Electric Circuits II - Philadelphia University

Electric Circuits II Resonance

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Dr. Firas Obeidat

Dr. Firas Obeidat – Philadelphia University

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• 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(𝜔)=Z(𝜔) 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 θ=0, the voltage across each of the elements in the circuit can

be written as

I =𝑉

𝑚∠θ

𝑅∠0=

𝑉𝑚

𝑅 ∠θ

𝑉𝑅 = 𝐼𝑅∠0𝑜

𝑉𝐿 = 𝐼𝑋𝐿∠90𝑜

𝑉𝐶 = 𝐼𝑋𝐶∠ − 90𝑜

Because 𝑿𝑳 = 𝑿C , The magnitude of VL equals to the magnitude of VC but 180° out

of phase

The average power dissipated by the resistor and the reactive powers of the

inductor and capacitor as follows

𝑃𝑅 =1

2𝐼2𝑅 (w)

𝑄𝐿 =1

2𝐼2𝑋𝐿

(VAR)

𝑄𝐶 =1

2𝐼2𝑋𝐶

(VAR)


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Series Resonance

The frequency response of the circuit’s current magnitude is as in the figure

The magnitude of the current at resonance is

I =𝑉𝑚

𝑅

𝑃 =1

2𝐼2𝑅 =

1

2

𝑉𝑚

2

𝑅

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 𝜔1

and 𝜔2 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

𝑃 =1

2𝐼2𝑅 =

1

2

𝑉𝑚

|𝑍|

2𝑅 =

1

2

𝑉𝑚

2

𝑅2+ 𝜔𝐿−1

𝜔𝐶2

𝑅

At the half-power frequencies, the power must be

1

2×

1

2

𝑉𝑚2

𝑅=

1

2

𝑉𝑚2

𝑅2 + 𝜔𝐿 −1

𝜔𝐶2

𝑅 → 1

2𝑅=

𝑅

𝑅2 + 𝜔𝐿 −1

𝜔𝐶2

2𝑅2 = 𝑅2 +𝜔2𝐿𝐶 − 1

𝜔𝐶

2

→ 𝑅2 =𝜔2𝐿𝐶 − 1

𝜔𝐶

2

±𝑅 =𝜔2𝐿𝐶−1

𝜔𝐶 → ±𝑅𝜔𝐶 = 𝜔2𝐿𝐶 − 1

−𝑅𝜔𝐶 = 𝜔2𝐿𝐶 − 1 → 𝜔2𝐿𝐶 + 𝑅𝜔𝐶 − 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

+𝑅𝜔𝐶 = 𝜔2𝐿𝐶 − 1 → 𝜔2𝐿𝐶 − 𝑅𝜔𝐶 − 1 =0 To find 𝜔2 take +𝑹𝝎𝑪

To find 𝜔1 take −𝑹𝝎𝑪


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Series Resonance

𝜔1 is called the lower half power frequency and 𝜔2 is called upper half

power frequency.

𝝎𝒐 =𝟏

𝑳𝑪= 𝝎𝟏𝝎𝟐

𝜔1𝜔2 = −𝑅

2𝐿+

𝑅

2𝐿

2

+1

𝐿𝐶

𝑅

2𝐿+

𝑅

2𝐿

2

+1

𝐿𝐶=

1

𝐿𝐶

𝝎𝟏 = −𝑹

𝟐𝑳+

𝑹

𝟐𝑳

𝟐

+𝟏

𝑳𝑪 𝝎𝟐 =

𝑹

𝟐𝑳+

𝑹

𝟐𝑳

𝟐

+𝟏

𝑳𝑪

𝜔1 and 𝜔2 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 “sharpness” of the resonance in a resonant circuit is measured quantitatively

by the quality factor Q.

𝑸 = 𝟐𝝅Peak energy stored in the circuit

Energy dissipated by the circuit in one period at resonance

𝜔2 − 𝜔1 =𝑅

2𝐿+

𝑅

2𝐿

2+

1

𝐿𝐶− −

𝑅

2𝐿+

𝑅

2𝐿

2+

1

𝐿𝐶 =

𝑅

𝐿

𝑩 = 𝝎𝟐 − 𝝎𝟏 =𝑹

𝑳

The relationship between the bandwidth B and the quality factor Q is given by

𝑸 =𝝎𝒐𝑳

𝑹=

𝟏

𝝎𝒐𝑪𝑹

The quality factor of a resonant circuit is the ratio of its resonant frequency to its

bandwidth.

𝑸 =𝝎𝒐

𝑩


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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 ω1 and ω2 are

The cutoff frequencies can be written in terms of the center frequency and the

bandwidth as:

𝝎𝟏 = −𝑩

𝟐+

𝑩

𝟐

𝟐

+ 𝝎𝒐𝟐 𝝎𝟐 =

𝑩

𝟐+

𝑩

𝟐

𝟐

+ 𝝎𝒐𝟐

The cutoff frequencies can be written in terms of the quality factor and the center

frequency as:

𝝎𝟏 = 𝝎𝒐 −𝟏

𝟐𝑸+ 𝟏 +

𝟏

𝟐𝑸

𝟐

𝝎𝟐 = 𝝎𝒐

𝟏

𝟐𝑸+ 𝟏 +

𝟏

𝟐𝑸

𝟐

𝝎𝟏 ≅ 𝝎𝒐 −𝑩

𝟐 𝝎𝟐 ≅ 𝝎𝒐 +

𝑩

𝟐


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Series Resonance

Example: In the circuit R=2Ω, L= 1 mH and C=0.4 μF. (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 𝜔o, 𝜔1 and 𝜔2.


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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?.

𝑎) 𝑅 = 2Ω

𝐼 =𝑉𝑠

𝑅=

10∠0𝑜

2= 5∠0𝑜

𝐴

𝑣𝑅 = 𝑉𝑠 = 10∠0𝑜A

𝑣𝐿 = 𝑍𝐿𝐼 = 10∠90𝑜 × 5∠0𝑜= 50∠90𝑜

𝑣𝐶 = 𝑍𝐶𝐼 = 10∠ − 90𝑜 × 5∠0𝑜= 50∠ − 90𝑜

𝑏) 𝑄𝑠 =𝜔𝑜𝐿

𝑅=

10Ω

2Ω= 5

𝑐) 𝐵 =𝜔𝑜

𝑄𝑠

=2 × π × 5000

5= 6283.19

𝑑) 𝑃𝐻𝑃𝐹 =1

2𝑃𝑚𝑎𝑥 =

1

2×

1

2𝐼𝑚𝑎𝑥

2R

=1

2×

1

2× 52 × 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 Ω, what is the value of XL at resonance?

c. Find the inductance L and capacitance C of the circuit.

𝑎) 𝑄𝑆 =𝜔

𝑜

𝐵 =

2×π×4000

2513.27 =10

𝑐) 𝐿 =𝑋𝐿

2π𝑓𝑜

=100

2π × 4000𝐻𝑧= 3.98 𝑚𝐻

𝑏) 𝑄𝑆 =𝜔

𝑜𝐿

𝑅 → 𝜔𝑜𝐿 = 𝑋𝐿 = 𝑄𝑆𝑅 =10×10=100Ω

At resonant XL=XC.

𝐶 =1

2π𝑓𝑜𝑋𝐶

=1

2π × 4000𝐻𝑧 × 100Ω= 0.398 μ𝐹


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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 μF, find the value of L and R.

𝜔1 = 2𝜋𝑓1 = 2𝜋 × 1000 = 6283.19 𝑟𝑎𝑑/𝑠

𝜔2 = 2𝜋𝑓2 = 2𝜋 × 10000 = 62831.9 𝑟𝑎𝑑/𝑠

𝜔𝑜= 𝜔1𝜔2 = 6283.19 × 62831.9 = 19869.19 𝑟𝑎𝑑/𝑠

𝑓𝑜 = 3162.28 Hz

𝐵 = 𝜔2 − 𝜔1 = 62831.9 − 6283.19 = 56548.71

𝑄 =𝜔𝑜

𝐵=

19869.19

56548.71 = 0.3514

𝑅 =1

𝜔𝑜𝐶𝑄=

1

19869.19 × 1 × 10−6 × 0.3514= 143.24 Ω

𝐿 =1

𝜔𝑜2𝐶

=1

19869.192 × 1 × 10−6 = 2.533 𝑚𝐻


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Series Resonance Example: A series RLC network has R=2 kΩ, L=40 mH and C=1μF. 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=10Ω, 𝜔o=1000 rad/s and B=20 rad/s. then find circuit’s Q.

Example: Design a series RLC resonant circuit with 𝜔o=40 rad/s and B=10 rad/s.

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Electric Circuits II - Philadelphia University