Third Semester B.E. Degree Examination Electronic .... Operation of electronic switch in oscilloscope: The electronic switch is used in cathode ray oscilloscope to display two channels

Third 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: (06 Marks) (i) Gross error and systematic error (ii) Absolute error and relative error

Ans. i) Gross and Systematic errors: The gross errors occur mainly due to the carelessness/lack of experience of a human being involved in the measurement of any quantity. These errors are due to incorrect adjustment of measuring instruments mathematically and these errors cannot be recti� ed completely

However, Gross errors can be controlled by taking proper care while reading, calculating and recording a quantity under measurement; at least 3 to 4 repeated measurements of the same quantity should be taken.

The systematic errors occur mainly due to the defective parts and ageing of the meter and also environmental effects.

Systematic errors includes the friction in bearings, moving parts, variation in air gap, loading effects of the measuring instrument.

Another reason for systematic errors in an instrument is the environmental factors which are external to the instrument; they are temperature changes, thermal E.M.F, stray capacitance, etc.

These errors can be reduced by hermetically sealing the instrument and using a magnetic electrostatic shields.

(ii) Absolute errors and relative errors.

Absolute error may be de� ned as the difference between the true value and the measured value. Absolute er-ror represent the amount of physical error in a measurement. As an example, consider the true value of voltage measurement is, say, 5V and the measured value is 5.1 V. Then the absolute error is 5-5.1 V = –0.1V.

Relative error is de� ned as the ratio of the absolute error produced in a given measurement to the measured value. That is

Absolute errorRelative error

Measured value=

Considering the above example

0.1relative error 0.0196 1.96%

5.1= = =

(b) Explain the working of true RMS voltmeter, with a neat block diagram. (08 Marks) Ans. True RMS voltmeter: Any complex waveforms can be accurately measured using true rms - voltmeter, these

instruments sense the signal heating power and produces meter indication proportional to square or rms value of voltage. This heating power is ampli� ed and fed to thermocouple and then the measured output voltage is proportional to square of rms - value.

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Electronic Instrumentation June 2012 June 12 - ��

To measure true rms voltage two thermocouples connected forming a bridge network is used as shown in � gure. The input voltage to be measured is applied to the heater element of the thermocouple. The heating effect of heater is measured by thermocouple and generates corresponding voltage called ‘V1’.

The input voltage is ampli� ed and then given to the heater element of measuring thermocouple to produce enough heating so that 1 2V V= .

The two thermocouples (balancing and measuring) form a bridge network and when 1 2V V= the bridge is

balanced, output voltage ( )0 1 2V A V V= − sine ‘A’ is gain of dc ampli� er (1) 1 2V V= .

V1 = K × V2irms, Virms = rms value of the input

22 0V KV= , 1 2V V=

= K.V2irms = V0

2

0 irmsV V=

(c) Convert a basic D’ Arsonal movement with an internal resistance of 100� and a full scale de� ection of 10 mA into a multi range dc voltmeter with ranges from 0 – 5 V, 0 – 50V and 0-100V. (06 Marks)

10 mA

R =100M �

+

–

+

–

R4 R3 R2 R1

V1

V2 V3

V4

Ans

(i) V3 = (R3 + Rm) I

5V = (R3 + 100�) × 10 × 10–3


June 2012 Electronic InstrumentationJune 12 - ��

3

5V

10 10−× = R3 + 100

500� = R3 + 100�

R3 = 500� – 100�

3R 400= Ω

(ii) V2 = I (R2 + R3 + Rm)

50 V = 100mA (R2 + 400� + 100�)

50 V = 10 mA (R2 + 500�)

50V

100mA = R2 + 500�

5000� = R2 + 500�

2R 4500= Ω

(iii) Vl = I (R1 + R2 + R3 + Rm)

100V = 10 mA (R1 + 4500� + 400� + 100�)

100V

10mA = R1 + 5000�

10000� – 5000�

1R 5000= Ω

2. (a) A 4½ digit voltmeter is used for voltage measurement (07 Marks)

(i) Find its resolution

(ii) How would 12.98 V be displayed on 10 V range?

(iii) How would 0.6973 V be displayed 1 V and 10 V range?

Ans. Given data 4½ digit display

(i) Full digit display N = 4

Resolution = 4

10.0001

10=

(ii) There are 5 digit places in a 4½ digit display.

To display 12.98V on a 10V range it displays as 12.980.

(iii) In 1V range, 0.5573V will be displayed as 0.5573, since the resolution in 1V range is 1V × 0.0001 = 0.0001V

In 10V range the resolution is 10V × 0.0001 = 0.001V

Therefore 0.557V will be displayed instead of 0.5573V. .

(b) Explain the working principle of successive approximation digital voltmeter, with the help of block diagram. (07 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.

Sample and Hold Clock

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.

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.

(c) With a basic block diagram, explain the method used for digital measurement of time period. (6 Marks)

Ans. To get good accuracy for low frequency measurement, it is a practice to measure the time period rather than make direct frequency measurement. The � gure below shows a simple block diagram of time measurement.

AmplifierSchmitttrigger

Amplifier

AttenuatorInputsignal1

Flip Flop

Inputsignal2

AttenuatorSchmitttrigger

CrystallOscillator

Schmitttrigger

100÷

Display

decade counting display

The gating signal is derived from unknown input signal,that controls opening and closing of � ip-� op. The



number of pulses which occur during one period of unknown signal are counted and displayed by the decade counting assemblies. The displayed time value is taken to calculate the unknown frequency using the formula,

1F

T=

3. (a) Explain the working of dual trace oscilloscope, with a neat block diagram and necessary waveforms.

(10 Marks)

Ans. Dual-trace oscilloscopes are those that can perform (more or less) the same operations as that of a dual-beam oscilloscope, yet use only one electron gun and one set of vertical and horizontal de� ection plates for their operations. Figure shows the simpli� ed block diagram of a dual-trace oscilloscope.

Y1 ampli� er

Ampli� er Y-plate

X-plate

Beam

Y1 input

Y2 input

Electronic Switch

1

2Y2 ampli� er

Simpli� ed block diagram of a dual-trace CRO

In this scheme, the electron beam is switched between positions 1 and 2 of an electronic switch at a very rapid rate. The electronic switch is connected to the Y1 input, when it is in position 1 and to the Y2 input, when it is in position 2. Thus the switch connects inputs 1 and 2 to the vertical de� ection plates so that the same beam will be subjected to the de� ection by two different input voltages in an alternating fashion as shown below. This switching produces the waveforms corresponding to inputs 1 and 2, respectively. Since the switching is done at a fast rate, we observe that the two waveforms appear to be stationary and independent.



(b) With the help of basic block diagram and circuit diagram, explain the working principle of electronic switch. (08 Marks)

Ans. Operation of electronic switch in oscilloscope: 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 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

(c) Brie� y explain about the focus control knob available on the CRO panel. (02 Marks)

Ans. The focus knob actually controls the grid voltage of CRT, which in turn control the � ow of electrons passing through the focusing anodes. Focusing anodes works like a lens and focus the beam to a � ne point on the screen. This helps the viewer to observe a clear wave form on the screen.

4. (a) Describe the working of oscilloscope delayed time base system, with the help of block diagram and associated waveforms. (10 Marks)

Ans.

CRT

Delayed Time base



The delayed time base system is as shown in � g. (a). This feature increases the versatility of the instrument by making it possible only to magnify the required portion of an undelayed sweep, measure the rise time/jitter and many more applications.

When the delay time in CRO is not used then the initial part of the signal is lost in the display. In order to counter act this disadvantage the delay line of 2000 sec is used and the signal is indirectly applied to vertical depletion plates as shown in � g. (a).

This gives � nite time for the sweep generator to start at the Horizontal plates before any input signal is applied to CRO, and the signal has reached the vertical plates. The Trigger pulse has off time of t0 sec after which the input signal from input has passed through main ampli� er.

The sweep generator generally delay the Horizontal sweep to Horizontal Ampli� er only after t0 + 90 n sec so that the Horizontal sweep start earlier to the vertical signal. The vertical sweep starts at t0 +200 n sec and by this approach the input signal is not lost.

(b) Explain the basic operation of digital storage oscilloscope with the help of block schematic and associated waveforms. (10 Marks)

Ans.

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.



InputVerticalamplifier

S/Hcircuit

Triggercircuit

Controllogic D/A

converter

Horizontaldeflectionamplifier

Verticaldeflectionamplifier

A/Dconverter

D/Aconverter

MemoryInputsignal

CRT deflection plants

Fig. 4.3 Block diagram of DSO

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.

PART - B

5. (a) With the help of block diagram, explain the working of modren laboratory signal generator. (10 Marks)

Ans.

The above � gure shown is a +ve feed back system. It basically consists of ampli� er with gain (A) and feedback n/w whose feedback fraction is ‘ ’.

When the product of the loop gain ‘A�’ = 1 and total phase shift around the loop is 360° sustained oscillations are generated called barkhausen criteria.



Block diagram of standard signal generator

Fig. 4(a)

Signal generators are extensively used in testing of radio receivers and transmitters. Signal generator is basically a radio frequency signal generator generating sinusoidal signal whose frequency is greater than 1MHz. The block diagram is shown in � gure 4(a). The RF oscillator is used to generate oscillation depending on range and frequency selected. It is an LC-tank-circuit oscillator generating a stable RF signal and fed to wide - band ampli� er. The modulation may be done by a sine wave, square wave or triangular wave using the setting ‘ON’ the front - panel indicating carrier frequency for frequency modulation.

The output of the wide band ampli� er is a modulated signal where the output is fed to attenuator which helps in selecting required attenuation and output signal level is controlled for high frequency modulation we require isolate oscillator from output circuit by using buffer ampli� er.

(b) Explain the working principle of frequency synthesizer, with a neat block diagram. (10 Marks)

Ans: Figure 5.e 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 f

i 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 f

0. 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 f

i and f

0.

Phasedetector

Low – pass filter

VCO

Divide – by N

Nf0

fi

f0

Figure 5.e 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 Nf

0. This is because, with output frequency equal to Nf

0 the input frequency becomes Nf

0/N = f

0.

This principle is used in frequency synthesizer circuits lo produce frequencies of all values and ranges.

The theory of the frequency multiplier using PLL may be extended to synthesize (artificially produce) oscillations in any desire frequency range. Figure 5.f shows a typical PLL frequency synthesizer. It consists of a crystal oscillator that produces a fixed frequency of, say, I 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 f

0, which

is obtained by frequency division, as shown in Fig. 5.e.

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. By using several crystal oscillators of different frequencies, and several PLL units, we can produce frequencies in all ranges, values and amplitudes.

CrystalOsc

Frequencycounter

Phasedetector

Low – passfilter

VCO

Divide – by N

Nf0

fi

f0

Figure 5.f Phase–locked loop frequency synthesizer

6. (a) Mention the limitations of wheatstone’s bridge. Derive the balance equation for Kelvin’s double bridge. (10 Marks)

Ans. The limitations of Wheatstone’s bridge are :

1. The resistance of the leads and contacts becomes signi� cant in the low resistance measurement that introduces error.

2. In the measurement of high resistive values, the resistance presented by the Wheatsotone’s bridge becomes so large that the galvanometer will be insensitive due to imbalance.

3. The heating effect of the current that rise the temperature which in turn causes a change in the value of resistance in bridge arms. Excessive current cause a permanent change in the value of resistance that affects the measurement.

Kelvin’s double bridge is a modi� cation of Wheatstone’s bridge and provides increased accuracy in the measurement of low value resistance typically below 1. The term double bridge is used because the circuit contains a second set of ratio arms as shown in Fig. 6 (a). This second sets of arms, labeled RA and RB in the diagram, connects the galvanometer to a point P at the appropriate potential between M and N, and eliminates the effect of the yoke resistance RY. The initial condition is that the resistance ratio of RA and RB is the same as the ratio of R1 and R2. Unbalanced wheatstone’s bridge

L

M N

O

K

E

RB RARXR3

R1R2

RY

Fig. Kelvin’s double bridge



The galvanometer indicates zero, when the potential at point K equals the potential at P. That is EKL = ELMP, where

( )( )

( )( )( )

2 23

1 2 1 2

3

A B YKL x

A B Y

A B YBLMP

A B A B Y

R R RR RE E I R R

R R R R R R R

and

R R RRE I R

R R R R R

⎡ ⎤+= = + +⎢ ⎥+ + + +⎣ ⎦

⎡ ⎤⎧ ⎫+⎪ ⎪= +⎢ ⎥⎨ ⎬+ + +⎪ ⎪⎢ ⎥⎩ ⎭⎣ ⎦

Equating EKL

and ELMP

, we get

( )( ) ( )

( )( )

23 3

1 2

A B Y A B XBX

A B Y A B A B Y

R R R R R RR RI R R I R

R R R R R R R R R R

⎡ ⎤⎡ ⎤ ⎧ ⎫+ +⎪ ⎪+ + = +⎢ ⎥⎨ ⎬⎢ ⎥+ + + + + +⎪ ⎪⎢ ⎥⎣ ⎦ ⎩ ⎭⎣ ⎦

Simplifying we get

( )( ) ( )

23 3

1 2

A B Y B YX

A B Y A B Y

R R R R R RR R R

R R R R R R R R

⎡ ⎤++ + = +⎢ ⎥+ + + + +⎣ ⎦

Expanding the RHS term results

( )( ) ( )

1 3 1 23 3

2 2

A B Y B Yx

A B Y A B Y

R R R R R R R R RR R R

R R R R R R R R

++ + = + + ×

+ + + +

Solving for RX, we get

( ) ( )( )( )

1 3 1

2 2

A B YB Y B Yx

A B Y A B Y A B Y

R R RR R R R R R RR

R R R R R R R R R R R

+= + × + −

+ + + + + +

Therefore,

( )1 3 1

2 2

B Y Ax

A B Y B

R R R R R RR

R R R R R R

⎛ ⎞= + −⎜ ⎟+ + ⎝ ⎠

Using the assumed initial condition, 1

2

,A

B

R R

R R= we see that the above equation reduces to

13

2x

RR R

R=

(b) A capacitance comparison bridge is used to measure a capacitive impedance at a frequency of 2 kHz. The bridge constants at balance are C3 = 100 �F. R1 = 10 K�. R2 = 50k� and R3 = 100K�. Find the equivalent circuit of the unknown impedance. (04 Marks)

Ans. Rx = 2 3

1

R R 100k 50k

R 10k

Ω× Ω=Ω

R 500kx = Ω

Cx = C3

61

2

R 10k100 10

R 50k−= × ×



C 20 fx = μ

The equivalent circuit is capacitor 20�f in series with 500k�

(c) Derive an expression for frequency of the wein bridge circuit. (06 Marks)

Ans. Wein bridge in its basic form is designed to measure frequency but also can be used to measure the unknown capacitance. Figure shown a wein bridge, which is combination of a series RC in one arm and parallel RC in another arm.

Detector

R1

R2

R3 R

4

C1

C3

From figure, impedance of one arm is

1 11

= f

Z RC

−ω

The admittance of the parallel arm is

3 33

1Y = + ωj C

R

Using bridge equation,

Z1Z

4 = Z

2 Z

3

21 4 2 1 4 3

3

i.e, = and =Z

Z Z Z Z Z YY

2 4 1 31 3

3 41 4 42 3 1 4

3 1 3 1

1⎡ ⎤⎡ ⎤∴ = − + ω⎢ ⎥⎢ ⎥ω⎣ ⎦ ⎣ ⎦

= − + ω +ω

jR R R j C

C R

C RR R jRR j C R R

R C R C

3 41 4 43 1 4

3 1 1 3

j⎡ ⎤ ⎡ ⎤

= + − − ω⎢ ⎥ ⎢ ⎥ω⎣ ⎦ ⎣ ⎦

C RR R RC R R

R C C R

Equality the real and imaginary parts, we have

1 4 1 4 42 3 1 4

3 1 1 3

32 13 1

4 3 1 1 3

2

1 1 3 3

and 0R

1Therefore, and

1

= + − ω =ω

= + = ωω

∴ω =

R R C R RR C R R

R C C

CR RC R

R R C C R

C R R C



1 1 3 3

1 1 3 3

1or

Since 2 , weget

1

2

ω =

ω= π

=π

C R R C

f

fC R C R

In most Wein bridge circuit, the components are chosen such that R1 = R

2 = R and C

1 = C

2 = C, then the above

equation becomes.

1

2=

πf

RC

which is the general equation for frequency of the bridge circuit. 7. (a) Explain the construction and working of bonded resistance wire strain gauge and semiconductor strain

gauge. (10 Marks)

Ans. Bonded resistance wire strain gauge:

Dir

ectio

n of

st

rain

Leads

Fine wire

A � ne metallic wire element about 25�m in diameter when looped fourth on back carriers base, and cemented to the external member under going stress is called bonded wire strain gauge is as shown in the � gure.

The cemented carrier is a thin sheet of backlight the metallic wire is wound back and fourth on cemented base so that it is not mechanically damaged.

The metallic wire spreading should be uniform to as so permit uniform distribution of stress, the cemented base permits good transfer of strain from carrier to metallic wire, the tensile stress permits the wire to elongate and decrease in C/S area and increasing wire resistance.

Rρ= l

a

The strain gauge is connected to one arm of the bridge and bridge is activated. The other arms of the bridge comprises of a constant resistance. When stress is applied to the strain gauge the resistance increases the bridge is brought to unbalanced condition by measuring the current through the galvanometer, which is proportional to the strain on the strain gauge.

Semi conductor strain gauge:



When very high gauge factor is required of the order 50 then we use semiconductor stain gauges. The semi conductor gauge consists of Base material like silicon or germanium and gold leads are used to make contacts with base material and two electrodes are taken out from the base semi conductor strain gauge depends upon the piezo resistance that is the change value due to change in resistants of semiconductor in these gauges can be cascaded with op-amp which can act as pressure sensor but these gauges are very sensitive to change in temperature.

(b) With necessary sketches, explain the construction and working principle 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.

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) Mention the advantages and limitations of RTD. (04 Marks)

Ans. Advantages :

1. Linearity over a wide range

2. Operation at high temperature is possible

3. Wide operating range

4. Better stability at high temperature

Disadvantages:

1. Sensitivity is less

2. Affected by contact resistance, vibration and shock

3. Comparatively expensive with other temperature transducers

4. Requires 4-wire operation to eliminate errors due to lead resistance.

(b) Define the terms: (i) Seebeck effect. (ii) Peltier effect. (04 Marks)

Ans. Seebeck effect: A thermocouple consists of a pair of dissimilar metals wires joined together at one end, called sensing or hot end and terminated at another end, which is called reference or cold end. When a temperature difference exists between the hot and cold junctions, an emf is produced causing a current in the circuit. This thermoelectric effect is known as the Seebeck effect, named after the German Physicist Thomas seebeck. Thermocouple works on the principle of Seebeck effect.

Peltier effect: Even at the same temperature, conductors made up of different materials will have different densities of free-carriers. When two dissimilar conductors are joined together, electrons will diffuse across the junction from the conductor that has higher electron density. And the conductor, which loses electrons, will acquire positive potential with respect to other conductor. This phenomena is called Peltier effect.

(c) Explain how bolometer bridge can be used for the measurement of power. Also discuss the application of unbalanced bolometer bridge. (04 Marks)

Ans. 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 � g. 8(c).



R1

R1

R1

low RF supplyV

R

Edc Bias

/4 stub for Bolometer

Ground return

Tapered Section for Impedance Matching

Bolometer ElementRF Bypass Capacitor

RF Input

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 R

1, 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 V

2.

The unknown RF power is calculated using the relation,2 2

2 1

1

RF power ,4

−=

V V

R

since the power delivered to the bolometer element is 1

4 of the power fed to the bridge.

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. 8(c).

Schematic diagram of an unbalanced Bolometer Bridge is shown in Fig. 8 (d).

When there is no RF power input, the bridge is brought in to balance by adjusting the exciting source voltage and the balanced condition of zero will be indicated by the detector. The RF test power is applied now to the bolometer element and the resistance of the bolometer element changes. This results in to the unbalancing of the bridge, the amount of which is indicated by the detector and gives the magnitude of the RF power directly.



Detector

Zero Adjust

ac Exciting source

RF power (Bolometer Element)

R1 R1

R1

Fig 8 (d) Unbalanced bolometer bridge

The unbalanced method is the simplest means of measuring low RF power by realizing a direct reading bridge. However there is one disadvantage that the bolometer impedance getting changed since the resistance of the bolometer is a function of RF power level. This upset the impedance matching of the RF system.

(d) List the important features of LCD. (04 Marks)

Ans.

1. The electric � eld required to active LCDs is typically of the order of 104V/cm

2. NLC materials possess high sensitivity of the order of 1010� and so the current required for scattering hight in an NLC is very small in the range of 0.1 �A/cm2

3. Light source for re� ective LCD is only the ambient light and so the power requirement is only to cause tarbulence in the cell, that is very small, typically 1 �W/cm.

4. They are very slow devices; The turn On and OFF time are typically in the range of a few milli seconds and tens of milliseconds respectively.


Documents

Third Semester B.E. Degree Examination Electronic .... Operation of electronic switch in oscilloscope: The electronic switch is used in cathode ray oscilloscope to display two channels