Thermal Transfer - LED PCB
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Thermal Transfer - LED PCB

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

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

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    Thermal Transfer - LED PCB Thermal Transfer - LED PCB Presentation Transcript

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