Question

4. Assume a 15 cm diameter wafer has a cost of 12, contains 84 dies, and has 0.020 defects/cm2. Assume a 20 cm diameter wafer has a cost of 15, contains 100 dies, and has 0.031 defects/cm2 a) Find the yield for both wafers. b) Find the cost per die for both wafers. c) If the number of dies per wafer is increased by 10% and the defects per area unit increases by 15%, find the die area and yield. d) Assume a fabrication process improves the yield from 0.92 to 0.95. Find the defects per area unit for each version of the technology given a die area of 200 mm2.

Answer 1

4. a) Die area, Wafer area, T(4-5cm) = 2.1037 omos 84 Die count, 1 T1 cm² Die area, a I (10 cm2) 1 Yieldy Die areas (1+ Defect rate 1 2 (1 +(0.020)(0.5) (2.1037)) = 0.959216 1 = 0.909289 Yield, (1+0.31 (0.5)(10)) = by Cost per die, Cost per waters Dies per water * Yield 12 (84)(0.959216) = 0.148931. 15 -0.164964 Cost per die, (100) 0.909289)

St cy for 1" Water. Water area Die area new (1.1) (Die Count) 2.103745 lol cm2 -1.912495 cm 2 Yield New Die area 1+ (1.15 )(Defect rate) new ( 2 ! (+ (115)(0-020) -91249512 = 0.957411 Water & For and T ㅠ 26855993 cm2 Die area New t-1.0m2 ! Yield New 2.855993 2 1939 (i+ (1.15)(0.031) = 0.905462

di Since, the die area is 2 square cm. 1 Yield ('+(Defect rate) 2 cm2 2 1 1 (1+ Defect rate)? Defect rate Yield So, Defect rate Defect 10.92 1 old $ 0.042572 defects cm2 1 Defect rate -1 new 10.95 = 0.025978 defects cm2

Hex Decimal Binary (8 bits) (2 digits) 18 0001 0010 12 63 0011 llll 3F 46 0010 1110 ZE 227 1110 0011 E3 145 1001 0001 91 СІ 193 1100 0001 OA to 0000 1010 17 23 0001 0111 55 85 0101 0101 96 0110 150 1001