1. 3441 Industrial Instruments 1 Chapter 4 Thermal Sensors
Princess Sumaya University
Industrial Instruments 1
Princess Sumaya Univ. Electronic Engineering Dept.
3441 Industrial Instruments 1 Chapter 4 Thermal Sensors
Dr. Bassam Kahhaleh
Dr. Bassam Kahhaleh

2. 3441 - Industrial Instruments 1
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Industrial Instruments 1
Thermal Sensors
Objective
Understand how thermal sensors work and how to interface them.

3. 3441 - Industrial Instruments 1
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Thermal Sensors
Metal Resistance
Energy bands for solids:
Energy gap: required energy for the electron to become free W W W
Conduction
Conduction
Conduction
Valence
Valence
Valence
Semiconductors
Insulators
Metals

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Thermal Sensors
Metal Resistance
Nickel
Platinum 3
R(25°C)
R(T) 2 1
Temperature (°C) −

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Thermal Sensors
Metal Resistance
Linear Approximation
Quadratic Approximation R2 R1 T1 T T2

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Thermal Sensors
Resistance-Temperature Detectors (RTD)
Sensitivity
0: 0.004/°C ~ /°C
Response Time
0.5 ~ 5 seconds
Construction
Wire
Signal Conditioning
Bridge with lead compensation
Dissipation Constant (Self-heating)

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Thermal Sensors
RTD
Example
0= 0.005/°C
R = 500 Ω at 20°C
PD = 30 mW /°C
R1 = R2 = 500 Ω
VS = 10 V
RTD is at 0°C
Find R3
R 1
R 2
R 3 VS D
RTD a b c

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Thermal Sensors
Thermistors 30 20
Resistance (KΩ) 10
Temperature (°C) −

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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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Thermal Sensors
Thermistors
Example
R = 3.5 KΩ at 20°C
S = - 10% /°C
PD = 5 mW /°C
VO = 5 at 20°C
Self-heating error?
R 1
R TH
10 V
V D

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Thermal Sensors
Thermocouples
Heat Flow T2 T1 T2 T1
EMF I
Seebeck Effect
Peltier Effect

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Thermal Sensors
Thermocouples TR +
VTC − TM TR

13. Thermal Sensors Thermocouples Type Material Normal Range J
Princess Sumaya University
Industrial Instruments 1
Thermal Sensors
Thermocouples
Type
Material
Normal Range J
Iron-constantan
-190°C to 760°C T
Copper-constantan
-200°C to 371°C K
Chromel-alumel
-190°C to 1260°C E
Chromel-constantan
-100°C to 1260°C S
90%platinum, 10%rhodium-platinum
0°C to 1482°C R
87%platimum, 13%rhodium-platinum

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Thermal Sensors
Thermocouples
Type E 50
Type J
VTC (mV) 30
Type R 10
Temperature (°C) −

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Thermal Sensors
Thermocouples
0 °C Ref. 4 3
20 °C Ref.
VTC (mV) 2 1 −
Temperature (°C) −1

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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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Thermal Sensors
Thermocouples
Cold junction Compensation T K
KVTC +
Vout
Tref VC
Temperature
Sensor
Signal
Conditioning

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Thermal Sensors
Bimetal Strips
Thermal Expansion γ1 T0
γ2 < γ1
T > T0

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Thermal Sensors
Gas Thermometers
Constant Volume:
Liquid-Expansion Thermometers

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Thermal Sensors
Solid-State Temperature Sensors
~ 12 mV / K
PD = 5 mW / °C
At 293 K:
VT = V
I = (5 – 3.516) / 510 = A
P = * = 10.2 mW
ΔT = 10.2 / 5 = 2.04 °C
 Increase R
5 V
510 Ω
V T T

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Thermal Sensors
Solid-State Temperature Sensors
Example
J-type TC: ~ 50 μV / °C
SS sensor: 8 mV / °C
VO = °C
VTC (200°C) = 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 − +
2 K T − +
10 K
V out
2 K
11.59 K
320 K
Tref

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Thermal Sensors
Example
Turn-on alarm when T = 10 ± 0.5 °C
Turn-off alarm when T ≤ 8 °C
RTH = °C
= °C
Keep self-heating within ± 0.5 °C. Use ± 0.25 °C
P = (5 mW / °C)*(0.25 °C) = 1.25 mW
@ 10 °C, ITH = mA, VTH = 3.5 V
Using voltage divider with 5 V supply:
R = (5 – 3.5) / = 4.28 KΩ
@ 10 °C, VD = 1.5 V
@ 8 °C, VD = 1.41 V  Hysteresis = 0.09 V

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Thermal Sensors
Example
5 V
500 K
R TH
9 K
V D + −
Vout
5 V
4.28 K
2.327 K
Vref = 1.5 V
1 K

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Thermal Sensors
Example
Measured T = 50 °C ~ 80 °C (error ≤ ± 1 °C)
VO = 0 ~ 2 V
RTD: R(65°C) = 150 Ω
(65°C) = / °C
PD = 30 mW / °C
R(50°C) = 150 [ (50 – 65) ] = 141 Ω
R(80°C) = 150 [ (80 – 65) ] = 159 Ω
Keep self-heating within ± 1 °C
P = (30 mW / °C)*(1 °C) = 30 mW
@ 80 °C, IRTD = 13.7 mA, VRTD = 2.17 V

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Thermal Sensors
Example
5 V
10 K
220
220
10 K − + + −
V out
10 K
141
RTD
10 K
138 K
10 K
@ 50 °C: RRTD = 141 Ω, ΔV = 0, Vout = 0
@ 80 °C: RRTD = 159 Ω, ΔV = V, Vout = * 13.8 = 2 V

27. 3441 - Industrial Instruments 1
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Thermal Sensors
Example
Measured T = 500 ~ 600 °F (260 °C ~ °C)
VO = 0 ~ 5 V
J-type TC: Tref = 25 °C
VTC(260 °C) = mV
VTC(315.6 °C) = 15.9 mV
Vout = m VTC + V0
0 = m( ) + V0
5 = m( ) + V0
m = 1634, V0 = − 21, Use Gain = 100 * 16.34

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Thermal Sensors
Example
5 V
2.88 K
Vref = V
10 K
1 K
100 K
163.4 K
1 K T − +
10 K − +
1 K
V out
100 K
Tref = 25 °C

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Thermal Sensors
End of Chapter 4