http://en.wikipedia.org/wiki/Input_impedance




R% P7ASE S7&# "S%&$$A"R

A&M: To design and stud" te wor.ing of an RC ase Sift oscillator using BJT$

%"MP"!E! RE'&RED:

Sno %omponent uantity Spe/ifi/ation% Transistor % BJT BC%&,' Capacitor + &$&%)* 0(1# &$+, )*( Resistor + ''./ # %./# 7$2./# +,&/ 3 4RB# ($./+ Signal 5enerator % 3- CR6 % 37 4ecade Resistance Bo8 % 3

, Connecting wires 3 3

%&R%'& D&A)RAM+

%al/ulations+

heoreti/al #re5uen/y %al/ulations

f 9 % ? 0'[RCs!rt07@+11 were 9 Rc ? R

let R9 ($./ and C 9 &$&%)*





Rc 9 %& ./

f 9 % ? 0'[($ ./ &$&% )* s!rt07@+0%?($111

9%$-( .= 0appro81

"bser0ed #re5uen/y

2

f 2 - 3

"BSERVA&"!+

Teoretical*re!uenc" f t 0<=1

ractical*re!uenc" f p0<=1

Error 9 0f t 3 f p1 ? f t %&&

heory+

RC pase sift oscillator or simpl" RC oscillator is a t"pe of oscillator were a simple RCnetwor. 0resistor3capacitor1 networ. is used for gi;ing te re!uired pase sift to te feedbac.signal$ In LC oscillators li.e <artle" oscillator and Colpitts oscillator an LC networ. 0inductor3capacitor networ.1 is used for pro;iding te necessar" positi;e feedbac.$ Te main feature of anRC pase sift oscillator is te e8cellent fre!uenc" stabilit"$ Te RC oscillator can output a puresine wa;e on a wide range of loads$RC pase sift networ. is a simple resistor capacitor networ. tat can be used to gi;e a desired pase sift to a signal$ Te circuit diagram of a simple single stage RC networ. is sown in tefigure below$

Teroticall" in a simple RC circuit# te output ;oltage will lead te input ;oltage b" a paseangle \ 9&]$ An"wa" in practical case te pase angle will be someting below &] Nust because it is impossible to get a purel" ideal capacitor$ ase sift of a practical RC networ.depends on te ;alue of te capacitor# resistor and te operating fre!uenc"$ Let * be te operatingfre!uenc"# R be te resistance and C be te capacitance$ Ten te capaciti;e reactance Hc to tefre!uenc" * can be gi;en b" te e!uation Hc 9 % ? 0'[*C1$Te effecti;e impedance of te circuit





can be gi;en b" te e!uation F 9 ^0 R_ @ Hc_1$ Te pase angle of te RC networ. can bederi;ed as \ 9 tan3% 0Hc?R1$ Just b" ma.ing an RC networ. wit pase sift e!ual to 7&] andcascading tree of tem togeter te desired pase sift of %2&] can be attained$ Tis %2&] pasesift b" te RC networ. plus te %2&] pase sift made b" te transistor gi;es a total pase sift

of (7&] between te input and output wic is te necessar" condition for maintaining sustainedoscillations$

Pro/edure+

%$ Test te components$'$ Connections are made as per te gi;en circuit diagram$($ AdNust te dc source to proper ;alue$+$ Te dc input is gi;en to te circuit and te output fre!uenc" is noted and compared

wit te desired fre!uenc"

E,PE%ED )RAP7+

RES'$+

E,PER&ME! !" ?


http://www.circuitstoday.com/hartley-oscillator

http://www.circuitstoday.com/colpitts-oscillator




#E 7AR$E> A!D %"$P&S "S%&$$A"R

A&M: To design <artle" and colpitts oscillator for gi;en fre!uenc" using *ET

%"MP"!E!S RE'&RED:

S6


% *ET % B*D%%' Resistors - %$-.P#%.P0'1# 2' .P# %MP( Capacitors + &$%)*0'1# +, )*# ((&p*+ Connecting wires 3 3- Signal 5enerator % 3

7 Bread Board % 3, 4C power suppl" % 32 CR6 % 3

7AR$E> "S%&$$A"R+

%&R%'& D&A)RAM+





DES&)!+





%"$P&S "S%&$$A"R+

%&R%'& D&A)RAM+

DES&)!+





7E"R>+

7artley os/illator+

Te <artle" oscillator is an electronic oscillator circuit in wic te oscillation fre!uenc" is

determined b" a tuned circuit consisting of capacitors and inductors# tat is# an LC oscillator$ Tecircuit was in;ented in %%- b" American engineer Ralp <artle"$ Te distinguising feature of te <artle" oscillator is tat te tuned circuit consists of a single capacitor in parallel wit twoinductors in series 0or a single tapped inductor1# and te feedbac. signal needed for oscillation ista.en from te center connection of te two inductors

Te <artle" oscillator is distinguised b" a tan. circuit consisting of two series3connected coils 0or# often# a tapped coil1 in parallel wit a capacitor# wit an amplifier betweente relati;el" <ig impedance across te entire LC tan. and te relati;el" low ;oltage?ig

current point between te coils$ Te original %%- ;ersion used a triode as te amplif"ing de;icein Common plate 0catode follower1 configuration# wit tree batteries# and separate adNustablecoils$ Te most simple of implementations sown ere uses noting but a J*ET 0in Common3drain configuration1 and te LC tan. circuit 0ere te single winding is tapped1# plus a single batter"` it will wor.# but probabl" wit ig distortion and ig current drain 0tat could beimpro;ed b" adding resistance between te source and te coil tap1# and ser;es as anuncomplicated illustration of te <artle" oscillator operation:

• te output from te J*ETYs source 0emitter # if a BJT ad been used` cathode for a triode1as te same pase as te signal at its gate 0or base1 3 and rougl" te same ;oltage as itsinput 0wic is te ;oltage across te entire tan. circuit1# but te current is amplified # i$e$ it isacting as a current buffer or ;oltage3controlled ;oltage3source$

• tis low impedance output is ten fed into te coil tapping# effecti;el" intoan autotransformer tat will step up te ;oltage# re!uiring a relati;el" ig current0compared wit tat a;ailable at te top of te coil1$

• Dit te capacitor3coil resonance# all fre!uencies oter tan te tuned fre!uenc" willtend to be absorbed 0te tan. will appear as nearl" =ero oms near 4C due to te inductorYs

low reactance at low fre!uencies# and low again at ;er" ig fre!uencies due to tecapacitor1` te" will also sift te pase of te feedbac. from te =ero degrees needed for oscillation at all but te tuned fre!uenc"$

%olpitts os/illator+ Te Colpitts circuit# li.e oter LC oscillators# consists of a gain de;ice 0sucas a bipolar Nunction transistor # field effect transistor# operational amplifier# or ;acuum tube1 wit





its output connected to its input in a feedbac. loop containing a parallel LC circuit 0tuned circuit1wic functions as a bandpass filter to set te fre!uenc" of oscillation$

A Colpitts oscillator is te electrical dual of a <artle" oscillator # were te feedbac. signal is

ta.en from an inducti;e ;oltage di;ider consisting of two coils in series 0or a tapped coil1$ *ig$% sows te common3base Colpitts circuit$ L and te series com bination of C % and C ' form te parallel resonant tan. circuit wic determines te fre!uenc" of te oscillator$ Te ;oltageacross C ' is applied to te base3emitter Nunction of te transistor# as feedbac. to createoscillations$ *ig$ ' sows te common3collector ;ersion$ <ere te ;oltage across C % pro;idesfeedbac.$ Te f re!uenc" of oscillation is appro8imatel" te resonant fre!uenc" of te LC circuit#wic is te series combination of te two capacitors in parallel wit te inductor

PR"%'D'RE+

%$ Ma.e te connections as sown in circuit diagram$'$ 6bser;e te sinusoidal output ;oltage$($ Measure te fre!uenc" and compare wit te teoretical ;alues$

E,PE%ED )RAP7+

RES'$+ 4iode clipping circuits for pea. clipping and pea. detection is ;erified$


http://en.wikipedia.org/wiki/Electronic_oscillator


http://en.wikipedia.org/wiki/Electronic_circuit

http://en.wikipedia.org/wiki/Tuned_circuit

http://en.wikipedia.org/wiki/Capacitor


http://en.wikipedia.org/wiki/Inductor


http://en.wikipedia.org/wiki/Ralph_Hartley

http://en.wikipedia.org/wiki/Feedback

http://en.wikipedia.org/wiki/Tank_circuit


http://en.wikipedia.org/wiki/Coil_tap

http://en.wikipedia.org/wiki/High_impedance

http://en.wikipedia.org/wiki/Distortion

http://en.wikipedia.org/wiki/Reactance



http://en.wikipedia.org/wiki/Electronic_circuit




http://en.wikipedia.org/wiki/Ralph_Hartley

http://en.wikipedia.org/wiki/Feedback

http://en.wikipedia.org/wiki/Tank_circuit


http://en.wikipedia.org/wiki/Coil_tap

http://en.wikipedia.org/wiki/High_impedance

http://en.wikipedia.org/wiki/Triode

http://en.wikipedia.org/wiki/Common_plate

http://en.wikipedia.org/wiki/Junction_FET

http://en.wikipedia.org/wiki/Common-drain

http://en.wikipedia.org/wiki/Common-drain

http://en.wikipedia.org/wiki/Distortion

http://en.wikipedia.org/wiki/Bipolar_Junction_Transistor

http://en.wikipedia.org/wiki/Phase

http://en.wikipedia.org/wiki/Buffer_amplifier#Current_buffer

http://en.wikipedia.org/wiki/VCVS

http://en.wikipedia.org/wiki/Autotransformer

http://en.wikipedia.org/wiki/Resonance


http://en.wikipedia.org/wiki/Transistor

http://en.wikipedia.org/wiki/Vacuum_tube

http://en.wikipedia.org/wiki/Feedback_loop

http://en.wikipedia.org/wiki/LC_circuit


http://en.wikipedia.org/wiki/Bandpass_filter




E,PER&ME! !" C

%R>SA$ "S%&$$A"R

A&M: To test te performance of BJT3Cr"stal 6scillator for fre!uenc" f & %&&.<=

%"MP"!E!S RE'&RED:

S6


% Transistor % SL%&&' Resistors ( %&.P0'1 # %.P( Capacitors ' %)* # &$+,)*+ Connecting wires 3 3- Cr"stal % 3

7 Bread Board % 3, 4CB ( 32 4C power suppl" % 3 4RB % 3

%&R%'& D&A)RAM+


http://en.wikipedia.org/wiki/Hartley_oscillator




%A$%'$A&"!S+

2 f 21

T 2

7E"R> +

6ne of te most important features of an" oscillator is its fre!uenc" stabilit"# or in oter wordsits abilit" to pro;ide a constant fre!uenc" output under ;ar"ing load conditions$ Some of tefactors tat affect te fre!uenc" stabilit" of an oscillator generall" include: ;ariations intemperature# ;ariations in te load as well as canges to its 4C power suppl" ;oltage to name afew$ *re!uenc" stabilit" of te output signal can be impro;ed b" te proper selection of tecomponents used for te resonant feedbac. circuit including te amplifier but tere is a limit tote stabilit" tat can be obtained from normal LC and RC tan. circuits$ To obtain a ;er" igle;el of oscillator stabilit" a Kuart= Cr"stal is generall" used as te fre!uenc" determining de;ice

to produce anoter t"pes of oscillator circuit .nown generall" as a Kuart= Cr"stal 6scillator#0H61$ Te !uart= cr"stal used in a Kuart= Cr"stal 6scillator is a ;er" small# tin piece or wafer of cut !uart= wit te two parallel surfaces metallised to ma.e te re!uired electricalconnections$ Te p"sical si=e and tic.ness of a piece of !uart= cr"stal is tigtl" controlledsince it affects te final or fundamental fre!uenc" of oscillations$ Te fundamental fre!uenc" iscalled te cr"stals caracteristic fre!uenc"$





Kuart= Cr"stal E!ui;alent Model

Te e!ui;alent circuit for te !uart= cr"stal sows an RLC series circuit# wic represents temecanical ;ibrations of te cr"stal# in parallel wit a capacitance# Cp wic represents teelectrical connections to te cr"stal$ Kuart= cr"stal oscillators operate at parallel resonance#

and te e!ui;alent impedance of te cr"stal as a series resonance were Cs resonates witinductance# L and a parallel resonance were L resonates wit te series combinationof Cs and Cp as sown$

PR"%ED'RE+

%$ Connections are made as per te circuit diagram$'$ Te dc ;oltage of %7 is applied to te circuit$($ 6bser;e te output wa;eform and note te fre!uenc" of te wa;eform$+$ Te fre!uenc" of output wa;eform is compared wit te fre!uenc" of cr"stal0 bot

sould matc1$E,PE%ED )RAP7+

RES'$+

#re5uen/y heoreti/al+ 1M7; Pra/ti/al+





E,PER&ME! !"

D&"DE %$&PP&!) %&R%'&S

A&M: To stud" te diode clipping circuits for pea. clipping and pea. detection

%"MP"!E!S RE'&RED:

S6


% 4iode % I+&&,' Resistors % %&.P( Connecting wires 3 3+ Signal 5enerator % 3

- Bread Board % 37 4C power suppl" % 3, CR6 % 3

%&R%'& D&A)RAM+

-. !egati0e Series %lipper ith Positi0e Referen/e Voltage.abular %olumn





1. !egati0e Series %lipper ith !egati0e Referen/e Voltage.

9. Positi0e Series %lipper ith Positi0e Referen/e Voltage.

. Positi0e Series %lipper ith !egati0e Referen/e Voltage.


in ref o

in ref o

in ref o




?. !egati0e Parallel %lipper ith Positi0e Referen/e Voltage.

C. !egati0e Parallel %lipper ith !egati0e Referen/e Voltage.


in ref o

in ref o

in ref o




. Positi0e Parallel %lipper ith Positi0e Referen/e Voltage.

F. Positi0e Parallel %lipper ith !egati0e Referen/e Voltage

G. %lipping at o &ndependent le0els.

-=. %lipping at o &ndependent le0els : Sli/er<.


in ref o

in ref o

in ref% ref' o




7E"R>+

Te 4iode Clipper# also .nown as a 4iode Limiter# is a wa;e saping circuit tat ta.es an inputwa;eform and clips or cuts off its top alf# bottom alf or bot al;es togeter to produce anoutput wa;eform tat resembles a flattened ;ersion of te input$ *or e8ample# te alf3wa;e

rectifier is a clipper circuit# since all ;oltages below =ero are eliminated$

But 4iode Clipping Circuits can be used a ;ariet" of applications to modif" an inputwa;eform using signal and Scott." diodes or to pro;ide o;er3;oltage protection using Fener 4iodes to ensure tat te output ;oltage ne;er e8ceeds a certain le;el protecting te circuit fromig ;oltage spi.es$ Ten diode clipping circuits can be used in ;oltage limiting applications$

PR"%'D'RE +

%$ Connections are made as per te circuit diagram$

'$ Te input ;oltages of 2 pea. to pea. at '.<= is applied from te signal generator to tecircuit$

($ *or te gi;en input# ref is ;aried and te clipped wa;eform i$e output wa;eform can beobser;ed on te CR6 screen and amplitude of o is noted down in te tabular column$

+$ eep te CR6 in te dual mode# connect te input in to Cannel % of CR6 and outputwa;eform is connected to Cannel ' of CR6# now .eep CR6 in H3 mode and obser;ete transfer caracteristics$

E,PE%ED )RAP7+

Vin


in ref% ref' o

http://amazon.in/s/?field-keywords=Zener+Diodes+and+Their+Applications






-. !egati0e Series %lipper ith Positi0e Referen/e Voltage.

1. !egati0e Series %lipper ith !egati0e Referen/e Voltage.

9. Positi0e Series %lipper ith Positi0e Referen/e Voltage.

. Positi0e Series %lipper ith !egati0e Referen/e Voltage.





?. !egati0e Parallel %lipper ith Positi0e Referen/e Voltage.

C. !egati0e Parallel %lipper ith !egati0e Referen/e Voltage.

. Positi0e Parallel %lipper ith Positi0e Referen/e Voltage.

F. Positi0e Parallel %lipper ith !egati0e Referen/e Voltage





G. %lipping at o &ndependent le0els.

-=. %lipping at o &ndependent le0els : Sli/er<.

RES'$+ 4iode clipping circuits for pea. clipping and pea. detection is ;erified$

E,PER&ME! !" F





%$AMPER %&R%'&S

A&M: To stud" te diode clamping circuits for positi;e and negati;e clamping$

%"MP"!E!S RE'&RED:

S6


% 4iode % I+&&,' Resistors % %&&.P( Capacitor % &$%)*+ Connecting wires 3 3- Signal 5enerator % 3

7 Bread Board % 3, 4C power suppl" % 3

2 CR6 % 3

%&R%'& D&A)RAM+

-. !egati0e %lamper ith Positi0e Referen/e Voltage. abular %olumn

D

Vin

Vref

CR)p- p/#0

1 23445

R1

100kΩ

4

C

67

4. 78F

1. !egati0e %lamper ith !egati0e Referen/e Voltage.

D

Vin

Vref

CR)p- p/#0

1 23445 R1

100kΩ

C

67

4. 78F

4


in ref o

in ref o




9. Positi0e %lamper ith Positi0e Referen/e Voltage.

D

Vin

Vref

CR")p- p/#0

1 23445 R100kΩ

C

6 4. 78F

7

4

. Positi0e %lamper ith !egati0e Referen/e Voltage.

D

Vin

Vref

CR)p - p/#0

1 23445 R1

100kΩ

C

6 4. 78F

4

7

7E"R>+

A clamping circuit is used to place eiter te positi;e or negati;e pea. of a signal at a desiredle;el$ Te dc component is simpl" added or subtracted to?from te input signal$ Te clamper isalso referred to as an IC restorer and ac signal le;el sifter$

In some cases# li.e a T recei;er# wen te signal passes troug te capaciti;e couplingnetwor.# it loses its dc component$ Tis is wen te clamper circuit is used so as to re3establiste te dc component into te signal input$ Toug te dc component tat is lost in transmissionis not te same as tat introduced troug a clamping circuit# te necessit" to establis tee8tremit" of te positi;e or negati;e signal e8cursion at some reference le;el is important$ A

clamp circuit adds te positi;e or negati;e dc component to te input signal so as to pus it eiter on te positi;e side or on te negati;e side$ Te circuit will be called a positi;e clamper # wente signal is pused upward b" te circuit$ Te circuit will be called a negati;e clamper# wente signal is pused downward b" te circuit$


in ref o

in ref o




Te important points regarding clamping circuits are:

0i1 Te sape of te wa;eform will be te same# but its le;el is sifted eiter upward or downward#

0ii1 Tere will be no cange in te pea.3to3pea. or rms ;alue of te wa;eform due to teclamping circuit$ Tus# te input wa;eform and output wa;eform will a;e te same pea.3to3 pea. ;alue tat is# 'ma8$ Tis is sown in te figure abo;e$ It must also be noted tat samereadings will be obtained in te ac ;oltmeter for te input ;oltage and te clamped output;oltage$

0iii1 Tere will be a cange in te pea. and a;erage ;alues of te wa;eform$ In te figure sownabo;e# te input wa;eform as a pea. ;alue of ma8 and a;erage ;alue o;er a complete c"cle is=ero$ Te clamped output ;aries from ' ma8 and & 0or & and 3'ma81$ Tus ts pea. ;alue of teclamped output is 'ma8 and a;erage ;alue is ma8$

0i;1 Te ;alues of te resistor R and capacitor C affect te wa;eform$

0;1 Te ;alues for te resistor R and capacitor C sould be determined from te time constante!uation of te circuit# t 9 RC$ Te ;alues must be large enoug to ma.e sure tat te ;oltageacross te capacitor C does not cange significantl" during te time inter;al te diode is non3conducting$ In a good clamper circuit# te circuit time constant t 9 RC sould be at least tentimes te time period of te input signal ;oltage$

PR"%ED'RE+

%$ Connections are made as per te circuit diagram$'$ Te input ;oltages of 2 pea. to pea. at '.<= is applied from te signal generator to te

circuit$($ *or te gi;en input# ref is ;aried and te clamped wa;eform i$e output wa;eform can be

obser;ed on te CR6 screen and clamped le;el is noted down in te tabular column$. eep te CR6 in te dual mode# connect te input in to Cannel % of CR6 and output

wa;eform is connected to Cannel ' of CR6

E,PE%ED )RAP7





-. Positi0e %lamper ith !egati0e Referen/e Voltage.

1. Positi0e %lamper ith Positi0e Referen/e Voltage.

9. !egati0e %lamper ith !gai0e Referen/e Voltage.

. !egati0e %lamper ith Positi0e Referen/e Voltage.





RES'$+ 4iode clamping circuits for positi;e and negati;e clamping are ;erified$

E,PER&ME! !" G





%$ASS B P'S7 P'$$ P"8ER AMP$&#&ER

A&M: Testing of a transformer less class B pus pull power amplifier and determination of itscon;ersion efficienc"

%"MP"!E!S RE'&RED:

S6


% Transistor ' SL%&&#C%&&' Resistors % %.P( Capacitors % %)*+ Connecting wires 3 3- *unction 5enerator % 3

7 Bread Board % 3

, 4c Ammeter % mA

%&R%'& D&A)RAM+

%A$'%'$A&"!S+

Collector current IC 9 33333

4C Current Idc 9 'QIc

9 3333333

6utput oltage o0p3p1 9





RL 9

cc 9 2

i0dc1 9ccQ Idc

9 33333333

o0ac1 9Vo ( p− p)8×RL

2

9 33333333333

9ɳ

Po(ac) Pi (dc) Q %&&

9 3333333333333

7E"R>+

6ne of te main disad;antages of te Class B amplifier circuit abo;e is tat it uses balancedcenter3tapped transformers in its design# ma.ing it e8pensi;e to construct$ <owe;er# tere isanoter t"pe of Class B amplifier called a Complementar"3S"mmetr" Class B Amplifier tatdoes not use transformers in its design terefore# it is transformerless using insteadcomplementar" or matcing pairs of power transistors$ As transformers are not needed tis

ma.es te amplifier circuit muc smaller for te same amount of output# also tere are no stra"magnetic effects or transformer distortion to effect te !ualit" of te output signal$ Te maindisad;antage of class B t"pe pus3pull amplifiers is tat te" suffer from an effect .nowncommonl" as Crosso;er 4istortion$ A simple wa" to eliminate crosso;er distortion in a Class Bamplifier is to add two small ;oltage sources to te circuit to bias bot te transistors at a pointsligtl" abo;e teir cut3off point$ Tis ten would gi;e us wat is commonl" called an Class ABAmplifier circuit$ <owe;er# it is impractical to add additional ;oltage sources to te amplifier circuit so 3Nunctions are used to pro;ide te additional bias in te form of silicon diodes$

PR"%'D'RE+

%$ Connections are made as per te circuit diagram$'$ Te suppl" ;oltages are .ept at 2 and te input pea. to pea. ;oltage .ept at %- in

CR6$($ Te resulting Idc is noted from ammeter$+$ Te output o0p3p1 is raised to matc input and resulting power are calculated and also

efficienc"$





E,PE%ED )RAP7+

RES'$+


http://www.electronics-tutorials.ws/amplifier/amp_7.html




E,PER&ME! !" -=

RE%&#&ERS

A&M: To test te alf wa;e# full wa;e rectifiers wit and witout filter and determining te ripplefactor# regulation and efficienc"$

%"MP"!E!S RE'&RED:

S6


% Resistors % %&.P

' Connecting wires 3 3( Capacitor % %&& )( Bread Board % 3+ 4c Ammeter % mA- 4iode % I+&&,7 Transformer % 3, Multimeter % 3

%&R%'& D&A)RAM+

-.A< 7alf 8a0e Re/tifier 8ithout #ilter





-.B< 7alf 8a0e Re/tifier ith filter

AB'$A&"! %"$"'M!+

Vsrms

:in 0olts<

Idc

0in mA1

dc*L

0in 1

Isrms

0in mA1

Vrrms

:in 0olts<

Ripple*actor

Efficienc"

oltageRegulation

DitoutfilterDit filter

%&R%'& D&A)RAM+

1.A <#'$$ 8AVE RE%&#&ER 8&7"' #&$ER+





1.B <#'$$ 8AVE RE%&#&ER 8&7 #&$ER+

AB'$A

&"! %"$"'M!+

Vsrms

:in 0olts<

Idc

0in mA1

dc*L

0in 1

Isrms

0in mA1

Vrrms

:in 0olts<

Ripple*actor

Efficienc"

oltageRegulation


%&R%'& D&A)RAM+

9.A< #'$$ 8AVE BR&D)E RE%&#&ER 8&7"' #&$ER+

9.B< #'$$ 8AVE BR&D)E RE%&#&ER 8&7 #&$ER+





AB'$A&"! %"$"'M!+

Vsrms

:in 0olts<

Idc

0in mA1

dc*L

0in 1

Isrms

0in mA1

Vrrms

:in 0olts<

Ripple*actor

Efficienc"

oltageRegulation


7E"R>+

A rectifier is an electrical de;ice tat con;erts alternating current 0AC1# wic periodicall" re;erses direction# to direct current 04C1# wic flows in onl" one direction$ Te process is .nown as rectification$ In alf wa;e rectification of a single3pase suppl"# eiter te positi;e or negati;e alf of te AC wa;e is passed# wile te oter alf is bloc.ed$ Because onl"

one alf of te input wa;eform reaces te output# mean ;oltage is lower$ <alf3wa;e rectificationre!uires a single diode in a single3pase suppl"# or tree in a tree3pase suppl"$ Rectifiers "ielda unidirectional but pulsating direct current` alf3wa;e rectifiers produce far more ripple tanfull3wa;e rectifiers# and muc more filtering is needed to eliminate armonics of te ACfre!uenc" from te output$

In a *ull Da;e Rectifier circuit two diodes are now used# one for eac alf of te c"cle$A multiple winding transformer is used wose secondar" winding is split e!uall" into two al;eswit a common centre tapped connection# 0C1$ Tis configuration results in eac diodeconducting in turn wen its anode terminal is positi;e wit respect to te transformer centre

point C producing an output during bot alf3c"cles twice tat for te alf wa;e rectifier so it is%&& efficient





Te full wa;e rectifier circuit consists of two power diodes connected to a single load

resistance 0RL1 wit eac diode ta.ing it in turn to suppl" current to te load$ Den point A of te transformer is positi;e wit respect to point C# diode 4% conducts in te forward direction asindicated b" te arrows$Den point B is positi;e 0in te negati;e alf of te c"cle1 wit respectto point C# diode 4' conducts in te forward direction and te current flowing trougresistor R is in te same direction for bot alf3c"cles$ As te output ;oltage across teresistor R is te pasor sum of te two wa;eforms combined# tis t"pe of full wa;e rectifier circuit is also .nown as a bi3pase circuit$ As te spaces between eac alf3wa;e de;eloped b"eac diode is now being filled in b" te oter diode te a;erage 4C output ;oltage across teload resistor is now double tat of te single alf3wa;e rectifier circuit and isabout &$7(,ma8 of te pea. ;oltage# assuming no losses$

Anoter t"pe of circuit tat produces te same output wa;eform as te full wa;e rectifiercircuit abo;e# is tat of te *ull Da;e Bridge Rectifier$ Tis t"pe of single pase rectifier usesfour indi;idual rectif"ing diodes connected in a closed loop bridge configuration to producete desired output$ Te main ad;antage of tis bridge circuit is tat it does not re!uire a specialcentre tapped transformer# tereb" reducing its si=e and cost$ Te single secondar" winding isconnected to one side of te diode bridge networ. and te load to te oter side of te circuit$

PR"%ED'RE +

%1 Connections are made as per te circuit diagram$

1< All te ;alues of srms# Isrms# rrms# dc*L# dcL are measured and noted down in te tabularcolumn.

(1 Te Efficienc"# oltage regulation and ripple factor are calculated$+1 Te same readings as to be noted down wit filter$

E,PE%ED )RAP7+

-. 7alf a0e re/tifier ithout filter


http://en.wikipedia.org/wiki/Electric_power_conversion

http://en.wikipedia.org/wiki/Alternating_current

http://en.wikipedia.org/wiki/Direct_current

http://en.wikipedia.org/wiki/Single-phase_electric_power

http://en.wikipedia.org/wiki/Three-phase_electric_power

http://en.wikipedia.org/wiki/Electric_power_conversion

http://en.wikipedia.org/wiki/Alternating_current

http://en.wikipedia.org/wiki/Direct_current

http://en.wikipedia.org/wiki/Single-phase_electric_power

http://en.wikipedia.org/wiki/Three-phase_electric_power

http://www.electronics-tutorials.ws/transformer/multiple-winding-transformers.html




1. 7alf a0e re/tifier ith filter

9. #ull a0e bridge re/tifier ithout filter

. #ull a0e bridge re/tifier ith filter





RES'$+ Ripple factor# Regulation and efficienc" of alf wa;e rectifier# full wa;e center taprectifier and full wa;e bridge rectifier are ;erified$

E,PER&ME! !" --

VER&#&%A&"! "# 7EVE!&!HS A!D MA,&M'M P"8ER RA!S#ER

7E"REM #"R D% %&R%'&S

A&M:

a1 erification of Te;enins teorem b" comparing te ;alue of load current 0 IL 1 witgi;en networ. and wit Te;enins networ.$

b1 erif" te ma8imum power transfer teorem for te gi;en networ.$

%"MP"!E!S RE'&RED:

S6


% Resistors - %&./ 0+1# +$,/#' Connecting wires 3 3( Bread Board % 3+ 4c Ammeter % mA- ower Suppl" % 3

7 4RB % 3

%&R%'& D&A)RAM+





7EV&!&!HS E'&VA$E! %&R%'&+

RES&SA! E'&VA$E! %&R%'&+

7E"RA&%A$ %A$%'$A&"!S+

"BSERVA&"!+

V7

:V<

R 7

:@I<

&$

:mA<Teoreticalractical

PR"%ED'RE " #&!D V7 +

%$ 6pen circuit te load branc$'$ eep all te acti;e sources as te" are in te networ.$9. *ind te open circuit ;oltage across te open circuited load branc$





PR"%ED'RE " #&!D R 7

-. 6pen circuit te load branc$1. Reduce all te sources to =ero$9. *ind te e!ui;alent resistance ta.ing into te open circuited load branc$

b<MaJimum Poer ransfer heorem.

%&R%'& D&A)RAM+

"BSERVA&"!S+

$oad Resistan/eR L 0/1

%urrentIL 0mA1

PoerP2 IL

1 R L 0D1

7E"R>+

Te;enins Teorem states tat An" linear circuit containing se;eral ;oltages and resistancescan be replaced b" Nust a Single oltage in series wit a Single Resistor$ In oter words# it is





possible to simplif" an" Linear circuit# no matter ow comple8# to an e!ui;alent circuit wit Nust a single ;oltage source in series wit a resistance connected to a load as sown below$Te;enins Teorem is especiall" useful in te Circuit Anal"sis of power or batter" s"stems andoter interconnected circuits were it will a;e an effect on te adNoining part of te circuit$

Te MaJimum Poer ransfer heorem is anoter useful Circuit Anal"sis metod toensure tat te ma8imum amount of power will be dissipated in te load resistance wen te;alue of te load resistance is e8actl" e!ual to te resistance of te power source$ Terelationsip between te load impedance and te internal impedance of te energ" source willgi;e te power in te load$ te impedance of te load will ;ar" from an open3circuit state to asort3circuit state resulting in te power being absorbed b" te load becoming dependent on teimpedance of te actual power source$ Ten for te load resistance to absorb te ma8imum power possible it as to be Matced to te impedance of te power source and tis forms te basis of MaJimum Poer ransfer$

PR"%ED'RE+

%$ Test te components$'$ Connections are made as per te circuit diagram$($ Load resistance is ;aried in steps and corresponding load current ;alues are noted down$

+$ ower deli;ered to te load is calculated and note it down in te tabular column$-$ A grap of load resistance ;ersus ower is plotted and ;erified$

E,PE%ED )RAP7+

RES'$+ Te;enins teorem and ma8imum power transfer teorem are ;erified$DEPARME! "# E$E%R"!&%S A!D %"MM'!&%A&"!S(A&! %"$$E)E "# E!)&!EER&!)* BE$)A'M




E,PER&ME! !" -1

RES"!A!%E

A&M: To design a series and parallel RLC circuit and obtain its caracteristics

%"MP"!E!S RE'&RED:

S

6


% Signal 5enerator % 3

' atc Cords? Connecting Dires 3 3( 4c Ammeter % mA+ 4ecade Resistance Bo8 % 3- 4ecade Inductance Bo8 %7 4ecade Capacitance Bo8 % 3

%&R%'& D&A)RAM+

Series Resonan/e


http://amazon.in/s/?field-keywords=Circuit+Analysis+For+Dummies

http://amazon.in/s/?field-keywords=Circuit+Analysis+For+Dummies




Parallel Resonan/e

DES&)!+

#or Series Resonan/e

Let Resonant fre!uenc" f & 9 %.<= Assume te ;alue of te Capacitance C

Resonant fre!uenc" f & for series RLC is gi;en b"

f & 91

2 π √ LC

f '& 91

4 π 2 LC

L 91

4 π 2

f 20C

L 9 $$$$$$$$$$$$$$$$$$$





#or Parallel Resonan/e

Let Resonant fre!uenc" f & 9 %.<=

Assume te ;alue of Inductance L

Resonant fre!uenc" f & is gi;en b"

f & 91

2 π √ 1

LC − R

2

L2

Substitute te ;alue of L and f & in te abo;e e!uation and find te ;alue of C

C 9 $$$$$$$$$$$$$$$$$$$"SERVA&"!S+

#or Series and Parallel Resonan/e

;i933333333

#re5uen/y:7;<

%urrent:mA<

7E"R> +

In a series RLC circuit tere becomes a fre!uenc" point were te inducti;e reactance of teinductor becomes e!ual in ;alue to te capaciti;e reactance of te capacitor$ In oter words#HL 9 HC$ Te point at wic tis occurs is called te Resonant *re!uenc" point# 0 hr 1 of te circuit# and as we are anal"sing a series RLC circuit tis resonance fre!uenc" producesa Series Resonance$ Series Resonance circuits are one of te most important circuits usedelectrical and electronic circuits$ Te" can be found in ;arious forms suc as in AC mains filters#noise filters and also in radio and tele;ision tuning circuits producing a ;er" selecti;e tuning

circuit for te recei;ing of te different fre!uenc" cannels$In man" wa"s a parallel resonance circuit is e8actl" te same as te series resonance

circuit$ Bot are (3element networ.s tat contain two reacti;e components ma.ing tem asecond3order circuit# bot are influenced b" ;ariations in te suppl" fre!uenc" and bot a;e afre!uenc" point were teir two reacti;e components cancel eac oter out influencing tecaracteristics of te circuit$ Bot circuits a;e a resonant fre!uenc" point$ Te difference tis





time owe;er# is tat a parallel resonance circuit is influenced b" te currents flowing trougeac parallel branc witin te parallel LC tan. circuit$ A tan. circuit is a parallel combinationof L and C tat is used in filter networ.s to eiter select or reNect AC fre!uencies$ A parallelcircuit containing a resistance# R# an inductance# L and a capacitance# C will produce aparallelresonance 0also called anti3resonance1 circuit wen te resultant current troug te parallelcombination is in pase wit te suppl" ;oltage$ At resonance tere will be a large circulatingcurrent between te inductor and te capacitor due to te energ" of te oscillations# ten parallelcircuits produce current resonance$

A parallel resonant circuit stores te circuit energ" in te magnetic field of te inductor and te electric field of te capacitor$ Tis energ" is constantl" being transferred bac. and fort between te inductor and te capacitor wic results in =ero current and energ" being drawnfrom te suppl"$ Tis is because te corresponding instantaneous ;alues of I L and IC will alwa"s

be e!ual and opposite and terefore te current drawn from te suppl" is te ;ector addition of tese two currents and te current flowing in IR $

In te solution of AC parallel resonance circuits we .now tat te suppl" ;oltage is common for all brances# so tis can be ta.en as our reference ;ector$ Eac parallel branc must be treatedseparatel" as wit series circuits so tat te total suppl" current ta.en b" te parallel circuit is te;ector addition of te indi;idual branc currents$

PR"%ED'RE+

%$ Connections are made as per te circuit diagram$'$ Te input ;oltages of 2 pea. to pea. at %.<= is applied from te signal generator to te

circuit$($ Set te designed R#L Z C ;alues decade bo8es$+$ Te fre!uenc" is ;aried at regular inter;als of time Z te corresponding current is noted

down$-$ *rom te abo;e reading a grap of fre!uenc" ;ersus current is plotted $

E,PE%ED )RAP7+

SERIES RES6AT CIRCOIT ARALLEL RES6AT CIRCOIT





RES'$+


aec_manual_vin (1).docx

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