1. THERMAL TRANSFER
CHARACTERISTICS ON HEATER
Jorge Hernandez PhD. Stundent
DIMES -UNICAL

2. Heat Transfer
• Temperature conduction is performed from a
hot body to a cold body.
• Thermal Equilibrium happens in two bodies
have the same temperature.

3. Thermal Transfer
Variables

4. Difference Temperature
• Contact surface of two bodies experiences a
difference in temperature.
• While thermal equilibrium come to happen,
the difference go to zero.

5. Material thermal characteristic
• In previous example, heat was transferred
from water through the metal to water.
• The rate of heat transfer depends on material
through heat is transferred.
• Heat transfer coefficient (k) express thermal
conduction characteristic of a material.
• Large value of K means thermal conductor
and a lower value means thermal isulators.

6. Material thermal characteristic
• k is determined experimentally and its unit is
𝑊
𝑚. °𝐶
Source:
http://www.roymech.co.uk/Related/Thermos/Thermos_HeatTransfer.html
Material k Material k
Aluminum 237 Porcelain 1.05
Brass 110 Wood 0.13
Copper 398 Water 0.58
Gold 315 Air 0.024

7. Material body characteristics
• Area (A)
– Wider areas conduct more heat and this means
greater heat dissipation. Not good for our
project.
• Thickness (d)
– Thin walls conduct heat in faster way. This affect
the rate thermal transfer.

8. Mathematical
Equation

9. Mathematical Equation
• Variables:
– Temperature difference: (T1 – T2)
– Thermal Conductivity Coefficient of the Material: k
– Area: A
– Thickness: d
(1) 𝑃𝑟𝑟𝑟𝑟 =
𝑄
𝑡
= 𝜎𝑡𝑡𝑡 (𝑇𝑇 − 𝑇𝑇) [
𝐽𝐽𝐽𝐽𝐽𝐽
𝑠
= 𝑊]
(2) 𝜎𝑡𝑡𝑡 =
𝑘 .𝐴
𝑑
[
𝑊
°𝐶
] Average Thermal Conductance
(3) 𝑅𝑡𝑡𝑡 =
1
𝜎 𝑡𝑡𝑡
[
°𝐶
𝑊
] Average Thermal Resistance
Information extracted from reference [1], assuming that areas are flat and regular.

10. Application
Problem
Case

11. Heater V1

12. Practical Considerations
– Contact area of SafeFET
(heat source size) is smaller
than the opposite side area
and even the DUT contact
area.
– Location of heat source is
important to note because
of thermal distribution on
sink.
– Also, thermal resistance of
electrical isulator must be
considered.

13. Mathematical Resolution
• Calculation of Total
Thermal Resistance
based on reference [1]
and [2].
1. Calculate 𝑅𝑡𝑡𝑡.
2. Calculate 𝑅𝑅𝑅𝑐
3. Calculate 𝑅𝑅𝑅𝑅.

14. Calculation 𝑅𝑡𝑡𝑡 according to [1]
𝐴 = 0.018 𝑥 0.022 [𝑚2
]
𝐴 = 3.96𝑥10−4
[𝑚2
]
𝑑 = 0.003 [𝑚]
𝑘 = 110 [
𝑊
𝑚. °𝐶
]
From (2) and (3):
𝑅𝑡𝑡𝑡 =
𝑑
𝑘. 𝐴
=
0.003
(110)(3.96𝑥10−4)
°𝐶
𝑊
𝑹 𝒕𝒕𝒕 = 𝟔𝟔. 𝟖𝟖𝟖𝟏𝟏−𝟑[
°𝑪
𝑾
]

15. Equations for 𝑅𝑡𝑡𝑡
• In [2], Spreading Thermal Resistance (𝑅𝑡𝑡𝑡) is stablished
in function of 𝑅𝑡𝑡𝑡 and the difference between size of
the heating areas in each side of the sink.
• 𝑅𝑡𝑡𝑡 is an additional quantity that is needed for
determining the maximum heat sink temperature.
• Also, [2] considers the
location of heating source
on the sink surface. 𝑪 𝒇 [
𝟏
𝒎
]
• Ideally, this should be in
the center of the same.
𝑪 𝒇 = 𝟏

16. Equations for 𝑅𝑡𝑡𝑡 and 𝑅𝑡𝑡𝑡
(4) 𝑅𝑡𝑡𝑡 = 𝐶𝑓 𝑥
𝐴 𝑝. 𝐴 𝑠
𝑘 . 𝜋. 𝐴 𝑝. 𝐴 𝑠
𝑥
λ. 𝑘. 𝐴 𝑝. 𝑅𝑡𝑡𝑡 + tanh(λ. 𝑑)
1 + λ. 𝑘. 𝐴 𝑝. 𝑅𝑡𝑡𝑡. tanh(λ. 𝑑)
[
°𝐶
𝑊
]
Where:
λ=
𝜋
3
2�
𝐴 𝑝
+
1
𝐴 𝑠
• 𝐴 𝑝: Footprint area of the heat sink base-plate
• 𝐴 𝑠: Contact area of the heat source
(5) 𝑅𝑡𝑡𝑡= 𝑅𝑡𝑡𝑡 + 𝑅𝑡𝑡𝑡

17. Calculation of 𝑅𝑡𝑡𝑡 according to [2]
𝐴 𝑝 = 0.018 𝑥 0.022 = 3.96𝑥10−4[𝑚2]
𝐴 𝑠 = 0.0104 𝑥 0.0101 = 105.04𝑥10−6
[𝑚2
]
λ =
𝜋
3
2�
𝐴 𝑝
+
1
𝐴 𝑠
= 377.39 [
1
𝑚
]

18. Calculation of 𝑅𝑡𝑡𝑡 according to [2]
• From (4) 𝑎𝑛𝑑 assuming that heat source is placed in the center of
the base-plate:
𝑅𝑡𝑡𝑡 = 𝐶𝑓 𝑥
𝐴 𝑝. 𝐴 𝑠
𝑘 . 𝜋. 𝐴 𝑝. 𝐴 𝑠
𝑥
λ. 𝑘. 𝐴 𝑝. 𝑅𝑡𝑡𝑡 + tanh(λ. 𝑑)
1 + λ. 𝑘. 𝐴 𝑝. 𝑅𝑡𝑡𝑡. tanh(λ. 𝑑)
[
°𝐶
𝑊
]
tanh 377.39 𝑥 0.003 = 0.81176
λ. 𝑘. 𝐴 𝑝. 𝑅𝑡𝑡𝑡 = 377.39 𝑥 110 𝑥 3.96𝑥10−4 𝑥 68.87𝑥10−3 = 1.1322
𝑅𝑡𝑡𝑡 = 1 𝑥
3.96𝑥10−4. 105.04𝑥10−6
110 . 𝜋(3.96𝑥10−4)(105.04𝑥10−6)
𝑥
1.944
1.919
[
°𝐶
𝑊
]
𝑹 𝒕𝒕𝒕 = 𝟓. 𝟐𝟐𝟏𝟏−𝟑
[
°𝑪
𝑾
]

19. Calculation of 𝑅𝑡𝑡𝑡 according to [2]
• From (5):
𝑅𝑡𝑡𝑡 = 𝑅𝑡𝑡𝑡 + 𝑅𝑡𝑡𝑡
𝑅𝑡𝑡𝑡 = 68.87𝑥10−3
+ 5.2𝑥10−3
[
°𝐶
𝑊
]
𝑹 𝒕𝒕𝒕 = 𝟕𝟕. 𝟎𝟎𝒙𝟏𝟏−𝟑 [
°𝑪
𝑾
]
Note.- Line graph shows the Cf variations in function of the distance from the
center of the heat sink to the heat source placed along the center line at y=0
and -37.5 < x < 37.5 [mm]. Cf is case dependent.

20. THERMAL
TRANSFER
SAFEFET - DUT

21. How does the heat flow?
• Considerations:
– Contact temperature
between case SafeFET
and right side of sink is
𝑇𝐶𝐶.
– Contact temperature
between case DUT and
left side of sink is 𝑇𝐶𝐶.
– Thermal resistence of
the electrical isulator
must be considered.

22. Case study extracted from Project
• Both SafeFET and DUT are heat
sources, but in practice SafeFET
deliveres more heat than DUT.
• All considerations are referenced
to regime work.
– SafeFET (DUT 2)
• 𝑇𝑗𝑗 = 156.82 °𝐶
• 𝑃𝐷𝐷 = 6.04 𝑊
– DUT 2
• 𝑃 𝐷𝐷 = 0.14 𝑚𝑚
• 𝑇𝐶𝐶 = 150 °𝐶 (measured with LM35
and It will be demonstrated
mathematically)

23. Analog focus for resolution

24. Case Temperature on SafeFET
Adapting (1) for this situation:
𝑇𝑗𝑗 − 𝑇𝐶𝐶 = 𝑅𝑡𝑡𝑡𝑡𝑡 ∗ 𝑃𝑆
Where, from datasheet SafeFET :
𝑅𝑡𝑡𝑡𝑡𝑡 = 0.6[
°𝐶
𝑊
] : Thermal resistance junction-case of SafeFET
𝑇𝑗𝑗 = 156.82 °𝐶
It is important to note that because both devices are heat sources due to
their power consumption:
𝑃𝑆 = 𝑃𝐷𝐷 + 𝑃 𝐷𝐷
𝑃𝑆 = 6.04014 [𝑊]
Therefore:
156.82 °𝐶 − 𝑇𝐶𝐶 = 0.6
°𝐶
𝑊
∗ 6.04014 [𝑊]
𝑻 𝑪𝑪 = 𝟏𝟏𝟏. 𝟐 °𝑪

25. Transfer Temperature between heat
sink sides
Adapting (1) for modified value of thermal resistance of the heat sink:
𝑃𝑆 = 𝜎𝑡𝑡𝑡 (𝑇𝐶𝐶 − 𝑇𝑆𝑆) [𝑊]
Where: 𝜎𝑡𝑡𝑡 =
1
𝑅 𝑡𝑡𝑡
= 13.5
𝑊
°𝐶
𝑇𝐶𝐶 = 153.2 °𝐶
𝑇𝑆𝑆 ≡ Temperature in side contact between heat sink and insulator
𝑃𝑆 = 𝑃𝐷𝐷 + 𝑃 𝐷𝐷 = 6.04𝑊 + 0.14𝑚𝑚 = 6.04014𝑊
𝑇𝑆𝑆 = 𝑇𝐶𝐶 −
𝑃𝑟𝑟𝑟𝑟
𝜎𝑡𝑡𝑡
𝑻 𝑺𝑺 = 𝟏𝟏𝟏. 𝟕𝟕 °𝑪

26. Case Temperature on DUT transferred
through insulator
Adapting (1) for this situation:
𝑃𝑠 = 𝜎𝑡𝑡𝑖 (𝑇𝑆𝑆 − 𝑇𝐶𝐶) [𝑊]
Where:
𝑇𝑆𝑆 = 152.75 °𝐶
𝑇𝐶𝐶 ≡ Case temperature of DUT
𝑷 𝑺 = 𝑷 𝑫𝑫 + 𝑷 𝑫𝑫 = 𝟔. 𝟎𝟎𝟎𝟎𝟎𝟎
Whit:
𝜎𝑡𝑡𝑖 =
𝑘𝑖 𝑥 𝐴𝑖
𝑡𝑖
= 2.62
𝑊
°𝐶
From the vendor site:
𝑘𝑖 = 3.5 [
𝑊
𝑚.°𝐶
]
𝐴𝑖 = 0.015 𝑥 0.019 = 285𝑥10−6 [𝑚2]
𝑡𝑖 = 381𝑥10−6
[𝑚]
Therefore:
𝑻 𝑪𝑪 = 𝑻 𝑺𝑺 −
𝑷 𝑺
𝜎𝑡𝑡𝑡
= 𝟏𝟏𝟏. 𝟒𝟒 °𝑪

27. Junction Temperature on DUT
Adapting (1) for this situation, and considering that heat
source is the SafeFET:
𝑇𝐶𝐷 − 𝑇𝑗𝑗 = 𝑅𝑡𝑡𝑡𝑡𝑡 ∗ 𝑃𝑆
Where:
𝑇𝐶𝐶 = 150.44 °𝐶
𝑇𝑗𝑗 ≡ Junction temperature of DUT
𝑷 𝑺 = 𝑷 𝑫𝑫 + 𝑷 𝑫𝑫 = 𝟔. 𝟎𝟎𝟎𝟎𝟎𝟎
𝑅𝑡𝑡𝑗𝑗𝑗 = 0.5
°𝐶
°𝑊
Therefore:
𝑻𝒋𝒋 = 𝑻 𝑪𝑪 − 𝑷 𝒔. 𝑹 𝒕𝒕𝒕𝒕𝒕 = 𝟏𝟏𝟏. 𝟒𝟒 °𝑪

28. Conclusions
• Difference of junction temperatures for this case study analyzed.
∆𝑇𝑗𝑗𝑗 = 𝑇𝑗𝑗 − 𝑇𝑗𝑗 = 156.82 °𝐶 − 147.42 °𝐶
∆𝑻𝒋𝒋𝒋 = 𝟗. 𝟒 °𝑪
• Error between temperature setpoint and 𝑇𝑗𝑗
%𝐸𝐸𝐸𝐸𝐸 =
150°𝐶 − 147.42°𝐶
150°𝐶
∗ 100 = 𝟏. 𝟕𝟕 %
• Thermal Resistance of Heat Sink
𝑹𝒕𝒕𝒕 = 𝟔𝟔. 𝟖𝟖𝟖𝟏𝟏−𝟑
[
°𝑪
𝑾
]
• Total Thermal Resistance of Heat Sink according location of heat source
criteria.
𝑹𝒕𝒕𝒕 = 𝟕𝟕. 𝟎𝟎𝟎𝟏𝟏−𝟑 [
°𝑪
𝑾
]
• Temperature aquisition of the system was adjusted with an LM35, which
performs a sensing of its case temperature, in this way this mathematic
process shows this right adjustment with the case temperature value on
DUT:
𝑻 𝑪𝑪𝑴 = 𝟏𝟏𝟏. 𝟒𝟒 °𝑪 (Mathematical value)
𝑻 𝑪𝑪𝑬 = 𝟏𝟏𝟏. 𝟎 °𝑪 (Experimental value)

29. References
• The Physics Classroom
http://www.physicsclassroom.com/class/ther
malP/u18l1f.cfm
• Calculating spreading resistance in heat sinks
http://www.electronics-
cooling.com/1998/01/calculating-spreading-
resistance-in-heat-sinks/

30. ¿QUESTIONS?