HW4 solution

# HW4 solution - V Sol/MTION Name CHE/ME 109 Homework 4...

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Unformatted text preview: V) Sol/MTION Name CHE/ME 109 Homework 4 Silicon chip 0.5 m Ceramic substrate Problem 1. A low power silicon chip (kchip = 110 W/mOC) is attached to a ceramic substrate. The chip is 6 mm x 6 mm x 0.5 mm. The chip is bonded to the ceramic with an epoxy (layer is approximately .2mm thick) that has thermal conductivity (kepoxy) of 9.3 W/mOC and the ceramic substrate has thermal conductivity (kbase) of 9.3 W/mOC. The top face of the chip is in air at room temperature (Too=250C) and the convective heat transfer coefficient (h) is 140 W/mZOC- Subswvlvt \§ lnnwx +l/HC'K a. Calculate the temperature of the upper surface of the chip if 90% of the heat is removed by convection from the upper surface of the chip. o.9*3W = hA(TS - To.) = (140 W/m2 OC)(.oo6*.oo6m)(TS - 25) 2.7 = 14o*3.6x10-5*( TS - 25) => Ts = 561°C b. Calculate the AT across the silicon chip. - AT . . . Q z —kA— => this 1s not correct, because of the heat generation, one must Ax write the differential equation and solve for the temperatures. Given the high value of k for the chip and the very high surface temperatures, which incidentally are too high for any real silicon chip to sustain, the temperature drop across the chip is negligible. c. Diagram the thermal resistance network for this problem. \ TV T5 ‘ x), (1. Using the thermal resistance concept, calculate the thermal resistances in the epoxy and the ceramic substrate. Determine the temperature at the base of the ceramic substrate on the opposite face from the silicon chip. * The temperature at the base of the silicon chip is approximately 561°C and the q = -.3W/[3.6x10'5m2] = -8333 W/m2. Then the resistances can be calculated ‘ __ , A m :- gk'g’“ ' . R‘s ' marl. RWY ‘13 w. I 9252290er mt r 4 L Q (RhﬁL: X(O- (1:20“ H, turf a “\$3333 1 Tswt~\$é>l I’lﬁxro“? Plexi gl &% uwer \ Copper Tmnsistor plane \ Problem 2. Six identical power transistors with aluminum casing are attached on one side of a 1.2 cm-thick 2o—cm x 30-cm copper plate (k = 386 W/mOC) by screws that exert an average pressure of 10 MPa. The base area of each transistor is 9 cm2, and each transistor is placed at the center of 10-cm x 10—cm section of the plate. The interface roughness is estimated to be about 1.4 pm. All transistors are covered by a thick Plexiglas layer, which is a poor conductor of heat, and thus all the heat generated at the junction of the transistor must be dissipated to the ambient at 150C through the back surface of the copper plate. The combined convection/ radiation heat transfer coefﬁcient at the back surface can be taken to be 30 W/m2 0C. If the case temperature of the transistor is not to exceed 85°C, determine the maximum power each transistor can dissipate safely, and the temperature jump at the case—plate interface. .C’OOZM 0 Properties The therinal conductivity of copper is given to be k = 386 W/m-°C. The contact conductance at the interface of copper-aluminum plates for the case of 1.3-1.4 pm roughness and 10 MPa pressure is hc = 49,000 W/m2-°C (Table 32). Analysis The contact area between the case and 'the plate is given to be 9 cmzt and the plate area f0r each transistor is 100 cm3. The thermal resistance network of this problem consists of three resistances in series (contact, plate, and convection) which are determined to be . 1 1 l R = = A = 0.0227 °C/ w “mm hcA‘. (49,000 W/ m2.°C)(9 x10" m2) Rpm =L=——————————0'0l2"' , =0.0031°0/w [04 (386 W/m.°C)(0.0l m') l 1 l Roonvection = = 3.333 °C/ W ,hoA = (30 W/m2.°C)(0.0l m2) The total thermal resistance is then Rm, = 12mm +‘R,,,mc + Rconvection = 0.0227 + 0.0031+3.333 = 3.359 °c / w Note that the thermal resistance of copper plate is very small and can be ignored all together. Then the rate of heat transfer is determined to be Q: AT z (85—15)C =20'8w 3.359°C/w Therefore, the power transistor should not be operated at power levels greater than 20.8 W if the case temperature is not to exceed 85°C. The temperature jump/at the interface is determined from ('1 K0“ K34; Kantia- ATMface = kam, = (20.8 W)(0.0227 °C/W) = 0.47°C ‘ <- '- which is not very large. Therefore, even if we eliminate the themial contact resistance at the interface completely, we will lower the operating temperature of the transistor in this case by less than l°C. 0.05 cm 2‘ I Copper filling Epoxy board Problem 3. Consider 15 cm x 20 cm epoxy glass laminate (k = 0.2 W/mOC) whose thickness is 0.13 cm. In order to reduce the thermal resistance across its thlckness, cylindrical copper ﬁllings (k = 417 W/mOC) of 0.05 cm in diameter to be planted throughout the board, with a center-to-center distanceof 0.15 cm. Calculate the value of the thermal resistance of the board with and without the copper ﬁllings. Aland : ’UJS cm X 70-2 m z i 0503. /i.— ] pg llVl (£006) ‘ (— i )1 32 A = (my? \3 32 “ ’OOOSW‘ " ‘32 6”") m we“ 7r ’ I 4' 3W, “ - 0‘s mz’ Q'L’ZNO I ﬂagoxy “ ’ Acoﬁxar: ’ 0113? m2 _. 0‘ CC Mal WSLS‘l—HW : ,DOll I waefz EA 3 m1 “:3; ;,(022<w‘ "‘ ‘I r We ‘ 2-2?th— Problem 4. A 40—W power transistor is to be cooled by attaching it to one o commercially available heat sinks shown in Table 3-4-in the textbook. Select a heat sink that will allow the case-temperature of the transistor not to exceed 80°C in the ambient air at 350C. I v AT . T - Q - Q ' V :- ‘l'anStS‘Hr 0V RmSe/‘ambien‘l' “xﬂm'o‘m-f 4 0 w l l Thy. flu/rle rCSis+anot air‘m WJr sink V mm“ W Wow {429%. RM Tqbu 37+ H5 5030 in Véf‘h‘cal postb’on/ H6 éH‘S in VQf‘ch) {JOSllﬁm/i CKV‘Q W only 2 oF‘h W15, Problem 5. A hot surface at 100°C is to be cooled by attaching 3—cm-long, 0.25- cm-diameter aluminum pin fins (k=237 W/mOC) to it, with a center—to—center distance of 0.6 cm. The temperature of the surrounding medium is 300C, and the heat transfer coefficient on the surfaces is 35 W/m2 oC. Determine the rate of heat transfer from the surface for a 1-m x 1—m section of the plate. Also determine the overall effectiveness of the fins. Ti) : [OODC pins. long) 07-6 cm ciia Top = 50°C ) =35 \_/\L distal/met ~= ,Qcm aggari‘ smug AH“ ; D). 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HW4 solution - V Sol/MTION Name CHE/ME 109 Homework 4...

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