6-electronic instrumentation dec.013-jan.014 · 1.c. explain in detail the working of true rms...

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sDec.2013/Jan.2014Third Semester B.E. Degree Examination

Electronic InstrumentationTime: 3 hrs. Max. Marks: 100

Note: 1. Answer any FIVE full questions, selecting at least two questions from each part.

PART - A

1. a. De� ne the following terms as applied to an instrument: i) Accuracy and Precision ii) Random Error iii) Absolute and relative error iv) Gross error v) Systematic error (06 Marks)

Ans: i) Accuracy is de� ned as the degree of exactness (closeness) or conformity to the value of quantity under measurement. Precision is defined as the degree of agreement within a group of measurement (repeatability of measurement).

ii) Random errors occur due to reasons that are unknown, and occur at random. Since the reason for them cannot be accounted for, they cannot be eliminated completely.

iii) Absolute error may be de� ned as the difference between the true value and measured value. Absolute error represents the physical error in the measurement. Relative error is de� ned as the ratio of the absolute error produced in a given measurement to the measured value.

iv) Gross errors occur mainly due to the carelessness/lack of experience of the person involved in the measurement of any quantity.

v) Systematic error occurs mainly due to the defective parts and ageing of the instrument and also environmental effects. They may be due to the friction in the bearings, moving parts, variation in air gap, loading effect etc.

1. b. Determine the value of the multiplier resistance on the 50V range of a dc voltmeter that uses a 250mA meter movement with an internal resistance of 100�. (06 Marks)

Ans: Given Ifsd = 250 mA, Internal resistance = 100 �, Range = 50VSensitivity of the meter movement is

6fsd

1 1S 4k / V

I 250 10−= = = Ω×

The value of multiplier resistance RS is found by the relationRS = S � Range – internal resistance = 4000 � 50 – 100 = 199900� = 199.99 k�

1.c. Explain in detail the working of true RMS voltmeter and give the difference between peak responding and average responding voltmeters. (08 Marks)

Ans: True RMS - Reading AC voltmeter: A DC meter can be converted into a true RMS voltmeter.For converting a given DC meter to read the RMS values, we use the form factor, which is the ratio of the RMS value of the wave to its DC (average) value.

6-Electronic Instrumentation Dec.013-Jan.014.indd 1016-Electronic Instrumentation Dec.013-Jan.014.indd 101 05/08/2014 12:49:37 PM05/08/2014 12:49:37 PM

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For a sinusoidal input waveform, the form factor is 1.11. We multiply the rectified DC with 1.11 to get the corresponding peak value of the input AC wave. The scale of the simple AC voltmeter can then be converted (and calibrated) to read peak values. For obtaining the RMS value of a non-sinusoidal waveform, corrections have to be made, as its form factor will be different from the standard sinusoidal value of 1.11.Figure shows the construction of a directly calibrated true - RMS reading voltmeter. The voltage to be measured is applied through an operational amplifier OA 1 to the heating element H1, of the main thermocouple unit TM. H1, in turn, produces a heat proportional to the input AC. The thermocouple TM senses this heat and produces a voltage that is proportional to it. It can be seen that this voltage represents the true RMS value of the input AC voltage, since it is proportional to the power dissipated in H1.The output voltage of TM is amplified by OA 2, which drives an ordinary moving-coil voltmeter M. To remove errors due to common-mode signals, such as ambient temperature, we use a feedback line from the output of OA 2 to drive heating element 2 of thermocouple TD, which senses the error produced and applies a corrective voltage to OA 2.Since this meter senses the heating effect and produces proportional voltages, this can indeed be directly calibrated to read the true RMS value of any applied input AC.

M

OA2

OA1

TM

TD

Heating elementH1

H2

Fig. True RMS reading voltmeter

ac input voltage

The difference between peak responding and average responding meters: The peak responding show the peak value (maximum value) of a signal while the average responding meters show the average value of the signal. This can be done by calibrating the meter for the required value after introducing the necessary changes in the meter circuits.

2. a. Explain in detail the working of successive approximation type digital voltmeter and give the output for a 4 bit control word. (08 marks)

Ans: Successive Approximation DVM is based on the principle of simple weighing technique used in practice. The basic block diagram of a successive approximation DVM is shown.

When a +ve start pulse is applied to multi-vibrator which activates the control circuit, SAR is cleared to “00000000” and VCout of DAC is 0V.

if Vin > VCout then comparator output is positive during � rst clock pulse of the counter and D7 = 1 and all

other registers to ‘0’ and VCout jumps to refV

2.

Now VCout > Vin then comparator output is negative during second clock pulse of the counter and D7 = 0, D6

= 1 and all other register to ‘0’ and VCout jumps to refV

4.


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s Similarly, the rest of bits beginning from D7 to D0 are set and tested. Therefore in 8 clock cycles the measurement is completed and the content in SAR is the actual digital output.

Sample and Hold Clock

2. b. A 4 ½ digit DVM is used for voltage measurements. (i) Find its resolution (ii) How would 67.50V be displayed on a 5V range (iii) How would 0.716V be displayed on 1 V and 10 V ranges. (06 Marks)

Ans: i) Number of full digits on a 4 ½ digit multimeter = 4In general, resolution of a DVM with an n-digit display is given by

n

1Resolution

10=

Substituting n = 4, we get

4

1Resolution 0.0001

10= =

ii) Using the 4 ½ digit display, there would be 5 digits on the display panel, with the most digit being a 0 or a 1.

Thus 67.50V would be displayed as 067.50iii) Resolution on 1 V range = 0.0001�1V = 0.0001V (100�V) It means any reading up to 4th decimal can be displayed Hence, 0.716 would be displayed as 0.7160 in 1V range.

Similarly, Resolution in 10 V range = 0.0001�10V= 0.001V (1mV). Hence decimal upto 3rd decimal places can be displayed. So, 0.716 would be displayed as 00.716.

2. c. Explain in detail the working of digital multimeter. (06 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



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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.

acattenuator

acattenuator

ohmsconverter

acConverter

Digital Display

A/Dconvertor

Precisionreference

Interface

BCDOutput

Shunt

High

Input

Low

ac

bc

Fig: 2(a) Basic block diagram of a DMMThe 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.

3. a. Explain in detail the generation of time base signal for the horizontal de� ecting plates. (05 Marks)

Ans: Sweep or time base generator is a part of horizontal de� ecting 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).

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. 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.



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+VBB

VoCT

RTR2

R1

B2

B1

Synchpulseinput

EVP

VBB

V

VV

TSTr

t

Circuit diagram Wave form in output

Ts � Sweep timeTr � Retrace time

Fig. 3(a) : Continuous sweep

Circuit diagram for the triggered sweep and the corresponding output waveforms are shown in Fig.3(b).

+VBB

VoCT

CCRT

R3

R4R1

R2

VD

Trigger Input D

Tr ThTs

Output

Trigger input pulse

V0

VD

t

Circuit diagram Wave form

Ts �Sweep TimeTr �Retrace TimeTh �Hold Time

Fig: 3(b) Triggered sweep wave formResistances 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. What is an electronic switch and explain in detail its various modes of operation (05 Marks)

Ans: The electronic switch is used in cathode ray oscilloscope to display two channels at a time. The electronic switch basically consists of two separate gain control and gate stages. These gate stages are alternating biased to cut-off by a square wave signal applied to the gate stage. This technique allows only one gate is in



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a condition to pass its signal at any given time. The outputs of these gate stages are directly coupled through a capacitor to VDF of CRT.

To vertical de� ection of CRT

3. c. List the various control knobs on the front panel of CRO (04 Marks)Ans: Typical front - panel controls of a basic oscilloscope (2 channel) is explained below:

1. On - off switch. - To switch on or off the scope.2. INTENS. This is the intensity control connected to the grid G to control the beam intensity and hence

the brightness of the screen spot. Make the intensity control just bright enough for clear visibility.3. FOCUS Control allows you to obtain a clearly defined line on the screen.4. POSITION knob allows you to adjust the vertical position of the waveform on the screen. (There is

one of these for each channel).5. AMPL/DIV. is a control of the Y (ie. vertical) amplitude of the signal on the screen. It is a rotary

switch individual for each channel.6. AC/DC switch. This should be left in the DC position unless you cannot get a signal on-screen

otherwise. If it is kept in AC position, DC voltage will be blocked.7. A&B/ADD switch. This allows you to display both input channels separately or to combine them into

one. 8. +/- switch. This allows you to observe the waveform on the display start from its +ve half or -ve half

cycle.9. X POSITION Only one control knobe for both the channels, that allow you to adjust the horizontal

position of the signals on the screen.10 LEVEL This allows you to determine the trigger level; ie. the point of the waveform at which the

ramp voltage will begin in timebase mode.11. Time/Div This selector controls the frequency at which the beam sweeps horizontally across the

screen in timebase mode, as well as whether the oscilloscope is in timebase mode or xy mode. It helps to measure the time period of the waveform and hence frequency.

12. A/B selector. This allows you to choose which signal to use for triggering (channel 1 or 2).

13. -/+ press button switch will force the ramp signal to synchronise its starting time to either the decreasing or increasing part of the unknown signal you are studying.

14. INT/EXT This will determine whether the ramp will be synchronised to the signal chosen by the A/B switch or by whatever signal is applied to the EXT. SYNC. input :



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s15. External trigger input: External trigger signal is applied to this terminal if EXT is selected (refer the above part 14)

3. d. An electrically de� ected CRT has a � nal anode voltage of 2000V and parallel de� ecting plates 1.5cm apart. If the screen is 50 cm from the center of the de� ecting plates, � nd (i) the beam speed (ii) de� ection sensitivity of the tube (iii) De� ection factor. (06 Marks)

Ans: GivenFinal anode voltage, VA = 2000 V,Distance between plates, d = 1.5 cm = 0.015 mDistance from center of deflection plates to screen, L = 50 cm = 0.5 m

i) Then, beam speed A2eVV

m=

where e= charge of electron (1.6�10–19C), m= mass of electron (9.1�10– 31 kg)

196

31

2 1.6 10 20000Therefore, V 26.52 10 m

9.1 10

−

−

× × ×= = ××

ii) De� ection sensitivity, d

A

LS

2dV

l=

where ld = effective length of the de� ection plates in meters and let it be 1cm = 0.01 m.

60.5 0.01

Therefore, S 83.3 10 m / V2 0.015 2000

−×= = ×× ×

iii) De� ection factor, ( )

6

1G Volt / meter

s

1G 12000V / m

83.3 10−

=

= =×

4. a. Explain in detail the working of sampling oscilloscope with necessary waveforms (10 Marks)Ans:

Fig. 4.a Sampling of high-frequency signal into a low-frequency signal

v

v

t

t

1

2

3 4 56

7

Sampling Pulses Sampling Pulses

Sampling Pulses

Reconstructed output wave Higher frequency input wave



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Consider an ordinary oscilloscope being applied with a high-frequency signal at its vertical input. With a high-frequency being applied across the vertical deflection plates, the writing speed of the electron beam increases. The immediate result of higher writing speed is a reduction in image intensity on the CRT screen. In order to obtain sufficient image brilliance, the electron beam must be^accelerated to a higher velocity so that more kinetic energy applied to the screen to produce normal image brightness. An increase in electron-beam velocity is easily achieved by raising the voltage on the accelerating anodes. A beam with higher velocity also needs a greater deflection potential to maintain the deflection sensitivity. This immediately places higher demands on the vertical amplifier.In the sampling oscilloscope, the input waveform is reconstructed from many samples taken during recurrent cycles of the input waveform and thus overcomes the frequency limitations of conventional CRTs. The technique is illustrated by the waveforms indicated in Fig. 4 (a).Figure 4 (a) shows a higher frequency input wave being sampled by a group of sampling pulses. The principle involves in taking one sample each from each recurrent full cycle of the applied wave. As illustrated in the figure, sample 1 is taken from the first cycle, sample 2 is taken from the second cycle, sample 3 is taken from the third cycle, and so on.It may also be noted that the samples are taken from different points on the applied wave in such a fashion that, when regrouped, these points will form a single wave, which is replica of one cycle of the applied wave, as shown in Fig. 4 (a), where the reconstructed wave is indicated by thick dotted lines.

Sampling gate

VoltageComparator

RampGenerator

Blocking oscillator

Vertical Ampli� er

StaircaseGenerator

Attenuator

Inputsignal

Trigger Input

To horizontal de� ection plate

To vertical de� ection plate

Fig. 4(b) black diagram of the sampling circuitry in a sampling scope

Figure 4 (b) shows a simplified block diagram of the sampling circuitry used in a sampling oscilloscope. The input waveform to be observed is applied to the sampling gate. Sampling pulses momentarily bias the diodes of the balanced sampling gate in the forward direction, thereby briefly connecting the gate input capacitance to the test point. These capacitances are slightly charged toward the voltage level of the input circuit. 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 the vertical amplifier, allowing the sweep triggering to be done by the input signal. When a trigger pulse is received, the avalanche blocking oscillator (so called because it uses avalanche transistors) starts an exactly linear ramp voltage, which is applied to a 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 the sampling gate. At this moment, a sample of the input voltage -is taken, amplified, and applied to the vertical deflection plates.Advantages over Real time CRO1. Greater band width2. Larger dynamic range



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s3. Very low signal to noise ratio4. Better frequency response

4. b. Explain in detail the working of digital storage oscilloscope and list the advantages of DSO. (10 Marks)

Ans: Digital storage oscilloscope:

InputVerticalamplifier

S/Hcircuit

Triggercircuit

Controllogic D/A

converter

Horizontaldeflectionamplifier

Verticaldeflectionamplifier

A/Dconverter

D/Aconverter

MemoryInputsignal

CRT deflection plants

In a digital storage Oscilloscope, the waveform to be displayed and stored is converted into binary digits (1s and 0s), stored in a random access memory, and retrieved for display on screen. The stored wave form may be continuously displayed by repeatedly scanning the stored waveform and, therefore, a conventional oscilloscope tube can be used for the display.

The stored data can be displayed inde� nitely as long as power is applied to the memory. The digitized waveform can be analyzed by either the oscilloscope itself or by using a digital computer connected to it.

Figure 4.3 shows the block diagram of a digital storage oscilloscope. The input is ampli� ed and attenuated with input ampli� ers as in any oscilloscope. The digital storage oscilloscope uses the same types of input circuitry as a conventional oscilloscope and can operate in a conventional mode, bypassing the digitizing and storing features.

As shown in the � gure, the input signal ampli� ed by the vertical ampli� er-attenuator combination is applied to an analog-to-digital converter, which then drives a random-access memory (RAM). This temporarily stores the digitized input data.

A control logic circuit is used to control the operations of the ADC and the memory. The output of the memory is applied to a digital-to-analog (DA) converter, which in turn is used to drive the vertical de� ection ampli� er and vertical de� ection plates.

The control logic also drives the horizontal-sweep DAC and the horizontal de� ection ampli� er. The combined action of the de� ection plates, as in the conventional oscilloscope produces display on the screen.



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Advantages of digital storage oscilloscope

1. They can be used to observe fast and slow phenomena alike. Can be used to analyze minute details in any type of waveform.

2. Data can be stored permanently for viewing at convenience.

3. Real-time analysis is also possible.

4. Can be directly connected to digital computers for analysis of waveforms.

5. Waveforms appearing only once (like a transient) can be observed and analyzed easily.

6. With large memory capacity, a lot of data can be stored in a DSO.

7. Several channels are possible in modern DSO’s which help in analyzing waveform at different parts of a system simultaneously.

8. Can be used to pre-trigger view.

9. Compatible with modern � at-panel displays using TFT’s

10. Analyses can be displayed on the screen itself by computational ability of the scope.

11. Brighter and bigger display.

12. Different colors employed to distinguish multiple traces.

13. Much higher resolutions possible.

14. Detection of peak signal possible.

15. Waveform zooming in and zooming out is possible for � ner examinations.

16. Allows automation.

17. Three dimensional imaging easy.

PART - B

5. a. Explain in detail the working 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.

Frequency – control

circuit

Constant – current source

Schmitttrigger

Wave – shaper

Constant – current source

Integrator

I

I

Square

Triangular

Sine

Fig: 5. a block diagram a typical function generatorThe generation of basic signal is done by two constant sources driving an integrator. The upper current



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ssource is used to charge the integrator capacitor at a uniform rate, which produces a positive going ramp according to the relation. V = –C idt. 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

+V

–V

V

b

c

a

t

t

t

Shaped sine wave

Original wave

Fig: b Triangular, square, and shaped sine wavesThe 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. Explain in detail the working of square and pulse generator. (08 Marks)

Ans: The block diagram of a typical square-wave and pulse generator circuit, shown in Fig. 5(b), which consists of two constant current (C-C) sources. The switch first connects capacitor C to the top C-C source, which charges it to a predetermined value, and then connects it to the bottom C-C source, which discharges it. The charging and discharging of C produces a triangular wave, which is squared by the Schmitt-trigger circuit as shown in Fig 5 (c).The charging and discharging can be made asymmetric to give different duty cycles of the square-wave output. The output may be a square wave or a trigger wave. Typical square-wave generators (of the type described above) are usually available with outputs in the range of 1 Hz to 10 MHz. They also produce square waves, which are needed to drive logic gates.



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C-C source

C-C source

Switch

Amp 1

Amp 2

Frequencycontrol

Schmitttrigger

Discharging of C

C

Charging of C Square wave

Triggerpulses

Fig. 5 (b) : Block diagram of square-wave and pulse generator circuit

t

t

+V

V

b

a

Fig 5(c) Outputs of square-wave and pulse generator

5. c. Explain in detail the working of sine and square wave generator. (04 Marks)Ans: Figure 5(d) shows the block diagram of a sine- and square-wave generator. It consists of a variable frequency

Wien-bridge oscillator that drives an isolation amplifier. The output of this amplifier is amplified further and attenuated to give a sine-wave output, as shown. The sine-wave is squared using a Schmitt-trigger circuit to give a square-wave output. This generator is used to give audio-frequency sine and square-wave outputs.

Wien-bridge oscillator

Isolation ampli� er

Ampli� er

Schmitttrigger

Sine wave

Square wave

Fig 5(d) Block diagram of a sine- and square-wave generator6. a. Explain in detail the working of wein bridge oscillator and � nd the parallel R and C that causes the

wein bridge to null with the following components values: R1 = 27k�, R2 = 22k�, C1 = 5�F, R4 = 100k�, and the operating frequency is 2.2 kHz. (12 Marks)



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sAns: i) Wein bridge oscillator is an RC oscillator and is used to generate sinusoidal oscillations in the audio frequency range. The following figure (a) shows the circuit of wein bridge oscillator. It consists of an Op-amp non-inverting amplifier in the forward path and a lead-lag network in the feedback path. Series R1C1 network is a lead network and the parallel R2C2 network is the lag network. The name wein bridge is due to the fact that the feedback resistor R3 and R4 of Op-amp amplifier and lead-lag network forms a bridge as shown in the following figure (b). At the oscillator frequency, the lead-lag network is designed to introduce 0o phase shift. The Op-amp non-inverting amplifier introduces 0o phase shift. Hence, the total phase shift around the loop is zero. The expression for the frequency of oscillation is obtained from the balancing condition of the bridge.

C1

C2

R1

+

–+VCC

R3

Vo

a

Lead - Lag network

b

dc

R4

R2

C2

R3R1

R2

R4

d

c

b

C1

Fig (a) Wein bridge Oscillator Fig (b) Wein bridge

ii) Given R1 = 27k�, R2 = 22k�, C1 = 5�F, R4 = 100k�, f = 2.2kHz

Then � = 2�f = 2 � 3.14 � 2200 = 13.83 krad/s

Parallel Resistance

( ) ( )( )

43 1 2 2

2 1 1

22 6

R 1R R

R R C

100k 127k

22k 13.83k 27k 5 10

4.545 27k 0.0077 122.7k

−

⎛ ⎞= +⎜ ⎟ω⎝ ⎠

⎛ ⎞⎜ ⎟= +⎜ ⎟× × ×⎝ ⎠

= + = Ω

Parallel Capacitance

( ) ( ) ( )6

2 13 22 2 2 2 2 6

4 1 1

6 6

R C 22k 5 10C

R 1 R C 100k 1 13.83k 27k 5 10

11.1 10 0.315 10 0.315pF

1 3485875.7

−

−

− −

⎛ ⎞⎛ ⎞ ×⎜ ⎟= =⎜ ⎟+ ω⎝ ⎠ ⎜ ⎟+ × × ×⎝ ⎠

⎛ ⎞= × = × =⎜ ⎟⎝ ⎠+

Parallel Resistance R3 = 122.7k� and Parallel Capacitance C3 = 0.315pF

6. b. For the Wheatstone’s bridge shown in Fig Q6 (b), the galvanometer has a current sensitivity of 12mm/�A. The internal resistance of galvanometer is 200�. Calculate the de� ection of the galvanometer caused due to 5� unbalance in the arm AD. (08 Marks)



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S BA

D R 3 = 10

00�

R4 = 2000�

R 1 = 20

0 �R

2 = 100 �

C

10V

R5 =

200

�

Fig Q6 (b)Ans: Given R1 = 200 �, R2 = 100 �, R3 = 1000 �, R4 = 2005 �, RG = 200 �, V = 10V.

To find the current through the galvanometer, we have to find the Thevenin’s equivalent circuit and the circuit may be modified as shown in Fig (a).

BA

10V

CD

R1 R2

R3R4

Fig (a) Circuit to � nd the Thevenin’s equivalentReferring Fig. (a), Thevenin’s Voltage between the points C and D is

32th

1 2 4 3

3

RRV V

R R R R

100 100010

200 100 2005 1000

5.5 10 V−

⎛ ⎞= −⎜ ⎟+ +⎝ ⎠

⎛ ⎞= −⎜ ⎟+ +⎝ ⎠

= ×

Thevenin’s Equivalent Resistance is

4 3 1 2th

4 3 1 2

R R R RR

R R R R

2005 1000 200 100

2005 1000 200 100

600

⎛ ⎞= +⎜ ⎟+ +⎝ ⎠

× ×⎛ ⎞= −⎜ ⎟+ +⎝ ⎠

= Ω

Referring circuit in Fig. (b), the current through the Galvanometer is3

thG

th G

V 5.5 10I 6.9 A

R R 600 200

−×= = = μ+ +

Deflection = Sensitivity � Galvanometer current



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s� = S � IG = 12 � 6.9 = 82.8 mm

Rth = 600�

Rg = 200� GVRh

5.5mv

Fig. (b) Thevenin's equivalent7. a. Explain in detail the working of resistive position transducer and also calculate the output voltage

when wiper is 10cm from extreme end for the applied voltage of 5V and if the resistive position transducer uses a shaft with a stroke of 50cm. The total resistance of the potentiometer is 5k�. (10 Marks)

Ans: it is a common requirement in industrial measurement is to able to sense the position of an object or the distance it has moved, which is possible by the use of a displacement transducer. A resistive position transducer is an electromechanical device containing a resistance element that is contacted by a movable slider and motion of the slider results in a resistance change. Fig. (a) shows the basic construction of this type of transducer and fig. (b) be illustrates the working principle or measurement technique.

Wiper

Conducting strip

Shift

B A W

Vo

W

A

B

Vt

R1

R2

Fig.(a) Resistive Position Transducer Construction Fig (b) Measurement techniqueIt has a resistive element with a sliding contact or wiper linked to the object being monitored or measured. It can be seen that the resistance between the slider and one end of the resistance element depends on the position of the object. The output voltage depends on the wiper position that is applied to a voltmeter calibrated in centimeters for visual display.Referring fig. (b), it may be seen that the output voltage is a certain fraction of the applied voltage depending on the position of the wiper.

o 2

t 1 2

v RThat is,

v R R=

+

Given size of the shaft stroke = 50 cm Position of the wiper = 10 cmTotal resistance of the Pot = 5 k� Applied voltage Vt = 5 VReferring Fig. (c) and given data,



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wiper Vo = ?

A

B

VtR1

R2

Fig (c)

2

o 2

t 1 2

10R 5000 1000 1k

50

V R

V R R

= × = Ω = Ω

=+

Therefore,

2o t

1 2

o

RV V

R R

1000V 5 1V

5000

= ×+

= × =

7. b. Explain the construction, principle and operation of LVDT. (10 Marks)Ans: LVDT

Figure shows the construction of the linear variable differential transformer (LVDT). The differential transformer consists of single primary winding and two secondary windings wound on a hallow cylindrical former. The two secondary windings have equal number of turns and are placed on both sides of the primary winding. The primary winding is excited by an ac source.

A movable soft iron core slides in & out the hallow former effecting the magnetic coupling between primary and two secondary windings.



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s When the core is at the normal position (exactly at the middle of the former) the secondary voltages induced are equal and hence the output voltage is the difference of these two voltages, Vo = E1 – E2 = 0 .

0V 0V= When the core is moved to the bottom more � ux links ‘S1’ than ‘S2’the output voltage is E1, the output

voltage 0 1 2V E E= − . when the moved to bottom most, the output voltage is very negligible (almost zero).

2 0 1E 0 so V E= → ≈

Similarly, when the core is moved in the opposite position Vo = E2 =E1. At the extreme end, E1 = 0

0 2V E=

Thus, we find that, as the position of the core changes within the former, the voltages induced in the individual secondary coils differ; this produces an output voltage that is linearly proportional to the position of the core; hence the name linear variable differential transformer.The transfer curve of the LVDT is shown in Fig. 7 (c). The transfer characteristic shows a fairly linear operation of the LVDT.

V0 = V

02 – V

01

–Xmax

+Xmax

V0 = V

01 – V

02

Position of the core (–X)

Position of the core (+X)

Fig. 7. c Transfer characteristic of LVDT

8. a. Write a short note on signal conditioning system. (06 Marks)Ans: Signal conditioning circuits Signal - conditioner: Signal conditioning equipment is required to perform basic linear processes such

as attenuation, ampli� cation and mathematical operations like integration and differentiation. They can also be used to perform non-linear process such as � ltering, multiplication, sampling and modulation/demodulation. These functions require proper selection of components and faithful reproduction of the

required � nal output. The signal conditioning / data acquisition system is an excitation system/ampli� cation system for passive transducers, or it can be an ampli� cation system for active transducer.

Simpli� er

Fig. DC signal conditioning system For a passive transducer excitation is essentially required. The passive transducer like strain gauge

thermometers, capacitive transducers and potentiometers to be excited from an external dc source. The



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active transducers like thermocouples, photo diode piezoelectric transducers does not require any external dc source for excitation. But these transducers are connected to ampli� ers to amplify the low level input signal.

A simple dc signal conditioning system is as shown in � gure. The dc bridge is excited by a dc source and stain gauge is connected to one arm of the wheastone’s - bridge. The bridge can be basically balanced using a potentiometer and output of calibration is applied to dc ampli� er the essential requirements of dc-ampli� er are as follows. 1. The input differential ampli� er should have a high CMRR. 2. It should have a large long term stability 3. If should be easy to calibrate The main draw back in using dc amplifier is the problem of drift. Hence at low frequency this spurious signal appear at the output of conditioning system as data. In order to avoid this spurious signal, LPF is used at the output of dc amplifier. In order to overcome the problem of drift an AC signal conditioning system is used..

8. b. Explain the working of Piezo Electric Transducer with circuit diagram. (10 Marks)Ans: Piezo means press. Then piezo electricity means press and generate electricity. It is found that when certain

crystaline materials such as quartz, Rochelle salt and Barium titanate placed under stress, they generated electricity. This property is used in the construction of piezoelectric transducer, where a crystal is placed between a solid base and the force summing member as shown in Fig 8 (a).An externally applied force entering the transducer through its pressure part applies pressure to the top of the crystal, which produces an emf across the crystal proportional to the magnitude of the external pressure. Equivalent electrical circuit of the crystal is also shown in Fig. 8 (a). The basic expression for the generated emf is given by

Crystal (Barium Titanate Rochelle Salt Quartz)

Base

output

Fig. 8.a Peizo Electrical Transducer

C1

R1

L1

Cp

Force - summing

Equivalent Circuit of Crystal

P

= QE

C

where E = generated electricity Q = generated charge CP = Shunt capacitanceFor a pizeo electric element under pressure, part of energy is converted to an electrical potential that appears on opposite faces of the element, analogous to a charge on the plates of a capacitor. The rest of the applied energy is converted to mechanical energy, analogues to compressions of a spring. The piezo electric element returns to its original shape and losses its charge, when the external pressure is removed. From these relationship, it is derived that



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Mechanical energy converted into electrical energy =

Applied mechanical energy

or

Electrical energy converted into Mechanical energy =

Applied electrical energy

K

K

The piezo electric effect is reversible. That means, these crystals exhibits the property of electrostriction when an electric field is applied across the crystal, mechanical stress is produced. An alternating voltage is applied to the crystal causes it to vibrate at its natural frequency. This property is used in high frequency accelerometers.Disadvantages:

1. Can not measure static conditions since voltage is generated only when external pressure is applied.

2. Output is affected by temperature changes.

8. c. Compare LED displays and LCD displays. (04 Marks)Ans: Comparison between LED displays and LCD displays.

LED LCD1. Small and compact 1. Medium and compact2. Small size 2. Small size3. Bright displayed 3. Good contrast 4. Low power 4. Low- power5. high speed 5. Medium speed


6-electronic instrumentation dec.013-jan.014 · 1.c. explain in detail the working of true rms...

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