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To: AK2004 who wrote (28372)	2/13/2001 10:24:14 PM
From: pgerassi	Read Replies (1) \| Respond to of 275872

Dear Albert: Whoops! Actual equation is y = 100 * exp(-(defects per wafer (fpw))/(die per wafer(dpw))) * constant(other factors). The formula I used is for yield in %. It makes assumptions the the defects are randomly spread and constant over the entire area. If you further simplify by using the first order approximation of exp(-x) = 1 - x where x is close to zero, the yield in good die per wafer gdpw = dpw * (1 - fpw / dpw). Thus, gdpw = dpw - fpw. Since flaws per wafer are constant, then the number of good die are about equal to possible die minus the flaws. Assuming a good yield of 90% for P3 at say 250 P3 die per wafer, fpw would be 25. Since P4 has 40% of possible P3 die, the result is 100 - 25 or 75 good P4 die or 225 good P3 die. Thus, the equation now delivers gp4dpw = p4dpw - (1 - p3y) * p3dpw. Assuming that P4 die per wafer is P3 die per wafer / k where k is the ratio of dies per wafer between the P3 die and P4 die, the equation now becomes gp4dpw = p3dpw / k - (1 - p3y) * p3dpw. Simplifying further, gp4dpw = p4dpw * ( 1 / k - 1 + p3y ) * k. Thus the yield of P4 die does not depend on the number of die on the wafer for either P3 or P4 but only on the size ratio and the yield of P3 ratio (for a first order approx.), or p4y = gp4dpw / p4dpw = (p3y - (1 - 1 / k)) * k, or p4y = p3y * k + 1 - k, or p4y = 1 - k * (1 - p3y). Thus to get a P4 yield of 40% with a 2.5 times poss die ratio, P3 yield must be roughly 76%. To get a P4 yield of 50%, P3 yield must be roughly 80% and so on. A P3 yield of 90%, makes for a P4 yield of roughly 75%. To get good die per wafer then gp4dpw = p4y * p4dpw, or gp4dpw = (1 - k * (1 - p3y)) * p3dpw * k. Thus one of the terms seems to be related at the square of the die size ratio. That is what I think the original post refers to. Pete