Ph5.Chapter.resonantCircuits

8/14/2019 Ph5.Chapter.resonantCircuits

1/21

PHYSICS MATHEMATICS AND ASTRONOMY DIVISION

CALIFORNIA INSTITUTE OF TECHNOLOGY

Sophomore Physics Laboratory (PH005/105)

Analog ElectronicsResonant Circuits

CopyrightcVirgnio de Oliveira Sannibale, 2003

(Revision December 2012)


2/21

DRAFT

Chapter 2

Resonant Circuits

2.1 Introduction

Resonators, one of the most useful and used device, are essentially phys-ical systems that present a more or less pronounced peak in their transferfunction.

In general, their performance is measured by a dimensionless param-eter named quality factorQ, which characterizes the sharpness of the res-

onant peak. The higher the quality factor the sharper is the peak and thebetter is the resonator.

Quite often, the major issues of building a resonator are to obtain veryhigh quality factors and good stability. For example, mechanical oscil-lators made of fused silica fibers under load, can achieve quality factorsabove 108 in the acoustic band[?]. Very high quality factors in electronicscan be achieved using the mechanical resonances of piezoelectric materialssuch as quartz. Lasers and resonant cavities made of mirrors can be usedto build resonators in the optical frequency range. The same principle can

be applied in the microwave range. Thermal stabilization is always a keyingredient to obtain high stability.

Resonators made with electronic passive components, reaching qualityfactors values up to 10-100 or more, are quite easy to realize. In the nextsections we will study two typical resonant circuits, the LCR series andLCR parallel circuits.

37


3/21

DRAFT

38 CHAPTER 2. RESONANT CIRCUITS

VL

VR

VC

C

Vi

R

L

Figure 2.1: LCR series circuit.

2.2 The LCR Series Resonant Circuit

Figure 2.1 shows the so called LCR series resonant circuit. Depending onvoltage difference, we are considering as the circuit output ( the capacitor,the resistor, or the inductor), this circuit shows a different behavior. Letsstudy in the frequency and in the time domain the response of this passivecircuit for each one of the possible outputs.

2.2.1 Frequency Response with Capacitor Voltage Differ-ence as Circuit Output

Considering the voltage differenceVCacross the capacitor to be the circuitoutput, we will have

Vin =

R+jL+

1

jC

I,

VC = 1

jCI,

and the transfer function will be

HC() = 1

jRC 2 LC+ 1.


4/21

DRAFT

2.2. THE LCR SERIES RESONANT CIRCUIT 39

For sake of simplicity, it is convenient to define the two following quan-

tities20 =

1

LC, Q=

1

R

L

C=0

L

R

The parameter Q is the quality factor of the circuit, and the angularfrequency0is the resonant frequency of the circuit ifR = 0.

Considering the previous definitions, and after some algebra, HC()becomes

HC() = 20

20 2 +j0Q. (2.1)

Computing the magnitude and phase ofHC

(), we obtain

|HC()| = 2020 2

2+0Q

2 ,

arg [HC()] = arctan

1

Q

0

20 2

.

The magnitude has maximum for

2C =20 1

1

2Q2 ,and the maximum is

|HC(C)| = Q1 1

4Q2

.

IfQ 1 thenC 0, and |HC(C)| Q.Far from resonanceC, the approximate behavior of|HC()| is

C |HC()| 1 ,

C |HC()| 2

02 .

Figure 2.2 shows the magnitude and phase ofHC(). In this case, thecircuit is a low pass filter of the second order because of the asymptoticslope 1/2.


5/21

DRAFT


Bode Diagram

Frequency (rad/sec)

Phase(deg)

Magnitude(dB)

40

30

20

10

0

10

20

104

105

106

180

135

90

45

0

Figure 2.2: Transfer functionHC()of the LCR series resonant circuit witha resonant angular frequencyC 10.7krad/s.

2.2.2 Frequency Response with Inductor Voltage Difference

as Circuit Output

Lets considering now the voltage differenceVLacross the inductor as thecircuit output. In this case, we have

HL() = 2 LC

jRC 2LC+ 1.

Using the definition ofQ, and0 and after some algebra, HL() be-comes

HL() = 2

20 2 +j0Q(2.2)


6/21

DRAFT


Computing the magnitude and phase ofHL(), we obtain

|HL()| = 2

20 22

+0Q

2arg [HL()] = arctan

1

Q

0

20 2

The magnitude has a maximum for

2

L

=2

0

1

1 12Q2,

and the maximum is

|HL(L)| = Q1 1

4Q2

.

Again, ifQ 1 thenL 0, and |HL(L)| Q.Far from resonanceL, the approximate behavior of|HL()| is

L

|HL()

| 2

20 L |HL()| 1

Figure 2.3 shows the magnitude and phase ofHL(). In this case thecircuit is a second order high pass filter.

2.2.3 Frequency Response with the Resistor Voltage Dif-ference as Circuit Output

Considering the voltage difference across the resistor as the circuit output,

we will haveHR() =

jRC

12LC+jRC .

Using the definition ofQ and0, and after some algebra, HR() be-comes


7/21

DRAFT


Bode Diagram

Frequency (rad/sec)

Phase(deg)

Magnitude(dB)

60

40

20

0

20

40

104

105

106

0

45

90

135

180

Figure 2.3: Transfer functionHL()of the LCR series resonant circuit witha resonant angular frequencyL 10.7krad/s.

HR() =j0Q

20 2 +j0Q. (2.3)

Computing the magnitude and phase ofHR(), we obtain

|HR()| =

0Q

20 22

+0Q

2arg [HR()] = arctan

Q20 20


8/21

DRAFT


The magnitude has maximum for

2R =20,

and the maximum is

|HR(R)| = 1 .Far from the resonanceR, the approximate behavior of|HR()| is

R |HR()| 1Q

0

R |HR()| 0

Figure 2.4 shows the magnitude and phase ofHR(). In this case thecircuit is a first order band pass filter.

2.2.4 Transient Response

The equation that describes the LCR series circuit response in the timedomain is

vi = Ri +Ldi

dt+

1

C

t

0

i(t)dt , (2.4)

wherei(t)is the current flowing through the circuit andvi(t)is the inputvoltage.

Supposing that

vi(t) =

v0, t > 0

0, t 0 ,

and differentiating both side of eq. 2.4, we obtain the linear differentialequation

Rdi

dt+L

d2i

dt2+

1

Ci = 0, t > 0

or, considering the definition of0, andQ,

d2i

dt2+

0Q

di

dt+ 20i = 0.


9/21

DRAFT


Bode Diagram

Frequency (rad/sec)

Phase(deg)

Magnitude(dB)

104

105

106

90

45

0

45

90

50

40

30

20

10

0

Figure 2.4: Transfer functionHR()of the LCR series resonant circuit withresonant angular frequencyR 10.7krad/s.

The solutions of the characteristic polynomial equation associated withthe differential equation are

1,2 = 12

0Q

1

1 4Q2

.

As usual, we will have three different solutions depending on the dis-criminant value

= 1

4Q2 .

Under-damped Case: discriminant less than zero(Q > 1/2)

In this case we have two complex conjugate roots and the differential equa-tion solution is the typical exponential ring down


10/21


11/21

DRAFT


Is

VoCR

L

r

Figure 2.5: The tank circuit.

2.3.1 LCR Circuit Frequency Response

Using Thvenin theorem for the current source and R, the LCR parallelcircuit considering the equivalent circuit as shown in figure 2.6 where thecurrent source and the resistor R have been replaced with the Thvenincircuit.

Considering that the currentIof the current source can be written as

I = Vi

R,

I = Y Vo = 1R

+ 1

r+jL

+jCVo,we have

ViR

=

1

R+

1

r+jL+ jC

Vo (2.5)

Defining the following complex quantity as

1

r()+

1

jL()=

1

r+jL, (2.6)

and

R=R || r,eq. 2.5 becomes

ViR

=

1

R+

1

jL+jC

Vo


12/21

DRAFT

2.3. THE TANK CIRCUIT OR LCR PARALLEL CIRCUIT. 47

Vo

sV

r

L

C

R

+

Figure 2.6: The tank circuit with the current source and the resistance Rreplaced with the Thvenin equivalent circuit.

After some algebra, we will have

VoVi

= jL

R 2CLR+jLR

R . (2.7)

Generalizing the definition of0, andQ

0 = 1

L()C

, Q= R()

C

L() ,

and substituting in eq. 2.7 we finally obtain

H() = j0 /Q

(0 )2 2 +j0 /QR

R

Lets find the implicitly defined functions r, L. Using the term con-taining the inductanceL in eq. 2.6, we obtain

1

r+jL=

1

r

1 +

L

r

2

+

1

jL

1 +

rL

2

.

and finally

r() =r

1 +

L

r

2, L() = L

1 +

rL

2


13/21

DRAFT


100

80

60

40

20

0

Magnitude(dB)

101

102

103

104

105

106

107

90

45

0

45

90

Phase(deg)

Bode Diagram

Frequency (rad/sec)

Figure 2.7: Typical bode plot of a LCR parallel circuit with resonant angu-lar frequency near the acoustic band. As expected , ifr is not zero, then

the magnitude doesnt go to zero for =0.

2.3.2 Transfer Function

From the solution of the LCR parallel circuit we have

|H()| =0Q

0

2 2

2+

0Q

2

|R|R

arg(H()) = arctan

Q20 20

,

whose bode plots are shown in figure 2.7.


14/21

DRAFT


2.3.3 Simplest Case

Ifr =0, then we will have much simpler expressions for the thank circuitformulas, i.e.

0 = 1

LC, Q= R

C

L ,

and

H() = j0/Q

20 2 +j0/Q.

The magnitude and the phase will be

|H()| =0

Q20 2

2+0

Q

2 ,

arg(H()) = arctan

Q20 20

.

2.3.4 High Frequency Approximation

For high frequency

r/L, we have

r() r

L

r

2, L L

and0becomes

0 1LC

.

Evaluating the several defined quantities at0, we will have

r(0)

L

rC

,

R(0) LRRCr+L

Q(0) LRRCr+L

C

L


15/21

DRAFT


0 0.2 0.4 0.6 0.8 1 1.2

x 103

0.03

0.02

0.01

0

0.01

0.02

0.03

0.04

Step Response

Time (sec)

Amplitude

Figure 2.8: Typical step response of a LCR parallel circuit near the acousticband.

2.3.5 LCR Parallel Circuit Transient Response

Lets briefly analyze the response to a step of the LCR parallel circuit forthe under-damped case.

If we define the following quantity

= 1

2Q,

called damping coefficient, and if 0 < < 1, then we will have at thecircuit output

v(t) =v02

1 e0t cos

1 2 0t+0

+v1 .


16/21

DRAFT


The voltage outputv(t)is a damped sinusoid with angular frequency1

2

0 and time constant = 1/0. The DC offsetv1 depends onthe inductor resistancer and the initial step.Figure 2.8, a typical step response of the LCR circuit shows a ring-down

with a DC offset.


17/21

DRAFT


2.4 Laboratory Experiment

Real inductors have not negligible resistance. To build a LCR series circuitwith a highest quality factor it is indeed necessary to minimize the resis-tance of the circuit by mounting in series the inductor and the capacitoronly. Typical effective resistance of the inductors used in the laboratory isabout 10 to 80 at resonance .

Because of the internal resistance of the function generator (the bestscenario gives 50) is then comparable at some frequencies to LCRload, we will expect that the approximation of ideal generator will be nolonger valid.

Moreover, harmonic distortion of the function generator will be quite

evident in the LCR series circuit because of the dependence of the load onthe frequency.

An estimation of a ring-down time constant can be obtained as fol-lows. From the ring-down equation we have that after a timet = theamplitude is reduced by a factor 1/3 (e 1/2.718). This means that wecan easily estimateby just measuring the time needed to reduce the am-plitude down to about 1/3 of its initial value. A similar but quite coarseway is to count how many periodsn the amplitude takes to decrease to1/3 of its initial value. Then the estimation will be

Tn =

n

res

,

whereT, andres are respectively the period and the frequency of the os-cillation. Considering thatQ = resthen

Q n .The quality factor can also be estimated from the frequency response

considering that

Q=res

,

where is the Full Width at Half Maximum (FWHM) of the peak reso-nance.

2.4.1 Pre-laboratory Exercises

It is suggested to read the appendix about the electromagentic noise tocomplete the pre-lab problems and the laboratory procedure.


18/21

DRAFT

2.4. LABORATORY EXPERIMENT 53

1. Determine the capacitanceCof a LCR series circuit necessary to have

a resonant frequencyC =20kHz ifL = 10mH, andR = 10

. Then,calculateQ,,0,(=2)of the circuit.

2. Find the LCR series input impedanceZi and plot its magnitude in alogarithmic scale. Determine at what frequency is the minimum of|Zi| .

3. Supposing that the internal resistance of the function generator isRs = 50, and using the previous values for L,C, and R, calculatethe circuit input voltage attenuation at the frequency of|Zi| mini-mum and at twice that frequency.

4. Determine the capacitance C of a tank circuit necessary to have aresonant frequency C = 20 kHz if L = 10mH, R = 10k, andr = 10. Use the high frequency approximation. Then, calculate Q,,0, of the circuit.

5. Estimate the time constantof the ring-down in figure 2.8. Suppos-ing thatR = 10k, estimaterfrom figure 2.7.

6. Calculate the maximum frequency of the EM field isolated by a Fara-day cage with a dimension d = 10mm (hint: consult the proper ap-pendix).

2.4.2 Procedure

1. Build a LCR series circuit with a resonant frequency of around 20kHz,using inductance, capacitance, and resistance values calculated inthe pre-lab problems. Then, do the following steps:

(a) Using the oscilloscope and knowing the expected magnitudeand phase values at the resonant frequency C, find Cand com-pare it with the theoretical value computed using the compo-nents measured values.

(b) Verify the circuit transfer functionHC()using the data acqui-sition system and the proper software.

(c) Estimate the quality factor Q of the circuit from the transferfunction measurement and compare it with the theoretical value.


19/21

DRAFT


(d) Explain why the input voltageVi changes in amplitude if we

change frequency.(e) Considering the harmonic distortion of the function generator,

explain why the frequency spectrum of the input signal changesquite drastically when we approach the resonance C.

(f) Download the simulation file from the ph5/105 website for theLCR series circuit, input the proper components values, run theAC response simulation, and find the discrepancies beteweenyour measurement and the simulation.

(g) Modify the simulated circuit to qualitatively account for theeventual notch measured between 100 kHz and 1 MHz (hint:use the inductor model specified in the appendix consideringthe extra capacitor only).

2. Build a LCR parallel circuit with a resonant frequency around 20kHz,using inductance, capacitance, and resistance values calculated inthe pre-lab problems. Then, do the following steps:

(a) Using the oscilloscope and knowing the expected magnitudeand phase values at the resonant frequency C, find Cand com-pare it with the theoretical value computed using the compo-nents measured values.

(b) Verify the circuit transfer functionHC()using the data acqui-sition system and the proper software.

(c) Estimate the quality factor Q of the circuit from the step re-sponse.

(d) Download the simulation file from the ph5/105 website for theLCR parallel circuit, input the proper components values, runthe AC response simulation, and find the discrepancies bete-ween your measurement and the simulation.

(e) Modify the simulated circuit to qualitatively account for theeventual notch measured between 100 kHz and 1MHz (hint:use the capacitor model specified in the appendix consideringthe extra inductors only).


20/21

DRAFT

2.4. LABORATORY EXPERIMENT 55

3. Check the effect of the Faraday cage ( a metallic coffee can) using 10x

probe connected to the oscilloscope. Add a 1m long wire to increasethe antenna effect.Note the differences when the antenna is approached to the fluores-cent lights, and when you touch the antenna.Keeping the cage in the same position and without touching thecage, explain what you observe and coarsely estimate the amplitudeand frequency content of the picked-up signal in the following con-ditions:

(a) Antenna outside the cage,

(b) Antenna inside the cage,

(c) Antenna inside the cage with ground probe connected to thecage.


21/21

DRAFT


Ph5.Chapter.resonantCircuits

Documents

Transcript of Ph5.Chapter.resonantCircuits