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

To: kpf who wrote (219055)	12/5/2006 4:06:26 PM
From: Elmer Phud	Read Replies (1) \| Respond to of 275869

kpf I don't see why. All else being equal, for a reasonable approximation to yields I'd suggest rather to look at transistorcount than diesize. Not necessarily so. Die size is a much better predictor within a reasonable range of cache sizes and it includes the size taken by redundancy. because designs for manufacturability compensate for the above by means of more redundancy for higher transistorcounts. Edit: Synopsis: For a given design adding redundancy obviously increases diesize and yield. :) You talk redundancy but you didn't mention cache where it's used. Yes, that will lift yield but at a price. The amount of redundancy you add takes up die size too. There's no reason to add more than you reasonably need, which gets us back to defect density. The lower the DD, the less redundancy you need to add so the lower the die size and the higher the yield. Which makes one wonder if it's related to why AMD reduced the cache size....

To: kpf who wrote (219055)	12/5/2006 11:27:53 PM
From: misen	Read Replies (4) \| Respond to of 275869

With the smaller die size AMD should see better yield in terms of both percent and total die. All else being equal. I don't see why. All else being equal, for a reasonable approximation to yields I'd suggest rather to look at transistorcount than diesize. Here is why Elmer's statement is generally true. First, assume that the defect density on the 90 nm process and 65 nm process are equal. Second, assume that yields are defect limited. Third, assume that defects are randomly distributed across a wafer. With these assumptions, a wafer with 200 candidate die and 50 defects would have ~ 150 good die or ~ 75% yield (yield would be somewhat higher as some die would have 2 or more defects of the 50). With a 50% reduction in die size, there should be about 400 candidate die => ~ 350 good die or 87.5% yield. In other words, a 50% reduction in die size should give ~ a 50% reduction in fail rate (percentage of failing die) This is the basis for the simple Poisson model. You may argue with the assumptions above, but the assumptions are close enough to reality that the example represents real behavior. Misen