Series and Parallel Resonances - Electrical engineeringjbornema/ELEC300/300-05 - Series,...

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1 Series Resonance The series resonant circuit. Consider the series RLC circuit shown in the frequency domain. The input impedance is 1 1 s s R j L R j L j C C V Z H I Resonance occurs when the imaginary part of the transfer function is zero, or 1 Im 0 L C Z The value of that satisfies this condition is called the resonant frequency . Thus, the resonance condition is Then 1 rad/sec o LC Series and Parallel Resonances o R Z

Transcript of Series and Parallel Resonances - Electrical engineeringjbornema/ELEC300/300-05 - Series,...

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Series Resonance

The series resonant circuit.

Consider the series RLC circuit shown in the frequency domain. The input impedance is

1 1s

s

R j L R j Lj C C

VZ HI

Resonance occurs when the imaginary part of the transfer function is zero, or

1Im 0LC

Z

The value of that satisfies this condition is called the resonant frequency . Thus, the resonance condition is

Then

1 rad/seco LC

Series and Parallel Resonances

o R Z

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Series ResonanceNote that 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. (Therefore, an ideal series resonance circuit with R0 has zero impedance at resonance.)

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() = Z() is minimum.

4. The inductor voltage and capacitor voltage can be much higher than the source voltage.

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Series Resonance The frequency response of the circuit’s current magnitude is.

The average power dissipated by the RLC circuit is

Hence, 1 and 2 are called the half-power frequencies.

221 1

2 2V

P I RR

21

2m

oVPR

1 212 oP P P

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Series ResonanceSince the impedance is lowest at resonance, the half-power frequencies are obtained by setting the magnitude of Z equal √2 R

22 1 2R L R

C

22 2

22

2 2

1 2

1 0

1 1 0

1 1 0

R L RC

L RC

L R L RC C

LC RC LC RC

2

1

2

2

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12 2

1 for >02 2

RC RC LCLC

R RL L LC

R RL L LC

2

2

Similarly

1 for >02 2R RL L LC

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Series Resonance2 2

1 2

2 2

2

1 12 2 2 2

12 2

1o

R R R RL L LC L L LC

R RL L LC

LC

1 2

2 1

2 2

Resonant Frequency Bandwidth

1 12 2 2 2

o

B

R R R R RL L LC L L LC L

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Series Resonance The “sharpness” of the resonance in a (single) resonant circuit is measured

quantitatively by the quality factor Q. At resonance, the reactive energy in the circuit oscillates between the inductor and the capacitor.

The quality factor relates the maximum or peak energy stored to the energy dissipated in the circuit per cycle of oscillation:

In the series RLC circuit, the peak energy stored is LI2/2, while the energy dissipated in one period is (I2R/2)(1/fo ). Hence,

Peak energy stored in the circuit2Energy dissipated by the circuit in one period at resonance

Q

2

2

1 1 12 = where / o o

o o

LI L LQRI f R RC R C LC

At resonance:

1

mL o m

mC m

o

VV L QVRVV QVR C

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Series Resonance

1 2

22 1

1 1

1=

oo

o

oo

L LQR RC R C

LCRB RCL Q

The quality factor of a single resonant circuit is the ratio of its resonant frequency to its bandwidth.The higher the circuit Q, the smaller the bandwidth.

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Series Resonance The selectivity of a single RLC circuit is its ability to respond to a certain

frequency and discriminate against all other frequencies ( filter applications).

If the band of frequencies to be selected or rejected is narrow, the quality factor of the single resonant circuit must be high. If the band of frequencies is wide, the quality factor is low.

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 (Q ≥ 10), the half-power frequencies are, for all practical purposes, symmetrical around the resonant frequency and can be approximated as

222

1,21

2 2 2 2

2

oo

o

R R BL L LC Q

B

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Example 1

2 7.958kHzo o of f

In the circuit, let R = 2 , L = 1 mH, and C = 0.4F.a) Find the resonant frequency and the half-power frequencies.b) Calculate the bandwidth and the quality factor.c) Determine the amplitude of the current at 0, 1, and 2.

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(b) Method 1:Example 1

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Example 1(b) Method 2:

1 2

20 V(c) At , 10 A220 V At , 7.071A

2 22

mo

m

VIR

VIR

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Parallel Resonance

The parallel resonant circuit.

1 1 1 1s

s

j C j CR j L R L

IY HV

The parallel RLC circuit is the dual of the series RLC circuit when voltage is exchanged with current and impedance with admittance.

Resonance occurs when the imaginary part of the transfer function is zero, or

1Im 0CL

Y

The value of that satisfies this condition is called the resonant frequency . Thus, the resonance condition is

1 rad/seco LC Then 1 ,o o R

R Y Z

(Therefore, an ideal parallel resonance circuit with R has infinite impedance at resonance.)

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Parallel Resonance

The voltage amplitude versus frequency for the parallel resonant circuit.

2

1

2

2

1 1 12 2

1 1 12 2

RC LC RC

RC LC RC

1 2

22 1

1=

1

oo

o

oo

R RQ RCL L C

LCLB

RC Q R

1,2 2oB

At resonance: ,mL m C m o m

o

I RI QI I I R C QIL

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The characteristics of resonant RLC circuits

1 1or or o

o

L LR RC R C

or or oo

R RRCL L C

Summary of the characteristics of RLC circuits.

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Example 20

1 2

Let 20cos( ) V.a) Calculate , and .b) Find and .c) Determine the power dissipated in the 45-k resistor at resonance.

sv tQ B

0

0

1 1a) Assume 10 : 4.082krad/s60mH 1μF

1μF12k 45k 38.6860mH

4.082krad/s 105.56 rad/s38.68

QLC

CQ RL

BQ

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Example 2

1 0

2 0

b) Since 10 : 4.03krad/s2

4.135krad/s2

BQ

B

2 245k

c) At resonance, an ideal parallel circuit has infinite impedance. Thus20V 35.09mA

12k 45k1 1 (35.09mA) 45k 2.77 mW2 2

I

P I R

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Example 3

2 1.27 kHzo o of f

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Example 3

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Example 4

2 15.9kHzo o of f

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PR

Practical ConsiderationsIn practical resonant circuits, the highest losses are usually attributed to the inductor and are represented as a series resistor. How can this be translated into a parallel resonant circuit?

SR ( , , ) ?P SR f R L C

2

1( ) 1

1 ( )

S

S

S

S

Z jj C

R j LR j L

j R C j LC

2

1( ) 1 1

1 ( )

P

P

P

Z jj C

R j Lj L

Lj j LCR

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Practical Considerations1. This is the ideal (lossless)

parallel resonant circuit.

2. In practice, we assume that the resonance frequencies of the two circuits are identical and that the Q’s are greater than 10. Then

( ) ( ) if 0 and .S P S PZ j Z j R R

2 21oo PS P P

S o S S

LL R LQ Q RR L R LC R

21 oP

S S

ZLRR C R

where oLZC

Example: Let L=20mH, C=5nF and RS=40, then

1 1100krad/s , 50 10 100koo S P

S S

L LQ RR R CLC

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Example 5Determine the resonant frequency.

2

00 012

0

2 20 0 0

1 1 1 2 20.1 0.110 2 2 10 4 4

2At resonance: Im 0 0.1 0 ( 0)4 4

0.4 0.4 2 0 4 2 rad/s 1 2.236 rad/s

in

in

jY j jj

Y

LC

0

0

2 2 2 (series resonance)22 10 0.1 2 (parallel resonance)

1 1 1 1 (not a good resonant circuit - losses are too high!)

Ls

C p

eqeq L C

LQR

Q R C

QQ Q Q

sRpR

Why is there a difference of more than 10 % in resonant frequency?

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sRpR

LC

A More Practical Example2mH, 20 , 2nF, 500ks pL R C R

811

2 2 2

00 012 2 2

0

22 2 2 50 0 2

102.5 10

5 00 0

1 1 1

Im 0 0 ( 0)

1 4.999 10 rad/s

1 5 10 rad/s , 79.6 kHz2

sin

p s p s

ins

ss

R j LY j C j CR R j L R R L

LY CR L

RL C L R CLC L

LC f

0 050 , 500

1 1 1 45.45

L s C p

eqeq L C

Q L R Q R C

QQ Q Q

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Relationship between and Q

svCv

2

111

1 ( )

C

s

j CR j L j C

j RC j LC

VHV

02

0 02

1 1

1 ( )RCLC j jLC

H

2

0 0

1Recall quadratic pole:1 2 / /j j

H

1 12 sinceRC C LR QL Q R CLC

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Q