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

Answer:

1) $Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959$

$Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909$

2) $Cost/die_1 = \frac{12}{84 x 0.959}=0.149$

$Cost/die_2 = \frac{15}{100 x 0.909}=0.165$

3) $Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2$

$Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2$

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

$Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957$

$Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905$

4) $DR_{old}=\frac{1}{\sqrt{0.92}} -1$=0.0426 defects/cm^2

$DR_{new}=\frac{1}{\sqrt{0.95}} -1$=0.0260defects/cm^2

Step-by-step explanation:

Part 1

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

$Area= \frac{W area}{Number count}$

$Area_1 = \frac{\pi (7.5cm)^2}{84}=2.104 cm^2$

$Area_2 = \frac{\pi (10cm)^2}{100}=3.1415 cm^2$

Now we can use the yield equation given by:

$Yield=\frac{1}{(1+ DR\frac{Area}{2})^2}$

And replacing we got:

$Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959$

$Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909$

Part 2

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

$Cost/die = \frac{Cost per day_i}{Number count_i x Yield_i}$

And replacing we got:

$Cost/die_1 = \frac{12}{84 x 0.959}=0.149$

$Cost/die_2 = \frac{15}{100 x 0.909}=0.165$

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:

$Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2$

$Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2$

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

$Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957$

$Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905$

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:

$Yield= \frac{1}{(1+DR\frac{2cm^2}{2})^2}=\frac{1}{1+(DR)^2}$

And if we solve for the Defect rate we got:

$DR= \frac{1}{\sqrt{Yield}}-1$

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

$DR_{old}=\frac{1}{\sqrt{0.92}} -1$=0.0426 defects/cm^2

And for the new defect rate we got:

$DR_{new}=\frac{1}{\sqrt{0.95}} -1$=0.0260defects/cm^2