SI
SI
discoversearch

We've detected that you're using an ad content blocking browser plug-in or feature. Ads provide a critical source of revenue to the continued operation of Silicon Investor.  We ask that you disable ad blocking while on Silicon Investor in the best interests of our community.  If you are not using an ad blocker but are still receiving this message, make sure your browser's tracking protection is set to the 'standard' level.
Technology Stocks : Advanced Micro Devices - Moderated (AMD)
AMD 203.14-0.8%Jan 9 9:30 AM EST
 Public ReplyPrvt ReplyMark as Last ReadFilePrevious 10Next 10PreviousNext  
To: dougSF30 who wrote (133646)9/23/2004 11:06:30 PM
From: Pravin KamdarRead Replies (3) of 275872
 
Doug,

The point is that one would expect the die temperature of an 84mm^2 part dissipating the same power as a 144mm^2 part to be considerably higher.

Therefore, the fact that the 90nm "die-like" temperature is only up a little bit, *and* that is with the 90nm part being overvolted by 5%, suggests that the 90nm part is dissipating LESS power than the 130nm Newcastle.

That doesn't seem logical to me. Consider a 'perfect' heat sink attached to both the 130nm and the 90 nm chips through a perfect thermal conductor between heatsink and die. In this limit, both would be at surrounding ambient. Now, consider a 'good' heatsink with low thermal resistivity contact between the die and the heatsink. I would expect that this combination would keep the 90 nm die almost as cool as the 130nm die. Although the power density in the 90nm chip will be higher, its temperature depends on how fast the energy is extracted.

Pravin
Report TOU ViolationShare This Post
 Public ReplyPrvt ReplyMark as Last ReadFilePrevious 10Next 10PreviousNext  

HomeMailHotSubjectMarksPeopleMarksKeepersSettings
Terms Of UseContact UsCopyright/IP PolicyPrivacy PolicyAbout UsFAQAdvertise on SI
© 2026 Knight Sac Media.  Data provided by Twelve Data, Alpha Vantage, and CityFALCON News