**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 |