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Unformatted text preview: Chapter 15 Cooling of Electronic Equipment 1555 A plastic DIP with 16 leads is cooled by forced air. Using data supplied by the manufacturer, the junction temperature is to be determined. Assumptions Steady operating conditions exist. Analysis The junctiontoambient thermal resistance of the device with 16 leads corresponding to an air velocity of 300 m/min is determined from Fig.1523 to be R junction ambient = 50 C / W Then the junction temperature becomes C 125 = + = = = C/W) W)(50 (2 + C 25 ambient junction ambient junction ambient junction ambient junction R Q T T R T T Q When the fan fails the total thermal resistance is determined from Fig.1523 by reading the value for zero air velocity (the intersection point of the curve with the vertical axis) to be R junction ambient = 70 C / W which yields = C/W) W)(70 (2 + C 25 C 165 = + = = ambient junction ambient junction ambient junction ambient junction R Q T T R T T Q 1556 A PCB with copper cladding is given. The percentages of heat conduction along the copper and epoxy layers as well as the effective thermal conductivity of the PCB are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat conduction along the PCB is onedimensional since heat transfer from side surfaces is negligible. 3 The thermal properties of epoxy and copper layers are constant. Analysis Heat conduction along a layer is proportional to the thermal conductivitythickness product (kt) which is determined for each layer and the entire PCB to be ( ) ( ) . ( ) ( . ) . ( ) ( ) . . . kt kt kt kt kt copper epoxy PCB copper epoxy = = = = = + = + = 386 0 02316 0 26 0 00013 0 02316 000013 0 02329 W / m. C)(0.06 10 m W/ C W / m. C)(0.5 10 m W/ C ( ) W/ C33 Therefore the percentages of heat conduction along the epoxy board are 0.6% 2245 = = = 0056 . C W/ 02316 . C W/ 00013 . ) ( ) ( PCB epoxy epoxy kt kt f and 99.4% = = )% 6 . 100 ( copper f Then the effective thermal conductivity becomes C W/m. 41.6 = + = + + = m 10 0.5) + (0.06 C W/ ) 00013 . 02316 . ( ) ( ) ( 3 copper epoxy copper epoxy eff t t kt kt k 1516 Air 25 C 300 m/min 2 W Q PCB 12 cm 12 cm Epoxy t = 0.5 mm Copper t = 0.06 mm Chapter 15 Cooling of Electronic Equipment 1557 "!PROBLEM 15057" "GIVEN" length=0.12 "[m]" width=0.12 "[m]" "t_copper=0.06 [mm], parameter to be varied" t_epoxy=0.5 "[mm]" k_copper=386 "[W/mC]" k_epoxy=0.26 "[W/mC]" "ANALYSIS" kt_copper=k_copper*t_copper*Convert(mm, m) kt_epoxy=k_epoxy*t_epoxy*Convert(mm, m) kt_PCB=kt_copper+kt_epoxy f_copper=kt_copper/kt_PCB*Convert(, %) f_epoxy=100f_copper k_eff=(kt_epoxy+kt_copper)/((t_epoxy+t_copper)*Convert(mm, m)) T copper [mm] f copper [%] k eff [W/mC] 0.02 98.34 15.1 0.025 98.67 18.63 0.03 98.89 22.09 0.035 99.05 25.5 0.04 99.17 28.83 0.045 99.26 32.11 0.05 99.33 35.33 0.055 99.39 38.49 0.06 99.44 41.59 0.065 99.48 44.64 0.07 99.52 47.63 0.075 99.55 50.57 0.08 99.58 53.47 0.085 99.61 56.31 0.09 99.63 59.1 0.095 99.65 61.85 0.1 99.66 64.55 1517 Chapter 15...
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 Spring '06
 Rajadas

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