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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.
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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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Lesson 900: Attenuators
- 2 -
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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Lesson 900: Attenuators
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
3 2 approximately 2 Vpp
4 1 approximately 0.8 Vpp
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Lesson 900: Attenuators
- 4 -
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
1 3 approximately 1 Vpp
2 4 approximately 0.5 Vpp
3 2 approximately 2 Vpp
4 1 approximately 0.8 Vpp
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Lesson 900: Attenuators
5
Q4 Which is the value of the attenuation (in dB) ?
SET
A B
1 4 approximately 1 dB
2 3 approximately 3.9 dB3 2 approximately 2 dB
4 1 approximately 6 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
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
901.1 THEORETICAL HINTS
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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Lesson 901: Series and Parallel R-L-C Circuits
- 7 -
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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Lesson 901: Series and Parallel R-L-C Circuits
- 8 -
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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Lesson 901: Series and Parallel R-L-C Circuits
- 9 -
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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Lesson 901: Series and Parallel R-L-C Circuits
- 10 -
Fig. 901.5 Universal resonance curves
901.2 EXERCISES
MCM20 Disconnect all the jumpers
SIS1 Set all the switches to OFF
SIS2 Enter the lesson code number: 901
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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Lesson 901: Series and Parallel R-L-C Circuits
- 11 -
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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Lesson 901: Series and Parallel R-L-C Circuits
- 12 -
- 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
1 3 approximately 500 kHz
2 1 approximately 70 kHz
3 4 approximately 150 kHz
4 2 approximately 700 kHz
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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Lesson 901: Series and Parallel R-L-C Circuits
- 13 -
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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Lesson 901: Series and Parallel R-L-C Circuits
- 14 -
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
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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
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.
902.1 THEORETICAL HINTS
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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Lesson 902: Coupled Circuits
- 16 -
)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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Lesson 902: Coupled Circuits
- 17 -
902.2 EXERCISES
MCM20 Disconnect all the jumpers
SIS1 Set all the switches to OFF
SIS2 Enter the lesson code number: 901
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)
connect the X axis of the oscilloscope to TP1 (X AXIS). Connect the
Y axis (probe 10:1) between TP39 and TP40
adjust the center frequency of the VCO and the Sweep amplitude
(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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Lesson 902: Coupled Circuits
- 18 -
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
SIS1 Set the switch SW9 to ON
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
SIS1 Set the switch SW9 to OFF
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Lesson 902: Coupled Circuits
- 19 -
902.2.2 Capacitive coupling. Frequency response with wobbulator
Arrange the section TUNED CIRCUITS & COUPLING as it is
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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Lesson 902: Coupled Circuits
- 20 -
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
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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
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
Frequency-meter Multimeter
903.1 THEORETICAL HINTS
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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Lesson 903: Impedance Transformation with Transformer
- 22 -
Fig. 903.1 Impedance transformation through autotransformer
903.2 EXERCISES
MCM20 Disconnect all the jumpers
SIS1 Set all the switches to OFF
SIS2 Enter the lesson code number: 963
903.2.1 Impedance transformation with transformer
Arrange the section TUNED CIRCUITS & COUPLING as it is
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
set the oscilloscope to X-Y (X axis on 1 V/div; Y axis on 10 mV/div)
connect the X axis of the oscilloscope to TP1 (X AXIS). Connect the
Y axis (probe 10:1) between TP39 and TP40
adjust the center frequency of the VCO and the Sweep amplitude
(DEPTH), to obtain a curve similar to that shown in the fig. 903.3
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Lesson 903: Impedance Transformation with Transformer
- 24 -
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
1 3 approximately 100 kHz
2 1 approximately 10 kHz
3 4 approximately 40 kHz
4 2 approximately 5 kHz
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
resistance 2 times as low
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
resistance 4 times as low
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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
Power Supply mod. PSU/EV
Module-holding base mod. MU/EV
Individual Control Unit SIS1, SIS2 or SIS3
904.1 THEORETICAL HINTS
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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Fig. 904.7 Constant-k low-pass filters:
a) "L" section
b) "T" section
c) "pi" section
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Fig. 904.8 Constant-k high-pass filters:
a) "L" section
b) "T" section
c) "pi" section
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Fig. 904.9 Constant-k band-pass filters :
a) "T" section
b) "pi" section
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Fig. 904.10 Constant-k band-rejection filters:
a) "T" section
b) "pi" section
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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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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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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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904.2 QUESTIONS
SIS2 Enter the lesson code number: 904
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
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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
Power Supply mod. PSU/EV
Module-holding base mod. MU/EV
Individual Control Unit SIS1, SIS2 or SIS3
Testing module mod MCM20/EV
Oscilloscope
Frequency-meter Function generator
905.1 THEORETICAL HINTS
Refer to the previous lesson 904.
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905.2 EXERCISES
MCM20 Disconnect all the jumpers
SIS1 Set all the switches to OFF
SIS2 Enter the lesson code number: 905
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
1 3 approximately 700 kHz
2 1 approximately 600 kHz
3 4 approximately 800 kHz
4 2 approximately 500 kHz
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Lesson 905: Filters of discrete components II
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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
Assemble the filter connecting the jumpers as it is shown in the
fig. 905.3
measure the filter passband.
Fig. 905.3 pi low-pass filter
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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
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. 905.5)
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
1 3 approximately 500 kHz
2 1 approximately 600 kHz
3 4 approximately 800 kHz
4 2 approximately 700 kHz
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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905.2.4 Constant-k pi high-pass filter
Assemble the filter connecting the jumpers as it is shown in the
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
connect the oscilloscope to the filter output (between TP11 and
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
2 3 at approximately 470 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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Q4 Which is the value of the lower cutoff frequency ?SET
A B
1 3 approximately 500 kHz
2 1 approximately 600 kHz
3 4 approximately 450 kHz4 2 approximately 750 kHz
Q5 Which is the value of the upper cutoff frequency ?SET
A B
1 3 approximately 500 kHz
2 1 approximately 650 kHz
3 4 approximately 400 kHz
4 2 approximately 750 kHz
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
1 4 at approximately 530 kHz
2 3 at approximately 430 kHz
3 2 at approximately 580 kHz
4 1 at approximately 610 kHz
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
deviations (in dB) of output amplitude from the reference, on the
table.
Q7 Which is the value of the upper cutoff frequency ?SET
A B
1 3 approximately 530 kHz2 1 approximately 600 kHz
3 4 approximately 620 kHz
4 2 approximately 750 kHz
Q8 Which is the value of the lower cutoff frequency ?SET
A B
1 3 approximately 500 kHz
2 1 approximately 350 kHz
3 4 approximately 400 kHz
4 2 approximately 450 kHz
Fig. 905.9 Band-rejection filter
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Lesson 905: Filters of discrete components II
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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
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 starting from 400 KHz, increase 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
Fig. 905.11 M-derivedpi low-pass filter
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Q9 Which is the value of cutoff frequency at -3 dB ?
SET
A B
1 4 approximately 700 kHz2 3 approximately 600 kHz
3 2 approximately 500 kHz
4 1 approximately 400 kHz
Q10 Which is the frequency value corresponding to a (theoretical) infiniteattenuation ?
SET
A B1 2 approximately 750 kHz
2 1 approximately 650 kHz
3 4 approximately 520 kHz
4 3 approximately 480 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
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Lesson 906: Ceramic filters I
- 50 -
LESSON 906: CERAMIC FILTERS
Objectives Examining the operation of ceramic filters
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
Oscilloscope
Frequency-meter
Multimeter
906.1 THEORETICAL HINTS
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
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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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Fig. 906.3 Frequency response of a ceramic filter
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906.2 EXERCISES
MCM20 Disconnect all the jumpers
SIS1 Set all the switches to OFF
SIS2 Enter the lesson code number: 906
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)
adjust the center frequency of the VCO and the Sweep amplitude
(DEPTH), to display the response curve of the filter, on the
oscilloscope
Fig. 906.4
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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
SIS1 Set the switch SW4 to ON
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
SIS1 Set the switch SW4 to OFF
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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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
1 4 approximately 10 kHz
2 1 approximately 20 kHz
3 2 approximately 455 kHz
4 3 approximately 4 kHz
Fig. 906.5
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Lesson 907: Quartz - Crystal filters
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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
Power Supply mod. PSU/EV
Module-holding base mod. MU/EV
Individual Control Unit SIS1, SIS2 or SIS3
Testing module mod MCM20/EV
Oscilloscope / frequency-meter
907.1 THEORETICAL HINTS
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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Lesson 907: Quartz - Crystal filters
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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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Lesson 907: Quartz - Crystal filters
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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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Fig. 907.5 a) reactance and attenuation when Cv>Co b) reactance and attenuation when Cv
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Lesson 907: Quartz - Crystal filters
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Fig. 907.6 Lattice filter
Fig. 907.7 Bridge structure of a lattice filter
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Lesson 907: Quartz - Crystal filters
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Fig. 907.8 Lattice filter : a) reactance
Fig. 907.9 Half-lattice filter
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Lesson 907: Quartz - Crystal filters
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907.2 EXERCISES
MCM20 Disconnect all the jumpers
SIS1 Set all the switches to OFF
SIS2 Enter the lesson code number: 907
907.2.1 Response curve of the filter detected by wobbulator
Arrange the circuits as it is shown in the fig. 907.10
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 (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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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
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LESSON 908: IMPEDANCE MATCH I
Objectives
Describing the characteristics of impedance-matching networks
Necessary equipment
Power Supply mod. PSU/EV
Module-holding base mod. MU/EV
Individual Control Unit SIS1, SIS2 or SIS3
908.1 THEORETICAL HINTS
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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Lesson 908: Impedance match I
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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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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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Lesson 908: Impedance match I
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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
Process of calculation
* Choose Q
Fig. 908.8
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Lesson 908: Impedance match I
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908.1.8 Matching network d (fig. 908.9)
Applicability
* Any condition of R1 and R2
Process of calculation
* 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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Lesson 908: Impedance match I
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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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Lesson 908: Impedance match I
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908.2 QUESTIONS
SIS2 Enter the lesson code number: 908
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
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Lesson 909: Impedance match II
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LESSON 909: IMPEDANCE MATCH II
Objectives
Assembling matching networks and carrying out measurements
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
Oscilloscope
Function generator with output impedance of 50 Frequency-meter
909.1 THEORETICAL HINTS
Refer to what explained in the previous lesson (Lesson 908).
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Lesson 909: Impedance match II
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909.2 EXERCISES
MCM20 Disconnect all the jumpers
SIS1 Set all the switches to OFF
SIS2 Enter the lesson code number: 909
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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Lesson 909: Impedance match II
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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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Lesson 909: Impedance match II
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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.
Q3 Which is the value of the power transferred to the load at the frequencyof 1000 kHz ?
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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Q4 Which is the value of the power transferred to the load at the frequencyof 640 kHz ?
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
3 1 at approximately 480 kHz
4 2 at approximately 650 kHz
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Lesson 924: RF Amplifier
- 78 -
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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Lesson 924: RF Amplifier
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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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Lesson 924: RF Amplifier
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924.2 EXERCISES
MCM20 Disconnect all the jumpers
SIS1 Set all the switches to OFF
SIS2 Enter the lesson code number: 924
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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Lesson 924: RF Amplifier
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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
3 4 between 80 and 100
4 3 between 25 and 40
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
- 83 -
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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