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To: combjelly who wrote (133671)	9/23/2004 11:41:33 PM
From: Pravin Kamdar	Read Replies (2) \| Respond to of 275872

Combjelly, Having thought about it for another minute, the rate of energy extraction must equal the rate of heat generation, or the temp would rise until meltdown. THe heat flow rate will be dependent on the temperature gradient between the die and the heatsink; die temperature and heatsink temperature being the boundary values. All else being equal, the temp gradient on the 90nm must be greater than that of the 130 nm (due to smaller area). This can occur in two ways: Keeping the case ambient the same would require a higher die temp. Lowering the ambient would give a lower die temp for the same gradient. My guess is that a good heat spreader between die and heatsink, and a slightly lower ambient requirement (better air flow) would give a die temp not much higher than the 130 nm part. Pravin

To: combjelly who wrote (133671)	9/24/2004 1:16:38 AM
From: heatsinker2	Respond to of 275872

Perhaps heatsinker2 should weigh in on this issue Hey, thanks for thinking of me. With an ideal interface, spreader and heatsink, there won't be any difference between 90 and 130. But I don't think we'll have a perfect spreader, so the 90nm chip should run a few degrees hotter. Also, there will be a slightly greater dT due to the chip-to-spreader interface. This means that, for equal heatsinking, the 90 nm chip won't be able to dissipate as much heat while maintaining the same Tj. But we don't really know much, do we? Who knows what that hkpc guy was actually measuring? And Hector has very clearly stated that the dissipation characteristics for 90nm are excellent :-)