Transcript Chapter 4 : Resonance Circuit

CHAPTER 4
RESONANCE CIRCUITS
Content
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Series Resonance
Parallel Resonance
Important Parameters
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Resonance Frequency, ωo
Half-power frequencies, ω1 and ω2
Bandwidth, 
Quality Factor, Q
Application
2
Introduction
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Resonance is a condition in an RLC
circuit in which the capacitive and
reactive reactance are equal in
magnitude, thereby resulting in a purely
resistive impedance.
Resonance circuits are useful for
constructing filters and used in many
application.
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Series Resonance Circuit
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At Resonance
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At resonance, the impedance consists
only resistive component R.
The value of current will be maximum
since the total impedance is minimum.
The voltage and current are in phase.
Maximum power occurs at resonance
since the power factor is unity.
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Series Resonance
Total impedance of series RLC Circuit is
ZTotal  R  jX L - jX C
Z Total  R  j(X L - X C )
At resonance
XL  XC
The impedance now reduce to
ZTotal  R
The current at resonance
Im 
Vs
V
 m
ZTotal
R
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Resonance Frequency
Resonance frequency is the frequency where the
condition of resonance occur.
Also known as center frequency.
Resonance frequency
1
ωo 
rad/s
LC
fo 
1
2 LC
Hz
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Half-power Frequency
Half-power frequencies is the frequency when the
magnitude of the output voltage or current is decrease
by the factor of 1 / 2 from its maximum value.
Also known as cutoff frequencies.
2
R
R   1 
ω1 
   
rad/s
2L
 2L   LC 
2
R
R   1 
ω2 
   
rad/s
2L
 2L   LC 
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Bandwidth, 
Bandwidth,  is define as the difference between the
two half power frequencies.
The width of the response curve is determine by the
bandwidth.
β  (ω2  ω1 )rad/s
R
β  rad/s
L
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Current Response Curve
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Voltage Response Curve
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Quality Factor (Q-Factor)
The ratio of resonance frequency to the bandwidth
o o L
Q


R
The “sharpness” of response curve could be measured
by the quality factor, Q.
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Q-Factor Vs Bandwidth
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Higher value of Q,
smaller the
bandwidth. (Higher
the selectivity)
Lower value of Q
larger the bandwidth.
(Lower the selectivity)
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High-Q
It is to be a high-Q circuit when its quality factor is
equal or greater than 10.
For a high-Q circuit (Q ≥ 10), the half-power
frequencies are, for all practical purposes, symmetrical
around the resonant frequency and can be
approximated as
1  o 

2
 2  o 

2
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Maximum Power Dissipated
The average power dissipated by the RLC circuit is
1 2
P(ωo )  I R
2
The maximum power dissipated at resonance where
I
Vm
R
Thus maximum power dissipated is
1 V2m
P(ωo ) 
2 R
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Power Dissipated at ω1 and ω2
At certain frequencies, where ω = ω1 and ω2, the
dissipated power is half of maximum power
1 (Vm / 2 ) 2 V 2 m
P(ω1 )  P(ω 2 ) 

2
R
4R
Hence, ω1 and ω2 are called half-power frequencies.
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Example 14.7
If R=2Ω, L=1mH and C=0.4 F, calculate
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Resonant frequency, ωo
Half power frequencies, ω1 and ω2
Bandwidth, 
Amplitude of current at ωo, ω1 and ω2.
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Solution
Resonant frequency
ωo 
1
LC

1
(1mH)(0.4μF)
 50 krad/s
Bandwidth

R
2

 2krad/s
L 1mH
Quality Factor
ωo
50krad/s
Q

 25

2krad/s
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Solution
Since Q ≥10 , we can regard this as high-Q circuit. Hence
ω1  ωo 

ω 2  ωo 
2

2
 49krad/s
 51krad/s
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Solution
Current, I at ω= ωo
Vm 20
I

 10A
R
2
Current, I at ω = ω1 , ω2
I
Vm
2R

20
2 (2)
 7.017 A
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Practice Problem 14.7
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A series connected circuit has R=4Ω
and L=25mH. Calculate
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Value of C that will produce a quality factor
of 50.
Find ω1 , ω2 and .
Determine average power dissipated at
ω = ωo , ω1 and ω2. Take Vm = 100V
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Solution
Value of C that will produce Q = 50
ωo L
QR (50)(4Ω)
Q
 ωo 

 8krad/s
R
L
25mH
1
1
1
ωo 
C 2 
2
LC
ωo L (8k) (25mH)
C  0.625 F
Bandwidth
R
4
 
 160 rad/s
L 25mH
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Solution
Since Q ≥10 , we can regard this as high-Q circuit. Hence
ω1  ωo 
ω 2  ωo 

2

2
 7920 rad/s
 8080 rad/s
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Solution
Power dissipated at ω = ωo
2
1 Vin
1 (100) 2
P(ω o ) 

 1.25kW
2 R
2 (4)
Power dissipated at ω = ω1 , ω2
2
1 Vin
(100) 2
P

 0.625kW
2 2R
4(4Ω)
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Parallel Resonance
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Parallel Resonance
The total admittance
YTotal  Y1  Y2  Y3
YTotal
YTotal
1
1
1
 

R (jL) (-j/C)
1 -j
 
 j ωC
R ωL
YTotal 
1
 j( ωC  1/ωL)
R
Resonance occur when
ωC 
1
ωL
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At Resonance
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At resonance, the impedance consists
only conductance G.
The value of current will be minimum
since the total admittance is minimum.
The voltage and current are in phase.
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Parameters in Parallel Circuit
Parallel resonant circuit has same parameters as the
series resonant circuit.
Resonance frequency
ωo 
1
LC
rad/s
Half-power frequencies
2
1
 1   1 
ω1 
 
 
rad/s
2RC
2RC
LC

 

2
1
 1   1 
ω2 
 
 
rad/s
2RC
2RC
LC

 

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Parameters in Parallel Circuit
Bandwidth
1
β  ω2  ω1 
RC
Quality Factor
ωo
Q
 o RC
β
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Example 14.8
If R=8kΩ, L=0.2mH and C=8F, calculate

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ωo
Q and 
ω1 and ω2
Power dissipated at ωo, ω1 and ω2.
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Solution
Resonant frequency
ωo 
1
LC

1
(0.2mH)(8μF)
 25krad/s
Bandwidth
1
1


 15.625rad/s
RC (8kΩ)(8μF )
Quality Factor
ωo
25krad/s
Q

 1600
 15.625rad/ s
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Solution
Since Q ≥10 , we can regard this as high-Q circuit. Hence
ω1  ωo 
ω 2  ωo 

2

2
 24992 rad/s
 25008 rad/s
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Solution
Power dissipated at ω = ωo
2
1 Vm
1 (10) 2
P(ω o ) 

 6.25mW
2 R
2 (8k)
Power dissipated at ω = ω1 , ω2
2
1 Vm
(10) 2
P

 3.125mW
2 2R 4(8kΩ)
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Practice Problem 14.8
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A parallel resonant circuit has R=100kΩ,
L=25mH and C=5nF. Calculate
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ωo
ω1 and ω2
Q

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Solution
Resonant frequency
ωo 
1
LC

1
(20mH)(5nF)
 100krad/s
Bandwidth
1
1


 2krad/s
RC (100k )(5nF)
Quality Factor
ωo
100krad/s
Q

 50

2krad/s
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Solution
Since Q ≥10 , we can regard this as high-Q circuit. Hence
ω1  ωo 

2
 99krad/s

ω2  ωo   101krad/s
2
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APPLICATION
PASSIVE FILTERS
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A filter is a circuit that is designed to pass
signals with desired frequencies and
reject or attenuates others
A filter is a Passive Filters if it consists
only passive elements which is R, L and C.
Filters that used resonant circuit
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Bandpass Filter
Bandstop Filter
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BANDPASS FILTER
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A bandpass filter is
designed to pass all
frequencies within
ω1  ωo  ω2
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BANDPASS FILTER
SERIES RLC CIRCUIT
ωo 
1
LC
R
R   1 
ω1 
   

2L
 2L   LC 
2
2
R
R   1 
ω2 
   

2L
 2L   LC 
β  ω 2  ω1 
R
L
ωo
L
Q

β
CR 2
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BANDPASS FILTER
PARALLEL RLC CIRCUIT
ωo 
1
LC
1
 1   1 
 
 

2RC
 2RC   LC 
2
ω1 
2
1
 1   1 
ω2 
 
 

2RC
 2RC   LC 
1
β  ω 2  ω1 
RC
ωo
R 2C
Q

β
L
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BANDSTOP FILTER

A bandstop or
bandreject filter is
designed to stop or
reject all
frequencies within
ω1  ωo  ω 2
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BANDSTOP FILTER
SERIES RLC CIRCUIT
ωo 
1
LC
R
R   1 
ω1 
   

2L
 2L   LC 
2
2
R
R   1 
ω2 
   

2L
 2L   LC 
β  ω 2  ω1 
R
L
ωo
L
Q

β
CR 2
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BANDSTOP FILTER
PARALLEL RLC CIRCUIT
ωo 
1
LC
1
 1   1 
 
 

2RC
 2RC   LC 
2
ω1 
2
1
 1   1 
ω2 
 
 

2RC
 2RC   LC 
1
β  ω 2  ω1 
RC
ωo
R 2C
Q

β
L
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