Thermal Transfer segment from LED PCB webinar addresses thermal conductivity and thermal resistance of LED PCB

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