Tutorial 5
Principles of Transducers and Temperature Measurement
Chapter 5 Principles of Transducers
Example 5.1 (Johnson 2006) A potentiometric displacement sensor is used to measure workpiece motion from 0 to 10 cm. The resistance changes linearly over this range from 0 to 1 kΩ.
Develop signal conditioning to provide a linear, 0- to 10-V output.
–15V
Sensor
250Ω
510Ω
Op-Amp
5.1V
Figure 5.1
SOLUTION
The key thing is to not lose the linearity of the resistance versus displacement. We cannot put the
varying resistance in a divider to produce a varying voltage because the voltage varies
nonlinearly with resistance. Remember though that the output voltage of an inverting amplifier
varies linearly with the feedback resistance. Therefore, let’s put the sensor in the feedback of a
simple inverting amplifier. Then we would have something like
R
Vout = − 2 V in
R1
We can now get rid of the pesky negative by using Vin as a constant negative voltage, say –5.1
volts from a zener diode. Then we pick R1 to give the desired output, 10 volts at 1 kΩ (10 cm),
1000
10 = −
( −5.1) so R1 = 510 Ω
R1
Example 5.2 The following figure shows a capacitive displacement sensor designed to monitor
small changes in work-piece position. The two metal cylinders are separated by a plastic
sheath/bearing of thickness 1 mm and dielectric constant at 1 kHz of 2.5. If the radius of 2.5 cm,
find the sensitivity in pF/m as the upper cylinder slides in and out of the lower cylinder. What is
the range of capacity if h varies from 1.0 to 2.0 cm? ( ε 0 = 8.854 pF/m)
SOLUTION
The capacity is given by the following equation
A
C = ε0ε r
d
1
Displacement
r
h
d
Figure 5.2
The effective area is the area of the shared cylindrical area, which has a radius, r, and height, h.
Thus, A = 2πrh, so the capacity can be expressed as
rh
C = 2πKε 0
d
The sensitivity with respect to the height, h, is defined by how C changes with h, that is, it is
given by the derivative
dC
r
= 2πKε0
dh
d
Substituting for the given values, we get
dC
2.5 ×10−2 m
= 2π ( 2.5 )( 8.85 pF/m )
= 3475 pF / m
dh
10−3 m
Since the function is linear with respect to h, we find the capacity range as Cmin = (3475
pF/m)(10–2m) = 34.75 pF to Cmax = (3475 pF/m)(2 × 10–2m) = 69.50 pF.
Example 5.3 An LVDT has a maximum core motion of ± 1.5 cm with a linearity of ± 0.3% over
that range. The transfer function is 23.8 mV/mm. If used to track work-piece motion from –1.2 to
+ 1.4 cm, what is the expected output voltage? What is the uncertainty in position determination
due to nonlinearity?
SOLUTION
Using the known transfer function, the output voltages can easily be found,
V(–1.2cm) = (23.8 mV/mm)(–12mm) = –285.6 mV
and
V(1.4cm) = (23.8 mV/mm)(14mm) = 333 mV
The linearity deviation shows up in deviations of the transfer function. Thus, the transfer fuction
has an uncertainty of
( ± 0.003)(23.8mV/mm) = ± 0.0714mV/mm
This means that a measured voltage, Vm (in mV), could be interpreted as a displacement that
ranges from Vm/23.73 to Vm/23.8 mm, which is approximately ± 0.3%, as expected. Thus, if the
sensor output was 33 mV, which is nominally 1.4 cm, the actual core position could range from
1.40329 to 1.39506 cm.
Example 5.4 The level of ethyl alcohol is to be measured from 0 to 5 m using a capacitive
system such as shown in the following figure.
2
C
h
Figure 5.3
The following specifications define the system:
for ethyl alcohol: ε r = 26 (for air, ε r = 1)
cylinder separation: d = 0.5 cm
plate area: A = 2πRL
where R = 5.75 cm = average radius, L = distance along cylinder axis
Find the range of capacity variation as the alcohol level varies from 0 to 5 m.
SOLUTION
The capacity is given by
A
C = ε0ε r
d
Therefore all we need to do is to find the capacity for the entire cylinder with no alcohol and then
multiply that by 26.
A = 2πRL = 2π(0.0575m)(5m) = 1.806 m2
Thus, for air,
C = (1)(8.85 pF/m)(1.806m2/0.005m) = 3196 pF ≈ 0.0032 µF
With the ethyl alcohol, the capacity becomes
C = 26(0.0032 µF) = 0.0832 µF
The range is 0.0032 to 0.0832 µF.
Exercise 5.1 The output voltage of a potentiometer-type resistance transducer is to be measured
by a recorder having an input resistance of 20 kΩ. If the error of measurement is not to exceed –
2% at 50% f.s.d, determine resistance value of the potentiometer. [1.633kΩ]
Exercise 5.2 The following is a typical specification for potentiometer-type resistance transducer.
Examine the specification and explain the meaning and significance of each item.
Type wire-wound resistance displacement potentiometer
Terminal resistance 10kΩ
Range 0-25mm
Resolution 0.4%
Power rating 0.25W
Maximum wiper current 15mA
Thermal drift 0.05% per oC
Life expectancy 108 cycles
Exercise 5.3 A linear variable differential transformer is excited with a 100 Hz 6V peak-to-peak
waveform. The input core motion is sinusoidal at 10 Hz and has a displacement amplitude of ± 3
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mm. If the l.v.d.t sensitivity is 2 V/mm, draw the waveforms of the excitation voltage, input
displacement and output voltage.
Exercise 5.4 The specifications for the l.v.d.t in Exercise 5.3 are as follows:
Linearity: 0.4%
Resolution: infinite
Residual voltage: 0.5%
Drift better than 0.1% per oC
Output impedance: 2.5kΩ
Response time: 1ms
Explain the meaning and significance of the specifications.
See attached sheet for an example for LVDT specifications.
Exercise 5.5
(a) Describe the principle of operation and construction details of the piezoelectric (quartz)
transducer.
(b) A quartz pressure transducer has a sensitivity of 80 pC/bar. If, when the input pressure is 3
bars, an output voltage of 1 V is produced, determine the capacitance of the device. [240 pF]
Chapter 6 Temperature measurement
Exercise 6.1 Explain the principle of operation of the thermocouples.
Exercise 6.2 Explain the principle of operation of the resistance temperature detectors.
Exercise 6.3 Explain the principle of operation of the thermistors.
Exercise 6.4 Explain the principle of operation of the radiation pyrometers.
Exercise 6.5 A type-K thermocouple is exposed to a temperature of 1200oC. If the indicator is
used as the cold junction and its temperature is 50oC. Use the following figure to calculate the
e.m.f indicated. [47mV]
80
Type E
70
60
Type K
50
40
30
Type T
20
Type S
10
0
200
400
600
800
1000
1200
1400
1600 1800 2000
Figure 6.1 Thermocouple characteristics (reference oC)
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Exercise 6.6 Using the above figure, determine (a) the sensitivity of the type-T thermocouple in
the range of 0oC to 300oC, (b) the sensitivities of the type-E and type-S thermocouples in the
range of 400oC to 1000oC. [0.5mV/oC] [0.08mC/oC] [0.01mV/oC]
Exercise 6.7 If the dynamic relationship between the measured temperature θ2 and the reference
temperature θ1 for a thermometer is given by
dθ 2
= k ( θ1 − θ2 ) where k = 0.2s-1
dt
determine the time constant and static sensitivity for the thermometer. [5s]
Exercise 6.8 A mercury thermometer used to measure the temperature of a liquid has dynamics
represented by the following equation:
dT
1
1
CT m +
Tm =
Ti
dt
RT
RT
where Tm (oC) is the temperature of mercury, and Ti (oC) a change in the temperature of the liquid,
RT and CT are constant. Find the time constant, sensitivity and the dynamic response of the
thermometer Tm. using these numerical values: Ti = 100 oC , RT = 131 oC/W, and CT = 0.56 J/oC.
Exercise 6.9 An RTD has α 0 = 0.005 /oC, R = 500 Ω and a dissipation constant of PD = 30
mW/oC at 20 oC. The RTD is used in a bridge circuit as shown in the following figure, with R1 =
R2 = 500 Ω, and R3 is a variable resistor used to null the bridge. If the supply is 10 V and the
RTD is placed in an ice bath at 0 oC:
a) Find the value of R3 to null the bridge [R3 = 454.5 Ω].
Power
supply
10V
R1
R3
V
R2
RTD
Figure 6.2 RTD with a bridge circuit for Exercise 6.9
b) Find the output voltage measured by a voltmeter (Rv = ∞ ) with the above value of R3 if the
temperature is 100 oC. Consider the effect of the self-heating and calculate the error of the RTD
at 100 oC.
Hints: An RTD is a resistance, therefore there is an I2R power dissipated by the device itself that
causes a light heating effect, a self-heating. This may also cause an erroneous reading or even
upset the environment in delicate measurement condition. Thus, the current through the RTD
must be kept sufficiently low and constant to avoid self-heating. Typically, a dissipation constant
is provided in RTD specifications. This number relates the power required to raise the RTD
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temperature by one degree of temperature. Thus, a 25-mW/oC dissipation constant shows that if
I2R power losses in the RTD equal 25 mW, the RTD will be heated by 1 oC.
The dissipation constant is usually specified under two conditions: free air and a well-stirred oil
bath. This is because of the difference in capacity of the medium to carry heat away from the
device. The self-heating temperature rise can be found from the power dissipated by the RTD,
and the dissipation constant from:
∆T =
where ∆T
P
PD
P
PD
= temperature rise because of self-heating in oC
= power dissipated in the RTD from the circuit in W
= dissipation constant of the RTD in W/oC
Exercise 6.10 (Thermistor)
A thermistor is to monitor room temperature. It has a resistance of 3.5 kΩ at 20 oC with a slope of
–10%/oC. The dissipation constant is PD = 5 mW/oC. It is supposed to use the thermistor in the
divider of the following figure to provide a voltage of 5.0 V at 20 oC. Evaluate the effects of selfheating [3.0 kΩ; VD = 4.6 V].
10 V
3.5 kΩ
Themistor
RTH
VD = 5 V
Figure 6.3 Thermistor in a divider circuit for Exercise 6.10
Exercise 6.11 (Thermocouples) Find the emf for a material with α = 50 µV/oC if the junction
temperatures are 20oC and 100oC.
Hint: The thermoelectric effect is expressed by the following equation:
ε=³
T2
T1
( QA − QB ) dT
where
ε
= emf (also called Seebeck emf) produced in volts
T1, T2 = junction temperatures in K
QA, QB = thermal transport constants of the two metals
This equation, which described the Seebect effect, shows that the emf produced is proportional to
the difference in temperature and, further, to the difference in the metallic thermal transport
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constants. Thus, if the metals are the same, the emf is zero, and if the temperature are the same,
the emf is zero.
In practice, it is found that the two constants, QA and QB, are nearly independent of temperature
and that an approximate linear relationship exists as
ε = α ( T2 − T1 )
where α
= constant in V/K
T1, T2 = junction temperatures in K
Exercise 6.12 A voltage of 23.72 mV is measured with a type K thermocouple at a 0oC reference.
Find the temperature of the measurement junction. [572.1oC]
Hints: use the attached sheet with the following interpolation equations:
Temperature:
ª T − TL º
TM = TL + « H
» ( VM − VL )
¬ VH − VL ¼
where VM = measured voltage that lies between a higher voltage, VH, and a lower voltage, VL,
which are in the tables. The temperatures corresponding to these voltages are TH and TL,
respectively.
Voltage:
ª V − VL º
VM = VL + « H
» ( TM − TL )
¬ TH − TL ¼
Exercise 6.13 Find the voltage of a type J thermocouple with a 0oC reference if the junction
temperature is –172oC. [–7.18oC]
Exercise 6.14 An RTD has α (20oC) = 0.004 /oC. If R = 106 Ω at 20oC, find the resistance at
25oC.
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