Electric Circuits II - Philadelphia University
Transcript of Electric Circuits II - Philadelphia University
Electric Circuits II Resonance
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Dr. Firas Obeidat
Dr. Firas Obeidat โ Philadelphia University
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โข Series Resonance
Table of Contents
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Series Resonance
Resonance is a condition in an RLC circuit in which the capacitive and inductive
reactances are equal in magnitude, thereby resulting in a purely resistive
impedance (voltage and current at the circuit input terminals are in phase).
Resonant circuits (series or parallel) are useful for constructing filters. They
are used in many applications such as selecting the desired stations in radio
and TV receivers.
For the series RLC circuit. The input
impedance in frequency domain is
Resonance results when the imaginary part of the transfer function is zero
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Series Resonance The value of that satisfies this condition is called the resonant frequency. Thus, the
resonance condition is
Since
or
At resonance:-
1. The impedance is purely resistive, thus Z=R, In other words, the LC series
combination acts like a short circuit, and the entire voltage is across R.
2. The voltage Vs and the current I are in phase, so that the power factor is unity.
3. The magnitude of the transfer function H(๐)=Z(๐) is minimum.
4. The maximum current for the circuit for an applied voltage Vs since Z is a
minimum value.
5. The inductor voltage and capacitor voltage can be much more than the source
voltage.
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Series Resonance
At resonance, when ฮธ=0, the voltage across each of the elements in the circuit can
be written as
I =๐
๐โ ฮธ
๐ โ 0=
๐๐
๐ โ ฮธ
๐๐ = ๐ผ๐ โ 0๐
๐๐ฟ = ๐ผ๐๐ฟโ 90๐
๐๐ถ = ๐ผ๐๐ถโ โ 90๐
Because ๐ฟ๐ณ = ๐ฟC , The magnitude of VL equals to the magnitude of VC but 180ยฐ out
of phase
The average power dissipated by the resistor and the reactive powers of the
inductor and capacitor as follows
๐๐ =1
2๐ผ2๐ (w)
๐๐ฟ =1
2๐ผ2๐๐ฟ
(VAR)
๐๐ถ =1
2๐ผ2๐๐ถ
(VAR)
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Series Resonance
The frequency response of the circuitโs current magnitude is as in the figure
The magnitude of the current at resonance is
I =๐๐
๐
๐ =1
2๐ผ2๐ =
1
2
๐๐
2
๐
The maximum power dissipated by the series resonant
circuit is given as
The bandwidth, BW, of the resonant circuit is the
difference between the frequencies at which the circuit
delivers half of the maximum power. The frequencies ๐1
and ๐2 are called the half-power frequencies, the cutoff
frequencies, or the band frequencies.
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Series Resonance
For the series resonant circuit the power at any frequency is determined as
๐ =1
2๐ผ2๐ =
1
2
๐๐
|๐|
2๐ =
1
2
๐๐
2
๐ 2+ ๐๐ฟโ1
๐๐ถ2
๐
At the half-power frequencies, the power must be
1
2ร
1
2
๐๐2
๐ =
1
2
๐๐2
๐ 2 + ๐๐ฟ โ1
๐๐ถ2
๐ โ 1
2๐ =
๐
๐ 2 + ๐๐ฟ โ1
๐๐ถ2
2๐ 2 = ๐ 2 +๐2๐ฟ๐ถ โ 1
๐๐ถ
2
โ ๐ 2 =๐2๐ฟ๐ถ โ 1
๐๐ถ
2
ยฑ๐ =๐2๐ฟ๐ถโ1
๐๐ถ โ ยฑ๐ ๐๐ถ = ๐2๐ฟ๐ถ โ 1
โ๐ ๐๐ถ = ๐2๐ฟ๐ถ โ 1 โ ๐2๐ฟ๐ถ + ๐ ๐๐ถ โ 1 =0
The solution of the above equation yields four values for the cutoff frequency.
Only two of these values are positive and have physical significance
+๐ ๐๐ถ = ๐2๐ฟ๐ถ โ 1 โ ๐2๐ฟ๐ถ โ ๐ ๐๐ถ โ 1 =0 To find ๐2 take +๐น๐๐ช
To find ๐1 take โ๐น๐๐ช
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Series Resonance
๐1 is called the lower half power frequency and ๐2 is called upper half
power frequency.
๐๐ =๐
๐ณ๐ช= ๐๐๐๐
๐1๐2 = โ๐
2๐ฟ+
๐
2๐ฟ
2
+1
๐ฟ๐ถ
๐
2๐ฟ+
๐
2๐ฟ
2
+1
๐ฟ๐ถ=
1
๐ฟ๐ถ
๐๐ = โ๐น
๐๐ณ+
๐น
๐๐ณ
๐
+๐
๐ณ๐ช ๐๐ =
๐น
๐๐ณ+
๐น
๐๐ณ
๐
+๐
๐ณ๐ช
๐1 and ๐2 are in general not symmetrical around the resonant frequency, because
the frequency response is not generally symmetrical
The height of the curve is determined by R. The width of the response curve
depends on the bandwidth B, which is defined as the difference between the two
half-power frequencies.
Solving the above two equations gives
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Series Resonance
The โsharpnessโ of the resonance in a resonant circuit is measured quantitatively
by the quality factor Q.
๐ธ = ๐๐ Peak energy stored in the circuit
Energy dissipated by the circuit in one period at resonance
๐2 โ ๐1 =๐
2๐ฟ+
๐
2๐ฟ
2+
1
๐ฟ๐ถโ โ
๐
2๐ฟ+
๐
2๐ฟ
2+
1
๐ฟ๐ถ =
๐
๐ฟ
๐ฉ = ๐๐ โ ๐๐ =๐น
๐ณ
The relationship between the bandwidth B and the quality factor Q is given by
๐ธ =๐๐๐ณ
๐น=
๐
๐๐๐ช๐น
The quality factor of a resonant circuit is the ratio of its resonant frequency to its
bandwidth.
๐ธ =๐๐
๐ฉ
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Series Resonance
A resonant circuit is designed to operate at or near its resonant frequency. It is
said to be a high-Q circuit when its quality factor is equal to or greater than 10.
For high-Q circuits, the approximate expressions for ฯ1 and ฯ2 are
The cutoff frequencies can be written in terms of the center frequency and the
bandwidth as:
๐๐ = โ๐ฉ
๐+
๐ฉ
๐
๐
+ ๐๐๐ ๐๐ =
๐ฉ
๐+
๐ฉ
๐
๐
+ ๐๐๐
The cutoff frequencies can be written in terms of the quality factor and the center
frequency as:
๐๐ = ๐๐ โ๐
๐๐ธ+ ๐ +
๐
๐๐ธ
๐
๐๐ = ๐๐
๐
๐๐ธ+ ๐ +
๐
๐๐ธ
๐
๐๐ โ ๐๐ โ๐ฉ
๐ ๐๐ โ ๐๐ +
๐ฉ
๐
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Series Resonance
Example: In the circuit R=2ฮฉ, L= 1 mH and C=0.4 ฮผF. (a) Find the resonant frequency
and the half-power frequencies. (b) Calculate the quality factor and bandwidth.
(c) Determine the amplitude of the current at ๐o, ๐1 and ๐2.
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Series Resonance
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Series Resonance Example:
a. For the series resonant circuit, find I, VR, VL, and
VC at resonance.
b. What is the Qs of the circuit?
c. If the resonant frequency is 5000 Hz, find the
bandwidth.
d. What is the power dissipated in the circuit at the
half-power frequencies?.
๐) ๐ = 2ฮฉ
๐ผ =๐๐
๐ =
10โ 0๐
2= 5โ 0๐
๐ด
๐ฃ๐ = ๐๐ = 10โ 0๐A
๐ฃ๐ฟ = ๐๐ฟ๐ผ = 10โ 90๐ ร 5โ 0๐= 50โ 90๐
๐ฃ๐ถ = ๐๐ถ๐ผ = 10โ โ 90๐ ร 5โ 0๐= 50โ โ 90๐
๐) ๐๐ =๐๐๐ฟ
๐ =
10ฮฉ
2ฮฉ= 5
๐) ๐ต =๐๐
๐๐
=2 ร ฯ ร 5000
5= 6283.19
๐) ๐๐ป๐๐น =1
2๐๐๐๐ฅ =
1
2ร
1
2๐ผ๐๐๐ฅ
2R
=1
2ร
1
2ร 52 ร 2 =12.5w
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Series Resonance Example: The bandwidth of a series resonant circuit is 2513.27.
a. If the resonant frequency is 4000 Hz, what is the value of Qs?
b. If R=10 ฮฉ, what is the value of XL at resonance?
c. Find the inductance L and capacitance C of the circuit.
๐) ๐๐ =๐
๐
๐ต =
2รฯร4000
2513.27 =10
๐) ๐ฟ =๐๐ฟ
2ฯ๐๐
=100
2ฯ ร 4000๐ป๐ง= 3.98 ๐๐ป
๐) ๐๐ =๐
๐๐ฟ
๐ โ ๐๐๐ฟ = ๐๐ฟ = ๐๐๐ =10ร10=100ฮฉ
At resonant XL=XC.
๐ถ =1
2ฯ๐๐๐๐ถ
=1
2ฯ ร 4000๐ป๐ง ร 100ฮฉ= 0.398 ฮผ๐น
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Series Resonance Example: A band pass circuit able to select inputs within the 1-10 kHz frequency band
(cutoff frequencies). If C=1 ฮผF, find the value of L and R.
๐1 = 2๐๐1 = 2๐ ร 1000 = 6283.19 ๐๐๐/๐
๐2 = 2๐๐2 = 2๐ ร 10000 = 62831.9 ๐๐๐/๐
๐๐= ๐1๐2 = 6283.19 ร 62831.9 = 19869.19 ๐๐๐/๐
๐๐ = 3162.28 Hz
๐ต = ๐2 โ ๐1 = 62831.9 โ 6283.19 = 56548.71
๐ =๐๐
๐ต=
19869.19
56548.71 = 0.3514
๐ =1
๐๐๐ถ๐=
1
19869.19 ร 1 ร 10โ6 ร 0.3514= 143.24 ฮฉ
๐ฟ =1
๐๐2๐ถ
=1
19869.192 ร 1 ร 10โ6 = 2.533 ๐๐ป
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Series Resonance Example: A series RLC network has R=2 kฮฉ, L=40 mH and C=1ฮผF. Calculate the
impedance at resonance and at one-fourth, one-half, twice, and four times the resonant
frequency.
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Series Resonance
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Series Resonance
Example: Design a series RLC resonant circuit with R=10ฮฉ, ๐o=1000 rad/s and B=20 rad/s. then find circuitโs Q.
Example: Design a series RLC resonant circuit with ๐o=40 rad/s and B=10 rad/s.
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