Teo Ec eBook-3

8/8/2019 Teo Ec eBook-3

1/136

EE-305.72 to 73 5

fr Frequency

Gain

Low

frequency

roll-off

High

frequencyroll-off

Resonant

rise

Flat

resp

onse

Frequency response of Transformer coupled amplifier


2/136

EE-305.72 to 73 6

Reactance of the coil w.r.t. Frequency

jXL

Frequency

0

XL=2 f L

Frequency

XL=2 f L


3/136

EE-305.72 to 73 7

At Low Frequencies

The output voltage of a TC amplifier is equal to the

product of the collector current and the reactance of

the primary winding of the coupling transformer.

Voltage gain rolls off (decreases)

Why?

Reactance of primary begins to fall, resulting in

decreased gain.


4/136

EE-305.72 to 73 8

Voltage gain rolls off

(decreases)

Why

The capacitance

between turns of

windings acts as a

bypass capacitor

Reduces the output

voltage and hence gain

At High Frequencies

Distributed

capacitance


5/136

EE-305.72 to 73 9

Peak gain and Flat Response

Peak gain

It results due to the resonance effect of inductance

and distributed capacitance

Resonant frequency

- The frequency at which the resonant occurs is called

resonant frequency fr Flat response

-Small as compared to that of RC coupled amplifier


6/136

EE-305.72 to 73 10

Transformer Impedance matching

2

1

2

1

V

V

n

n

=2

1

2

1

I

I

n

n

=

V1V2

I1 I2

V1/I1=RL=Effective input impedance

V2/I2=RL=Effective output impedance

2

2

1

1 Vn

nV = 2

1

2

1 In

nI =

2

2

2

1

1

2

1

I

V

n

n

I

V

=LLL RnR

n2

n1R2

2

=

=


7/136

EE-305.72 to 73 11

If we want to match a 20 speaker load to a amplifier

so that the effective load may be 8K,then the turns ratio

should be

LLL RnR

n2

n1R

2

2

=

=

Given

R'L=8000 ,

RL=20

n=20


8/136

EE-305.72 to 73 12

Advantages Of Transformer Coupled Amplifier

No signal power is lost in the collector or base resistors

Excellent impedance matching

Higher gain

D.C. isolation between first and second stages


9/136

EE-305.72 to 73 13

Out put

impedance is

several

hundred ohms

Input impedance

of speaker isonly a few ohms

impedance

matching

CE

Amplifier

Reflected

impedancehigh

Impedance matching


10/136

EE-305.72 to 73 14

Poor frequency response

The coupling transformer is bulky and costly

Introduces hum in the circuit

Not used for amplifying audio frequencies

Disadvantages


11/136

EE-305.72 to 73 15

Applications

Used for amplifying radio frequency signals

Final stage of a multistage amplifier

For amplifying the power level of the input signal


12/136

EE-305.72 to 73 16

Summary

We have discussed about the

Frequency response of transformer coupled amplifier

Factors affecting the frequency response

Advantages

Disadvantages

Applications


13/136

EE-305.72 to 73 17

QUIZ

1. The frequency response of transformercoupling is

(a) Good

(b) Very good

(c) Excellent

(d) Poor


14/136

EE-305.72 to 73 18

2. The final stage of a multistage amplifier uses

(a) RC coupling

(b) Transformer coupling

(c) Direct coupling

(d) Any of the above


15/136

EE-305.72 to 73 19

1. In Transistor amplifiers ,the type of transformerused for impedance matching is

(a) Step-Up

(b) Step-Down

(c) Same turns ratio

(d) None of the above


16/136

EE-305.72 to 73 20

2. Transformer coupling is generally employedwhen the load impedance is

(a) Large

(b) Very large

(c) small

(d) None of the above


17/136

EE-305.72 to 73 21

Frequently Asked Questions

1. Draw the frequency response of Transformer coupledamplifier and Explain.

3. Explain why the gain falls at high frequencies as well

as at low frequencies?

5. List the advantages and disadvantages of Transformercoupled amplifier

7. List the applications of transformer coupled amplifier?


18/136

Recap

Already we discussed about the

Multistage amplifiers and

Their necessity

EE-305.75 2


19/136

EE305.75 3

Objectives

After the completion of the period student will be

able to know

Gain and band width of an amplifier.

Decibel gain

Why the gain is expressed in decibels.

Gain of a multistage amplifier

Frequency response of an amplifier.


20/136

EE305.75 4

GAIN

Def: The ratio of the output electrical quantity to the

input of the amplifier is called gain Electrical quantities

are voltage, current and power accordingly gain can be

voltage gain, current gain or power gain.

VinVoutAV =

in

outI

IIA =

PinPoutAP=

Voltage gain Current gain Power gain


21/136

EE305.75 5

Block diagram of 2- Stage CE cascade amplifier

FIRST

STAGE

SECOND

STAGEV1

V2V3


22/136

EE305.75 6

Voltage Gain of a Multi Stage Amplifier

voltage gain of the first stage

is

voltage gain of the secondstage

voltage gain of the multi stage

amplifier is

From equation 1 & 2

1

21

V

VAV =

2

32

V

V

AV =

1

2

2

3

1

3

V

VX

V

V

V

VAV ==

21 VVV XAAA =

Eq no.1

Eq no.2

From Eqno.1&2


23/136

EE305.75 7

Voltage gain continued

Voltage gain of the multi stage amplifier is equal to the

product of the voltage gains of the individual stages.

For n stage cascaded amplifier.

Magnitude of the total voltage gain.AV=AV1XAV2XAVn

= 1 + 2 +------------- n

Total phase shift

Due to loading effect above condition is not satisfied


24/136

EE305.75 8

Decibel Gain

Def: Ten times the common logarithm of the ratio of output

power to the input power is known as decibel gain

in

t

P

P

Powergain

ou

10log10=

in

t

V

VnindBVolatgegai

ou

10log20=

in

t

I

InindBCurrentgai

ou

10log20=

1 bel =10 Decibels


25/136

EE305.75 9

Properties of power gain and voltage gain

Factor Power gain indecibels Voltage gain indecibels

X 1 0 dB 0 dB

X 2 +3 +6

X 10 +10 +20

X 0.5 -3 -6

X 0.1 -10 -20


26/136

EE305.75 10

Gain of multi stage amplifier in dB

Gain of multi stage (n stage) amplifier is the product of

the gains of the individual stages

A = A1 X A2 X A3-----An

Taking logarithm on both sides10log10A = 10log10 (A = A1 X A2 X A3----An )

= 10log10 (A1) + 10log10 (A2) ----+ 10log10 (An)

AdB = AdB1 +AdB2 + AdB3 +----+AdBn

The over all gain in dB of a multi stage amplifier is the

sum of the decibel voltage gains of the individual stages.


27/136

EE305.75 11

Why dB is used?

It permits gains to be directly added when a number of

stages are cascaded

Use of logarithms changes multiplication in to an

addition

It permits us to denote, both very small as well as very

large quantities of linear scale by conveniently small

figures

Ex: Voltage Gain = 0.000001 = -120dBVoltage Gain = 456000 = 56dB

It tallies with the natural response of our ears


28/136

EE305.75 12

Frequency Response

The curve between Gain and signal frequency of anamplifier is known as frequency response.

Vol

tagegain

Frequencyfr

Maximum gain


29/136

EE305.75 13

Band Width The range of frequencies over which the gain is equal to or

greater than 70.7% of the maximum gain is known as band

widthVoltage gain

Frequencyfrf1 f2

Am

0.707 Am

f1-lower cut-off

frequency

f2-upper cut-off

frequency

The gain bandwidth

product is consant BAND WIDTH=f2-f1


30/136

EE305.75 14

3 dB frequencies

Lower cut off frequency:

f1 Frequency at which the magnitude of the low

frequency range falls to 0.707 times of the maximum

gain .

Upper cut off frequency

f2 Frequency at which the magnitude of the voltage

gain in the high frequency range falls to 0.707 times

of the maximum gain .

f1 and f2 are also known as 3dB frequencies or half

power frequencies.


31/136

EE305.75 15

QUIZ

1. A gain of 1,000,000 of times in power isexpressed by

(a) 30dB

(b) 60dB

(c) 20dB

(d) 600dB


32/136

EE305.75 16

1. 1dB corresponds to -----------change in powerlevel

(a) 50%

(b) 35%

(c) 26%

(d) 22%


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EE305.75 17

1.The band width of a single stage amplifier is

--------that of a multi stage amplifier

(a) more than

(b) the same as

(c) less than

(d) data


34/136

EE305.75 18

1. The upper or lower cut off frequency is also

called--------- frequency

(a) Resonant

(b) Side Band

(c) 3dB

(d) none of the above


35/136

EE305.75 19

Assignment Problems

1. The input power to an amplifier is 15mW while output

power is 2W. Find the decibel gain of the amplifier

3. A multi stage amplifier consists of three stages. The

voltage gains of the stages are 30, 50, and 60. calculate

the over all gain in dB.

EE305.75


36/136

EE305.75 20

Frequently asked questions

1. Define gain and bandwidth of an amplifier ?

3. Define decibel gain and frequency response of an

amplifier?

5. Define lower and upper cutoff frequencies?

EE305.75

Objectives


37/136

Upon completion of this period , you would

be able to

Know what is parallel resonance.

Derive the expression for resonant frequency.

Understand the condition for resonance in

parallel LC circuit.

EC 303 . 61 2

Objectives

Parallel Resonance


38/136

EC 303 . 61 3

Parallel Resonance

The Resonance Phenomenon Occurs in

Parallel LC circuits also as in the case of Series LC

circuits

The parallel resonance occurs at a frequency when

the imaginary part of circuit impedance becomes

Zero

Resonant Frequency


39/136

EC 303 . 61 4

Resonant Frequency

Step1 : Find the expression for admittance of the circuit.

Step2 : Equate the susceptance part of it to zero.

Step3 : Solve for resonant frequency


40/136

EC 303 . 61 5

Y(admittance)

i

Capacitive admittance

CjyC=

Inductive admittance

Ljy

L

1=

Total admittance

LjCjyyy

LC

1

+=+=

Analysis

It is convenient use admittance method when solving

parallel Networks


41/136

EC 303 . 61 6

The above expression can be reduced to.

imaginary part of the admittance is known as the susceptance (B).

Real part of the admittance is called as conductance (G).

= LCjy 1

(1)

Analysis


42/136

EC 303 . 61 7

Conductance G = 0

Susceptance B =

L

C

1

Since Imaginary part is zero under Resonance conditions

Equate succeptance to 0 to find resonant frequency

= 0 (2)

LC

0

0

1


43/136

EC 303 . 61 8

By solving Equation (2) for0

LC

1

0=

f 2=As

LCf 21

0= (3)

Impedance at resonance


44/136

EC 303 . 61 9


=

LCjy

1

Admittance of parallel LC circuit

At resonant frequency f0

LC

1

= 0

Impedance is reciprocal of admittance.

Y = 0



45/136

EC 303 . 61 10

Impedance of parallel LC circuit at

resonance is infinite ( ).

If frequency increase above f0 or decrease

below f0 impedance will decrease.


Frequency versus admittance


46/136

EC 303 . 61 11

Frequency versus admittance

f0

Resonant frequency

Current at resonant frequency


47/136

Current at resonant frequency

EC 303 . 61 12

Impedance at resonant frequency Z = 0.

Current entering into the circuit = .z

v

At resonant frequency the Net current drawn by

parallel LC circuit is equal to zero .

Example1


48/136

EC 303 . 61 13

Example1

Q) Find the value of L at which the circuit shown resonates at

frequency of 1000 rad/sec ?

-12j1000rad/secLC

1

0 = (1)

C0

1 = 12 (2)

FC

3.831000*12

1

12

1

0

===


49/136

EC 303 . 61 14

( )

H

C

L

LC

LC

012.010*3.83*1000

11

1

11

622

0

2

0

0

===

=

=

Therefore the value of L required for the circuit

to be resonant = 0.012H .

Summary


50/136

y

Parallel LC circuit resonates when its succeptance is 0.

Resonant frequency .

Impedance at resonant frequency

Z0 =infinite.

At resonant frequency circuit will act as open circuit.

EC 303 . 61 15

LCf

2

1

0=


51/136

QUIZ


52/136

EC 303 . 61 17

QUIZ

1. Parallel LC circuit resonates when its ___ is zero

a) impedance

b) reactance

c) susceptance

d) none of the above

Ans : ( c )

1) At resonant frequency impedance of parallel


53/136

EC 303 . 61 18

1) At resonant frequency impedance of parallel

resonant circuit is

a) zero

b) infinite

c) low


Ans : ( b )

h d i h f f


54/136

3. Factors that determine the resonant frequency of

parallel LC circuit are

a) L,C values

b) coil resistance

c) only C value

d) only L value

Ans : ( a )

EC 303 . 61 19

4 ) At resonant frequency current entering into the


55/136

4 ) At resonant frequency current entering into the

parallel LC circuit is

a) Zero

b) medium

c) Infinite


Ans : ( a)

EC 303 . 61 20


56/136

EC 303 . 61 21


1. Define resonant frequency of parallel LC circuit ?

3. Derive the expression for resonant frequency of a parallel

LC circuit ?

5. Explain about variation of impedance with respect to

frequency in parallel resonant circuit ?

Objectives


57/136

Objectives

EC 303 . 62 2

On completion of this period , you would beable to

Understand the condition for resonance in parallel

RL-C circuit.

Derive the expression for resonant frequency .

Know the current in parallel RL-C circuit under

resonance.

Know the impedance of parallel RL-C circuit under

resonance.


58/136

EC 303 . 62 3

Resonance in parallel RL-C circuit


59/136

EC 303 . 62 4

A) When do you say parallel RL-C circuit is in

resonance ?

When the susceptance part of its admittance

is zero.

The frequency of excitation at which the

susceptance part of admittance is 0 is known

as resonant frequency f0.


60/136

EC 303 . 62 5

Steps to find the resonant frequency

Step1 : Find the expression for admittance of the circuit

Step2 :Equate the susceptance part of it to zero

Step3 :Solve for frequency which is nothing but resonant

frequency

h fi d


61/136

EC 303 . 62 6

Now execute these steps to find resonant

frequency

R

L

C

Y

Admittance of coil

LjRYL

+

=1

Admittance of C

CjYC =

i

Si lif Y b l i l i


62/136

EC 303 . 62 7

Simplify YL by multiplying numerator

and denominator with (R- j L)

=

+=

)(

)(

)(

1

LjR

LjRX

LjRYL

22 )(

)(

LR

LjR

+

2222

)()( LR

Lj

LR

R

+

+

= (1)


63/136

EC 303 . 62 8

Now the total admittance y

CL YYY +=

Real part in the above eq. is the conductance.

Imaginary part is the susceptance.

++

+= 2222 )()( LR

LCjLR

R

(2)

Equate susceptance part to zero to


64/136

EC 303 . 62 9

Equate susceptance part to zero to

find the resonant frequency

(3)


65/136

EC 303 . 62 10

C

LLR =+ 20

2 )(

Replace with and solve for0 0

2

2

0

1

L

R

LC=

002 f =

(4)


66/136

EC 303 . 62 11

2

2

0

12

L

R

LCf =

Therefore resonant frequency parallel RL-Ccircuit is

2

2

0

1

2

1

L

R

LCf = (5)


67/136

EC 303 . 62 12

Expression for f0can also be written as

=

L

CR

LC

f2

0 1

2

1

(6)

What will happen when >1 ?

Resonant frequency will become imaginary.

But frequency must be real and positive.


68/136

EC 303 . 62 13

Therefore the circuit to have a resonant

frequency

Q) how should be the component values ?

component values should be such that.

>1

Otherwise f0 will become imaginary.


69/136

IMPEDANCE AT RESONANT FREQUENCY

++

+=

2222 )()( LRLCj

LRR

EC 303 . 62 14

Admittance of the circuit from Eq.1

Y

At resonant frequency susceptance part is zero

2

0

2 )( LR

RY

+=


70/136

EC 303 . 62 15

Impedance is reciprocal of admittance

From Eq. (4)

L

CRY =

Impedance of parallel LR-C circuit at resonant

frequency

CR

LZ =

0

CURRENT IN THE CIRCUIT UNDER


71/136

EC 303 . 62 16

CURRENT IN THE CIRCUIT UNDER

RESONANCE

Applied voltage =

Impedance at f0

v

CR

LZ =0

Q)Now what is the current at resonant frequency ?

L

vCR

CRLv

Z

vI ===

0

0

Power factor of parallel LC circuit under


72/136

EC 303 . 62 17

Power factor of parallel LC circuit under

resonance

Impedance /admittance of parallel resonant circuit

Under resonance is pure resistive/conductive.

So the voltage and current are in phase.

Cosine of angle between voltage and current is

known as power factor.

A) Now what is the power factor of an


73/136

EC 303 . 62 18

A) Now what is the power factor of an

parallel LC circuit ?

Power factor = 1)0cos(cos ==


74/136

EC 303 . 62 19

VARIATION OF IMPEDANCE WITH FREQUENCY

Impedance

frequencyfo

Z0

XL>XCXC>XL

Impedance decreases

as frequency deviates

from f0

R

Differences between series and


75/136

Differences between series andparallel resonant circuits

LCf

2

1

0=

L

CR

LC

f2

01

2

1=

EC 303 . 62 20

Parameter series parallel

Resonant frequency

At f0

Impedance(Z0) R(minimum) L/CR(maximum)

Current V/R(very large) (VCR)/L(minimum)

Acts as short circuit Open circuit

power factor unity unity

A) For RL-c circuit shown in fig.find the


76/136

EC 303 . 62 21

) g

resonant frequency ?

10

0.1H

F1010v

Resonant frequency

= L

CR

LCf

2

0 12

1

On substitution of R,L and Cvalues in the above equation

0f =158.35Hz


77/136

summary

LR-C circuit formed by connecting an RL branch and

a capacitor in parallel.

Resonant frequency of the circuit is

Impedance at f0 ,Z0=L/CR.

Impedance is maximum at f0.EC 303 . 62 22

=

L

CR

LCf

2

0 12

1


78/136

EC 303 . 62 23

Impedance decreases as frequency deviates from f0.

Current in the circuit is minimum at f0.

Current increases as frequency deviates from f0.

power factor under resonance is unity.

Quiz) h ll h d f C


79/136

EC 303 . 62 24

1)What will happen to impedance of an RL-C circuit

as frequency increases beyond f0?

a) Increases

b) decreases

c) Unchanged

d)none

Ans : ( b )

2) what is the phase difference betweenl d i ll l RL C i i


80/136

voltage and current in parallel RL-C circuitunder resonance ?

a) 00

b) 900

c) 1350

d) 1800

Ans : ( a)

EC 303 . 62 25

3) To get minimum current in parallel RL-C circuitf f it ti t b


81/136

fequency of excitation must be

a) very high

b) very low

c) equal to f0

d) None

Ans : ( c )

EC 303 . 62 26

4) What will be the impedance of parallel RL-C


82/136

4) What will be the impedance of parallel RL Ccircuit when R=0 ?

a) Zero

b) Infinite

c) Low

d) None

Ans : ( b )

EC 303 . 62 27



83/136


1. Draw the parallel RL-C circuit and derive the

expression for its resonant frequency ?

2. With neat graph explain how impedance of

parallel RL-C circuit vary with frequency ?

3. Derive the expressions for impedance and current

of parallel RL-C circuit under resonance ?

4. Prove that the power factor of tank circuit under

resonance is unity.

EC 303 . 62 28

OBJECTIVES


84/136

EC 303.65 to 66 2

On completion of this period ,you would able to

solve:

problems on series resonance

problems on parallel resonance

RECAP


85/136

EC 303.65 to 66 3

Define resonance ?

Give formula of quality factor ?

What is the relation between Q, Bw, fo?

What is formula for resonant frequency ?

1.Determine the resonant frequency of

i t i it h i Fi


86/136

EC 303.65 to 66 4

series resonant circuit shown in Fig.

10

0.5mH

10 F

Vs

Resonant frequency of series

RLC circuitLC

f2

10=

What is the expression for resonant

frequency of series RLC circuit ?

Fig 1


87/136

EC 303.65 to 66 5

On Substitution of L and C values in the above Eq.

Hzf 22511010105.02

1

630 =

=

Therefore resonant frequency = 2251 Hz

2.Determine the value of inductive reactance


88/136

EC 303.65 to 66 6

of the circuit shown at resonance ?

jXL

50

-j25

What will be the net reactance of

RLC series circuit under resonance ?

Net reactance of RLC series

circuit under resonance X=0

VS

Fig 2

XXXNet reactance


89/136

EC 303.65 to 66 7

=

=

=

25

025

L

L

CL

X

X

XXXNet reactance

Therefore the inductive reactance of the circuit under

resonance = 25 ohms

What is the impedance of the circuit at resonance ?

As reactance is zero impedance of the circuit at

resonance Z0 = R= 50 ohms

3.For the circuit shown in fig3 .Find thefrequency at which maximum voltage


90/136

EC 303.65 to 66 8

frequency at which maximum voltage

appears across capacitor and also find the

maximum voltage across capacitor

0.1 H

10

50 F

50V

When will the voltage across

the capacitor be maximum ?

2

2

2

1

L

R

LC= At

Fig 3

On substitution of L,C,R values in the above Eq.


91/136

EC 303.65 to 66 9

, , q

Current flowing through the circuit at this frequency

A

CLR

VI

968.4

)6

105058.441

11.058.441(100

50

2)(2

=

+

=

+

=

sec/58.4411.02

100

10501.0

126 rad=

=


92/136

EC 303.65 to 66 10

Therefore maximum voltage across the capacitor

voltsXIV CC 1.225105017.447

11.46max=

==

4.In the circuit shown,the maximum current flowsthrough the circuit is 0 5ma determine the


93/136

EC 303.65 to 66 11

through the circuit is 0.5ma.determine the

resonant frequency,the bandwidth,and the

quality factor Q at resonance ?

0.1 H

R

5 F

When maximum current flows through RLC series circuit ?

5v

Continued


94/136

EC 303.65 to 66 12

What is the impedance of RLC at resonant frequency ?

At resonant frequency maximum currentflows through the RLC series circuit

It is Z0 = R

Determination of R value


95/136

EC 303.65 to 66 13

Maximum current

Determination of R value

0

0

50.1

V I mA

Z R

3

550

0.1 10R

50R

Determination of resonant frequency


96/136

EC 303.65 to 66 14

Determination of resonant frequency

What is the expression for resonant frequency of series

RLC circuit ?

It is

LC

f2

1

0=

06

1225

2 0.1 5 10 f Hz

0

0 0

225

2 2 225 14142 / sec

resonant frequency f Hz

and f rad

Determination of quality factor Q


97/136

EC 303.65 to 66 15

What is the expression for quality factor Q of series RLC

resonant circuit ?

It isR

LQ 0

0

=

0

14142 0.128

50Quality factor Q

0 28Q

Determination of bandwidth


98/136

EC 303.65 to 66 16

What is the expression for bandwidth of a series RLC

resonant circuit ?

It is Q

fBW 0=

HzBW 36.8028

225==

Therefore the bandwidth BW=80.36 Hz

5.Determine the lower and upper half-powerfrequencies, bandwidth and then quality


99/136

EC 303.65 to 66 17

factor Q of circuit shown in Fig 5.

0.1 H

R=10

10 F

Maximum current I0 flows

through the circuit at resonant

frequency

current falls to 0.707I0 at half

power frequencies f1 and f2

At resonant frequency

impedance Z0 = R

FIG 5

So what is the impedance of circuit at half power


100/136

EC 303.65 to 66 18

frequencies ?

At half power frequencies

RZ

R

VRVZVI

Z

VI

h

h

h

2

2222

00

=

=====

Therefore at half power frequenciesf1/f2 impedance of the circuit Zh= R2

DETERMINATION OF LOWER HALF POWERFREQUENCY f


101/136

EC 303.65 to 66 19

FREQUENCY f1

1impedance at lower half power frequency f

2

2

1 1

1

1 2 22

h Z R f L Rf C

2

2 2

1

1

12 2

2

R f L R

f C

At lower half power frequency XC

>XL

(1)11

12

2 f L R

f C

On Solving equation(1) for f1


102/136

EC 303.65 to 66 20

2

1

4

4

L

R R Cf

L

6

4 0.110 100

10 10151.39

4 0.1Hz

1lower half power frequency 151.39 f Hz

DETERMINATION OF UPPER HALF POWER

FREQUENCY f2


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EC 303.65 to 66 21

FREQUENCY f2

2impedance at f

2

2

2 2

2

1

2 22h Z R f L R f C

2

2 2

2

2

12 2

2 R f L R

f C

At upper half power frequency XL>XC

2

2

12

2 f L R

f C

On Solving equation(2) for f2


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EC 303.65 to 66 22

2

2

4

4

LR RC

fL

6

4 0.1

10 100 10 10 167.114 0.1

Hz

2upper half power frequency 167.11 f Hz

DETERMINATION OF RESONANT

FREQUENCY


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EC 303.65 to 66 23

FREQUENCY

Hzf

LCf

2.159

10101.02

1

21frequencyresonant

60

0

=

=

=

Therefore the resonant frequency of the

circuit f0 = 159.2 Hz

DETERMINATION OF QUALITY FACTOR Q


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What is the expression for bandwidth ?

Bandwidth BW = f2-f1

Therefore BW = 167.11- 151.39 = 15.72

What is the expression for Q in terms of f0 and BW ?

Q=f0/BW

Therefore quality factor Q=159.2/15.72=10.13

. or an ser es c rcu s own n g. ns an aneousexcitation v=70.7sin(600t) and current i=1.5sin(600t+


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EC 303.65 to 66 25

450) Find R and C values.Also find the frequency at

which the circuit is resonant

0.1 H

C

Continued


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0

0

455.1ithen07.70If=

= v

At the given frequency a leading phase of 450 is

introduced by the circuit

After studying the problem what is your comment on

phase introduced by the circuit

The instantaneous impedance at given frequency


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EC 303.65 to 66 27

The instantaneous impedance at given frequency

(1)

0

070.7 47.13 45

1.5 45vzi

33.33 33.33j 33.33 33.33 z R jX j

On comparison of real and imaginary parts

in the above equation


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EC 303.65 to 66 28

in the above equation

33.33R

600 0.1 60LX L

33.33 33.33 60 93.33C LX X 1

93.33C

XC

117.85

600 93.33C F

33.33C L X X X

DTERMINATION OF RESONANT

FREQUENCY


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EC 303.65 to 66 29

FREQUENCY

Hzf

LCf

12.1191085.171.02

1

2

1frequencyresonant

60

0

=

=

=

Therefore the frequency at which the circuitresonates =119.12 Hz

7.Find the resonant frequency of the

parallel circuit shown in fig.


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EC 303.65 to 66 30

parallel circuit shown in fig.

7

1mH

20 F

i

Given data

R= 7

L= 1mH

C= 20 F

2

2

0

1

2

1

L

R

LCf =

What is the expression for Resonant frequency ?

Fig 7

On substitution of R,L, and C values in the above

equation


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EC 303.65 to 66 31

equation

Hzf 15910

49

102010

1

2

16630=

=

Resonant frequency = 159Hz


114/136

On substitution of L and C values


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EC 303.65 to 66 33

Hz

LCf

6.7117

1001.010502

1

2

1

630

=

==

Therefore resonant frequency = 7117.6Hz

9.The circuit shown in fig. is under resonance atfrequency 500Hz. Q factor of L is 5.find the coil

i t d it l ?


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EC 303.65 to 66 34

resistance and capacitance values ?

RLC

i

L=0.1H

Given data

f0 = 500Hz

QL = 5

L = 0.1H

What is the expression for Q

factor of a coil ?

R

Lf

R

LQ 00

2==

Fig 8

On substitution f0 and L values


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EC 303.65 to 66 35

02 2 500 0.1

62.835

f L

R Q

we know that

0

1

LC

Therefore

R=62.83 and C 0.101F

22

0

1 1C 0.101 F

L 0.1 2 500

it is resonant at frequency 5000 rad/ sec .xL=6


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EC 303.65 to 66 36

RL=8 and RC=7

RL

RC

Given data

XL = 6

RL=8

RC=7

0 = 5000 rad/sec

What will happen to susceptance of the circuit atresonant frequency ?

It will become zero

Find the susceptance of the circuit and equate it to zero

and then solve it for X


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EC 303.65 to 66 37

and then solve it for XC

admittance of R-L branch

2 2

1 8 6 8 6

8 6 8 6 100L

j jY

j

admittance of R-C branch

2 271

7 7C

C

C C

jXY jX X

Total admittance of the circuit


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EC 303.65 to 66 38

Real part is conductance and the imaginary part is known

as susceptance

2 2 2 2

8 7 6

100 7 7 100C

C C

Xj

X X

2 2 2 2

8 6 7

100 100 7 7

C

L C

C C

XY Y Y j j

X X

Equate the susceptance part to zero to find XC


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EC 303.65 to 66 39

2 2 6 07 100C

C

X

X

2 2

6

7 100

C

C

X

X

2100 6 294C CX X

(1)2

6 100 294 0C CX X

On Solving equation (1)

2


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EC 303.65 to 66 40

26 100 294 0

C CX X

2100 100 4 6 29412.85

2 6C

X

0

112.85

CX

C

Therefore the value of C for resonance = 15.56F

0

1 115.56

12.85 5000 12.85C F

11.Determine the value of RL for which theparallel circuit shown in fig is resonant.given

R 4 X 5 d X 10


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EC 303.65 to 66 41

RC=4 XC=5 and XL=10 .

RL

RC

Given data

RC=4

XC=5

XL=10 .

Find the admittance of the circuit and then

equate the susceptance part to zero to find the

RL value

Fig 10


124/136

EQUATE THE SUSCEPTANCE TO ZERO


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EC 303.65 to 66 43

2

5 100

41 100LR

2

10 5

100 41L

R

25 500 410LR

Equating imaginary part in the above eq. To zero

2

18LR 18 (imaginary value)LR j

What is your comment on resistance value ?


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EC 303.65 to 66 44

A resistor can not posses an imaginary value

As you can not have a resistor with an

imaginary value we can conclude that no

value of RL can make the circuit resonant

Summary


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We have solved problems dealing with

The resonant frequency,bandwidth,and Q-factor of series

resonant circuits.

The resonant frequency Q-factor,and bandwidth of

different parallel resonant circuits


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EC 303.65 to 66 46

QUIZ

1.An RLC series circuit has fixed values of R L and

C values what should we do make the circuit


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EC 303.65 to 66 47

C values what should we do make the circuit

resonant ?

a. Increase input voltage

c. Decrease input voltage

e. Adjust the input frequency

g. None of the above

Ans : c

2.Quality factor of coil depends upon ___


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EC 303.65 to 66 48

a. Its inductance

c. Coil resistance

e. Operating frequency

g. All the above

Ans : d

3.Quality factor of a capacitor depends upon___


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EC 303.65 to 66 49

a. Capacitance value

c. Resistance value

e. Operating frequency

g. All the above

Ans : d

4.which one of the following does noteffect the resonant frequency of the tank

circuit ?


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EC 303.65 to 66 50

circuit ?

a. R value

c. L value

e. Magnitude of input excitation

g. C value

Ans : c

5. A fixed frequency series RLC circuitintroducing a leading phase. To make

th t t h t


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EC 303.65 to 66 51

that resonant we have to __

a. Reduce the capacitive reactance

c. Increase the inductive reactance

e. Either (a) or (b)


Ans : c

6. Input voltage to series RLC circuit underresonance is v=Vmsin(5000t+30

0).its

current i=


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EC 303.65 to 66 52

current i=____

(Vm/R )sin(5000t+30)

(Vm/R )sin(5000t+45)

Zero amps

(Vm*R )sin(5000t+30)

Ans : a

7. Resonant frequency of a tank circuit is150kHz and Q0=10.Bandwidth of tuned

i it i


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EC 303.65 to 66 53

circuit is___

a. 5 kHz

c. 10 kHz

e. 15 kHz


Ans : c

8. Current passing through a seriesresonant circuit at f0 is 1A.current at

lower half power frequency=


136/136

lower half power frequency=___

a. 0.5A

c. 0.707A

e. 1A

g. 0A

Ans : b

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