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Unformatted text preview: Chapter 15 Cooling of Electronic Equipment 15-140 A multilayer circuit board consisting of four layers of copper and three layers of glass-epoxy sandwiched together is considered. The magnitude and location of the maximum temperature that occurs in the PCB are to be determined. Assumptions 1 Steady operating conditions exist 2 Thermal properties are constant. 3 There is no direct heat dissipation from the surface of the PCB, and thus all the heat generated is conducted by the PCB to the heat sink. Analysis The effective thermal conductivity of the board is determined from C W/ 0.00039 = m)] C)(0.0005 W/m. 26 . [( 3 ) ( C W/ 0.1544 = m)] C)(0.0001 W/m. 386 [( 4 ) ( 2 2 1 1 = = epoxy copper t k t k C W/m. 5 . 81 ) m 0005 . ( 3 ) m 0001 . ( 4 C W/ ) 00039 . 1544 . ( ) ( ) ( 2 1 2 2 1 1 = + + = + + = t t t k t k k epoxy copper eff The maximum temperature will occur in the middle of the board which is farthest away from the heat sink. We consider half of the board because of symmetry, and divide the region in 1-cm thick strips, starting at the mid-plane. Then from Fouriers law, the temperature difference across a strip can be determined from Q k A T L T QL k A eff eff = = where L = 1 cm = 0.01 m (except it is 0.5 cm for the strip attached to the heat sink), and the heat transfer area for all the strips is A = = ( . 015 m)[4(0.0001 m) + 3(0.0005 m)] 0.000285 m 2 Then the temperature at the center of the board is determined by adding the temperature differences across all the strips as T QL k A Q Q Q Q Q Q Q Q L k A eff eff center heat sink 2 W m) W / m. C m ) C- = = + + + + + + + = + + + + + + + = ( / ) ( . . . . . / )( . ( . )( . . 1 2 3 4 5 6 7 8 2 15 3 4 5 6 7 5 9 105 1125 2 0 01 815 0 000285 205 and T T T center heat sink center heat sink C + C = + = = - 35 205 . 55.5 C Discussion This problem can also be solved approximately by using the average heat transfer rate for the entire half board, and treating it as a constant. The heat transfer rate in each half changes from 0 at the center to 22.5/2 = 11.25 W at the heat sink, with an average of 11.25/2 = 5.625 W. Then the center temperature becomes Q k A T T L T T Q L k A ave eff ave eff 2245- 2245 + = = 1 2 35 center heat sink 2 C + (5.625 W)(0.075 m) (81.5 W / m. C)(0.000285 m ) 53.2 C 15-74 Copper Glass-epoxy 15 cm 15 cm 11.25 W 10.5 W 9 W 7.5 W 6 W 4.5 W 3 W 1.5 W Heat sink Center 1 cm 1 cm 0.5 cm Chapter 15 Cooling of Electronic Equipment 15-141 A circuit board consisting of a single layer of glass-epoxy is considered. The magnitude and location of the maximum temperature that occurs in the PCB are to be determined....
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This note was uploaded on 08/25/2009 for the course AET AET432 taught by Professor Rajadas during the Spring '06 term at ASU.
- Spring '06