Thermal Transfer - LED PCB


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

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Thermal Transfer - LED PCB

  1. 1. Thermal Transfer Transfer Heat Common Callouts include Thermal Impedance / Resistance (°C in2/W) and Thermal Conductivity / (°W/m-K).
  2. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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