The document discusses thermal transfer principles, focusing on thermal impedance, resistance, and conductivity metrics such as °C in²/W and °W/m-K. It emphasizes the advantages of using Kapton®/polyimide materials in thermal management through low thermal impedance and high reliability, while also detailing the measurement techniques outlined in ASTM D5470 for evaluating material performance. Additionally, it provides examples and calculations related to thermal resistance and conductivity in various materials and situations.

1. Thermal Transfer
Transfer Heat
Common Callouts include Thermal Impedance / Resistance (°C in2/W) and Thermal Conductivity / (°W/m-K).

2. Why Kapton® / Polyimide?
Heat Resistance
Stability
Flexibility
Dielectric Properties
Density/Weight
CooLamTM MCPCB (metal core PCB) Offers:
• Very Low Thermal Impedance
• Excellent Reliability Performance
• Excellent Durability and Stability at High Temperature
• Uniform Thermal, Mechanical & Electrical Properties Under Environmental Stress
• Lead Free Solder and Wirebond Process Compatibility
• Halogen Free
• Meets UL 94 V-0
• Construction Variations to Meet Thermal Management Needs
• 3D Shapes

3. Thermal Measurement
Characterize the Thermal Performance of Materials
Objective:
Measuring thermal performance per ASTM D5470.
Equipment:
1) Steady State “Thermal Interface Material Tester” (TIM)
(Analysis Tech. Inc.)
Factors to Consider:
1) Reducing contact resistance between sample and test unit
2) Repeatability of measured values
3) Identify any equipment and/or material limitations

4. Power (Watts)
Basic Principles of ASTM D5470
Heat Source T1 What is Thermal Resistance ?
Thermal Resistance (Rth) is defined as the difference in
temperature between two closed isothermal surfaces
divided by the total heat flow between them.
T2
Rth = (T(A) – T(B)) Power
(surface A)
(surface B)
T3
Heat Sink ∆ C
T4 Watt (V*I)
* Present system does a good job of accounting for all heat and monitoring temperature but nothing is perfect.

5. Power (Watts)
Output from TIM unit:
Thermal Resistance (Rth): ∆ C
Heat Source T1 Watt
= (∆ Temp. power)
T2
Thermal Impedance (RthI): C - in2
Watt
Test Sample = (Thermal Resistance) x area (of test sample)
and
T3
Thermal Conductivity: Watt
Heat Sink m- C
= thickness (material) thermal impedance
T4
Basic Principles of ASTM D5470 (cont.)

6. Power (Watts)
• Total Thermal Resistance (Rth total) is the sum of
components and its thermal resistance value.
“thermal grease” Grease
Copper
Test Sample Dielectric
Aluminum
“thermal grease” Grease
A B C
ASTM D5470 Thermal Measurement Technique

7. ASTM D5470 Thermal Measurement Technique
Measured Thermal Impedance (C-in2/Watt):
RthI (grease_top)
“thermal
Grease”
+
RthI (sample)
Test
Sample Total RthI
+
RthI (grease_bottom)
“thermal
Grease”
Total RthI B A
RthI (sample) = Total RthI - (RthI (grease_top) + RthI (grease-bottom) )

8. Thermal Impedance
Electronics industry defines thermal impedance as the following:
thickness
impedance
thermal
k
This does not include an area through which the heat flows!
Thermal Impedance Thermal Resistance
tDiel tDiel
I R
kDiel kDielA
Units: Units:
m m2 K m K
I R 2
mK m W
W
W
mK W
8

9. Thermal Impedance
Calculating the thermal conductivity of a material
Plot Ra vs. thickness from TIM tester
1/k =thermal conductivity k = is the slope of the line
Fitted Line Plot
K-in^2/W = 0.01366 + 39.35 Thickness (in)
0.225 S 0.0010461
R-Sq 100.0%
R-Sq(adj) 100.0%
Thermal Impedance K-in^2/W
0.200
0.175
0.150 Thermal Conductivity LG:
0.25 W/M-K
0.125
0.100 = 1.0 in / 39.35 K-in2/W
0.075
= .006394 W/in-K x 39.4
in/M (convert to metric)
0.050
0.001 0.002 0.003 0.004 0.005
= 1.0 W/M-K
9 Thickness (in)LC
Data Plotted from multiple readings

10. Example problems
What is the thermal resistance through the thickness of a 0.3 meter by 0.3 meter piece of
stainless steel that is 1.5 cm thick? Assume the thermal conductivity of stainless steel is 15
W/mK.
k=15 W/mK 1.5cm
0.3m
t
R 0.3m
kA
t 0.015m
R
kA (15W mK)(0.3m)(0.3m)
R 0.011K W

11. Example Problems
This same piece of stainless steel now has a
heater attached to one side that generates
1.5cm
100 Watts of heat into the plate. Ttop
If the bottom of the plate measures 45°C k=15 W/mK 0.3m
(318K), what is the temperature of the top Tbottom 0.3m
side of the plate?
(Ttop Tbottom ) (Ttop Tbottom ) q Rthermal
q
Rthermal
Ttop q Rthermal Tbottom
Ttop 100 0.011K W 318K
W
11 Ttop 319.1K 46.1 C

12. Basic Question Room Temp = 30°C
Same LED and power input
MCPCB 1
LED
MCPCB 2 LED
Cu Cu
Dielectric Dielectric
Al/Cu Base Al/Cu Base
Heat Sink Heat Sink
Tjunction = 65°C Tjunction = 70°C
WHY???
12
Heat Transfer