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Technology Stocks : Advanced Micro Devices - Moderated (AMD) -- Ignore unavailable to you. Want to Upgrade?


To: Sarmad Y. Hermiz who wrote (219002)12/5/2006 2:34:27 PM
From: dougSF30Read Replies (1) | Respond to of 275872
 
Well, yes, that definition of "mature yields" is lame. It would even include the case where you get 1 more 65nm die from a 300mm wafer, on average, which would result in virtually no change in production.

I'd still like confirmation from a non-Inq source of the ~125mm^2 die.



To: Sarmad Y. Hermiz who wrote (219002)12/5/2006 2:36:14 PM
From: combjellyRead Replies (3) | Respond to of 275872
 
"The whole point of the shrink is to get more revenue per wafer."

Yes.

"If yield is poor; and the process cannot produce high bin parts,"

Don't know if either of these is true. Assuming the Inq has it right, then at a minimum they are getting more die per wafer than they are getting at 90nm. How much, we don't know. We also don't know if the process can or cannot bin high clocking parts. Just because it isn't being done that way doesn't mean it can't. AMD very well could be concentrating on maximum yields instead of high clocks. The manufacturer always has this option, but usually the focus is on performance. Which results in more dumpster filler. In addition, they seem to be focussed on lower power, so that is another trade off against max. clock rate.

Another thing we don't know is who the source of this article is. If it is Charlie with the info, then that is a lot more credible than if it is Fuad or Theo.



To: Sarmad Y. Hermiz who wrote (219002)12/5/2006 2:45:47 PM
From: Elmer PhudRead Replies (4) | Respond to of 275872
 
Sarmad

Since both are built on 12 inch wafers, or 300mm as this size is called in Europe, the 33% or so shrink means that AMD will get lower yield as a percentage of dice but it is a net gain in terms of compontents.

Wrong! With the smaller die size AMD should see better yield in terms of both percent and total die. All else being equal.

Unfortunately all things are not always equal...