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To: Joe NYC who wrote (107560)	8/14/2000 10:56:10 PM
From: semiconeng	Read Replies (2) \| Respond to of 186894

At half of your Fab25 estimate (or 1500 wpw) + 1000 wpw Fab, the 3.6 M would be acceptable (but not outstanding) yield. Joe Sorry Joe, the numbers don't add up. At current die size, each 8 inch wafer can contain greater than 250 die, so: T-Bird Die Size (Largest Size Athlon Die) = 117 mm2 tomshardware.de Area of circle (Wafer Surface Area) = pi * Radius squared 3.14 * (100mm) squared 3.14 * 10,000 = 31,400mm Wafer Area / Die Area = Total Die Per Wafer 31,400 / 117 = 268.37 Die Per Wafer But let's be fair, and count for partial die, and say 250 Die Per wafer. Total wafers per quarter: 1500 * 12 = 18,000 Wafers/Quarter F25 1000 * 12 = 12,000 Wafers/Quarter F30 ------------------------------------- 30,000 Wafers/Quarter 2 Fabs 30,000 * 250 Die Per Wafer = 7.5 Million Die/Quarter Assuming 100% Yield of course. So if AMD is claiming 3.6 Million, that would mean a yield of less than 50%. A yield of 48%, doesn't sound too good to me. SemiconEng

To: Joe NYC who wrote (107560)	8/14/2000 11:07:38 PM
From: Elmer	Read Replies (1) \| Respond to of 186894

Re: "At half of your Fab25 estimate (or 1500 wpw) + 1000 wpw Fab, the 3.6 M would be acceptable (but not outstanding) yield" It would come out to about 117 DPW and that's with a conservative estimate of 2500 total wafers per week and only ~25% of Fab25s capacity given to Athlon. A more realistic backend yield estimate of 97.5% would put their yield at ~114 DPW. Either way I would consider that to be poor yield for a die TBird's size. The number of wafer starts could easily be higher making the yield even lower. Sorry that this isn't what you want to hear but that's what the numbers say. EP