June/July 2013
Third Semester B.E. Degree Examination
Electronic Instrumentation
Time: 3 hrs.
Max. Marks: 100
Note: 1. Answer any FIVE full questions, selecting at least two questions from each part.
PART – A
1. a.
Explain with suitable example accuracy and precision.
(05 Marks)
Ans:
Accuracy refers to the degree of exactness (closeness) or conformity to the true value of quantity under
measurement. Precision refers to the degree of agreement within a group of measurement (repeatability of
measurement).
Accuracy: To illustrate the concept of accuracy, let us consider an example. Suppose that a digital voltmeter
has an accuracy of ±2% and the measured value is 10V. This means that the true value is somewhere
between 10±2%=9.98 V or 10.02 V. To find the true value, or the actual percentage of accuracy, the readings
can be compared with a standard meter.
Precision: Consider a voltmeter that shows consistently for a number of repeated measurements as 9.99V
for a true value of 10.00V. This is close to the true value as the observer can read the scale by estimation.
Although there are no deviations from the observed value, the error created by the limitation of the scale
reading is a precision error. On the other hand, assume that repeated measurements by the voltmeter
produced readings like 9.98V, 9.99V, 10.01, 9.99 etc (not consistent) for a true value of 10V for a number
of measurements. This indicates that the meter readings are not showing precision even though they are
within the prescribed range of accuracy.
The above examples illustrate that conformity is necessary but not a sufficient condition for precision.
Similarly precision is necessary but not a sufficient condition for accuracy.
1. b.
Permanent Magnet Moving Coil instrument (PMMC) with Full Scale Deection (FSD) of l00 PA
and coil resistance is 1 K is to be converted into a voltmeter. Determine the required multiplier
resistance if the voltmeter is to be measure 50 V at Mill scale. Also calculate the applied voltage when
the instrument indicates 0.8. 0.5 and 0.2 of FSD.
(08 Marks)
Ans:
For the given values, following circuit is developed for the required voltmeter, since a PMMC can be
converted in to a voltmeter by connecting a series resistance across it.
From the figure, the voltage across the meter and series resistance,
V = I (Rm + Rs) = Im (Rm + Rs)
Using the given values of Im = 100 PA, Rm = l k, V= 50V in the above equation results,
50 = 100 u 10–6 (1 u 10–3 + RS)
Therefore,
50
⎛
⎞
Rs = ⎜
− 1 × 10−3
⎝ 100 × 10−6 ⎟⎠
Rs = 499kΩ
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I
Voltmeter
RM
M
100 PA
V
RS
The Combined meter resistance is 499+1 = 500 k and Full Scale division (FSD) of PMMC is 100 PA.
Then at 0.8 of FSD,
Applied voltage
V = 0.8 u 100 u 10–6 u 500 u 10–3 = 40V.
At 0.5 of FSD,
V = 0.5 u100 u10–6 u500 u10–3 = 25V.
At 0.2 of FSD,
V = 0.2 u100 u–6 u 500 u10–3 = 10V.
1. c.
Explain with neat circuit diagram and wave forms full wave rectier type AC voltmeter. (07 Marks)
Ans:
One of the methods of measuring ac voltage is to use rectier types ac voltmeter that use a permanent
magnet moving coil (PMMC) movement along with a recties circuit. Fig. 1(a) shows the circuit of a
fullwave rectier type ac voltmeter, which consists of a multiplier, a bridge rectier, and amplier and a
PMMC movement.
R1
R2
RS
Multiplier
A
ac
input
R
PMMC
Fig 1(a): Full wave rectier type AC voltmeter
The ac input is fed to the rectifier through a multiplier resistor, the purpose of which is to limit the current.
The bridge rectifier converts the input ac in to a pulsating full wave dc. Since the movement coil has inertia,
a steady state deflection is shown by the meter. This steady state reflection is proportional to the average of
the current flowing through the meter as shown in the wave form in Fig.(b). The average value of current,
I av = I dc =
2Im
π
where Im is the maximum value of ac input current. However the meter can be calibrated to read the RMS
value of the ac sine wave input. The calibration factor is related to the form factor, which is the ratio of rms
value to the average value.
The output of the rectifier is applied to an amplifier before it is connected to the movement coil. The
purpose of amplifier is reduce the loading effect.
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June 13 - I
Iac
0
t
Idc
Irms
Iac
t
0
Fig. 1(b) Wave forms
2. a.
Explain basic operation of digital multimeter with neat block diagram.
(07 Marks)
Ans:
Digital multimeter is one of the most versatile instrument capable of measuring dc and ac voltage as well
as current and resistance. Over its analog counter parts, it offers high accuracy, high input impedance and
smaller in size.
The basic circuit of digital multimeter is a dc voltmeter only. Current is measured by converting it into
voltage by passing through a precision low shunt resistance.. Similar way the alternating current is converted
in to dc by employing rectifiers and filters. The meter having a precision low current source applied across
the unknown resistance for the purpose of measuring resistance. The output across this unknown resistance
is again a dc voltage that is digitized and read out as ohms.
Fig. 2(a), shows the basic block diagram of a digital multimeter.
ac
attenuator
ac
High
bc
Digital
Display
ac
Converter
ac
attenuator
A/D
convertor
ohms
converter
Input
Shunt
BCD
Output
Interface
Precision
reference
Low
Fig: 2(a) Basic block diagram of a DMM
The DMM can be a bench top, which is used mainly for stand alone operation and usual operation ready, or
it may be a system meter that provide a BCD output or some time micro processor based computing power,
depend upon the customer requirement.
2. b.
Suppose the converter can measure a maximum of 5 V. i.e. 5 V corresponds to the maximum count
of 111111111, if the test voltage is Vin = 1 V. Show the steps take place in the table format in the
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measurement for the successive approximation type Digital Volt Meter (DVM).
Ans:
(06 Marks)
Table below shows the steps take place in the measurement of voltage using the successive approximation
type digital voltmeter, considering a test voltage of 1V and the maximum count of 11111111 indicated 5V.
Vin = 1V
00110011
00110011
00110011
00110011
00110011
00110011
00110011
00110011
Operation
D7 set
D6 set
D5 set
D4 set
D3 set
D2 set
D1 set
D0 set
D7
1
0
0
0
0
0
0
0
D6
0
1
0
0
0
0
0
0
D5
0
0
1
1
1
1
1
1
D4
0
0
0
1
1
1
1
1
D3
0
0
0
0
1
0
0
0
D2
0
0
0
0
0
1
0
0
D1
0
0
0
0
0
0
1
1
D0 Compare
0
Vin < Vo
0
Vin < Vo
0
Vin > Vo
0
Vin > Vo
0
Vin < Vo
0
Vin < Vo
0
Vin > Vo
1
Vin > Vo
Output
D7 Reset
D6 Reset
D5 Set
D4 Set
D3 Reset
D2 Reset
D1 Set
D0 Set
Voltage
2.5
1.25
0.625
0.9375
0.9375
0.9375
0.97725
0.99785
2. c.
Explain with neat block diagram. Digital Frequency Meter.
(07 Marks)
Ans:
To measure the frequency of signal, which is unknown, the signal is converted into a train of pulses such
that one pulse represents each cycle of the signal. Then an electronic counter is used to count the number
of pulses occurring in a denite time interval. The number of count is the direct indication of the frequency
since each pulse represents a cycle.
Fig. 2(c) illustrates typical basic block diagram of a digital frequency meter.
Unknown
frequency
signal
Amplier
Schmitt
trigger
Start/stop
gate
Digital
Read out
0000
Fig. 2(c) : Block diagram of Digital frequency meter
The signal with the unknown frequency is applied to an amplifier. The amplifier output then fed to a schmitt
trigger circuit to convert the signal in to square wave form with fast and fall times, which is then passed
through a differentiation and clipping circuit so as to make the wave form into a train of pulses such that
each pulse represents one cycle of the signal. These pulses are fed to a start/stop gate with another input of
the gate being enable input. When enable input is made high (ON) the input pulses are passed through the
gate and are fed to the electronic counter. When the enable input is low (OFF) the counter stops counting
the pulses. The number of pules passed through the gate is displayed in the counter, in the time interval
between start and stop. If the enable input is a clock pulse of ON – time equals 1 seconds, then the counter
indicates the unknown frequency.
3. a.
Explain sweep or time base generator with neat circuit diagram and wave forms, for a continuous
sweep CRO and triggered sweep CRO.
(02 Marks)
Ans:
Sweep or time base generator is a part of horizontal deecting circuit of a CRO. There are two kinds of
sweep generators; continuous sweep and triggered sweep, depend upon the application.
i)
Continuous sweep CRO: Time base generator for a continuous sweep CRO, using a UJT and the
corresponding wave forms are shown Fig.3(a).
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+VBB
R2
Synch
pulse
input
Ts Sweep time
Tr Retrace time
V
RT
VBB
B2
VP
E
B1
CT
R1
Vo
VV
t
Tr
TS
Wave form in output
Circuit diagram
Fig. 3(a) : Continuous sweep
Initially, when the power is switched on, the UJT is OFF and the capacitor CT charge exponentially through
the variable resistance RT, which makes the emitter voltage, VE rises towards VBB and reaches VP as shown
in the wave form. This makes the emitter to Base 1 junction forward biased and the UJT triggers ON, which
in turn provides a low resistance path for the capacitor discharge that subsequently make the emitter voltage
to reach the minimum voltage resulting into the UJT OFF. Once the UJT becomes OFF, the capacitor
recharges and the cycle repeats. The charging period to increase the emitter voltage to reach VP is called
sweep time (TS) and that falls to minimum is called retrace time (Tr). RT is a used for continuous control of
frequency within a particular range and CT is varied in steps to change the range itself. For this reason, they
are called timing resistor and timing capacitor. The sync pulse enables the sweep frequency to be exactly
equal to the input (CRO) signal frequency, so that the signal is looked on the screen.
ii.
Trigger
Input
CC
Triggered sweep CRO: A triggered sweep is necessary, when short duration signals such as voice or
music signals are to be displayed. In triggered mode, the input signal is used to generate substantial
pulse that trigger the sweep, ensuring that the sweep is always synchronised with the input signal that
drives it.
Circuit diagram for the triggered sweep and the corresponding output waveforms are shown in Fig.3(b).
Ts Sweep Time
+VBB
Tr Retrace Time
Th Hold Time
V
R3
R2
RT
0
Trigger
D
VD
input pulse
R1
CT
R4
Vo
VD
Output
Tr
Circuit diagram
Ts
Th
t
Wave form
Fig: 3(b) Triggered sweep wave form
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Resistances R3 and R4 form a Voltage divider so as to make the voltage Vd at the cathode of the diode below
peak voltage VP such that UJT dosen't conduct in the normal condition. At the time of switching ON the
circuit, UTT is OFF and CT charges exponentially through RT towards VBB until the diode becomes forward
biased and conducts. Because of the action of diode, the capacitor voltage can not reach the peak voltage
required for UJT to conduct but is clamped at VD. However, a negative pulse of sufficient amplitude is fed
to the base – 2 of the UJT and the peak voltage is lowered for a moment, the UJT fires and this makes the
capacitor discharges rapidly through the UJT until the maintaining voltage of the UJT is reached. Once it
reaches this voltage, the UJT switches OFF and the CT charges towards VBB till it clamped at VD and the
cycle repeats.
3. b.
Explain dual trace–oscilloscope with neat block diagram.
(08 Marks)
Ans:
A dual trace oscilloscope is one which has only one cathode ray gun but display two input wave forms
simultaneously using an electronic switch that switches the signals to a single vertical amplier. Block
diagram of dual trace oscilloscope is shown in Fig. 3(c).
Electronic switch
Position
Input A
Preamplier and
Attenuator
To vertical
deection
plate
Position
Input B
Vertical
amplier
Pre–amplier and
Attenuator
From
Trigger
circuit
C
D
Q
Toggle Flip – op
Q
Fig: Block diagram of dual – trace oscilloscope
There are two methods to employ to generate two independent traces. In the first method, deflect the
oscilloscope and display the input 'A'. Then the oscilloscope is triggered and the input 'B' is displayed at
a different position on the oscilloscope. An electronic switch is used to switch between the two vertical
sources that are processed separately in two vertical amplifiers. The electronic switch is changed or alternate
between the two channels, each time the sweep generator triggered. This is called alternate sweep method.
In the second method, called chop method, the same vertical amplifier is switched from one channel to the
other at a very fast rate such that the display is created from small segments of the actual waveform. This
requires the chopping frequency be much greater than the input frequency inorder to prevent the display
from being unrecognizable.
The electronic circuit required to generate the chop method is same as those used to generate the alternate
method, except for the electronic switch is a high frequency clock rather than the trigger generator.
4. a.
Discuss need for delayed sweep in digital storage oscilloscope.
Ans:
In a regular CRO (one without a time – delay circuit incorporated in to its sweep circuit), the sweep
voltage will be in synchronism with the applied input voltage resulting in the display of steady wave form
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(04 Marks)
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June 2013
June 13 - on the screen. However, we can not observe various portions of the applied wave form with this – single sweep.
To observe each and every part of a given wave form in minute details, a delayed time base can be used. That is, to
analyze a given waveform, we must apply the time base at the appropriate point on the waveform to be observed.
Delayed sweep is a technique that adds a precise amount of time between the trigger point and the beginning
of the scope sweep. In the sweep mode, the start of the horizontal sweep can be delayed typically from a
few micro seconds to several seconds. This delayed sweep operation allows the user to view clearly a small
segment of the wave form by magnifying it. By this technique the versatality of the instrument is increased
so that better measurement of waveform jitter, rise time, pulse modulation are possible.
4. b.
Explain basic principle of sampling oscilloscope with neat diagram and wave forms.
(06 Marks)
Ans:
In an ordinary CRO, when we try to measure high frequency, the image brilliance reduces. It is because,
when the frequency of the vertical deection signal increases, the writing speed of electron beam increases.
In order to obtain sufcient image brilliance and to maintain normal image brightness the electron beam
must be accelerated to a higher velocity so that more kinetic energy is available for transfer of electron
to the screen. The higher electron beem velocity can be easily achieved by raising the accelerating anode
voltage.
A high velocity beam also requires a greater deflection potential to maintain the deflection sensitivity.
All these leads to higher demands on the vertical amplifier.
To make the matter simple, a sampling oscilloscope uses a different approach altogether to improve the high
frequency performance. In the sampling scope, the input waveform is reconstructed from many samples
taken during recurrent cycles of the input wave form and thus over come the frequency limitations of
conventional CRT. This sampling technique is illustrated in Fig 4(a). From the figure, it can be seen that
the high frequency input being sampled by a train of sampling pulses. The principle involves in taking
one sample each from each recurrent full cycle of the applied wave at a slightly later delayed position.
As illustrated in the figure, sample 1 is taken form the first cycle, sample 2 from cycle 2, sample 3 from
cycle 3 and so on but from a slightly advanced position or from different points on the applied wave, when
regrouped these points will form a single wave, which is the replica of the applied wave form as shown in
the figure 4(a).
2
Reconstructed output wave
Higher frequency input wave
1
3
5
6
7
t
Sampling points
V
Triggering pulse
t
V
Sampling pulses
Sampling pulses
t
Fig: 4(a) Sampling of high – frequency signal into a low – frequency signal
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A simplified block diagram of the sampling circuitry used in the sampling scope is shown in Fig. 4(b).
The input wave form to be observed is applied to the sampling gate. Sampling pulses momentarily bias the
diodes of the balanced sampling gate in the forward directions briefly connecting the gate input capacitance
to the test point. These capacitances are slightly charged towards the voltage level of the input circuit.
Input
signal
Trigger
Input
Blocking
oscillator
Ramp
Generator
Sampling
gate
Vertical
Amplier
Voltage
Comparator
Staircase
Generator
To vertical
deection plate
To horizontal
deection plate
Attenuator
Fig. 4(b) black diagram of the sampling circuitry in a sampling scope
The capacitor voltage is amplified by the vertical amplifier and applied to the vertical deflection plates.
Since the sampling must be synchronized with the input signal frequency, the signal is delayed in
vertical amplifier, allowing the sweep triggering to be done by the input signal. When the trigger pulse
is received, the avalanche blocking oscillator starts an exactly linear ramp voltage, which is applied to
voltage comparator. The voltage comparator compares the ramp voltage to the output voltage of a staircase
generator. When the two voltages are equal in amplitude, the staircase generator is allowed to advance one
step and simultaneously a sampling pulse is applied to tar sampling gate. At this movement, a sample of the
input voltage is taken, amplified and applied the vertical deflection plate.
4. c.
Explain two types of storage techniques used in storage oscilloscope with neat diagram. (10 Marks)
Ans:
The storage is accomplished using the principle of secondary emission. When the ordinary writing electron
beam passes a point on the phosphor surface, not only does it momentarily cause the phosphor to illuminate,
but the kinetic energy of the electron beam knocks other electrons loose from the phosphor surface.
This leaves a net positive charge. Storage oscilloscopes provided with one or more secondary electron
guns, called the "flood guns", that provide a steady flood of low–energy electrons traveling towards the
phosphor screen. Flood guns cover the entire screen uniformly. The electrons from the flood guns are more
strongly drawn to the areas of the phosphor screen where the writing gun has left a net positive charge; in
this way, the electrons from the flood guns re–illuminate the phosphor in these positively charged areas of
the phosphor screen.
If the energy of the flood gun electrons is properly balanced, each impinging flood gun electron knocks out
one secondary electron from the phosphor screen, thus preserving the net positive charge in the illuminated
areas of the phosphor screen. In this way, the image originally written by the writing gun can be maintained
for a long time – many seconds to a few minutes.
There are two types of storage techniques : One is "bistable storage" (binary — on/off — form of storage) and the
other is "variable persistence storage". In both the cases a special type of cathode ray tube (CRT) is necessary.
Bistable storage : In bistable storage oscilloscope, a bistable CRT is used, the basic construction of which
is shown in Fig. 4(c). The term bistable on the context of CRT is that it can display a stored wave form
at only one level of brightness; waveform is either displayed or not displayed and no variation in display
intensity is possible. The screen has a storage layer of phosphor material that is capable of secondary
emission and has a high insulation resistance between the particles. Between the glass of the CRT and the
storage layer, metal film is deposited. Another metal film known as "collimator" is deposited around the
neck of the CRT.
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June 2013
Flood gun
electrons
Collimator
June 13 - Glass face
plate
Write gun
Write beam
Flood
guns
Collimator
Phosphor Metal
storage layer lm
Fig. 4(c) Basic construction of bistable storage CRT
The "write gun", shown in the figure, is made up of the accelerating and deflecting anodes same as in a
conventional CRT. The two "flood guns" are basically cathodes, the purpose of which is to generate low–
energy electrons. The cathodes are at ground potential and the collimator may also be kept at ground level
or slightly positive. The metal film is kept at a potential of +lV to +3V with respect to ground. The clouds of
electrons emitted from the flood guns are attracted towards the metal film because of the positive potential
and flood the screen with electrons as depicted in the figure.
It should be noted that if the write gun is not activated, the phosphor layer will be affected by the low energy
electrons from the flood guns. These electrons do not have the energy to penetrate the phosphor layer and
are collected by the collimator, and in this circumstance there is no display occurs.
Suppose the write gun is energized and a wave form is applied to the oscilloscope input for a very brief
time period (like a transient). The beam of high–velocity electrons from the write gun is deflected across
the CRT screen. These electrons will strike the phosphor layer with such a high energy that it produce
secondary emission. The electrons produced by the secondary emission are collected by the "collimator"
and now, every point on the screen, where the secondary emission occurs becomes positively charged
because the electrons are lost from that point. Thus a positive path is traced on the storage layer in the shape
of the input wave form. Since the storage layer has high insulation properties, charge leakage is very less
and so the traced path will remain for several minutes.
The low energy electrons from the flood gun are attracted to the positive charge path and pass through
it to the metal film that is more positive. While passing through the storage layer, the electrons
cause the phosphor to continue glowing. In this way a short time signal like a transient is displayed
or a long time.
To erase the display, make the metal film negative that will repel back the flood gun electrons to the storage
layer, where they accumulate and return the traced path to the same potential as the surrounding material
and the wave form is wiped out.
Variable–Persistence Storage: A CRT capable of doing variable persistence storage operation is shown
in Fig 4(d). The major difference from the bistable storage CRT is that a fine wire storage mesh and a
collector mesh are included in Variable–Persistence CRT. The storage layer is deposited on the inside
surface of the storage mesh and the screen is just a conventional CRT screen with a low–persistence
phosphor backed by an aluminium film. The additional change is that control grids are added in front of
each flood gun.
The collector mesh has positive potential around 100V, and the voltage on the storage mesh is between
0 and –10V. The high energy electron beam from the "write gun" produces secondary emissions at the
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storage layer and creates a positive change path in the shape of the input wave form. The low energy flood
gun electrons are repelled from all parts of storage layer except from the positively charged path created
by the write beam. Now the flood gun electrons pass through the charged path and produce a trace on the
screen. The secondary emitted electrons and the flood gun electrons that do not pass through the storage
layer are attracted back to the collector mesh since it is kept at highly positive potential. Even though it is
low persistence phosphor screen, the flood gun electrons passing through the positively charged trace on
the storage layer continue to maintain a display on the screen for a very long time.
Flood
guns
Collimator
Metal
lm
Phosphor
layer
Glass face
plate
Write gun
Write beam
Control Collector Storage Storage
Mesh
mesh
grids
layer
Fig. 4(d) Basic construction of variable persistence storage CRT
The stored wave form can be erased by connecting the storage mesh to the same high positive potential
as the storage mesh for a brief period of time. By doing this, the flood gun electrons produce secondary
emission all over the storage layer so that written wave form is wiped out.
PART – B
5. a.
Explain with neat block diagram, operating principle of function generator.
(08 Marks)
Ans:
A function generator is used to generate a variety of wave form functions, whole frequencies are adjustable
over a wide range. The most common output wave forms are sine, triangular and square wave forms.
Fig. 5(a) shows a typical block diagram of a function generator. The basic wave form produced in a function
generator is a triangular wave and the sine and square waves are derived out of this triangular wave form.
Constant – current
source
Frequency
– control
circuit
Schmitt
trigger
Square
I
Triangular
Integrator
I
Constant – current
source
Wave –
shaper
Sine
Fig: 5. a block diagram a typical function generator
The generation of basic signal is done by two constant sources driving an integrator. The upper current
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source is used to charge the integrator capacitor at a uniform rate, which produces a positive going ramp
according to the relation. V = –C³idt. The capacitor after reaching the final value will discharge through the
lower constant current source that results into a negative going ramp. The two actions; the charging and
discharging of the capacitor will produce a triangular wave as shown in Fig. 5(b).
V
t
a
+V
t
b
+V
c
t
Original
wave
–V
Shaped sine
wave
Fig: b Triangular, square, and shaped sine waves
The triangular wave is fed to a schmitt – trigger (voltage comparator) circuit to produce a square wave form.
To get a sine wave, a wave shaping circuit using diodes are used. The wave shaping circuit will smoothen
the edge of the triangular wave to form into a sine wave as shown in the figure. The output circuit of the
function generator consists of two output amplifiers to provide simultaneous amplifications for individually
selected outputs of any of the wave form function.
5. b.
Elaborate with neat block diagram, conventional standard signal generator.
(06 Marks)
Ans:
A standard signal generator produces known and controllable voltages. It is extensively used in the testing
of radio transmitters and receivers.
Am
modulator
AF
oscillator
RF
oscillator
FM
modulator
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AM
Amplier
Attenuator
Output
FM
Fig: 5. c Standard signal generator
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Figure 5.c shows the block diagram of a standard signal generator, which some times is called the standard
AM/FM signal generator. It consists of an RF generator, an AF generator, an amplitude modulator, a
frequency modulator, an amplifier and an attenuator.
The radio–frequency (RF) oscillator is used to produce sine waves in the range of 1 MHz to 100 MHz.
Usually, an RF oscillator of the Hartley or Colpitt's type is used to produce the desired radio–frequency
sine waves. The output frequency of the oscillator can be varied by varying the tank–circuit inductance
using ferrite tuning.
The audio–frequency (AF) oscillator is of the Wien – bridge type and is used to produce sine waves in the
range of 10 Hz to 100 kHz. The amplitude– and frequency– modulator circuits are, respectively, used to
amplitude and frequency modulate the RF signal with AF signal. These modulations are taken out through
a band switch S to an amplifier – attenuator combination circuit.
The amplifier – attenuator combination circuit is used to control the amplitude of the output signal. Some times,
a test–signal terminal of 1 –volt, 1000–Hz sine wave is also incorporated on the equipment's panel board.
Standard AM/FM signal generator is used to test and measure the performances of radio circuits. They arc
used as laboratory standards for testing and repairing RF transmission systems. They also help in tuning
radio receivers for optimum performance in factories and repair shops. In military and police signaling
schemes, standard, AM/FM signal generator is one of the highly essential test equipment.
5. c.
Explain with neat block diagram and waveforms, frequency synthesizer in signal generators.
(06 Marks)
Ans:
Figure 5.d shows a frequency multiplier using a PLL (phase–locked loop). It consists of a PLL with a
divide–by–N counter connected in its feedback path. Let the input frequency be fi and the fedback
input f0. These two are compared in the phase detector, which produces an output voltage that is proportional
to the phase difference between fi and f0. The low pass filter removes the AC content in the voltage and
produces an almost pure DC voltage which drives a voltage–controlled oscillator (VCO). The VCO in turn
produces a frequency that is proportional to the input DC voltage. Thus we find that the output frequency
is proportional to the phase difference between fi and f0.
fi
f0
Phase
detector
Low – pass
lter
VCO
Nf0
Divided by N
Figure 5.d Phase–locked loop frequency multiplier
Now with the divide by N circuit introduced in the feedback path, we observe that the output frequency is
really Nf0. This is because, with output frequency equal to Nf0 the input frequency becomes Nf0/N = f0.
This principle is used in frequency synthesizer circuits to produce frequencies of all values and ranges.
Frequency Synthesizer using PLL
The theory of the frequency multiplier using PLL is extended to synthesize (artificially produce) oscillations
in any desired frequency range. Figure 5.e shows a typical PLL frequency synthesizer. It consists of a
crystal oscillator that produces a fixed frequency of, say, 1 MHz. This is divided in a frequency counter by
M, an integer of appropriate value. This forms the fi of the PLL, whose feedback frequency is, as before f0,
which is obtained by frequency division, as shown in Fig. 5.d.
Thus we observe that the frequency output of the PLL depends on the ratio N/M and by suitably choosing
this ratio, we can obtain several frequencies, which are all crystal controlled frequencies, and hence are
stable.
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Electronic Instrumentation
Frequency
counter
Crystal
Osc
June 13 - June 2013
Phase
detector
fi
Nf0
Low – pass
lter
VCO
f0
Divided by N
Figure 5.e Phase–locked loop frequency synthesizer
By using several crystal oscillators of different frequencies, and several PLL units, we can produce
frequencies in all ranges, values and amplitudes.
6. a.
Derive an expression for galvanometer current (Ig) when the wheatstones bridge is unbalanced.
(05 Marks)
Ans:
The circuit diagram of wheatstones bridge is shown in Fig.6(a). When the bridge is unbalanced current
ows through the galvanometer that causes a deection of its pointer. To determine the amount of deection
or the amount of galvanometer current, circuit analysis by the thevenins theorem may be used.
R1
V
R2
R1
V
G
R3
R2
a
R4
b
R3
Fig: 6(a) Wheatstone's bride
R4
Fig: 6(b) Unbalanced Wheatstone's bridge
Since we need to determine the current through the galvanometer, the thevenin's equivalent as seen by the
galvanometer is to be found.
Step 1: Disconnect the galvanometer from the bridge to determine the open circuit voltage between the
terminals a and b, as shown in fig 6(b).
Step 2: Using voltage divider equation,
At point Na, Va =
V × R3
R1 × R3
and at point b, Vb =
V × R4
R2 × R4
Step 3: The voltage between a and b is the difference between Va and Vb, which is the Thevenin's equivalent
voltage.
i.e., Vth = Vab = Va − Vb =
VR3
VR4
−
R1 × R3 R2 + R4
R4 ⎤
⎡ R3
−
∴Vab = V ⎢
⎥
⎣ R1 + R3 R2 + R4 ⎦
Step 4: To find thevenin's equivalent resistance, find the resistance looking into the terminals a and b by
short circuiting the voltages source V. This arrangement is shown in Fig. 6(c).
Rth
a
R1
R2
a
b
R3
R4
Fig. 6(c) Thevenin's resistance
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Vth
G
b
Fig. 6(d) Thevenin's equivalent
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June 13 - June 2013
Electronic Instrumentation
Step 5: The Thevenin's equivalent resistance is R1 || R3 +R2 || R4
R1R3
R2R4
i.e. Rth = R +R + R +R
1
3
2
4
The Thevenin's equivalent circuit for the bridge looking to the terminals a and b is shown in Fig. 6(d).
If the galvanometer is connected across the terminals a and b and consider the galvanometer resistance is
Rg, then the deflection current in the galvanometer is given by
Vth
Rth + Rg
Ig =
6. b.
An unbalanced wheat stones bridge is shown in Fig. Q6(b). Calculate current in the galvanometer.
(05 Marks)
1.5k:
R1
10V
3k:
R2
RG=250:
G
R3
5k:
R4
14k:
Fig. Q6. (b)
Ans:
Given R1 = 1.5 k, R2 = 3k, R3 = 5k, R4 = 14k, Rg = 250 and V = 10V
The Thevenin's equivalent voltage between the galvanometer terminals is the difference of voltage at these
points.
R3 ⎞
⎛ R4
−
i.e., Vth = V ⎜
⎝ R + R4 R1 + R3 ⎟⎠
5k ⎤
⎡ 14k
= 10 ⎢
−
⎣ 3k + 14k 1.5k + 5k ⎥⎦
= 0.5423V
The Thevenin's equivalent resistance is
RR
RR
Rth = 1 3 + 2 4
R1 + R3 R2 + R4
1.5k × 5k 3k × 14k
+
1.5k + 5k 3k + 14k
= 3.624kΩ
=
Rth = 3.624 k
Eth
0.5423v
Rg = 250
G
The equivalent circuit connected along with the galvanometer is shown in figure. The current through the
galvanometer is given by
Ig =
Vth
0.5423
=
= 140μA
Rth + R g 3.624k + 0.25k
6. c.
Derive an expression for Lx and Rx which is a series impedance in the Maxwell's bridge. And nd
series equivalent unknown impedance, when C1 = 0.01 Pf, R1 = 470 K, R2 = 5.1 K, R3 = 100K.
(10Marks)
Ans:
i) Fig. shows a Maxwell's bridge that is used to measure unknown value of impedance interms of known
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June 2013
June 13 - value of capacitance and resistance.
The general equation for the balanced bridge is Z1 Zx = Z2 Z3
C1
R2
R1
Detector
Lx
R3
Rx
Fig : Maxwell's bridge
i.e., Z x =
Z 2 Z3
= Z 2 Z 3Y1
Z1
.............(1)
Where Z1 = R1 in parallel with C,
i.e., Y1 =
1
1
=
Z1 R1 + jωC ,
Z 2 = R2 , Z 3 = R3
and Zx = Rz is series with Lx
= Rx + jZLx
where Zx is the unknown impedance
From equation (1), we have
⎡1
⎤
Rx + jωLx = R2 R3 ⎢ + jωC1 ⎥
⎣ R1
⎦
RR
Rx + jωLx = 2 3 + jωC1 R2 R3
R1
Equating real and imaginary terms, we get
Rx =
R2 R3
and Lx = C1 R2 R3
R1
ii) Given values are :
C1 = 0.01 PF,
R1 = 470k, R2 = 5.1 k, R3 = 100 k
Substituting these values in the derived expressions,
Rx =
R2 R3 5.1k × 10k
=
= 1.085kΩ
470k
R1
and Lx = R2 R3 C1 = 5.1 k u 100 k u 0.01 PF
= 5.1 H
The equivalent unknown impedance is a resistance of 1.085 k in series with inductance value of 1H as
shown below.
Rx = 1.085k
Lx = 5.1H
a
7. a.
List at least ve advantages of electrical transducer.
6-Electronic Instruments June-July 2013.indd 143
b
(05 Marks)
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June 13 - Ans:
7. b.
June 2013
Electronic Instrumentation
The advantages of electrical transducers are,
1.
Amplication and attenuation can be easily achieved.
2.
Remote indication and recording of the output data are possible.
3.
Output can be easily modied to meet requirements of the indicating and controlling unit.
4.
Power level requirement is very small.
5.
The output signals of similar transducers can be mixed or conditioned.
6.
Friction and mass inertia effect is almost nil.
7.
The electrical output can be easily processed and transmitted.
A displacement transducer with a shaft stroke of 3.0 inch, is applied to the circuit as shown in
Fig. Q7 (b) below. The total resistance of the potentiometer is 5 K. the applied voltage Vt is 5 V.
When the wiper is at 0.9 inch from B. What is the value of output voltage Vo.
(05 Marks)
A
Vt
R1
R2
wiper
Vo=?
B
Fig. Q7(b)
Ans:
Given size of shaft stroke = 3 inches
Position of wiper = 0.9 inch
Total refinance of Pot = 5k
Applied voltage Vt = 5V
Referring the figure and given data
0.9 in
× 5k
3.0 in
= 1.5Ω
R2 =
∴
Vo
R2
=
Vt R1 + R2
i.e. Vo =
R2
× Vt
R1 + R2
1.5k
× 5V = 1.5V
5k
Vo = 1.5V
=
A
R1
Vt
w
R2
Vo
B
B
w – wiper position
7. c.
Dene Gauge factor. Derive expression for gauge factor of bounded resistance wire strain gauge.
(10 Marks).
Ans:
Strain gauges works under the property called Piezo resistive. When a strain gauge undergoes tension or
compression, the length of the material changes, which results in the change of resistance of the material.
Gauge factor is a characteristic of a strain gauge, which is dened as the unit change in resistance per
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June 2013
June 13 - unit change in length and denoted by the letter k.
A tensile stress tends to elongate the wire resulting in increase its
length and decrease its cross – sectional area. The combined effect
will be increase in resistance that can be observed by the following
equation.
R=
ρl
A
Direction of
strain
Fine wire
A bonded strain gauge is shown in Figure.
Leads
where
Fig. Bonded resistance wire strain gauge
R = The resistance of the gauge material
U = Specific resistance of the material in – m
l = length of the conductor (material)
A = Area of cross section of the conductor wire
From the definition, the gauge factor can be deduced as
K=
ΔR / R
Δl / l
..........(1)
⎛ Δl ⎞
The ratio of change in length to original length ⎜ ⎟ is called as strain and denoted as V. Therefore we can
⎝ l ⎠
rewrite equation (1) as
ΔR / R
..........(2)
σ
Here V indicates strain in the lateral direction. We know that the resistance of a conductor with uniform
cross section is
K=
R=ρ
l
A
l
π
πd / 4
..........(3)
d2
4
Where d is the diameter of the conductor
=ρ
l
2
=ρ
When the conductor undergoes stress, due to the strain, the length of the conductor increase by l but
simultaneously decrease the diameter by d. Hence the resistance of the conductor after undergoes the
strain can be written as
RS = l
l + Δl
π
(d − Δd )2
4
=
l (l + Δl )
π 2
[d − 2d Δd + Δd 2 ]
4
Since d is small, d2 can be neglected. Therefore,
RS =
(
)
ρl 1 + Δl l
l (l + Δl )
=
π 2 ⎛ 2Δd ⎞ π 2 ⎡ 2Δd ⎤
d ⎜1 −
d 1−
⎟
4 ⎝
d ⎥⎦
d ⎠ 4 ⎢⎣
Now, the ratio of strain in the lateral direction to the strain in the axial direction is nothing but Poisson's
Δd / d
i.e. μ =
..........(5)
Δl / l
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June 13 - ∴
June 2013
Δd μΔl
=
d
l
..........(6)
Substituting the value of
Rs =
Electronic Instrumentation
(
ρl 1 + Δl l
)
Δd
in equation (4), we get
d
π 2 ⎡ 2μΔl ⎤
d 1−
4 ⎢⎣
l ⎥⎦
Rationalizing,
Δl ⎞
⎛
ρl ⎜1 + ⎟
⎝
l ⎠
RS =
π 2 ⎛ 2μΔl ⎞
d ⎜1 −
⎟
4 ⎝
l ⎠
Δl ⎞
⎛
⎜⎝1 + 2μ ⎟⎠
l
Δl ⎞
⎛
⎜⎝1 + 2μ ⎟⎠
l
⎡ 2μΔl 2Δl 2μΔl 2
+
+ 2
1+
ρl ⎢
l
l
l
∴ RS =
⎢
2
π 2⎢
Δl
d
1 − 4μ 2 2
4 ⎣⎢
l
⎤
⎥
⎥
⎥
⎦⎥
Neglecting higher powers of l, we have
ρl
π 2
d
4
ρl
=
π 2
d
4
ρl
RS =
π 2
d
4
RS =
⎡ 2μΔl Δl ⎤
⎢⎣1 + l + l ⎥⎦
⎡ Δl
⎤
⎢⎣1 + l (1 + 2μ )⎥⎦
+
ρl Δl
(1 + 2μ )
π 2 l
d
4
From equation (3), substituting R =
Rs = R + ΔR
where ΔR +
ρl
2
d
π
4
ρl
2
d
π
, the above equation becomes
4
Δl
(1 + 2μ )
l
Therefore the gauge factor will now becomes
Δl
(1 + 2μ )
k=
= l
Δl
Δ ll
l
= 1 + 2μ
ΔRR
∴ k = 1 + 2μ
8. a.
Explain Photo transistor. With neat diagram and output characteristics. How is it used as a
transducer?
(05 Marks)
Ans:
Photo – transistor is a semi conductor device, in which the collector current can be controlled by subjecting
base area to illumination. It should be noted that the photo – transistor has base connection like a BJT. The
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June 2013
June 13 - optically generated Electron-Hole pairs (EHPs) is the base ow through the base-emitter junction to produce the
base current that results in the ow of the collector current Ic. Basic construction of a photo-transistor is shown
in Fig. 8(a). Illumination of the central region (base) causes the release of EHPs. This lowers the barrier potential
across both the junctions, causing the ow of electrons from emitter region to the collector region through base
region. Fig. 8(b) shows a set of output characteristics for different level of illumination.
Illumination
IC(mA)
400 w/m2
300 w/m2
200 w/m2
100 w/m2
8
6
4
E
B
2
C
0
Symbol
(a) Construction
5
10 15 20 25
VCE (V)
(b) Output characteristic
Fig. 8 Photo transistor
Photo transistor can be used as a transducer in the operation of a relay circuit. A simple relay circuit is
shown below. The light incident on the photo transistor causes its current to increase, which results into
increase in the voltage drop across the resistor R that inturn increase input to the transistor, which drives
the relay.
VCC
R
Relay
coil
Fig: Relay circuit using photo transistor
8. b.
Ans:
8. c.
List at least ve classications of digital displays.
(05 Marks)
The display can be classied in number of ways as follows.
1. On the method of convertion from electrical signal into visible light
eg : CRTs, LCDs, LEDs
2. On the application
eg : Nixies, LEDs, Alpha numeric
3. On the size of display and physical dimension
eg : Symbolic display like LEDs
Control display like CRT, LED
Large screen Display like projection system
4. On the display format
eg: Segmental, dotmatrix
5. In terms of resolution and clarify of characters
eg : Simple single element indicator
Multi element displays
List out the requirement of a dummy load. And explain measurement of power by means of a
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June 13 - Ans:
June 2013
Electronic Instrumentation
bolometer bridge.
(10 Marks)
The requirements of a dummy load are
Ñ It should have the ability to dissipate the specied amount of power
Ñ It should have low reactances
Ñ It should have minimum skin effect
Ñ It should have a value equal to that of the actual load
An example of dummy load is the Bolometer element, which is a short ultra thin wire having a positive
temperature coefficient of resistance. Bolometer is a small temperature sensitive resistive element that is
used to measure RF power. The RF power to be measured heats the bolometer and causes change in its
electrical resistance, which is used as an indication of the magnitude of power. The bolometer is generally
used in a bridge network so that even a small change is resistance can be detected easily and corresponding
power can be measured. The bolometer bridge is shown in fig. 8(c).
RF Input
O/4 stub for
Bolometer
Ground return
Tapered Section for
Impedance Matching
Bolometer Element
RF Bypass Capacitor
R1
R1
R
V
R1
E
dc Bias
low RF
supply
Fig. 8(c) bolometer Bride
To measure the unknown RF power, a small value of known RF power indicated by a voltage V1 is
superimposed on the RF test power. The dc current form the dc source E is adjusted by varying the resistance
R that heats the bolometer element until its resistance equals the value of R1, which is the value required to
balance the bridge. Now the test power is turned OFF, which unbalances the bridge. Restore the balance by
increasing the AF voltage indicated by V2.
The unknown RF power is calculated using the relation,
RF power =
V22 − V12
,
4 R1
1
of the power fed to the bridge.
4
In a transmission system, where coaxial cable or wave guide is used, the bolometer should provide the
necessary impedance matching. This is done by using of a tapered section as shown in Fig.
since the power delivered to the bolometer element is
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