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 233.54-1.8%Nov 7 9:30 AM EST

 Public ReplyPrvt ReplyMark as Last ReadFilePrevious 10Next 10PreviousNext  
To: kpf who wrote (219055)12/5/2006 11:27:53 PM
From: misenRead Replies (4) of 275872
 
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
Report TOU ViolationShare This Post
 Public ReplyPrvt ReplyMark as Last ReadFilePrevious 10Next 10PreviousNext