manual de analogicas

8/12/2019 Manual de Analogicas

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COPYRIGHT BY ELETTRONICA VENETA & INEL SPA

ANALOG COMMUNICATIONS I

module MCM20/EV

Volume 1/2

THEORY AND EXERCISES

TEACHER/STUDENT handbook

31045 MOTTA DI LIVENZA (Treviso) ITALY

Via Postumia, 16


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MC2011E0.DOC


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Contents

CONTENTS

Lesson 900: ATTENUATORS 1

Lesson 901: SERIES and PARALLEL R-L-C CIRCUITS 6

Lesson 902: COUPLED CIRCUITS 15

Lesson 903: IMPEDANCE TRANSFORMATION WITH TRANSFORMER 21

Lesson 904: FILTERS OF DISCRETE COMPONENTS I 25

Lesson 905: FILTERS OF DISCRETE COMPONENTS II 41

Lesson 906: CERAMIC FILTERS 50

Lesson 907: QUARTZ-CRYSTAL FILTERS 56

Lesson 908: IMPEDANCE MATCH I 64

Lesson 909: IMPEDANCE MATCH II 72

Lesson 924: RF AMPLIFIER 78

Lesson 925: AM TRANSMITTER 84


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SAFETY RULES

Keep this handbook at hand for any further help.

After the packaging has been removed, set all accessories in order so

that they are not lost. Check that the equipment is integral and shows no

visible damage.

Before connecting the +/-12V power supply to the module, be sure that

the power cables are properly connected to the power supply.

This equipment must be employed only for the use it was conceived, i.e.

as educational equipment, and must be used under the direct supervision

of expert personnel. Any other use is not proper and therefore

dangerous. The manufacturer cannot be held responsible for eventual

damages due to inappropriate, wrong or unreasonable use.


5/92

Lesson 900: Attenuators

1

LESSON 900: ATTENUATORS

Objectives

Describing the characteristics of attenuators measurements of attenuation

Necessary equipment

Power Supply mod. PSU/EV

Module-holding base mod. MU/EV

Individual Control Unit SIS1, SIS2 or SIS3

Testing module mod MCM20/EV

Dual-trace oscilloscope

Function generator

900.1 THEORETICAL HINTS

Attenuators are quadripoles that are connected between a source and a

load, when the amplitude of the signal across the load is required to be

lower than the amplitude of the signal generated by the source (fig.

900.1). Attenuation is usually expressed in decibeland is defined by the

following formula:

AdB= 20 log10(Vout/Vin)

The impedance match between generator and load (with attenuatorenabled) can be obtained only if:

the input impedance Zi of the attenuator is equal to the generator

impedance ZG the output impedance Zo of the attenuator is equal to the load

impedance ZL.

Fig. 900.1


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An important group of attenuators is represented by the purely resistive

networks of T or pi structure (fig. 900.2). When input and output

impedances are equal (Zi=Zo=Z), the formulae for dimensioning

attenuators are:

T attenuator pi attenuator

R1 = R2 = Z [(K-1)/(K+1)] R1 = R3= Z [(K+1)/(K-1)]

R3 = 2Z K/(K-1) R2 = Z/2 (K-1)/K

AdB/20where: K = 10

This module includes four attenuators of 1-2-4-8 dB which have

input/output impedances of 600 . The used resistors (fig. 900.3) havevalues approaching the rated values calculated with the previousformulae.

Fig. 900.2

Fig. 900.3


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3

900.2 EXERCISES

MCM20 Disconnect all the jumpers

SIS1 Set all the switches to OFF

SIS2 Enter the lesson code number: 900

900.2.1 Measurements of attenuation

Carry out the connections as it is shown in the fig. 900.4 taking care

that:

- the generator must be connected to the attenuator input, if the

generator output impedance is 600 - if the generator has an output impedance of 500 , it must be

connected to the attenuator input, after crossing the resistor R23

(of approximately 550 ), so that a total resistance of 600 canbe obtained at the output of R23

Fig. 900.4

connect the probes of the oscilloscope to the attenuator input and

output (TP53 and TP59)

set the generator frequency to 1 kHz; adjust the amplitude, to obtain

1 Vpp at the attenuator input (TP53)

Q1 Which is the amplitude of the signal available at the attenuator output(TP59) ?

SET

A B

1 3 approximately 1 Vpp

2 4 approximately 0.5 Vpp




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Q2 Which is the value of the attenuation (in dB) ?

SET

A B

1 3 approximately 1 dB

2 1 approximately 3.9 dB3 4 approximately 2 dB

4 2 approximately 0.8 dB

repeat the same measurements on the other attenuators.

900.2.2 Cascade attenuators

Carry out the connections as it is shown in the fig. 900.5 taking care

that:

- the generator must be connected to the attenuator input, if its

output impedance has a value of 600 - if the generator has an output impedance of 50 , it must be

connected to the attenuator input, after crossing the resistor R23

(of approximately 550 ), so that a total resistance of 600 canbe obtained at the output of R23

Fig. 900.5

connect the probes of the oscilloscope to the attenuator input and

output (TP53 and TP59)

set the generator frequency to 1 kHz; adjust the amplitude, to obtain

1 Vpp at the attenuator input (TP53)

Q3 Which is the amplitude of the signal available at the attenuator output(TP59) ?

SET

A B






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5

Q4 Which is the value of the attenuation (in dB) ?

SET

A B


2 3 approximately 3.9 dB3 2 approximately 2 dB


repeat the same measurements on other combinations of attenuators.

900.3 QUESTIONS

Q5 If Vinand Voutare the input and output voltages of an attenuator, theattenuation (in dB) is expressed by the following formula:

SET

A B

1 3 AdB= 10 log10(Vout/Vin)2 1 AdB= 20 loge(Vout/Vin)3 2 AdB= 20 log10(Vout/Vin)4 5 AdB= 10 log10(Vin/Vout)5 4 AdB= 10 log20(Vout/Vin)

Q6 A resistive attenuator is characterized by the following parameters:

SET

A B

1 4 input impedance / attenuation / center frequency

2 3 output impedance / attenuation / cutoff frequency

3 2 input impedance / gain / output impedance

4 1 input impedance / attenuation / output impedance


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Lesson 901: Series and Parallel R-L-C Circuits

- 6 -

LESSON 901: SERIES and PARALLEL R-L-C- CIRCUITS

Objectives

Measuring the resonance frequency and the bandwidth of series andparallel R-L-C circuits

calculating the Q factor

plotting the voltage-vs-frequency and phase-vs-frequency curves of

series and parallel R-L-C circuits

Necessary equipment







901.1.1 Series R-L-C circuits

Consider a series R-L-C circuit powered by a generator of alternating

voltage (fig. 901.1). The total circuit impedance is expressed by the

following formula:

Zs= R + jL + 1/(jC) = R + jXL - jXC

When frequency varies, the voltages across L and C vary in opposite

directions: VL increases together with frequency, whereas VCdecreases. The respective reactances XL and XC undergo the same

variation. It can intuitively be inferred that at a certain frequency (fo)

both reactances have the same value. Furthermore, being of opposite

signs (an inductance and a capacitance provoke opposite phase shifts),

they tend to cancel out as frequency varies. In particular, at the

frequency fo their sum will be null, and the circuit becomes only

resistive. In fact, if XL

=XC

,the circuit impedance will be minimum

and only resistive; so it will be expressed as follows:

Zs= R

In this condition the circuit is resonant and fo is the resonance

frequency. Being XL and XC equal, the value of fo is determined

through the following formulae:

wL = 1/C

from which:

fo= 1/(2LC) = o/2 [Hz]


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Q factor In resonance conditions, the Q factor of the circuit is defined as the

result of the ratio between the stored (reactive) power and the (active)

power dissipated in the circuit. IfI is the current crossing the circuit,

Q is expressed as follows:

R

X

R

X

ICR

I

IR

ILQ CL

2

o

2

2

2o ==

=

=

The above parameter allows to qualify the behaviour of the resonant

circuit.

Resonance curves

The resonance curves (fig. 901.2) of a series R-L-C circuit represent the

module and phase of the current versus frequency. The sketch of these

curves depends on Q.

Fig. 901.1 Series R-L-C- circuit

Fig. 901.2 Resonance curves of a series R-L-C circuit


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Passband

Consider the resonance curve of the current module. The bandwidth is

the difference between the two frequencies f2 and f1, at which the

current decreases of 3 dB (corresponding to a value 0.707 as high as the

maximum value):

B = f2- f1 [Hz]

Bandwidth, resonance frequency and Q factor of a series R-L-C circuit

are linked to each other by the following relationship:

B = fo/Q = R/(2L) [Hz]

901.1.2 Parallel R-L-C circuits

When the R-L-C elements are connected in parallel to each other (fig.901.3), the resulting circuit has a dual behaviour with respect to the

series circuit. The total admittance of the circuit is expressed by:

Fig. 901.3 Parallel R-L-C- circuit

Y = 1/Zp= 1/R + 1/jL + jC = 1/R - j/XL+ j/XC

The same formulae of series circuits are still valid for parallel circuits,

therefore, in this case too, the resonance frequency is expressed as

follows:

fo = 1/(2LC) [Hz]

The value of Zp

depends on frequency. At the frequency fo

, the reactive

components cancel out, therefore the circuit becomes only resistive. The

circuit impedance is maximum and is expressed as follows:

Zp= R

Q factor is:

Q = R/XL= R/XC

The resonance curves (fig. 901.4) of a parallel R-L-C circuit represent

the module and phase of voltage versus frequency. The sketch of thesecurves depends on Q.


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Bandwidth, resonance frequency and Q factor of parallel R-L-C circuits

are linked with each other by the following relationship:

CR2

R

Q

fB o

== [Hz]

Fig. 901.4 Resonance curves of a parallel R-L-C circuit

901.1.3 Universal resonance curves

Resonant circuits with different values of resonance frequencies and Q

show different resonance curves, although they keep the same shape.

The responses of different resonant circuits can be represented with only

two curves (one for amplitude and the other for phase). These curves are

called universal resonance curves(fig. 901.5) and are constructed in the

following way:

series R-L-C circuits: in ordinate there are the values of the ratio

between the current I and its maximum value Io (amplitude diagram),

and of the phase shift between I and Io (phase diagram)

parallel R-L-C circuits: in ordinate there are the values of the ratio

between the voltage V and its maximum value Vo, and of the phase

shift between V and Vo

in abscissa there are the values of Q, where is the relativedeviation of frequency with respect to the resonance frequency:

o

o

f

ff=


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Fig. 901.5 Universal resonance curves

901.2 EXERCISES




901.2.1 Parallel R-L-C circuits:Resonance frequency

Arrange the section "TUNED CIRCUITS & COUPLING" as it is

shown in the fig. 901.6, in order to assemble a parallel R-L-C circuit.

Turn RV3 completely in clockwise direction (maximum resistance

on). Turn the COUPLINGknob to MIN

Fig. 901.6


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apply a sine wave with amplitude of approximately 1 Vpp and

frequency of approximately 400 kHz, between TP36 and TP38 (use

the VCO of the module, output TP4)

connect the oscilloscope (probes 10:1) to the input of the whole

circuit (between TP36 and TP38) and to the only terminals of the R-

L-C circuit (TP37)

Q1 How is this frequency named ?SET

A B

1 2 image frequency

2 1 series frequency

3 4 parallel frequency

4 3 resonance frequency

Q2 Which is the value measured ?SET

A B

1 4 approximately 500 khz

2 3 approximately 400 kHz

3 2 it depends on the position of the variable capacitor CV3

4 1 approximately 700 kHz.

calibrate CV3 to obtain a resonance frequency of approximately 650

kHz.

Parallel R-L-C circuits:Resonance curves Apply a signal with frequency of 650 kHz (corresponding to the

resonance frequency set before), between TP36 and TP38. Be Vmaxthe peak-to-peak voltage measured across the R-L-C circuit in these

conditions, and o the phase difference between the signal acrossthe R-L-C circuit and the input signal.

Q3 Which is the value of the phase difference between these two signals ?SET

A B

1 3 approximately 902 1 approximately 180

3 4 approximately 0

4 2 approximately -90

recvord the frequency value and the corresponding peak-to-peak

voltages and phase differences measured, on a table (fig.901.7)

increase the frequency from 580 to 720 kHz, by steps of 10 kHz;

repeat the previous measurements and record the data on the table.

Calculate the ratio Vo/Vmax

corresponding to each frequency

starting from the data of the table, plot two graphs:


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- the resonance curve of the amplitudes of the parallel R-L-C circuit

can be plotted with Vo/Vmaxon the Y axis and the frequency on

the X axis

- the resonance curve of the phases of the parallel R-L-C circuit can

be plotted with on the Y axis and the frequency on the X axis

Frequency

[kHz]

Output voltage Vo

[Vpp]

Vo/Vmax Phase difference

[]

580 ... ... ...

590 ... ... ...

... ... ... ...

... ... ... ...

650 Vmax ... 0

... ... ... ...

... ... ... ...

710 ... ... ...

720 ... ... ...

Fig, 901.7

Parallel R-L-C circuits:Passband

Considering the resonance curve of amplitudes, calculate the value of

the passband B of the circuit through the formula:

B = f2 - f1

where f2 and f1 are the frequencies at which the ratio Vo/Vmax

decreases of 0.707 times (3 dB), with respect to its maximum value.

Q4 Which is the value of the passband ?SET

A B





reduce the value of RV3 and check whether the resonance curve is

broadenedand the passband increases.

SIS1 Set the switch SW6 to ON

SIS2 PressINS


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Q5 Which effect can be noted in the circuit ?

SET

A B

1 3 the center frequency increases because the parallel

capacitance has decreased2 1 the center frequency increases because the parallel inductance

has increased

3 4 the passband increases because the parallel resistance has

increased

4 2 the passband increases because the parallel conductance has

increased

SIS1 Set the switch SW6 to OFF

Resonance curve detected with wobbulator

Arrange the circuits as it is shown in the fig. 901.8

prearrange the oscilloscope in X-Y (X axis on 1 V/div; Y axis on 20

mV/div)

connect the X axis of the oscilloscope to TP1 (X AXIS). Connect the

Y axis (probe 10:1) across the R-L-C circuit (between TP37 and

ground)

adjust the center frequency of the VCO and the Sweep amplitude

(DEPTH), so that the resonance curve of the circuit can be displayed

on the oscilloscope. This curve is similar to that shown in the fig.

901.9

vary the capacitance CV3 and note that the resonance frequency is

shifted. Vary the resistance RV3 and note how the passband varies.

Fig. 901.8

Fig, 901.9


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901.2.2 Series R-L-C circuits

Arrange the section "TUNED CIRCUITS & COUPLING" as it is

shown in the fig. 901.10, in order to construct a series R-L-C circuit.

Turn RV3 completely in anticlockwise direction (minimum

resistance on). Turn the COUPLINGknob to MIN carry out measurements similar to those concerning parallel R-L-C

circuits (resonance frequency, resonance curves, passband). In this

case, the minimum signal will correspond to the resonance frequency

across the R-L-C circuit. But the coil is not shielded (in fact it will

also be used in the test on mutual coupling), therefore there may be a

signal leak coupled directly to the output.

Fig. 901.10

901.3 QUESTIONS

Q6 Which is the value of the resonance frequency in a circuit with thefollowing values: L = 200 H; C = 330 pF; R = 10 K?

SET

A B

1 3 251.5 kHz

2 1 467.2 kHz

3 2 891.1 kHz

4 5 619.5 kHz

5 4 712.6 kHz

Q7 The bandwidth of a parallel resonant circuit depends on the values of

SET

A B

1 4 R and L

2 3 the resonance frequency

3 2 R and C

4 1 L and C


19/92

Lesson 902: Coupled Circuits

- 15 -

LESSON 902: COUPLED CIRCUITS

Objectives

Examining the operation of resonant circuits coupled inductively Examining the operation of resonant circuits coupled capacitively

Necessary equipment





Dual-trace oscilloscope.


902.1.1 Inductive coupling

Consider the diagram shown in the fig. 902.1 where the tuned circuit

L1-C1 is inductively coupled with another tuned circuit L2-C2. The

frequency response of this circuit (resulting from Vout/Vin as

frequency varies) depends remarkably on the coupling coefficient K

between the primary L1 and the secondary L2.

Fig. 902.1 Inductive coupling Fig. 902.2Frequency response

of tuned circuits

The coupling coefficientKis defined as follows:

2L1L

MK

2

= M = mutual induction

As K varies (i.e. as M varies), also the frequency response of the

circuit varies, as it is shown in the fig. 902.2. These curves show that

responses have two different peaks (although the primary and

secondary circuits are tuned at the same frequency), if K exceeds a

critical value, Kc, defined by:


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)QQ(

1Kc

21=

where Q1and Q2are the quality coefficients (of equal value) of the

primary and secondary circuits. If K=Kc, the response is flat at its top;whereas, if K


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902.2 EXERCISES




902.2.1 Inductive coupling Frequency response with wobbulator

Arrange the section TUNED CIRCUITS & COUPLING as it is

shown in the fig. 902.5, in order to obtain an inductive coupling of

two resonant circuits. Turn RV3 and RV4 completely in clockwise

direction (maximum resistance on). Turn the COUPLING knob to

MIN apply a wobbled signal ranging from 500 to 900 kHz, with an

amplitude of approximately 1 Vpp, between TP36 and TP38. Use the

VCO set as shown in the fig. 902.5, as wobbulator

set the oscilloscope to X-Y (X axis on 1 V/div; Y axis on 10 mV/div)


Y axis (probe 10:1) between TP39 and TP40


(DEPTH), to obtain a curve similar to that shown in the fig. 902.6a.

Adjust the capacitances CV3 and CV4, to obtain the same resonance

frequency of the two tuned circuits (a symmetrical peak of maximum

amplitude is obtained)

Fig. 902.5

Fig. 902.6


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Turn the COUPLINGknob to MAX. The resulting curve is like that

shown in the fig. 902.6b.

Q1 When CV3 and CV4 are varied:

SET

A B

1 2 the two peaks of the curve are superimposed

2 1 a third peak due to the series resonance frequency appears

3 4 a peak disappears for the band rejection

4 3 the two peaks are not superimposed

Q2 How can a curve with flat top, like that shown in the fig. 902.6c, beobtained ?

SET

A B

1 4 connecting RV3, increasing RV3/RV4 and the coupling

2 3 connecting and increasing RV3, and decreasing RV4 and the

coupling

3 2 connecting RV3, decreasing RV3/RV4 and the coupling

4 1 varying CV3 and CV4


SIS2 PressINS

Q3 Which effect can be noted in this measurement ?

SET

A B

1 3 the curve disappears because L15 is in short circuit2 1 the curve amplitude increases

3 4 the curve disappears because the VCO is not modulated

4 2 the curve disappears because L15 is open



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902.2.2 Capacitive coupling. Frequency response with wobbulator


shown in the fig. 902.7, in order to obtain a capacitive coupling of

two resonant circuits. Turn RV3 and RV4 completely in clockwise

direction (maximum resistance on). Turn the COUPLING knob toMIN

detect the frequency response of the circuit. Note that there are two

peaks.

Q4 How do the peaks of the curve var, if the inductive coupling is alsoinserted (COUPLINGknob turned to MAX) ?

SET

A B

1 2 they are reduced2 1 they are more accentuated

3 4 another peak (the third one) appears

4 3 they do not vary

Fig. 902.7


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902.3 QUESTIONS

Q5 The frequency response of two coupled circuits strictly depends on:

SET

A B1 3 the bandwidth of inductances

2 1 the mutual induction and on the value of capacitances

3 4 the coupling coefficient

4 2 the oscillation coefficient

Q6 Coupled circuits are commonly used to construct:

SET

A B

1 5 broad-band amplifiers2 3 oscillators

3 2 high-pass filters

4 1 band-pass filters

5 4 multivibrators

Q7 Two resonant circuits are usually coupled through:

SET

A B

1 3 anti-inductive or capacitive resistances

2 1 inductances and capacitances

3 5 mutual inductance and capacitances

4 2 mutual inductance and resistances

5 4 transformer


25/92

Lesson 903: Impedance Transformation with Transformer

- 21 -

LESSON 903: IMPEDANCE TRANSFORMATION with

TRANSFORMER

Objectives Examining the impedance matching technique with transformer and

autotransformer.

Necessary equipment






Frequency-meter Multimeter


The Q factorof a parallel resonant circuit depends on the value of the

equivalent resistance connected in parallel to the coil (Q=R/wL). As

the load resistance of the resonant circuit (RV4 - fig. 903.1) is also

considered in the calculation of the equivalent resistance, some

particular techniques must often be used to increase the equivalent

value of the load resistance across the coil, so that a higher Q can be

obtained.A typical system for obtaining this transformation consists of an

autotransformer (L16, with center tap - fig. 903.1). The winding

(n2+n2) acts as secondary circuit, for the winding n1, and as primary

circuit for the winding n2 which the load resistance RV4 is connected

to. The results of the equations concerning the transformer explain that

an equivalent load resistance Req can be measured across L16

(formed by the windings n2+n2), as it is shown in the following

formula:

4RV44RVn

nnqRe

2

2

22 =+=

Therefore a center-tap transformer like that shown in the fig. 903.1

enables to transform the impedance value according to the square of

the ratio between the total number of turns in the winding and the

number of turns in parallel to the impedance to be transformed.


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Fig. 903.1 Impedance transformation through autotransformer

903.2 EXERCISES




903.2.1 Impedance transformation with transformer


shown in the fig. 903.2, in order to obtain a transformer coupling

between the input signal and the load RV4. Turn RV4 completely in

clockwise direction (maximum resistance on). Turn the COUPLINGknob to MAX

Fig. 903.2

apply a wobbled signal ranging from 500 to 900 kHz, with an

amplitude of approximately 2 Vpp, between TP36 and TP38. Use the

VCO prearranged as shown in the fig. 903.2, as wobbulator



Y axis (probe 10:1) between TP39 and TP40


(DEPTH), to obtain a curve similar to that shown in the fig. 903.3


27/92


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Q3 Why ?

SET

A B

1 4 because the load resistance across the secondary circuit has

been increased by the autotransformer2 1 because the load resistance across the secondary circuit has

been reduced by the autotransformer

3 2 because the load resistance across the secondary circuit has

been halved by the autotransformer

4 3 because no variation occurred

Q4 Which is the value of the passband B2?

SETA B





set RV4 to approximately 1/4 of its value (47/4 12 k) carrying outa measurement between TP39 and TP40, after removing the jumper

Q5 Connect the jumper again and check whether:

SET

A B

1 4 the new bandwidthB3is larger thanB1, because it has a load

resistance 4 times as low

2 3 the new bandwidthB3is very nearB1, although it has a load


3 2 the new bandwidthB3is very nearB1, although it has a load

resistance 4 times as high4 1 the new bandwidthB3is very nearB1, although it has a load



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Lesson 904: Filters of Discrete Components I

- 25 -

LESSON 904: FILTERS of DISCRETE COMPONENTS I

Objectives

Describing the operation of constant-K, T and pi, high-pass, low-pass, band-pass, band-rejection passive filters

Describing the operation of M-derivedlow-pass passive filters.

Necessary equipment





904.1.1 Electric filters

Electric filters are quadripoles enabling the passage of certainfrequencies, whereas other frequencies are eliminated.

There are 4 types of filters:

low-pass filters: they allow the passage of all the frequencies below a

critical frequency fc, and attenuate the frequencies exceeding fc

high-pass filters: they allow the passage of all the frequencies

exceeding a critical frequency fc, and attenuate the frequencies lower

than fc

band-pass filters: they allow the passage of all the frequencies

included within two values f1 and f2, and attenuate all the other

frequencies

band-rejection filters: they attenuate all the frequencies included

within two values f1 and f2, and allow the passage of all the other

frequencies

The fig. 904.1 indicates the symbol and the ideal and actual frequency

responses of each type of filter.

904.1.2 Characteristic parameters

Refer to the fig. 904.2.

Insertion loss: it is the ratio between the amplitudes of the desiredfrequencies before and after the filter insertion

Rejection: it is the ratio between the amplitudes of the undesired

frequencies before and after the filter insertion

Cutoff frequency: it is the frequency leading to an attenuation of 3 dB

with respect to the insertion loss. Actually, consider the maximum

amplitude of the filter response: the 3 dB of attenuation and the

attenuation frequency (or frequencies) are determined with reference

to this value

Bandwidth: it is the band included between the lower (fc1) and upper

(fc2) cutoff frequencies of a band-pass or band-rejection filter. This


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parameter is also applied to low-pass filters, but in this case it

corresponds to the band included between zero frequency anf fc

Q factor: it is the ratio between the center frequency fo and the

bandwidth B of a band-pass or band-rejection filter:

Q = fo/B

Form factor: it is the ratio between the two bands at -60 dB and -6 dBF = B(-60 dB)/B(-6 dB)

Sometimes the band at -3 dB is taken as reference, so the result is:

F = B(-60 dB)/B(-3 dB)

Impedance: it is the value of input and output impedances of a filter.

Fig. 904.1

Fig. 904.2


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904.1.3 Constant-k filters

The simplest structure of a constant-k filter is shown in the fig. 904.3

(the meaning of constant-k will be explained later). The figure shows

an example of low-pass filter: only this type of filter will be examined

as the resulting data can easily be related to high-pass, band-pass andband-rejection filters. The following theoretical formulae of passive

quadripole networks can describe the main parameters of the circuit

shown in the fig. 904.3:

CL1ZCL1C

LZi 20

2 == input impedance

CL1

1Z

CL1

1

C

LZo

20

2

=

= output impedance

CLjCL1lnAi 2 += image attenuation constant

When: )(CL

10 c=

The value of Ai is 0, that is, the network does not provoke any

attenuation. When > c, Ai becomes: Ai = ln(LC-1)

In these conditions Ai exceeds zero, so the network provokes some

attenuation. As the circuit has a different behaviour at different

frequencies, this network may be used as a filter. The following value:

CL2

1

2fc c

=

=

is called cutoff frequency of the filter.

This network is called Constant-k filterbecause the product of the two

impedances (jL and 1/jC) is kept constant as frequency varies:

2KC

L

Cj

1Lj ==

The fig. 904.4 shows the typical sketches of Z1, Zo and Ai.


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Fig. 904.3 "L" low-pass constant-kfilter

Fig. 904.4 "L" low-pass constant-k filter:

a) real part of Zi and Zo

b) imaginary part of Zi and Zoc) imageattenuation constant Ai


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Exchanging L and C transforms this network from low-pass filter into

high-pass filter .

When two "L" constant-k sections are cascaded and comply with the

impedance match between them, the result is a "pi" or "T" filter, as it is

shown in the fig. 904.5. As the number of cascaded sections increases,also the slope of the filter response increases (fig. 904.6).

In actual applications, the source and load impedances of filters are

usually resistive and constant. The input and output impedances of

constant-k filters vary as frequency varies, although they are real

within the passband; therefore an impedance match can be obtained at

a single frequency.

The transmission characteristics shown in the fig. 904.4, obtained in

perfect matching conditions within the whole band, can only be

approximated when sources and loads are constant. Luckily in a lot ofactual applications, filter specifications are not very restrictive, so the

distortion caused by mismatching can be tolerated.

When impedance must be matched within all the operating band of the

filter,M-derived filters will be used (refer to the next paragraph).

As conclusion of the description of constant-k filters, the figures

904.7, 904.8, 904.9, 904.10 show:

the three "L", "T" and "pi" basic sections in low-pass and high-pass

configurations

the "T" and "pi" sections in band-pass and band-rejection

configurations

the input and output impedances of each section

the attenuation versus frequency (in perfect matching conditions)

the formulae for the calculation of filters.


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- 31 -

Fig. 904.7 Constant-k low-pass filters:

a) "L" section

b) "T" section

c) "pi" section


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- 32 -

Fig. 904.8 Constant-k high-pass filters:

a) "L" section

b) "T" section

c) "pi" section


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- 33 -

Fig. 904.9 Constant-k band-pass filters :

a) "T" section

b) "pi" section


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- 34 -

Fig. 904.10 Constant-k band-rejection filters:

a) "T" section

b) "pi" section


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- 35 -

904.1.4 M-derived filters

The constant-k filters examined in the previous paragraph have two

inconvenients:

impedance varies together with frequency

the out-of-band attenuation is not sufficient for many applications.The structure of M-derived filters is derived from that of constant-k

filters; they can:

minimize the impedance mismatching

improve the out-of-band attenuation.

Consider the circuit shown in the fig. 904.11a. The impedance Zi is:

1C)2L1L(11C

1LZi 2 +=

this is the same mathematical formula of input impedance for theconstant-k structure. This means that the M-derived structure can be

chosen to be perfectly matched to the constant-k structure.

Furthermore, the impedance Zm of anM-derived filter can be adjusted

to obtain a constant trend, and hence a good matching to a load or a

source being only resistive (fig. 904.12).

Actually the output impedance Zm depends on a value "M" being

expressed by: M=L1/L. The fig. 904.11b shows the sketch of Zm

versus different values of "M": note that when M=0.6, impedance is

constant enough and its value is Z0= (L/C).

Fig. 904.11 a) M-derived"L" low-pass filter

b) Zm versus "M"


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- 36 -

Fig. 904.12 Use of M-derived sections for impedance match

When the input or output element of a constant-k filter is a parallel

capacitance (as, for instance, in "pi" low-pass filters), the M-derived

output section corresponds to the structure shown in the fig. 904.13a.

Its impedance Zm', depending on "M", is shown in the fig. 904.13b.

Both structures of the figs. 904.11 and 904.13 have the sameattenuation constant whose trend versus frequency is shown in the fig.

904.14. Note that the attenuation becomes infinite when f=f:

212

1

CbLa2

1

1C2L2

1f

=

=

=

Of course "L" sections of M-derived filters can be used to assemble

"T" and "pi" filters (fig. 904.15). The figures 904.16 and 904.17 show

the low-pass configurations and the calculation formulae. Exchanging

L and C transforms low-pass filters into high-pass filters.

Fig. 904.13 a) M-derived"L" low-pass filter

b) Zm versus "M"


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- 37 -

Fig. 904.14 Attenuation Ai of an M-derived filter

Fig. 904.15 M-derivedlow-pass filters

a) "pi" section

b) "T" section


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Fig. 904.16 "L" and "pi" M-derivedlow-pass filters


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Fig. 904.17 "L" and "T" M-derivedlow-pass filters


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- 40 -

904.2 QUESTIONS


Q1 Electric filters are:

SET

A B

1 4 quadripoles which amplify all the frequencies

2 3 bipolar circuits which amplify only certain frequencies

3 2 quadripoles which allow the passage of certain frequencies

and elliminate the other frequencies

4 1 quadripoles which attenuate all the frequencies

Q2 The cutoff frequency of a filter is:

SET

A B

1 3 the frequency at which the filter starts self-oscillating

2 1 the frequency at which the filter output deviates of 3 dB from

the maximum (low-pass and band-pass) or minimum (high-

pass and band-rejection) output

3 4 the frequency at which the filter output deviates of 12 dB

from the maximum (low-pass and band-pass) or minimum

(high-pass and band-rejection) output

4 2 the double of passband

Q3 A band-pass filter with center frequency of 800 kHz has a Q of 40. Howmuch is its bandwidth ?

SET

A B

1 4 840 kHz

2 3 80 kHz

3 2 0.05 kHz4 1 20 kHz

Q4 The input and output impedances of a constant-k filter:

SET

A B

1 3 do not depend on frequency

2 1 depend on the product LC

3 4 depend on frequency

4 2 depend on amplitude


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Lesson 905: Filters of discrete components II

- 41 -

LESSON 905: FILTERS of DISCRETE COMPONENTS II

Objiectives

Examining the operation of constant-k T and pi, high-pass, low-pass, band-pass and band-rejection passive filters

examining the operation of M-derived low-pass passive filters

Necessary equipment





Oscilloscope

Frequency-meter Function generator


Refer to the previous lesson 904.


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- 42 -

905.2 EXERCISES




905.2.1 Constant-k T low-pass filter

Assemble the filter connecting the jumpers as it is shown in the

fig. 905.1.

Fig. 905.1 "T" low-pass filter

connect a generator with output impedance of 50 , to the filterinput (between TP5 and TP6), and supply a signal with amplitude

of approximately 2 Vpp and frequency of approximately 100

kHz

connect the oscilloscope to the filter output (between TP7 and

TP8), and record the measured amplitude on a table: this value

will be the reference (fig. 905.2)

starting from 400 kHz increase the generator frequency by steps of

10 kHz, whereas amplitude does not vary. Record the deviations

(in dB) of the output amplitude from the reference, on the table

Q1 Which is the value of cutoff frequency at -3 dB ?

SET

A B






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- 43 -

Frequency

[kHz]

Output voltage Vo

[Vpp]Vo/Vref Vo

20log Vref

100 Vref 1 0 dB 1

450 ... ... ...

460 ... ... ...

... ... ... ...

... ... ... ...

900 ... ... ...

1000 ... ... ...

Fig. 905.2 Frequency response of a low-pass filter

905.2.2 Constant-k pi low-pass filter


fig. 905.3

measure the filter passband.

Fig. 905.3 pi low-pass filter


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- 44 -

905.2.3 Constant-k T high-pass filter

Assemble the T low-pass filter connecting the jumpers as it is

shown in the fig. 905.4

connect a generator with output impedance of 50 , to the filterinput (between TP5 and TP6), and supply a signal with amplitudeof approximately 2 Vpp and frequency of approximately 1000

kHz



will be the reference (fig. 905.5)

starting from 600 kHz, reduce the generator frequency by steps

of 10 kHz, whereas the amplitude does not vary. Record the

deviations (in dB) of output amplitude from the reference, on the

table.

Q2 Which is the value of cutoff frequency at -3 dB ?SET

A B





Fig. 905.4 "T" high-pass filter

Frequency

[kHz]

Output voltage Vo

[Vpp]Vo/Vref Vo

20log Vref

1000 Vref 1 0 dB

600 ... ... ...

590 ... ... ...

... ... ... ...

200 ... ... ...

100 ... ... ...

Fig. 905.5 Frequency response of a high-pass filter


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- 45 -

905.2.4 Constant-k pi high-pass filter


fig. 905.6

measure the filter passband.

Fig. 905.6 pi high-pass filter

905.2.5 Constant-k band-pass filter

Use the filter shown in the fig. 905.7. Connect a generator with

output impedance of 500 , to the filter input (between TP9 and

TP10) and supply a signal with amplitude of approximately 2 Vppand frequency of approximately 400 kHz


TP12) and vary the frequency to obtain the maximum output.

Q3 At which frequency is the maximum output obtained ?

SET

A B

1 4 at approximately 500 kHz


3 2 at approximately 570 kHz4 1 at approximately 600 kHz

record the measured amplitude on a table: this value will be the

reference (fig.905.8)

vary the generator frequency by steps of 10 kHz, whereas the

amplitude does not vary. Record the deviations (in dB) of output

amplitude from the reference, on the table.


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- 46 -

Q4 Which is the value of the lower cutoff frequency ?SET

A B



3 4 approximately 450 kHz4 2 approximately 750 kHz

Q5 Which is the value of the upper cutoff frequency ?SET

A B





Fig. 905.7 Band-pass filter

Frequency

[kHz]

Output voltage Vo

[Vpp]Vo/Vref Vo

20log Vref

350 ... ... ...

360 ... ... ...

... ... ... ...

... ... ... ...

??? Vref 1 0 dB

... ... ... ...

... ... ... ...

690 ... ... ...

700 ... ... ...

Fig. 905.8 Frequency response of a band-pass filter


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905.2.5 Constant-k band-rejection filter

Use the filter shown in the fig. 905.9. Connect a generator with

output impedance of 50 , to the filter input (between TP16 andTP17), and supply a signal with amplitude of approximately 2

Vpp and frequency of approximately 1000 kHz

connect the oscilloscope to the filter output (between TP18 andTP19), and vary the frequency to obtain the minimum output

Q6 At which frequency is the minimum output obtained ?SET

A B





supply a signal having an amplitude of approximately 2 Vpp and afrequency of approximately 1000 kHz, and record the measured

amplitude on a table: this value will be the reference (fig. 905.10)

starting from 700 kHz reduce the generator frequency by steps of

10 kHz, whereas the amplitude does not vary. Record the


table.

Q7 Which is the value of the upper cutoff frequency ?SET

A B




Q8 Which is the value of the lower cutoff frequency ?SET

A B





Fig. 905.9 Band-rejection filter


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- 48 -

Frequency

[kHz]

Output voltage Vo

[Vpp]Vo/Vref Vo

20log Vref

350 ... ... ...

360 ... ... ...

... ... ... ...

... ... ... ...

690 ... ... ...

700 ... ... ...

1000 Vref 1 0 dB

Fig. 905.10 Frequency response of a band-rejection filter

905.2.6M-derived low-pass filter

Assemble the m-derived low-pass filter connecting the jumpers as

it is shown in the fig. 905.11

connect a generator with output impedance of 50 , to the filteroutput (between TP5 and TP6), and supply a signal having an

amplitude of approximately 2 Vpp and a frequency of

approximately 100 kHz



will be the reference starting from 400 KHz, increase the generator frequency by steps

of 10 kHz, whereas the amplitude does not vary. Record the


table

Fig. 905.11 M-derivedpi low-pass filter


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- 49 -

Q9 Which is the value of cutoff frequency at -3 dB ?

SET

A B




Q10 Which is the frequency value corresponding to a (theoretical) infiniteattenuation ?

SET

A B1 2 approximately 750 kHz




determine the almost constant trend of the filter impedance within

the passband carrying out the following operations:

- connect the oscilloscope after the resistor R1 and check

whether the signal level is kept rather constant, as frequency

varies up to approximately 500 kHz: these conditions, that is,

impedance matching between source (generator of 50 +resistance R1 of 100 ) and filter input (constant impedanceof 150 ), are shown in the fig. 905.12;

- but, if this test is repeated on constant-kfilters, there should be

a level varying together with frequency, because the filter input

impedance varies.

Fig. 905.12 Matching between source and filter


54/92

Lesson 906: Ceramic filters I

- 50 -

LESSON 906: CERAMIC FILTERS

Objectives Examining the operation of ceramic filters

Necessary equipment





Oscilloscope

Frequency-meter

Multimeter


Ceramic filters are band-pass filters that use piezoeletric ceramic

material as eletrical-mechanical transducer and mechanical resonator.

The figure 906.1 shows the symbols of the filter sections with 2 and 3

terminals. This figure also shows the equivalent circuit of the 2-terminal

filter, as well as the trend of its impedance versus frequency. In a lot of

cases, this type of filter consists of 2 sections coupled through a smallcapacitor (fig. 906.2).

Some important parameters of ceramic filters are the input impedance,

the output impedance and the coupling capacitance between two

sections. The fig. 906.3 shows the frequency response of a ceramic filter

at 455 kHz (this is the filter used in the tests) versus the coupling

capacitance and the input and output impedance match.


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Lesson 906: Ceramic filters

- 51 -

a) 2-terminal filter b) 3-terminal filter

c) equivalent electric circuit d) impedance

Fig.906.1 Ceramic filter

Fig. 906.2 Ceramic filter with coupled sections


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- 52 -

Fig. 906.3 Frequency response of a ceramic filter


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- 53 -

906.2 EXERCISES




906.2.1 Response curve of the filter detected with wobbulator

Applying a signal of variable frequency to the input and detecting the

amplitude of the output signal determine the response curve (or

frequency response) of a common quadripole. The quadripole

attenuation, measured at different frequencies, is expressed as follows:

A = Vo/Vi

or (in dB):

AdB= 20log(Vo/Vi)

Representing the attenuation versus frequency determines the response

curve of the quadripole.

Prearrange the circuits as it is shown in the fig. 906.4

set RV1 to the minimum (anticlockwise direction) and RV2 to the

maximum (clockwise direction)

set the oscilloscope to X-Y (X axis on 1 V/div; Y axis on 20

mV/div)

connect the X axis of the oscilloscope to TP1 (X AXIS). Connect

the Y axis (probe 10:1) to the filter output (TP35)


(DEPTH), to display the response curve of the filter, on the

oscilloscope

Fig. 906.4


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- 54 -

Q1 To obtain a steep and flat reponse curve it is necessary:

SET

A B

1 3 to increase RV2, to reduce RV1 slightly, to increase CV22 1 to increase RV1 slightly, to reduce RV2, to increase CV2 up

to its maximum

3 4 to increase RV1 slightly, to reduce RV2, to set CV2 to its

intermediate position

4 2 to increase RV1, to reduce RV2, to set CV2 to its minimum


SIS2 PressINS

Q2 Which effect can be detected in the response curve ?

SET

A B

1 3 it is cancelled out because CV2 is open

2 1 two peaks appear because CV2 is open3 4 two peaks appear because CV2 is in short circuit

4 2 two peaks appear because CV2 is increased


vary the coupling capacitance CV2 at will and check the 3

following conditions: the response curve is narrowed; the response

is flat; the response shows two resonance peaks.

906.2.2 Measurement of the filter response by steps

adjust RV1 and RV2 to approximately 3 k (check with amultimeter)

set CV2 to its intermediate position

apply a signal with frequency of 455 kHz (corresponding to the

center frequency of the filter), to the filter

be Vo and Vi the peak-to-peak voltages, measured at the filter

output and input. The filter attenuation Aat 455 kHz is expressed

by the following formulae:A=Vo/Vi; andAdB= 20log(Vo/Vi)[in dB]


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- 55 -

repeat this measurement varying the frequency from 455 to 465

kHz, by steps of 1 kHz; calculate Adb in correspondence with

each ffrequency value and record the resulting data on a table like

that shown in the fig. 906.5

starting from the data of this table plot a graph with AdBon the Y

(vertical) axis and the frequency on the X (horizontal) axis: theresult is the curve of frequency response of the filter

calculate the filter passband;

B = f2 - f1

where f2 and f1 are the frequencies at which the attenuation AdBhas a 3 dB increase with respect to its minimum value.

Q3 Which is the value of bandwidth at -3 dB ?

SET

A B





Fig. 906.5


60/92

Lesson 907: Quartz - Crystal filters

- 56 -

LESSON 907: QUARTZ-CRYSTAL FILTERS

Objectives

Describing the quartz characteristics

examining the operation of a band-pass filter constructed with a

quartz crystal

Necessary equipment





Oscilloscope / frequency-meter


907.1.1 Properties of quartz crystals

Quartz is a piezoelectric material; when properly cut and assembled, it

vibrates at high frequencies; there are also some frequencies at which

it vibrates more intensely.

Quartz symbol and its equivalent electric circuit are shown in the fig.

907.1.

Fig. 907.1 a) electrical symbol of quartz

b) equivalent circuit

It is characterized by a series resonance frequencyfs and a slightly

higherparallel resonance frequency fp; Q is very high (since R is very

low) and can exceed 106. In the diagram of the fig. 907.1, Co

represents the static capacitance formed by the quartz armatures and

the case. As Co is much higher than C, it can be omitted in the

calculation of theseries resonance frequencyfs:

)LC(2

1fs

=


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- 57 -

The value of theparallel resonance frequencyfpis slightly higher than

fsand is determined as follows:

CoC

CoCL

2

1fp

+

=

The distance between fp and fs can be approximated by the following

formula:

Co2

Cfsfsfpf

=

The fig. 907.2 shows the trend of quartz reactance versus frequency: it

has negative values (capacitive reactance) up to fs and after fp,

whereas it is characterized by positive values (inductive reactance)between fsand fp. Impedance is only resistive; it is almost null in fs,

almost infinite in fp.

Fig. 907.2 Quartz reactance Fig, 907.3 Reactance of

versus frequency "quartz + inductance"

When an inductance in connected in parallel to a quartz crystal, its

effect on the total reactance is double (fig. 907.3): the parallel resonance frequency slightly increases passing from fp

to fp1, whereas the series resonance frequency does not vary

another parallel resonance frequency (fp2) whose value is lower

than fsis introduced.

907.1.2Crystal-Gatefilter

The simplest quartz-crystal band-pass filter is shown in the fig. 907.4

and is called Crystal-Gatefilter.

The circuit is a bridge. The voltage applied to the quartz crystal Q1 isphase-shifted of 180 with respect to that applied to CV1, because of


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- 58 -

the center tap of the secondary winding of TR1. When the reactance of

CV1-C11 is equal to the quartz reactance, the circuit does not suffer

any phase shift any longer and the voltages applied at the output of the

two branches of the bridge are phase-shifted of 180 and so they cancel

out.

As it is shown in the fig. 907.5, this equality surely occurs because the

quartz capacitive reactance is below the series resonance and varies

rapidly together with frequency, whereas the reactance of CV1 varies

more slowly. When both these reactances intersect, the outputs of the

two branches of the bridge cancel out and the filter output is

theoretically null. On the contrary, if the quartz reactance is inductive

and its value is equal to that of CV1, the filter output will reach its

maximum value because an additional phase shift of 180 is provoked

by the two branches and, in this case, their outputs are added instead of

cancelling out. In the other points the output value and the attenuation

are included within these two limits.

The fig. 907.5 shows that the passband is extended from the frequency

f1 (slightly higher than fs of quartz crystal) up to the frequency f2(slightly lower than fp). Actually, the frequency f1 is higher than fs,

when C2>Co, and lower, when C2


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- 59 -

Fig. 907.5 a) reactance and attenuation when Cv>Co b) reactance and attenuation when Cv


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- 60 -

Fig. 907.6 Lattice filter

Fig. 907.7 Bridge structure of a lattice filter


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- 61 -

Fig. 907.8 Lattice filter : a) reactance

Fig. 907.9 Half-lattice filter


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- 62 -

907.2 EXERCISES




907.2.1 Response curve of the filter detected by wobbulator

Arrange the circuits as it is shown in the fig. 907.10



Y axis (probe 10:1) to the filter output (TP15) adjust the center frequency of the VCO and the Sweep amplitude

(DEPTH) to obtain the response curve of the filter.

Fig. 907.10

Q1 The shape of the response curve shows that:

SET

A B

1 3 the filter allows the passage of all the frequencies, excepting a

narrow band

2 1 the filter eliminates all the frequencies, excepting a verynarrow band. The top of the curve is flat and broad.

3 4 the filter eliminates all the frequencies, excepting a very

narrow band. The top of the curve is very narrow and it is not

uniform

4 2 the filter eliminates all the frequencies, excepting a very

narrow band. The top of the curve is characterized by two

clear peaks


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- 63 -

Q2 Vary CV1. How does the filter response vary ?

SET

A B

1 4 the center frequency varies2 3 the maximum gain is shifted between the two ends of the

resonance peak

3 1 the maximum amplitude varies

4 2 the maximum attenuation is shifted between the two ends of

the resonance peak

907.3 QUESTIONS

Q3 Crystal-gatefilters are not used in many applications because:

SET

A B

1 3 their passband is too narrow

2 4 their passband is too broad

3 1 they have a flat response within the passband

4 2 they need 4 quartz crystals

Q4 The shape of the response curve of a ceramic filter depends on:

SET

A B

1 4 the coupling capacitance

2 3 the impedance match

3 1 the coupling capacitance and the impedance match

4 2 the coupling inductance

Q5 A lattice filter consists of:

SET

A B

1 4 four equal quartz crystals

2 3 two equal quartz crystals

3 1 four quartz crystals being equal two by two

4 2 only one quartz crystal


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Lesson 908: Impedance match I

- 64 -

LESSON 908: IMPEDANCE MATCH I

Objectives

Describing the characteristics of impedance-matching networks

Necessary equipment





908.1.1 Power transfer

Consider a generator having an internal impedance ZG=r+jx and a

load ZL=R+jX (fig. 908.1). The maximum power transfer from

generator to load occurs when ZGand ZLare complex conjugates, that

is:

r = R x = -X

In these conditions the power P on the load is:

R4vP

2

=

Actually, the required conditions seldom occur in a lot of applications,

therefore a matching between generator and load is necessary (fig.

908.2).

Fig. 908.1 Power transferfrom generator to load


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- 65 -

Fig. 908.2 Impedance matching between generator and load

908.1.2 Matching networks

An impedance matching network consists only of reactances, so that all

the power applied to the input terminals is supplied to the load (fig.

902.3).

This network can be dimensioned so that its input impedance Zi is

complex conjugate of the generator output impedance Zg: the result is

the maximum power transfer between generator and network.

Furthermore, the output impedance Zo may be complex conjugate of the

load impedance ZL, so that the maximum power can also be transferred

between network and load.

Various types of matching networks will be explained in the following

paragraphs.

Fig. 908.3 Impedance matching between generator and load

908.1.3 Two-impedance matching networks

The simplest network for matching two resistive impedances R1and

R2is the reactive L network shown in the fig. 908.4.

The perfect matching between R1and R2occurs only at the frequency

fo; at this frequency the reactances X1 and X2 have the following

values:

1n

1

1R2R1R

2R

1R1X == m


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- 66 -

n

1n1R

2R1R

2R1R2X

=

= mm

X1and X2have opposite signs, therefore, when X1is inductive, X2is

capacitive, and vice versa. The parallel reactance (X1

) must be

connected on the side of the higher resistance.

The trend of this L network may be of low-pass type, if X1 is

capacitive and X2inductive, or of high-pass type, in the opposite case;

of course, in this case too, the perfect matching conditions occur only

at the frequency fo.

The fig. 908.5 shows the normalized frequency response of the circuit

of the fig. 908.4, for both the high-pass and low-pass configurations.

The Q of this circuit at fois:

Qo = n 1

This equation shows that there is only one possible value of Q for a

certain ratio of transformation n. Therefore, when the values of R1and R2are fixed, the Q of the network is univocally set and it cannot

be reduced to broaden the matching range, nor increased to increase

the network filtering power.

Fig. 908.5 Normalized response of low-pass and high-pass L network

Fig. 908.4 Matching

L network


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- 68 -

908.1.6 Matching networkb (fig. 908.7)

Applicability

* R1 > R2

Process of calculation

* Choose Q

Fig. 908.7

908.1.7 Matching network c (fig. 908.8)

Applicability

* R2 > R1


* Choose Q

Fig. 908.8


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- 69 -

908.1.8 Matching network d (fig. 908.9)

Applicability

* Any condition of R1 and R2


* Choose Q

Fig. 908.9

908.1.9 Choosing standards

As already mentioned, each application will need the network with the

most realistic values of its components. For this reason the following

considerations may be very useful:

1. the network "a" may be applied only when R2>R1; when R2approaches R1the value of C1tends to infinite

2. the network "b" may be applied when R1>R23. the network "c" may be applied when R2>R14. the network "d" may be applied in any condition of R

1and R

2.

908.1.10 Matching impedances not only resistive

If one of the output impedances has a reactive component, this

component can be considered as a part of the matching network. For

example, match the impedance:

Z1= Rs+ jXCs

to the only resistive impedance R2, through the network "a" (fig.

908.10). According to the formulae of the network "a" and considering

X'Land XCs we can also write:


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- 70 -

X'L+ XCs= QR1

from which:

X'L = QR1- XCs = Q

R1+ XCs= XL+ XCs

Fig. 908.10

Therefore, if the impedance Z1 is not only resistive, but if it also

includes a reactive part due to a capacitive component (this is the most

frequent case), the absolute value of the "external" reactance must be

added to the value calculated for XL.


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- 71 -

908.2 QUESTIONS


Q1 Impedance-matching networks are used:

SET

A B

1 3 to transfer the minimum power from the generator to the load

2 4 to transfer the maximum power from the load to the generator

3 1 to transfer the maximum power between two equal

impedances

4 2 to transfer the maximum power from the generator to the load

Q2 Matching networks are constructed with:

SET

A B

1 4 resistive components

2 3 the reactive components R and C

3 2 the reactive components L and C

4 1 the reactive components L and R

Q3 Furthermore matching networks:

SET

A B

1 2 amplify the signal

2 1 filter the signal

3 4 multiply the frequency

4 3 match the inductance

Q4 Matching networks are used:

SET

A B

1 3 to match only resistive impedances

2 4 to increase the load impedance

3 1 to match different impedances

4 2 to reduce the generator impedance


76/92

Lesson 909: Impedance match II

- 72 -

LESSON 909: IMPEDANCE MATCH II

Objectives

Assembling matching networks and carrying out measurements

Necessary equipment





Oscilloscope

Function generator with output impedance of 50 Frequency-meter


Refer to what explained in the previous lesson (Lesson 908).


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909.2 EXERCISES




909.2.1 Two-impedance matching network. Low-pass configuration. Q=2

Match an impedance of 50 to another impedance of 10 at thefrequency fo= 680 kHz, through a two-impedance network in low-

pass configuration.

The used network is shown in the fig. 909.1. The calculationformulae give the following results:

L 4.7 H C 9.4 nF

Fig. 909.1 Fig. 909.2

Before measuring the power transferred from the generator (50 ) tothe load (10 ), measure the reference power supplied by thegenerator, as follows:

- prearrange the generator for a frequency of approximately 680 kHz

and connect to the resistance of 50 of the module (between TP20and TP21, fig. 909.2): this means loading the generator with animpedance equal to its internal impedance; in these conditions the

generator transfers the maximum power to the load

- adjust the generator amplitude to obtain a load voltage of 0.1 Veff(283 mVpp)

- calculate the power supplied to the load as follows:

W20050

01.0

R

effVPef

L

2

===

Project data

Solution

Measurements


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- the power Pref is the maximum power transferred from the

generator to the load, because this power transfer occurs in perfect

matching conditions

disconnect the resistance of 50 and connect the generator to thematching network (TP22-TP23) which the load resistance R10 of

10 has already been connected to (fig. 909.3)

Fig. 909.3

measure the effective voltage across the load of 10 and calculatethe power transferred from the generator (50 ) to the load (10 )through the matching L-C network:

10

efVP

2

=

When the matching network is perfect and has no losses, the power

P should be equal to Pref; but actually it does not happen because of

the losses due to L and C. vary the frequency and note how the power on the load varies:

record the resulting data on a table (the values indicated in the

following table are only approximate)

f

[kHz]

V10

[mVeff]

P= Veff/10

[w]

400 38 144

500 41 170

680 ?? ???

1000 29 84

Pref= 200 W


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Q1 Which is the value of the power transferred to the load at the frequencyof 680 kHz ?

SET

A B1 3 approximately 220 W2 4 approximately 120 W3 1 approximately 190 W4 2 approximately 100 W

Q2 The measurements carried out show that this network also acts as:

SET

A B1 4 resistive attenuator

2 3 narrow-band filter

3 2 high-pass filter

4 1 low-pass filter

909.2.2 Three-impedance matching network. Q=4

The used network is shown in the fig. 909.4: the impedance match is

between 50 and 10 at the frequency fo= 636 kHz, with Q=4.

Fig. 909.4

Before measuring the power transferred from the generator (50 ) tothe load (10 ), measure the reference power supplied by thegenerator, as follows:

- prearrange the generator for a frequency of approximately 640 kHz

and connect to the resistance of 50 of the module (between TP20and TP21, fig. 909.5): this means loading the generator with an

impedance equal to its internal impedance; in these conditions the

generator transfers the maximum power to the load

- adjust the generator amplitude to obtain a load voltage of 0.1 Veff(283 mVpp)

- calculate the power supplied to the load as follows:


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W20050

01.0

R

effVefPr

L

2

===

- the power Pref is the maximum power transferred from the

generator to the load, because this power transfer occurs in perfectmatching conditions

disconnect the resistance of 50 and connect the generator to thematching network (TP26-TP27) which the load resistance R11 of

10 has already been connected to (fig. 909.6)

Fig. 909.6

measure the effective voltage across the load of 10 and calculatethe power transferred from the generator (50 ) to the load (10 )through the matching network:

10

effVP

2

=

When the matching network is perfect and has no losses, the power

P should be equal to Pref; but actually it does not happen because of

the losses due to L and C.

vary the frequency and note how the power on the load varies: recordthe resulting data on a table.


SET

A B

1 3 approximately 90 W2 4 approximately 180 W

3 1 approximately 40 W4 2 approximately 120 W

Fig. 909.5


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SET

A B1 4 approximately 130 W2 3 approximately 170 W3 1 approximately 240 W4 2 approximately 80 W

909.2.3 Three-impedance matching network. Q=2

The used network is shown in the fig. 909.7: the impedance match is

between 50 and 200 .

Fig. 909.7

Measure the power transferred to the load in the frequency band

ranging from 300 to 1000 kHz

Q5 Which frequency does the maximum power transfer occur at ?

SET

A B

1 4 at approximately 720 kHz2 4 at approximately 370 kHz




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Lesson 924: RF Amplifier

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LESSON 924: RF AMPLIFIER

Objectives

Examine the operation of an RF amplifier in class B and the power

matching network

calibrate the matching network

measure the output power.

Material

basic unit (power supply mod. PSU/EV, module-holder unit mod.

MU/EV, individual control unit mod.SIS1/SIS2/SIS3)

experiment module mod.MCM20/EV

dual trace oscilloscope

frequency-meter.

924.1 THEORETICAL NOTIONS

The RF power amplifiers are commonly used in radio transmitters to

amplifier the signal before sending it to the transmission antenna.

If the signal must be amplified without introducing excessive

distortions, the so called linear amplifiers are used, which operate inclass A, AR or B, and which efficiency goes from the 20 to the 65%

about. Amplifiers in class C with higher efficiency cannot be used, as

the output amplitude of such amplifier does not vary linearly with the

input. The class C is used when there isnt an amplitude variation in the

modulated signal to be transmitted, as in case of FM signal.

To amplify an AM signal, linear amplifiers are required; a modulator in

c/ass C, ifthe modulation is carried out directly on the final power stage

and no further amplification is necessary.

The examined circuit, reported in fig.924. 1, is an amplifier in class B

with low-pass matching network toward the load.

The section consisting of the transistors T1 and T2 is used to introduce

the amplitude modulation, not used in this lesson and described in the

next chapter 3.

The RF signal is applied to the base of T3, which biasing is obtained

with the diode Dl. On the anode of D1, directly biased by R20, there are

about 0.7V, which correspond to the voltage drop of T3. In this way,

only the signals with amplitude over 0V can take T3 in active zone, andare amplified in this way. In other words, only the positive half-waves of


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a sine signal are amplified, as scheduled in the operation of an

amplifiers in class B.

The impedance L17 prevents the RF input signal to be short-circuited to

ground with the diode Dl. The impedance L18 prevents the RF output

signal is short-circuited to ground with the power supply.

If the output of the amplifier (collector of T3) is directly sent to the load,

the transmitted signal would be highly distorted. The distortion is

removed by means of the band pass filter (C28-C29-CV5-L19-C30)

centered to the work frequency; this is also used as matching network

between the amplifier and the load, consisting of the resistance R21 or

the antenna.

fig.924.1 RF Amplifier


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924.2 EXERCISES




Matching network calibration

Set the circuit as in fig.924.2 (jumper J23 connected)

across the input of the amplifier (TP43), apply a sine signal with

1MHz-frequency and about 0.5Vpp-amplitude. This signal can betaken by the VCO (TP4)

connect the oscilloscope (probes 10:1) to the input of the amplifier

and to the load R21 (TP43 and TP46)

adjust the variable capacity CV5and the coil L19 to obtain the max.

output

fig.924.2


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Q1 How much is the output signal?SET

AB

1 4 about 5Vpp

2 3 about 1Vpp

3 1 about 30Vpp4 2 about l5Vpp

Amplifier power gain

Adjust the amplitude of the input signal to about lVpp. Calculate the

voltage gain of the amplifier (G = Vout/Vin)

Q2 How much is the gain?SET

AB

1 2 between 5 and 10

2 1 between 0.5 and 1



increase the amplitude of the input signal to obtain the max. output

calculate the output power of the amplifier with the following

relation:

Pout = (Vrms)2/R21

where R21 is the amplifier load and Vrms is the effective voltage on the

same load

Q3 How much is the power?SET

AB

1 4 between 5 and 10 mW

2 1 between 0.5 and 1 W3 2 between 150 and 250 mW

4 3 between 30 and 40 mW

connect one probe of the oscilloscope (10:1) to the transistor base

T3 (TP44): note that there is a distortion of the positive half-waves

(over about 0.7V) of the input signal, due to the low impedance

shown by the transistor when in conduction. This shows that its

operation is in class B


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Lesson 925: AM Transmitter

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LESSON 925: AM TRANSMITTER

Objectives

Examine the operation of the AM modulator

check the antenna operation

examine the wave-forms of the transmitter

Material

basic unit (power supply mod. PSU/EV, module-holder unit mod.

MU/EV, Individual control unit mod.SIS1/SIS2/SIS3)

experiment module mod.MCM20/EV

dual trace oscilloscope

frequency-meter.

925.1 THEORETICAL NOTIONS

An AM transmitter generally consists of the following stages (fig.

925.1):

an RF oscillator, which generates the carrier for the modulator

the amplitude modulator, which receives the carrier and the

modulating signal, and supplies the modulated signal

the RF amplifier, which amplifies the signal supplied by the

modulator.

Sometimes, as in the circuit under test, the modulation and the

amplification are carried out by the same stage

a matching network between the amplifier and the load (antenna),

to transfer the max. power to the load

the transmitting antenna.

fig.925.1


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Lesson 925: AM Transmitter

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Refer to the complete electrical diagram of fig.925.2.The 1MHz-carrier

signal is applied to the amplifier in class B (transistor T3), which carries

out the amplitude modulation, too. The power supply voltage of T3 is

varied by the modulating signal, across the low frequency amplifier

consisting of T1 and T2. This causes a variation of the carrier amplitude

on the collector of T3, carrying out so the amplitude modulation.

As the amplifier-modulator operates in class B, the output signal is

much distorted. The next matching network has two functions:

it transfer the max. power from the generator (collector of T3) to the

load (antenna)

it filters the signal, removing the distortions and supplying a proper

modulated signal.

The antenna used in the experiments is a ferrite antenna commonly used

in AM radioreceivers. It is practically a transformer, which windings are

wired around a ferrite bar. The primary receives the RF signal from the

transmitter, while the secondary is tuned to the transmission frequency

with a parallel capacity.

fig.925.2


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