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

7 / 29 Princess Sumaya University 3441 - Industrial Instruments 1

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 

8 / 29 Princess Sumaya University 3441 - Industrial Instruments 1

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

13 / 29 Princess Sumaya University 3441 - Industrial Instruments 1

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

 19 / 29

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

26 / 29 Princess Sumaya University 3441 - Industrial Instruments 1

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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28 / 29 Princess Sumaya University 3441 - Industrial Instruments 1

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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End of Chapter 4