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Technology Stocks : Advanced Micro Devices - Moderated (AMD)

AMD 259.65

+2.3%

Jan 23 9:30 AM EST

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To: pgerassi who wrote (21393)	12/3/2000 10:38:57 PM
From: Ali Chen	Read Replies (3) of 275872

Pete, <If the wafer is 1mm thick and the thermal conductivity of common silicon is .8W/cmC, The wafer will cause a thermal drop of (45.7W/sqcm)(.1cm)/(.8W/cmC) or 5.7C.> <Most good heatsinks are rated ... In this case, the drop is (55W)(.35 C/W) or 19.3C.> <This places almost all of the thermal resistance in the bulk die itself. Here a 40% increase in conductivity will drop the temperatures 30%> From what you wrote one can conclude that for the more or less realistic and manufacturable conditions, the die is responsible for 25% of the total resistance. Making the die even infinitely conductive would save you those 5.7C, or only 25% of total temperature difference. How did you arrive to 30% savings is beyond my comprehension ;) :) I reiterate my opinion on Insaneonic: this is BS. The 30% number is highly misleading, to say very softly. Reminds me the RMBS. Regards, - Ali

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