Lecture 18
Parallel Resonant Circuit with Non-ideal Circuit Elements
The following items are covered:
•
•
•
•
Conversion of a practical tank circuit, with coil resistance, to an
equivalent parallel resonant circuit with ideal elements
Derivation of the expression for resonant frequency for the
practical tank circuit
Analysis of the effect of approximating the practical tank circuit
Solve numerical example with the practical tank circuit
Parallel Resonance with Practical Tank circuit
The circuit for a practical tank circuit is shown in Figure 18-1 Rl
represents the resistance of the coil, having inductance L.
Rl
XC
XL
Fig. 18-1 Practical tank circuit
The resistance Rl can no longer be included in a simple series or
parallel combination with the source resistance and any other
resistance for design purposes. Even though Rl is usually relatively
small in value compared with other resistance and reactance
levels of the network, it does have an important impact on the
parallel resonant conditions, as will be demonstrated through
numerical example.
First, our effort will be to find a parallel network equivalent of the
series R-L branch of Fig.18-1.
Now we will learn that a resistance Rl in series with XL, as shown in
Fig. 18-2 can be converted to an equivalent resistance Rp in
parallel with XLp.
Rl
Rp
Xl
=
Rl2 + X L2
RL
2
2
X Lp = Rl + X L
XL
Fig.18-2 Parallel equivalent of two impedances in series
The expressions of Rp and XLp in terms of Rl and XL can be written
as,
Rl2 + X L2
Rp =
RL
Rl2 + X L2
X Lp =
XL
Proof: optional
The admittance of the series Rl and XL branch can be written as,
1
YR− L = 1 =
Z R− L Rl + jX L
R
X
= 2 l 2−j 2 L 2
Rl + X L
Rl + X L
The admittance of the parallel Rp and Xlp branch is
YRp − X = 1 + 1
Rp jX Lp
Lp
Equate the two admittances and the result follows.
Redrawing the network of Fig.18-1 with the equivalent of Fig. 18-2
and a practical current source having an internal resistance Rs will
result in the network of Fig. 18-3
ZT
I
+
Rp
Rs
X Lp
XC
VP
-
YT
Source
Fig. 18-3 The equivalent of a practical tank circuit with current
source I and its internal resistance Rs
If we define the parallel combination of Rs and Rp by the notation,
R = Rs Rp
the network of Fig. 18-4 will result which has the same format as
the ideal configuration of Fig. 17-. (11 check exact number from
chapter 17 for figure 11)
ZT
I
R
X Lp
XC
YT
Fig.18-4 Equivalent R,L,C parallel resonant circuit
Recall that for series resonance the resonant frequency was the
frequency at which the impedance was a minimum, the current a
maximum, the input impedance purely resistive and the network
had a unity power factor.
For parallel networks, since the resistance Rp in our equivalent
model is frequency dependent, the frequency at which maximum
Vp (=VC) is obtained is not the same as required for the unity power
factor characteristic.
The condition for resonance of a practical tank circuit
From Fig. 18-4, the parallel circuit is in resonance (unity power
factor condition) when,
X p = XC
Rl2 + X L2
= XC
XL
Rl2
+ X L = XC
XL
Rl2
+ω p L = 1
ω pL
ω pC
This gives,
ωp =
1 − ( Rl )2
LC L
fp can be obtained from the relationship,
For coils having
ω p = 2π f p .
Qcoil 1, it can be shown that the second term
under the square root is small, giving,
ωp 1
LC
Xp in this case is,
Rl2
Xp =
+ XL
XL
XL
and
Rl2 X L2
Rp = +
Rl Rl
ω 2p L2
= RL +
Rl
2
1 L
LC Rl
L
Rl C
Example
Given that the resonant frequency for the following parallel
resonant network with a practical tank circuit is fp = 0.04 MHz.
I
Rs
40k Ω
10Ω
Rl
C
L
1mH
f p = 0.04MHz
Fig. 18-5 An exercise problem with practical tank circuit
a) Determine the quality factor of the coil, Ql
b) Determine the total impedance ZT seen by the source
at resonance
c) Find the value of capacitance, C
d) Find the quality factor of the circuit, Qp
e) Calculate the band width
f) Find the voltage across the capacitor.
Solution
a)
b)
Ql =
X L 2π f p L 2π (40000)(0.001)
=
=
= 25.12
Rl
Rl
10
Rl2 + X L2 102 + 251.22
Rp =
=
= 6.32k Ω
Rl
10
ZT = Rp Rl = 5.457k Ω
c) Since Ql ≥ 10,
fp 1
⇒ C = 21 2 = 15.9nF
4π f p L
2π LC
d) Since Ql ≥ 10,
Qp RT ZT 5450
= 21.68
=
=
X L X L 251.2
Alternatively, the exact value can be found from
Qp =
e)
RT RT
=
X p XC
Band width=
f) The voltage V
f p 40000
=
= 1.85KHz
Qp 21.68
= IZT = 0.002(5450) = 10.9V
Series-Parallel Resonant Circuits
A circuit is said to be in resonance if the total impedance exhibited
by the circuit is purely resistive. Under this condition the power
absorbed by the inductive elements are supplied by the capacitors.
The circuit then has a unity power at resonance.
The concept of resonance is explained for a series-parallel circuit
with an example.
Example
The values of the parameters in the following circuit are:
L=1H ,
4
R = 2Ω and L = 1 F . Find the resonant frequency of the series8
parallel circuit of Fig. 18-6.
L
R
C
Fig. 18-6 Example of series-parallel resonant circuit.
Solution
The impedance of the circuit as seen by the source is
2( 8 )
Z = jω + jω
4 2+ 8
jω
= jω + 16
4 8 + j(2ω )
− j 2ω )
= jω + 16(8
4 82 + (2ω )2
= 128 2 + jω − j32ω 2
4 64 + 4ω
64+4ω
At resonance the imaginary part of Z is zero,
Or,
4ω 2 + 64 −128 = 0
Solve for the positive value of
Summary
ω =4
The parallel resonant circuit with a practical tank circuit was
considered in this lecture. The following items were covered:
• Conversion of a series circuit (R,X) to its parallel
equivalent
• Derivation of the exact resonant frequency of a
practical tank circuit with non-zero coil resistance
• Comparison with resonant frequency expressions for
ideal parallel resonant circuit
• Derivation of the expressions for Rp and Xp when
quality factor of the coil is large
• Solved an example with a practical tank circuit
The class will end with a quiz on parallel resonant circuit
Practice Quiz:
I = 80mA∠0ο
Rl
ZT
P
Rs
450k Ω
2Ω
IL
XL
IC
XC
30Ω
Fig.18-7 Problem for quiz
For the network of Fig. 18-7,
A. If the coil is converted to a resistance Rp in parallel with XLp, the
value of Xp is (30Ω)
B. The value of XC at resonance is (30Ω)
C. The total value of impedance ZTp at resonance is (225Ω)
D. The value of current IL through the coil is
(0.6∠− 86.190 )
F. The value of current IC through the capacitor
(0.6∠900 )
G. If the resonant frequency is 20,000Hz, the value of L is
(0.239mH)
H. If the resonant frequency is 20,000Hz, the value of C is
(265.26nF)
I. The quality factor, (Qp) for the above parameters is (2.32)
J. The bandwidth for the circuit is (2.67kHz)