Question

This background applies to the next several questions. 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.

Find the yield for both wafers. 2. Find the cost per die for both wafers. 3. 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. 4. 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
200mm2

Answer 1

1)
`Yield_1= (1)/((1+ 0.02 (1)/(2) 2.104)^2)=0.959`

`Yield_2= (1)/((1+ 0.031 (1)/(2) 3.1415)^2)=0.909`

2)
`Cost/die_1 = (12)/(84 x 0.959)=0.149`

`Cost/die_2 = (15)/(100 x 0.909)=0.165`

3)
`Area_1 = (1.1 \pi (7.5cm)^2)/(84)=(2.104 cm^2)/(1.1)=1.913 cm^2`

`Area_2 = (1.1 \pi (10cm)^2)/(100)=(3.1415 cm^2)/(1.1)=2.856 cm^2`

And for the new yield we need to take in count the increase of 15% for the area and we got this:

`Yield_1= (1)/((1+(1.15) 0.02 (1)/(2) 1.913)^2)=0.957`

`Yield_2= (1)/((1+(1.15) 0.031 (1)/(2) 2.856)^2)=0.905`

4)
`DR_(old)=(1)/(√(0.92)) -1`=0.0426 defects/cm^2

`DR_(new)=(1)/(√(0.95)) -1`=0.0260defects/cm^2

Explanation:

Part 1

For this part first we need to find the die areas with the following formula:

`Area= (W area)/(Number count)`

`Area_1 = (\pi (7.5cm)^2)/(84)=2.104 cm^2`

`Area_2 = (\pi (10cm)^2)/(100)=3.1415 cm^2`

Now we can use the yield equation given by:

`Yield=(1)/((1+ DR(Area)/(2))^2)`

And replacing we got:

`Yield_1= (1)/((1+ 0.02 (1)/(2) 2.104)^2)=0.959`

`Yield_2= (1)/((1+ 0.031 (1)/(2) 3.1415)^2)=0.909`

Part 2

For this part we can use the formula for cost per die like this:

`Cost/die = (Cost per day_i)/(Number count_i x Yield_i)`

And replacing we got:

`Cost/die_1 = (12)/(84 x 0.959)=0.149`

`Cost/die_2 = (15)/(100 x 0.909)=0.165`

Part 3

For this case we just need to calculate the new area and the new yield with the same formulas for part a, adn we got:

`Area_1 = (1.1 \pi (7.5cm)^2)/(84)=(2.104 cm^2)/(1.1)=1.913 cm^2`

`Area_2 = (1.1 \pi (10cm)^2)/(100)=(3.1415 cm^2)/(1.1)=2.856 cm^2`

And for the new yield we need to take in count the increase of 15% for the area and we got this:

`Yield_1= (1)/((1+(1.15) 0.02 (1)/(2) 1.913)^2)=0.957`

`Yield_2= (1)/((1+(1.15) 0.031 (1)/(2) 2.856)^2)=0.905`

Part 4

First we can convert the area to cm^2 and we got 2 cm^2 the yield would be on this case given by:

`Yield= (1)/((1+DR(2cm^2)/(2))^2)=(1)/(1+(DR)^2)`

And if we solve for the Defect rate we got:

`DR= (1)/(√(Yield))-1`

Now we can find the previous and new defect rate like this:

`DR_(old)=(1)/(√(0.92)) -1`=0.0426 defects/cm^2

And for the new defect rate we got:

`DR_(new)=(1)/(√(0.95)) -1`=0.0260defects/cm^2