2. 9.2 BASIC ELECTRICAL AND ELECTRONICS ENGINEERIM
9.1 Introduction to Transducers
nich
Most of the instruments present in our day-to-day life deals with physical quantities uhii
any
are non-electrical quantity. These quantities can not be recorded and measured with an.
electric or clectronic measurement devices. It is required to convert non-clectrical
1antity into electrical quantity so that it can be measurcd, recordcd and analysed by the
instruments.Transducers are the devices whichconvert one form ofenergy into another
form of energy. The device which is used for converting the non-electrical quantity into
clcctrical quantity is called electrical transducer
e
When the transducer converts one form of mechanical quantity into another form of
mechanical quantity,it iscalled as mechanicaltransducer. For example, Bourdon tube
converts pressure into displacement In electrical transducers, the input quantity canhe
force.displacement, pressure, level, fow, stress, temperature etc. and the output electrical
quantity may be current, voltage or frequency
9.2 Classification of Transducers B
The electrical transducers are classified according to their application area,outputsignal
nature, structures, method of energy conservation etc. The various classification of
transducers are described in this section.
9.2.1 On the Basis of Transduction Form Used
According to the principle used in transduction, transducers are classified as following
(a) Capacitive Transducers: The transducer is called capacitive transducer, if the
output of transduceris obtained as
change into the capacitance. In a
capacitor, two
conductors are separated by an insulator called dielectric. So, to bring the change in
the capacitor either the distancé between two plates or the dielectric is changed
(b) Resistive Transducers: The transducer is called resistive transducer, if the outpu
of a transducer changes with the resistance of transducer. Examples of resisuv
transducers are
potentiometer, photoconductive cell, strain gauge,
thermometer etc. resistance
(c) Inductive Transducers: The transducer is called inductive transducer, if theoup
tput
of a transducer changes with the inductance of transducer. Examples of inducu
tive
transducers are differential transformer, magnet restriction gauge, reluctanco p
etc ick-up
(d) Voltage and Current
Generating Transducers: The transducers which pro
ce
their output in
voltage or current are called voltage or current generating traisu
3. INSTRUMENTATION 9.3
e Examples for voltage and current generating transducers are
piezoclectric, clectron-
tube, photocmissive cell, ionization chamber etc.
9.2.2 U
9.2.2 Primary and Secondary Transducers
Primary transducers are used to convert non-electrical input quantity into intermediate
quantity. This intermediate quantity is then converted into the final electricat quantity by
thesecondary transducers.)To understand both the transducers, let us take an example
ofthe combination ofthe Bourdon tube and LVDT. Figure 9.1 shows the arrangement
where bourdon tube acts as a primary transducer and LVDT acts as a secondary
transducer.
Pulley
Bourdon
Output
0 voltage
tube
LVDT
n m T Fixed
end
Supts Pressure
Figure 9.1 Example of primary and secondary transducer
Bourdon tube senses the pressure and converts the pressure into the displacement of its
free end. Now, linear variable differential transformer (LVDT) acting as a secondary
transducer has its core moved by the displacement of the free end which produces an
output voltage. This output voltage is proportional to the movement of the core which is
proportional to the displacement of the free end, which in turn is proportional to the
vnes pressure.
Hence, there are two stages of transduction, firstly the pressure is converted into
displacement by Bourdon tube which acts as primary transducer then the displacement
is converted into voltage which acts as secondary transduce
9.2.3 Passive and Active Transducers
These type of transducers are classificd according to whether they are passive or active.
stab (a) Passive Transducers: Passive transducers are those transducers which do not
generate any clectrical signal by themselves and derive the power required for
transduction from an auxiliary power source. They are also termed as "extremely
4. BASIC ELECTRICAL AND ELECTRONICS ENGINEEP
RING
s
9.4
LVDTs, variable capacitive transducers ctc. as they cannot transduce witho
presence of the auxiliary power source.
powered transducers". Examples of passive transducers are potentiometers, RVDT
hout
(b) Active Transducers: Active transducers are those transducers which do not require
the auxiliary power source to produce their output and are sclt
generating tyDe of
transducers. These transducers do not require any external source power for thei.
operation since they develop their own voltage or current output. Examples of active
transducers are thermocouple, photovoltaic cells, Tachogenerators, accelerometere
piezoelectric crystals etc.
9.2.4 On the Basis of Nature of Output
Transducers can be classified on the basis of the output of the transducers which can be
continuous function of time or in discrete steps
(a) Analog Transducers: These type of transducers convert the input physical quantity
into an analog output_which is a continuous function of time. Examples of analog
transducers are LVDT, thermistor, strain gauge, thermocouple etc. which produce
continuous output.
fb) Digital Transducers: Thesetypeoftransducers convertthe input physical quantity
into digital output which is in the form ofpulses. For the cach value sensed, a unique
code is generated at the output.
9.2.5 Onthe Basisof Type of Output
Transducers can be classificd on the basis of type of the output produced which can be
electrical or non-electricalquantity
(a) Transducers: As discussed earlier, transducers are the devices which convert a non-
electrical quantity into an electrical quantity.
(b) Inverse Transducers: Inverse transducers are those transducers which convert
electrical quantity into non-electrical quantity. It is used at the output side and hence
called output transducers Apiezoelectriccrystal is an inverse transducer becaus
when a voltage is applied across itssurface,it produces mechanical displacementdu
tochange in its dimensions. Similarly, PMMC (Permanent Magnet Moving Coll)
inversetransducer as the current carried by its coil is converted into a force WIu
ch
causes translational or rotational motion (displacement). Loudspeaker 1s
aiso
an
inverse transducer which converts electrical signal into sound signal. Many
indicating and recording instruments such as ammeter and voltmeter are
transducers which convert current or voltage into mechanical displacement.
data
verse
following section, some
commonly used transducers are briefly discussed.
5. 9.5
INSTRUMENTATION
9.3 Thermocouple
Thermocouple is a temperature transducer which converts thermal energy into electrical
energy, Generally, it is used as a primary transducer and for the measurement of high
temperatures
9.3.1 Working Principle of Thermocouple
Thermocouple works on the principle of thermoelectric effect. This was first observed
by Thomas Seebeck in 1821 and is known as seebeck effect. It says that if two
dissimilar metals having different work functions are joined together to form a closed
circuit,there will be two junctions where they meet each other. If one of these junction
is heated and other junction is kept at lower temperature,then
across the two junctions. The emf generated is dependent upon the difference of
an emf is generated
temperature of the two junctions.
Voltmeter
Measuring
junction (hot)
Reference
Junction (cold)
o Heat source
Two dissimilar
metal wires
Figure 9.2 Thermocouple
u The emf generated in a thermocouple can be approximately expressed as
E = a(A0) + b(AG9 9.1)
A0 difference in the temperature between the measuring
(hot) junction and reference (cold) junction, °C
where
a, b =
constants
tatese Generally, a 1s
Vo
Generally, a is very large as compared to b and it is usually neglected and hence equation
becomes
ionue
E =a(A6)
wwwwii
6. 9.6 BASIC ELECTRICALAND ELECTRONICS ENGINEERING
a
9.2)
40 cE
Hence, the equation (9.2) shows that generated voltage is proportional to the difference
in temperatures of two junctions.
932 Construction of Thermocouple
Thermocouple ismadeupoftwo wires of dissimilar metals joined together to form two
junctions as shown in figure 9.3(a).
Metal A(+) Hot
Junction
Voltmeter
Metal B(-)
Cold junction
(a) (b)
Figure 9.3 (a) Basic thermocouple circuit and (b) Practical thermocouple circuit
Out of the two junctions Jj and J, J is kept at constant referencetemperatureand
hence called cold junction.Junction Jjis subjected to the temperature changes to be
measured which is called as hot
junction.The two wires of the thermocouple are
generallytwisted and welded together. To measure higher temperature,the wire used
should be heavier. But if the size of the wireincreases, the response time of the
thermocouple increases. So, the size of the wire is increased considering above two
conditions. The thermocouples can be
made from a
number of different metal
combinations like
Platinum-rhodium, copper-constantan, iron-constantan, chromel-
constantan etc
The combination of metals used to form
thermocouple depends upon the
temperatur
range required to
measure which is shown in table 9.1. Platinum is used to form
thermocouples having high sensitivity. Copper-constantan thermocouple gives highest
output with maximum sensitivity. Iron-constantan is most widely usd in industrial
applications which proved to be inexpensive. Tungsten and Molybdenum are used to
higher temperature measurements
AadyuwwM- Phaw
Totm-tonstehka -)0-160
M
Cotc te
7. 9.7
JNSTRUMENTATION
Table 9.1 Thermocouples and temperatur ranges
Material used
Copper-constantan
Temperaturerange
270°C to 400°C
-210°C to1200°C
270°Cto 1370°C
Iron-constantan
Chromel-Alumel
Chromelconstantan -270°C to 1000°C
Platinum-rhodium 0°C to 1400°C
Tungsten-Molybdenum 0°C to 2700°C
Tungsten-Rhenium 0°C to 2600°C
The two metals or wires of thermocouples must be insulated from each other at cold
unction. According to temperature ranges, suitable insulating materials are used like
cotton, glass, ceramics etc. Millivoltmeter is used to read the difference between hot and
cold junctions voltages which is calibrated with temperature reading
9.3.3 S Applications of Thermocouple
Thermocouple can be used for various purposes which are listed below:
(a) They can be applied for temperature measurements at various points.
(b) Theyareparticularly used for measuringthesurfacetemperature. For such purposes,
a special junction is developed consisting of two parallel flat flexible metal strips with
a welded junction at their centre. They can be held pressed against the test surface
by strips across a carrier.
(c) Thermocouples are used for indicating rapid change in temperatures and used for
localizedandaccessible positions because of their small heat capacity and very small
bulk
(d) They can be embedded into various parts of a device or machine to measure local
temperature (heating) of different part of the machines. For example, winding
temperature ofmachine
34 Advantages and Disadvantages of Thermocouple
The advantages of thermocouple are following
a Thermocouple is rugged in construction.
t0) It is cheaper and simpler to use than resistance thermometer
(C) It can be used for wide temperature range from 250C to 2700°C.
aThey are suitable for recording rapid changes in temperature as they follow
temperature changes with a small time lag.
8. 9.8 BASIC ELECTRICAL AND ELECTRONICS ENGINEERING
(e) Its calibration can be casily checked.
)They have high response speed compared to filled system thermometers.
(g They have good measurement accuracy and reproducibility,
(h) Long transmission distance for temperature measurement are possible using
extension leads and compensating cables.
The disadvantages of thermocouple are following:
a)They cannot be used in temperature spans of less than about 33°C as there is
relativelysmallchangein junction voltages with temperature
t5) Stray voltage pickups is possible.
(c) The temperature voltage relationship is non-linear.
(d In many applications, amplifiers are required.
e )
For any deviation, it is required to hold cold junction temperature constant or
compensate it.
(For control applications, they require expensive accessories.
Example 9.1 A thermocouple is made of iron and constantan. Find the emf developed
C difference of temperature between the junctions. Given that the thermoelectric emf
of iron and constantan against platinum are +18 and 42 uvPC difference of temperature.
Solution: Emf of iron w.r.t. platinum
EF18 x 10 VPC
Emf of constantan w.rt. platinum
EC 42 x
10 VPC
Emf of platinum w.r.t. constantan
Ep"-EC42 x 106 VPC
Emf of iron-constantan junction per C difference of temperatures
EpeEfe+EP (by law of intermediate metal)
18 x 10 +42 x 10
60 VC Ans.
Example 9.2 A copper constantan thermocouple has a = 39.5 VC and = 0.005
VPC. Determine the emf developed by the thermocouple when its hot junction isat
200°C and cold junction is kept in ice.
a -
39.5 uVPC = 39.5 x 10 VPC
B 0.005 VPC =0.005x 10 uVrC
Solution: Given
9. INSTRUMENTATION 9.9
Temperature of hot junction,
T=
200 + 273=473 K
Temperature of cold junction,
To-0+ 273 273 K
emf developed by the thermocouple
E =a (AT) +B (AT)
39.5 x10 (473 273) +0.005x10 (473 273)
0.0081 V Ans.
9.4 Resistance Temperature Detector (RTD)
Resistance temperature detector is a primary electrical transducer which is also called as
the resistance thermometer and used to measure the change in the temperature. It can
be used for measurement of small temperature difference as well as for wide ranges of
temperature.
9.4.1 Working Principle of RTDD
RTD works on the principle that when the temperature changes the resistance of the
conductor changes. The electrical resistance of the certain metals increases
proportionally to the rise of the temperature and RTD utilises this property for the
measurement of temperature. The variation of resistance R with the temperature T(°K)
can given by the following relationship for most of the metals as
R =
Ro(1 + aT + aT + ... +alT" + 9.3)
where Ro = resistance at temperature T =
0°K
a1, a2, , am constants
Generally, a2, a3, an are small. Therefore, higher order terms can be neglected
without much loss of accuracy. Equation (9.3) can also be expressed approximately as
R=Ro(1+ayA0)
where Ro =
temperature at 6,°C
R temperature at 0°C
aoreSistancetemperature coefficient at 0°C
Hence by measuring the resistance of the known material, the temperature of the wire
can be obtained. Platinum, nickel and copper are the examples of the metals used for
measuring temperature.
Generally, all the metallic conductors have a positive temperature cocfficient so that
10. 9.10 BASIC ELECTRICAL AND ELECTRONICS ENGINEERING
their resistance increases with the increase in temperature. In a temperature sensing
element, it is desired to have large value of a so that small change in temperature brings
a substantial change in the resistance of element. It helps in measuring small changes in
Temperature The change in resistance (AR) can be measured with the help of
whcatstone bridge. The temperature sensing clement (X) is placed in the medium whose
temperature is to be measured and connected to the leads of Wheatstone bridge as
shown in figure 9.4. The Wheatstone bridge consists four arms in which A, B and C are
constant known resistances whose values do not change with temperature and fourth arm
1s
Sensing element X having high temperature coefficient of resistance. Now in balanced
condition, no current flows through galvanometer and the ratio of resistance is given by
X (9.4)
B
When the temperature of the medium changes the resistance of X changes and bridge
becomes unbalanced which is shown by galvanometer and will give deflection which can
be calibrated with the suitable temperature scale.
B
w
A
Galvanometer
oe
ww
C gt
w
Power
Figure 9.4 RTD bridge circuit
94.2 Construction of RTD
Resistance elements used for sensing the temperature are
generally long, spring like wires
enclosed in a metal sheath. Porcelain insulator is used to surround resistance element to
prevent short circuit between wire and sheath. Connecting leads are attached to each
side of wire. Sheath is used to detect the temperature of the medium in which it is placed
The wire inside the sheath change its resistance according to the temperature of the
sheath. This change in resistance can be directly calibrated to indicate the temperature
of the medium. The construction of the RTD is shown in figure 9.5.
11. INSTRUMENTATION
9.11
Connecting
leads
ebnibootbes d lesiole
ZZZA}ITZZIZIIIZZZIZZIIIITIZILL
-Element dsle pl bs ielaoecus l (
Mounting
threads
Lead support
bolls
-Sheathet
IIZIITIITIITIIIIIZIIZZZZZTL
Figure 9.5 Resistance temperature detector
The sensing element should have the following properties to measure the temperature
effectively:
(a) It should exhibit relatively large change in resistance for a given change in
temperature.
on(b) It should not undergo permanent change with time.
(c) It should have lincar change in resistance with the change in temperature.
(d) It should be small in size so that less heat is required to raise its temperature and this
makes it suitable for measurement of rapid variations in temperature.
The relationship of resistance of most of the metals with temperature is linear initially and
then increases exponentially. Copper, nickel and platinum exhibit good sensitivity and
reproducibility for temperature measurement purposes Copper has the highest
temperature cocficient among these metais with the most linear dependence. It has low
resistivity and requires large size of the resistance element for reasonable sensitivity and
hence it is not used generally. Nickel and nickel alloy wires can be uscd in the
temperature range of 100 to 450 K as it is less expensive and has higher temperature
coctficient. Platinum wire is mostly used in laboratories and industries for the
measurements of high accuracy. It has inherent property of accuracy and reproducibility
which makes it best choice for many applications. It is recommended as an international
12. 9.12 BASIC ELECTRICAL AND ELECTRONICS ENGINEERING
standard for temperature measurement because of its stable and reproducible resistance
to temperature variation.
9.4.3 Applications of RTD B
RTD is used for the following applications
(a) For temperature control in the textile industry, chemical industry and food industry.
(b) For automatic temperature control in ovens, solder pots and testing chambers.
9.4.4 Advantages and Disadvantages of RTD
The advantages of RTD are following
(a) RTD measures the temperature with high accuracy.
(b) They can be easily installed and replaced.
(c) They have good reproducibility and are fast in response.
(d) RTDs can be used for wide temperature range from 200°C to 650c
(e) They are smaller in size and have stability over long periods of timne.
(f) It does not require temperature compensation.
Although, it also has following disadvantages:
(a) They are costly in comparison to others.
(b) They have large bulb size as compared to thermocouple.
(c) External power source is required for the operation of bridge circuit.
(d) It gives inaccurate results due to current flowing through the bridge circuit and hence
heats the resistance element.
9.4.5 Comparison between RTD and Thermocouple
Table 9.2
RTD
S.No. Parameters
1. Measurement Itmeasuresthe temperature lt measures the temperature
Thermocouple
directly. difference between hot andd
coldjunction.
Its accuracy is not affected Its accuracy changes with
byambienttemperature.
Higher
Lower
Slower
AccuracCy
Sensitivity
Range
Response
Installation
ambienttemperature.
Lower
Higher
Faster
Not easier
Easier
13. I N S T R U M E N T A T I O N
9.13
Example 9.3 A platinum resistance thermometer has a resistance of 1402 at 25°C.
Determine its resistance at 75°C. The temperature coefficient of resistance of platinum
at 25° C is 0.004 UPC. In case the resistance of the thermometer is found to be 1502,
determine the temperature.
Solution: Resistance at 25°C,
Ri = 100 Q
Temperature cocfficient of resistance at 25° C.
a 0.004 Q/QPC.
Resistance at 75° C, R2 =
R [1 +a (-0)
R 100 [l +0.004 (75 -
23)1speoo d aob
= 120 2
Let the temperature corresponding to the resistance of 150S2 be 0,° C.
then R3 R [1 +a (3 - 6,)
150 = 100 [1 +0.004 (03 - 25)]
(150 1
100
+25 150° C Ans.
or
0.004
Example 9,4 A copper resistor having resistance of 182 at 20° C is used to indicate the
temperature of bearings of a machine. Determine the limiting value of resistance if the
maximum temperature of the bearing does not to exceed 175°C. The temperature
coefficient of resistance of copper at 20°C is 0.0045 /PC.
Solution: Resistance at 20° C,
R = 182
Temperature coefficient of resistance at 20° C, a =0.0045
Initial temperature 6 = 20°C.
Limiting temperature 6 =
175°C
Limiting resistance R2 =
R1 [l+a (2 -
01)1
=
18 [1 +0,0045 (175 20)]
Ans.
30.33 2
Example 9.5 A thermistor has a temperature cocfficient of resistance of 0.05 over a
temperature range of 25°C to 50°C. Determine the resistance of thermistor at 40°C, if the
resistance of the thermistor at 25°C is 1002.
14. 9.14 BASIC ELECTRICAL AND ELECTRONICS ENGINEERING
Solution: Resistance at 25°C,
R2s 1002
Temperature coefficient of resistance over a temperature range of 25°C to 50°C is
a =0.05
Resistance at 40°C, R4o=R2s (1 +a(2 61)
R4o100 [1 0.05 (40 25)]= 1752
Example 9.6 A thermistor has a resistance of 4,000 2 at 0°C and 8002 at 40°C. The
Ans.
resistance temperature relationship is given by an expression,
R = RoaeAT
determine the constants a and 8. Determine the range of resistance to be measured in
case the temperature rises from 50°C to 100°C.
Solution: At temperature of 0°C:
Absolute temperature, To 273 K
Resistance, Ro=4,0002
4000 a x 4000 e273
)
At temperature of 40°C:
Absolute temperature, T1 273 + 40 313° K
en Resistance, Ri = 8002
slogior@
800= a x 4000 e313
By dividing the equation () by (1i), we get
)
nereitso
4000 ax4000x eB/273
-
800 ax4000x eB313
eB/273
or =
B/313
or e 273313 = 5
s13= 5
or
take antilog
B=3,438
15. uSTRUMENTATION 8 9.15
Putting in cquation (), we get
3438
4000 a x 4000x e 73
3438
1=ae 273
a =3.39 x 10
Absolute temperature at 50°C = 273 + 50 323°K
Resistance at 50°C 3.39 x 10x 4000x e
e438/323
E568.72 Ans.
Res1stance at 100°C =3.39 x 10 x 4000x e35513
Ans.
136.52
Resistance range is from 136.52 to 568.72
9.5 Strain Gauge N
Strain gauge is a passive resistive transducer which converts mechanical displacement
into change of resistance. This type of transducer is used for the measurement of
displacement, stress and strain etc.
9.5.1 Working Principle of Strain Gauge
Strain gauges work on the principle that when a metal conductor is strained (stretched
or compressed) it resistance changes. When a metal conductor is stretched or
compressed its diameter and length change as a result there is change in resistance.
Some metals also exhibit piezo-resistive effect. The piezo-resistive effect is change of
resistivity (p) due to strain. The strain gauges also exhibit piezo-resistive effect.
L+AL
L
-F F D-AD D
F D
Tensile
force
(b) After being strained
(a) Before being strained
Figure 9.6 Change in the dimensions of a strain gauge element when subjected
to a tensile-force
16. 9.16 BASIC ELECIRICAL AND ELECTRONICS ENGINEERING
The deformation of the strain gauge causes a change in resistance of the strain gauge
which is usually converted into voltage signal. One, two or four similar strain gauges
are connected as arms of Wheatstone bridge to measure the change in resistance.
Ihe output of the strain gauge bridge is a measure of the strain senscd by cach strain
gauge 010
Now, let us consider a strain gauge made up of a wire of length L, area A, diameter
D and the resistivity p.
Let the tensile stress S be applied to the wire which causes increase in
length of the
wire and decrease in area of the wire and change in resistivity.
stere
Let AL =
change in length
AA change in arca
AD =
change in diameter
and AR= change in resistance nisie
The resistance of conductor is given as
3narnoosigetb lsoinGiosrn 2ooo
to momo0aso od 1ot bozu p 3rtdaien gvieasq s 21 9gMBg nisie
og alt onieieT 1o 9gnsrdo olm
A 310 BDe bas 229112 nomOoslgail***.
differentiate cquation (9.5) with respect to S, we get anbhoW L.e.e
bodotota)bunite2ro30bao to dR p dL plL dAL dp1o how 20gusg e
10 boroaue 1ooubsoo Ion ds AdS A2ds
301
to oe Dividing equation (9.6) by R. we get
1osfto oviaizo-oorg dR
AdS (bo2esiqmo9 10g9.6)
9gEdo iignol bru 1913/mctb 21 boe21qtt09
vibeveo-osoig tididro oalsaletorm omo2
1dR P dl pl. dA L do ub(o) irieies
RdS AR dS AR dS AR dS
=
..(9.7)
pL
Putting R=
in cquation (9.7), we get
A
slenol
1 dR
P d pl dA L dp
Rds
icsie gnisd votl A
(d)
legnbd 1olsf (s)
boliojdiznode nonuoly sgusg9
RdS
R ds Lds Ads
-+
S 9gngl d.e swgi....(9.8)
AdS p ds
03
1 dR
1d1dAd
17. INSTRUMENTATION
9.17
From the cquation (9.8). it is clear that per unit change in resistance dR IS due to
R
following
nesof ro
d
L loo1.
) per unit change in length
SIST
(i) per unit change in area =
(1)
dp
(1)per unit change in resistivity=
P
Now, Area is given by A= S
(L) where r radius of wire
diameter
.
ey obupo tnoit
2
A= mD/2),
(.0
(9.9)
or
gusgeTe to sostere 9g to 9ulsr oifh 2origoitnoposvods ol
Differentiate equation (9.9) with respect to S, we getnotog3uehog sal
dA (2D) ondde brotrecel cgn...(9.10)
Dividing cquation (9.10) by A, we get
1 dA T D dD
(01 9 A dS 2 A dS
TD dD
2 D ds
1 dA 2 dD
(9.11)
A dS D ds
dA
Putting from cquation (9.11) into (9.8), weget
PuttingAdS
18. 9.18 BASIC ELECTRICAL AND ELECTRONICS ENGINEERN
1 dL 2 dD dp t
1 dR
R dS L dS D dS p ds
9.
Now, Poisson's ratio
dD/D
Lateral strain
dlLIL
Longitudin al strain
dD
.9.13)
F
D
Multiply equation (9.12) by dS, we get
dR d dD dp
L D P 9.14)
R
dD
Putting from equation (9.13) into (9.14), we get
D
dR
d 2a d
RL (9.15)
The above equation gives the value of per unit change in resistance of a strain gauge.
The important performance parameter of a stran gauge is gauge factor.
Gauge Factor: This is defined as the ratio of per unit change in resistance to per unit
OE2 change in length and it is abbreviated by G
dRIR
G dLI L
dR G=Ge
R
....(9.16)
where
& strain=
dl
Dividing cquation (9.15) by We get
dR/R
dLIT +20+p/p
dL/L spemot
19. INSTRUMENTATION 9.19
G=1+2a+plp
...(9.17)
G 20
Resistance
dplp
Resistance
Resistance
change due to
change of length
change due
to change in area change due
to piezores1stive
effect
Equation (9.17) is valid when piezo-resistive effect exists. However, if a strain gauge do
not have piezo-resistive property, the third term of the equation reduces to zero.
Example 9.7 A resistance strain gauge with a gauge factor of 4 is fastened to a steel
member which is subjected to a strain of 2 x 100. If the original value of the resistance
is 1502, calculate the value of change in resistance.
Solution: Given R=1502
AL
strain= =2 x106
L
gauge factor = 4
We know that
AR/R
gauge factor =
AL/L
ARIR
4 AL/L
AR
4
R
AL
AR =4xRx
L
AR=
4x 150x 2 x 100
AR= 1200 Q Ans.
Example 9.8 Calculate the gauge factor if a 1.5 mm diameter conductor and 24 mm long
changes length by 1 mm and diameter by 0.02 mm under a compressive foree of 10
Amicrostrain. Find the new resistance ofthe wire ifthe nominal resistance is 100 Q
20. 9.20 BASIC ELECTRICAL AND ELECTRONICS ENGINEERIM
Solution: Given D= 1.5 mm
AD=-0.02 mm
L 24 mm
AL =1 mm
sulers R= 100 2
abgo
Now, we know that rtoof gtollo ogrrie
Gauge factor, G ae ALIL nof bilsy ai (i9)noiaup
Gauge factor, G=1+2a+P
oote s o1 bonstestato 10t 9 G-1+2 le somsteeot A T.e sfqmez8
Ifwe neglect the Ap then torol biidd on ogoiq odereg-osoiq ovsrd1on
la o botoojdue 2idoirly1odmom
eAD/Dseo1ooulsu seb otsluoleo 2021 2i
orid oihulo
9ons7ez1or to sis ipuigno odi
Now,
ALIL
0.02
5
101961 gusg
24
O= +0.32
So, the gauge factor G-1 +20
G=1 +2 x 0.32
G-1.64
ARIR
Now, finally we know that
GyLIL
AL
AR GT
R
AR RxG, x
24 303 A
AR 100x1.64x g olhotsiole)8.2slqgmex
24 brs itnvd digrel gnste
AR=6.84 Q iA
Ans.
21. lusTRUMENTATION RAR EEn 9.21
Example 9.9 A resistance wire strain gauge uses a soft iron wire of small diameter with
the gauge factor is 2. Neglecting the piczoclectric resistive effect. Calculate the Poisson
O000
ratio.
Solution: Given G 2
We know the equation of gauge factor
91A
O = 1+2o+
4plpo
93101,
E
If piezoelectric effect is neglected then
solbobnod 2 to1o10st9gus8 G-142 eoi sonpteieo A 11.esigmez
M 081to 22311e s otbrajdualou e lost2
d to oulsr ort m
m Gr 2-19 orf ohaliole) mVo 00
eilooteto ioieslo to 2uluborm onfT
Qb 9ortsieies
Poisson ratio0,
2 2 2e011e boiiqqs ordt o
O 0.5 Ans.
Example 9.10 A single strain gauge is bonded to a beam 0.5 m long and has a cross
section arca 6 cm,. Young modulus for steel is 200 GN/m*. The strain gauge has
unstrained resistance of 2502 and gauge factor of 4. When a load is applied, the
resistance of gauge changes by 0.0132. Calculate the change in length of the steel beam
and the amount of force applied to the beam.
AR 0.013 2
R 250 2gos
Solution: Given
L= 0.5 m
d tovig2t9011stete1 9gUsg tn 9gnsi0
G= 4
Young's modulus =
200
ARIR
G LIL
We know that
AL ARIR
LT G
0.013/250
0.5
AL =
4
AL6.5 x 10m Ans.
22. 9.22
EERING
BASIC ELECTRICAL AND ELECTRONICS ENGINEEPI
Now Stress Young's modulus Strain
200x10 x6.5 x 10
0.5
=2.6 x 10 N/m o otepe pdt roolo
Force =Stress Area
2.6x 10° x 6 x 10
Force= 1560 N Ans.
Example 9.11 A resistance wire strain gauge with a gauge factor of 2 is bonded to0a
steel structure subjected to a stress of 180 MN/m. The modulus of elasticity of steel is
200 GNm*. Calculate the percentage change in the value of the gauge resistance due
to the applied stress.
Solution: Given S=180x10
E= 200 x 10
G-2dt2afborr geo so13 noilo
By Hook's law, we have
180x10 gor oiskoe
900 x10-6
200x10
2.0
Change in gauge resistance is given by
AR/R
G NLIL
AR
-G,
R
2 x 900x 10
R
AR
= 1,8 x 10-3
AR
0.18%%
R
