teo ec ebook-3
TRANSCRIPT
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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
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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
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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.
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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
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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
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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
=
=
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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
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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
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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
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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
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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
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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
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EE-305.72 to 73 17
QUIZ
1. The frequency response of transformercoupling is
(a) Good
(b) Very good
(c) Excellent
(d) Poor
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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
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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
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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
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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?
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Recap
Already we discussed about the
Multistage amplifiers and
Their necessity
EE-305.75 2
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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.
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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
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EE305.75 5
Block diagram of 2- Stage CE cascade amplifier
FIRST
STAGE
SECOND
STAGEV1
V2V3
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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
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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
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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
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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
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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.
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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
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EE305.75 12
Frequency Response
The curve between Gain and signal frequency of anamplifier is known as frequency response.
Vol
tagegain
Frequencyfr
Maximum gain
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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EC 303 . 61 8
By solving Equation (2) for0
LC
1
0=
f 2=As
LCf 21
0= (3)
Impedance at resonance
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EC 303 . 61 9
Impedance at resonance
=
LCjy
1
Admittance of parallel LC circuit
At resonant frequency f0
LC
1
= 0
Impedance is reciprocal of admittance.
Y = 0
Impedance at resonance
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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.
Impedance at resonance
Frequency versus admittance
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EC 303 . 61 11
Frequency versus admittance
f0
Resonant frequency
Current at resonant frequency
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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
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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
===
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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
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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=
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QUIZ
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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
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EC 303 . 61 18
1) At resonant frequency impedance of parallel
resonant circuit is
a) zero
b) infinite
c) low
d) none of the above
Ans : ( b )
h d i h f f
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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
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4 ) At resonant frequency current entering into the
parallel LC circuit is
a) Zero
b) medium
c) Infinite
d) none of the above
Ans : ( a)
EC 303 . 61 20
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EC 303 . 61 21
Frequently asked questions
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
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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.
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EC 303 . 62 3
Resonance in parallel RL-C circuit
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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.
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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
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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
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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)
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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
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EC 303 . 62 9
Equate susceptance part to zero to
find the resonant frequency
(3)
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EC 303 . 62 10
C
LLR =+ 20
2 )(
Replace with and solve for0 0
2
2
0
1
L
R
LC=
002 f =
(4)
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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)
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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.
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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.
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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
+=
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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
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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
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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
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EC 303 . 62 18
A) Now what is the power factor of an
parallel LC circuit ?
Power factor = 1)0cos(cos ==
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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
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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
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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
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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
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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
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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
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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
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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
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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
Frequently asked questions
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Frequently asked questions
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
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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
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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
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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
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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
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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
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=
=
=
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
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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.
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, , 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=
=
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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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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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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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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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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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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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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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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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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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equation
Hzf 15910
49
102010
1
2
16630=
=
Resonant frequency = 159Hz
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On substitution of L and C values
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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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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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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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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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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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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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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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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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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
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EQUATE THE SUSCEPTANCE TO ZERO
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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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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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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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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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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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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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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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that resonant we have to __
a. Reduce the capacitive reactance
c. Increase the inductive reactance
e. Either (a) or (b)
g. None of the above
Ans : c
6. Input voltage to series RLC circuit underresonance is v=Vmsin(5000t+30
0).its
current i=
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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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circuit is___
a. 5 kHz
c. 10 kHz
e. 15 kHz
g. None of the above
Ans : c
8. Current passing through a seriesresonant circuit at f0 is 1A.current at
lower half power frequency=
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lower half power frequency=___
a. 0.5A
c. 0.707A
e. 1A
g. 0A
Ans : b