Princess Sumaya Univ. Electronic Engineering Dept. Industrial Instruments 1 Chapter 4 Thermal Sensors Dr. Bassam Kahhaleh.
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Princess Sumaya Univ.
Electronic Engineering Dept.
3441 Industrial Instruments 1 Chapter 4
Thermal Sensors
Dr. Bassam Kahhaleh
2 / 29 Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Objective
Understand how thermal sensors work and how to interface them.
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Metal Resistance
Energy bands for solids:
Energy gap: required energy for the electron to become free 3 / 29
W
Conduction Valence
Metals W
Conduction Valence
Semiconductors W
Conduction Valence
Insulators
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Thermal Sensors
Metal Resistance Nickel 3 Platinum 2 1
R
l A R
(
T
)
R
( 25 ) (
T
( 25 ) )
−100 0 100 200 300 400 500 600 Temperature ( °C)
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Metal Resistance
Linear Approximation
R
(
T
)
R
(
T
0 ) [ 1 0
T
] 0 1
R
(
T
0 )
R
2
T
2
T R
1 1
R 2 R 1 T 1
Quadratic Approximation
R
(
T
)
R
(
T
0 ) [ 1 1
T
2 (
T
) 2 ]
T T 2
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6 / 29 Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Resistance-Temperature Detectors (RTD)
Sensitivity
0 : 0.004/ °C ~ 0.005 /°C
Response Time
0.5 ~ 5 seconds
Construction
Wire
Signal Conditioning
Bridge with lead compensation
Dissipation Constant (Self-heating)
T
P P D
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Thermal Sensors
RTD
Example
0 = 0.005/ °C R = 500 Ω at 20°C P D = 30 mW / °C R 1 = R 2 = 500 Ω V S = 10 V RTD is at 0 °C Find R 3
V S a R 1 R 3 D R 2 b c R T D
R
(
T
) 500 [ 1 0 .
005 ( 0 20 )] 450
I P
10 500 450 0 .
( 0 .
011 ) 2 * 450 011
A
0 .
054
W
T
0 .
054 0 .
030 1 .
8
C R
3 500 [ 1 0 .
005 ( 1 .
8 20 )] 454 .
5
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Thermal Sensors
Thermistors 30 20 10 −20 0 20 40 60 80 100 Temperature ( °C)
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Thermistors
Sensitivity
~ 10% / °C
Response Time
0.5 ~ 10 seconds
Construction
Discs, beads, rods … etc
Signal Conditioning
Divider circuit, Bridge
Dissipation Constant (Self-heating)
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Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Thermistors
Example
R = 3.5 K Ω at 20°C S = - 10% / °C P D = 5 mW / °C V O = 5 at 20 °C Self-heating error?
R
1 3 .
50
K
P
( 5 ) 2 3 .
5
K
T
7 .
1
mW
7 .
1 1 .
42
C
5
R 1 R TH 10 V
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V D
R TH
3 .
5
K
1 .
42
C
10 % * 3 .
5
K
C
3
K
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Thermocouples
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T 2 EMF
E
(
T
2
T
1 )
Seebeck Effect T 1 T 1 Heat Flow T 2 I Peltier Effect
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Thermocouples T R T M + V TC −
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T R
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Thermal Sensors
Thermocouples Type
J T K E S R
Material
Iron-constantan Copper-constantan Chromel-alumel Chromel-constantan 90%platinum, 10%rhodium-platinum 87%platimum, 13%rhodium-platinum
Normal Range
-190 °C to 760°C -200 °C to 371°C -190 °C to 1260°C -100 °C to 1260°C 0 °C to 1482°C 0 °C to 1482°C
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Thermocouples Type E 50 Type J
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30 Type R −200 10 Temperature ( °C) 0 200 400 600 800 1000 1200
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Thermocouples 1 0 −1 4 3 2 0 °C Ref.
20 °C Ref.
−20 0 20 40 60 80 Temperature ( °C)
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Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Thermocouples
Sensitivity
Type J: 0.05mV / °C Type R: 0.006mV / °C
Response Time
0.01 ~ 20 seconds
Construction
Welded junction
Signal Conditioning
High-gain differential amplifier, with high CMRR
Reference Compensation (cold junction comp.)
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Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Thermocouples
Cold junction Compensation T T ref K
KV TC V C
+
Temperature Sensor Signal Conditioning
V out
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Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Bimetal Strips
Thermal Expansion
l
l
0 1
T
γ
1
γ
2 <
γ
1
T 0
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T > T 0
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Gas Thermometers Constant Volume:
p
1
T
1
p
2
T
2
Liquid-Expansion Thermometers
V
(
T
)
V
(
T
0 ) 1
T
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Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Solid-State Temperature Sensors ~ 12 mV / K P D = 5 mW / °C At 293 K: V T = 3.516 V I = (5 – 3.516) / 510 = 0.0029 A P = 3.516 * 0.0029 = 10.2 mW ΔT = 10.2 / 5 = 2.04 °C
Increase R T 510 Ω 5 V
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V T
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Solid-State Temperature Sensors
Example
J-type TC: ~ 50 μV / °C SS sensor: 8 mV / °C V O = 2 V @ 200 °C V TC (200 °C) = 10.78 mV TC Gain = 8 mV / 50 μV = 160 Total Gain = 2000 / 10.78
= 185.5
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Thermal Sensors
Solid-State Temperature Sensors
Example
−8 mV / °C 10 K 11.59 K SS 320 K T 2 K − + 10 K − + V out 2 K 11.59 K 320 K T ref
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Example
Turn-on alarm when T = 10 ± 0.5 °C Turn off alarm when T ≤ 8 °C R TH = 10 K Ω @ 10 °C = 11 K Ω @ 8 °C Keep self-heating within ± 0.5 °C. Use ± 0.25 °C P = (5 mW / °C)*(0.25 °C) = 1.25 mW @ 10 °C, I TH = 0.354 mA, V TH = 3.5 V Using voltage divider with 5 V supply: R = (5 – 3.5) / 0.354 = 4.28 KΩ @ 10 °C, V D @ 8 °C, V D = 1.5 V = 1.41 V Hysteresis = 0.09 V 23 / 29
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Example
5 V 500 K R TH V D 9 K 4.28 K 5 V + − V out 2.327 K V ref = 1.5 V 1 K
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Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Example
Measured T = 50 °C ~ 80 °C (error ≤ ± 1 °C) V O = 0 ~ 2 V RTD: R(65 °C) = 150 Ω (65 °C) = 0.004 / °C P D = 30 mW / °C R(50 °C) = 150 [ 1 + 0.004 (50 – 65) ] = 141 Ω R(80 °C) = 150 [ 1 + 0.004 (80 – 65) ] = 159 Ω Keep self-heating within ± 1 °C P = (30 mW / °C)*(1 °C) = 30 mW @ 80 °C, I RTD = 13.7 mA, V RTD = 2.17 V 25 / 29
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Thermal Sensors
Example
5 V 10 K 220 220 10 K − + + − 10 K 141 RTD 10 K 138 K 10 K V out @ 50 °C: R RTD = 141 Ω, ΔV = 0, V out = 0 @ 80 °C: R RTD = 159 Ω, ΔV = 0.1447 V, V out = 0.1447 * 13.8 = 2 V
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
Example
Measured T = 500 ~ 600 °F (260 °C ~ 315.6 °C) V O = 0 ~ 5 V J-type TC: T ref = 25 °C V TC (260 °C) = 12.84 mV V TC (315.6 °C) = 15.9 mV V out = m V TC + V 0 0 = m(0.01284) + V 0 5 = m(0.01590) + V 0 m = 1634, V 0 = − 21, Use Gain = 100 * 16.34
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Thermal Sensors
Example
5 V 2.88 K V ref = 1.289 V 10 K 1 K T T ref = 25 °C 1 K 1 K 100 K − + 100 K 10 K − + 163.4 K V out
Princess Sumaya University 3441 - Industrial Instruments 1
Thermal Sensors
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