Transcript Parallel Resonance ET 242 Circuit Analysis II Electrical and Telecommunication

ET 242 Circuit Analysis II
Parallel Resonance
Electrical and Telecommunication
Engineering Technology
Professor Jang
Acknowledgement
I want to express my gratitude to Prentice Hall giving me the permission
to use instructor’s material for developing this module. I would like to
thank the Department of Electrical and Telecommunications Engineering
Technology of NYCCT for giving me support to commence and complete
this module. I hope this module is helpful to enhance our students’
academic performance.
OUTLINES
 Introduction to Parallel Resonance
 Parallel Resonance Circuit
 Unity Power Factor (fp)
 Selectivity Curve
 Effect of QL ≥ 10
 Examples
Key Words: Resonance, Unity Power Factor, Selective Curve, Quality Factor
ET 242 Circuit Analysis II – Parallel Resonance
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Parallel Resonance Circuit - Introduction
The basic format of the series resonant
circuit is a series R-L-C combination in
series with an applied voltage source. The
parallel resonant circuit has the basic
configuration in Fig. 20.21, a parallel R-LC combination in parallel with an applied
current source.
Figure 20.21 Ideal parallel resonant network.
If the practical equivalent in Fig. 20.22 had
the format in Fig. 20.21, the analysis would
be as direct and lucid as that experience for
series resonance. However, in the practical
world, the internal resistance of the coil must
be placed in series with the inductor, as
shown in Fig.20.22.
Figure 20.22 Practical parallel L-C network.
ET 242 Circuit Analysis II – Parallel Resonance
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The first effort is to find a parallel
network equivalent for the series R-L
branch in Fig.20.22 using the technique
in earlier section. That is
Z R  L  Rl  jX L
and YR  L 

with
1
Z RL

R
X
1
 2 l 2 j 2 L 2
Rl  jX L Rl  X L
Rl  X L
1
1

Rl2  X L2
 Rl2  X L2
j 
Rl
 XL
Rl2  X L2
X Lp 
Xl




1
1

R p jX LP
as shown in Fig .20.23.
Figure 20.23 Equivalent parallel network for a series R-L combination.
Parallel Resonant Circuit – Unity Power Factor, fp
For the network in Fig .20.25,
1
1
1
1
1
1
YT 


 

Z 1 Z 2 Z 3 R jX L p  jX C
 1 
1

  j
 XL 
R
 p
 1
1
and YT   j 
 XC
R

 1
j 
 XC
1

X Lp







ET162 Circuit Analysis – Parallel Resonance
Figure 20.25 Substituting R = Rs//Rp for the network in Fig. 20.24.
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For unity power factor, the reactive component
must be zero as defined by
1
1

0
X C X Lp
1
1

X C X Lp
Therefore,
and
X Lp  X C
Substituting for X L p yields
Rl2  X L2
 XC
XL
The resonant frequency, f p , can be det er min ed as follow :
Rl2 C
fp 
1
L
2 LC
1
or
f p  fs
Rl2 C
1
L
Where fp is the resonant frequency
of a parallel resonant circuit (for Fp
= 1) and fs is the resonant
frequency as determined by XL =
XC for series resonance. Note that
unlike a series resonant circuit, the
resonant frequency fp is a function
of resistance (in this case Rl).
(20.31)
Parallel Resonant Circuit – Maximum Impedance, fm
At f = fp the input impedance of a parallel resonant
circuit will be near its maximum value but not quite its
maximum value due to the frequency dependence of
Rp. The frequency at which impedance occurs is
defined by fm and is slightly more than fp, as
demonstrated in Fig. 20.26. Figure 20.26 ZT versus frequency
for the parallel resonant circuit.
ET 242 Circuit Analysis II – Parallel Resonance
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The frequency fm is determined by differentiating the general equation for ZT with respect to
frequency and then determining the frequency at which the resulting equation is equal to zero.
The resulting equation, however, is the following:
1  Rl2 C 

f m  f s 1  
4  L 
Note the similarities with Eq. (20.31). Since square root factor of Eq. (20.32) is always more than
the similar factor of Eq. (20.31), fm is always closer to fs and more than fp. In general,
fs > fm > fp
Once fm is determined, the network in Fig. 20.25 can be used to determine the magnitude and
phase angle of the total impedance at the resonance condition simply by substituting f = fm and
performing the required calculations. That is
ZTm = R // XLp // XC
f =f m
Parallel Resonant Circuit – Selectivity Curve
Since the current I of the current source is constant for
any value of ZT or frequency, the voltage across the
parallel circuit will have the same shape as the total
impedance ZT, as shown in Fig. 20.27. For parallel
circuit, the resonance curve of interest in VC derives
from electronic considerations that often place the
capacitor at the input to another stage of a network.
ET 242 Circuit Analysis II – Parallel Resonance
Figure 20.27 Defining the shape of the Vp(f) curve.
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Since the voltage across parallel elements is the same,
VC = Vp = IZT
The resonant value of VC is therefore determined by the value of ZTm and magnitude of the
current source I. The quality factor of the parallel resonant circuit continues to be determined
as following;
Rs // R p Rs // R p
R
Qp 


X L p  X C at resonance
X Lp
X Lp
XC
For the ideal current source (Rs = ∞ Ω) or when Rs is sufficiently large compared to Rp, we can
make the following approximation:
XL
Qp 
 Ql
Rl
Rs  R p
In general, the bandwidth is still related to the resonant frequency and the quality factor by
fr
BW  f 2  f1 
Qp
The cutoff frequencies f1 and f2 can be determined using the equivalent network and the quality
factor by
f1 
1
4C
1
 
R
1
4C 


L 
R2
ET 242 Circuit Analysis II – Parallel Resonance
and
f2 
1
4C
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1
 
R
1
4C 


L 
R2
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The effect of Rl, L, and C on the shape of the
parallel resonance curve, as shown in Fig.
20.28 for the input impedance, is quite
similar to their effect on the series resonance
curve. Whether or not Rl is zero, the parallel
resonant circuit frequently appears in a
network schematic as shown in Fig. 20.28.
At resonance, an increase in Rl or decrease
in the ratio L/R results in a decrease in the
resonant impedance, with a corresponding
Figure 20.28
increase in the current.
Effect of R1, L, and, C on the parallel resonance curve.
Parallel Resonant Circuit – Effect of QL ≥ 10
The analysis of parallel resonant circuits is significantly more complex than encountered for
series circuits. However, this is not the case since, for the majority of parallel resonant
circuits, the quality factor of the coil Ql is sufficiently large to permit a number of
approximations that simplify the required analysis.
Effect of QL ≥ 10 – Inductive Resistance, XLp
X Lp  X L
Ql 10
ET 242 Circuit Analysis II – Parallel Resonance
and
X L  XC
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Ql 10
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Effect of QL ≥ 10 – Resonant Frequency, fp
(Unity Power Factor)
f p  fs
1
1 2
Ql
Ql  10
f p  fs 
and
1
2 LC
Ql  10
Effect of QL ≥ 10 – Resonant Frequency, fm
(Maximum VC)
fm  fs
Rp
1 1
1   2
4  Ql




R p  Q Rl
2
l
Ql  10
Ql  10
and
f p  fs 
1
2 LC
Ql  10
XL
substituting Ql 
Rl
ET 242 Circuit Analysis II – Parallel Resonance
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In total
f p  fm  fs
L
then R p 
Rl C
Ql  10
Ql  10
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Z Tp  Rs // R p  Rs // Ql2 Rl
Qp
Rs // Ql2 Rl
R
Qp 

X Lp
XL
BW
Rl
1  Rl
1 
BW  f 2  f 1 
  
 and BW  f 2  f 1 
Q p 2π  L Rs C 
2ππ
Ql 10
and Ql 
XL
then Z Tp  Ql2 Rl
Rl
ZTp
and
Q p  Ql
Ql 10, Rs  R p
Ql  10, Rs  R p
fp
IL and IC
Rs  Ω
A portion of Fig. 20.30 is reproduced in Fig. 20.31, with IT defined as shown
and
I C  Ql I T
Ql  10
I L  Ql I T
Ql  10
Figure 20.31 Establishing the relationship
between IC and IL and current IT.
ET 242 Circuit Analysis II – Parallel Resonance
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Ex. 20-6
Given the parallel network in Fig. 20.32 composed of “ideal” elements:
a. Determine the resonant frequency fp.
b. Find the total impedance at resonance
c. Calculate the quality factor, bandwidth, and cutoff frequencies f1 and f2 of the system.
d. Find the voltage VC at resonance.
e. Determine the currents IL and IC at resonance.
a. The fact that R1 is zero ohms results in
a very high Ql ( X L / Rl ), permitting the
use of the following equation for f p :
f p  fs 
1
2 LC

1
2 (1mH )(1F )
 5.03 kHz
b. For the parallel reactive elements :
(X 90 )(X C   90 )
Z L //Z C  L
 j(X L  X C )
Figure 20.32
Example 20.6.
but X L X C at resonance, resulting in a zero in the
denominator of the equation and a very high impedance
that can be approximated by an open circuit. Therefore,
Z T p  Rs //Z L //Z C  Rs  10 kΩ
d . VC  IZ Tp  (10mA)(10k)  100V
e. I L 
VC
VL
100V
100V



 3.16 A
X L 2f p L 2 (5.03 kHz)(1 mH ) 31.6
VC
100V

 3.16 A ( Q p I )
X
31
.
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
C Circuit Analysis II – Power for AC Circuits
ET 242
IC 
c. Q p 
BW 
1
4C
f1 

Rs
Rs
10k


 316.41
X Lp
2f p L 2 (5.03k)(1mH )
fp
Qp

1
 
R
5.03kHz
 15.90 Hz
316.41
1
4C 


L 
R2
 1
1


4 (1F ) 10k
1
4(1F ) 


1mH 
(10k) 2
 5.03 kHz
1 1
 
4C  R
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f2 
1
4C 

  5.04 kHz
L 
R2
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Ex. 20-7
For the parallel resonant circuit in Fig. 20.33 with Rs = ∞ Ω:
a. Determine fp, fm, and fp, and compare their levels.
b. Calculate the maximum impedance and the magnitude of the voltage VC at fm.
c. Determine the quality factor Qp.
d. Calculate the bandwidth.
e. Compare the above results with those obtained using the equations associated with
Ql ≥ 10.
a.
fs 
1
2 LC

1
2 (0.3mH )(100 F )
 29.06 kHz
1  Rl2 C 
1  (20) 2 (100nF ) 

(
29
.
06
kHz
)
1





4 L 
4
0.3mH

 25.58 kHz
fm  fs 1
 (20) 2 (100nF ) 
Rl2 C
f p  fs 1
 (29.06kHz) 1  

L
0.3mH


 27.06 kHz
c. Rs   ; therefore
Qp 

Rs // R p
X Lp

Rp
X Lp
X
 Ql  L
Rl
2 (29.06kHz)(0.3mH ) 51

 2.55
20
20 
ET 242 Circuit Analysis II – Parallel Resonance
Figure 20.33 Example 20.7.
b. Z Tm  ( R1  jX L ) //  jX C at f  f m
X L  2f m L  2 (28.58kHz)(0.3mH )  53.87 
XC 
1
1

 55.69 
2f m C 2 (28.58kHz)(100nF )
Rl  jX L  20   j 53.87   57.4669.63
(57.4669.63)(55.69  90)
159.34  15.17
 IZ Tm  (2mA)(159.34 )  318.68 mV
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Z Tm 
VCmax
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d . BW 
fp
Qp

27.06kHz
 10.61 Hz
2.55
e. For Ql  10, f m  f p  f s  29.06 kHz
Q p  Ql 
2f s L 2 (29.06kHz)(0.3mH )

 2.74 (versus 2.55 above)
Rl
20
Z T P  Ql2 Rl  (2.74) 2  20  150.150 (versus 159.34  15.17 above)
VCmax  IZ T p  (2mA)(150.15)  300.3 mV
BW 
fp
Qp

(versus 318.68 mV above)
29.06kHz
 10.61 kHz (versus 10.61 kHz above)
2.74
Ex. 20-8
For the network in Fig. 20.34 with fp provided:
a. Determine Ql. b. Determine Rp.
c. Calculate ZTp.
d. Find C at resonance.
e. Find Qp.
f. Calculate the BW and cutoff frequencies.
a. Ql 
X L 2f p L

Rl
Rl
2 (0.04MHz )(1mH )
 25.12
10
b. Ql  10. Therefore,

R p  Ql2 Rl  (25.12) 2 (10)  6.31k
c. Z T p  Rs // R p  40 k // 6.31 k  5.45 k
ET 242 Circuit Analysis II – Parallel Resonance
Figure 20.34 Example 20.8.
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d . Ql  10. Therefore,
fp 
and
C
1
2 LC
1
4 2 f 2 L
1
 15.83 nF
4 2 (0.04 MHz ) 2 (1mH )
e. Ql  10. Therefore,

Qp 

Z Tp
XL
Rs // Ql2 Rl

2f p L
5.45k
 21.68
2 (0.04 MHz ) 2 (1mH )
Ex. 20-10
a.
f1 
c.

1 1
1 4C 




2
4C  R
L 
R
 1
1
1
4(15.9mF ) 




4 (15.9mF )  5.45k
1mH 
(5.45k) 2
 39 kHz
1 1
f2 
 
4C  R
1 4C 


L 
R2
 5.005  10 6 [183.486  10 6  7.977  10 3 ]
 40.84 kHz
Repeat Example 20.9, but ignore the effects of Rs, and compare results.
f p is the same, 318.31 kHz.
b. For Rs   
Q p  Ql  100 (versus 4.76)
c. BW 
fp
Qp

318.31kHz
 3.18 kHz (versus 66.87 kHz)
100
d . Z T p  R p  1 M (versus 47.62 k)
V p  IZ T p  (2 mA)(1 M)  2000 V
ET 242 Circuit Analysis II – Series Resonance
(versus 95.24 V )
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Figure 20.35 Example 20.9.
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