Question

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 .

1.10.1 Find the yield for both wafers.

1.10.2 Find the cost per die for both wafers.

1.10.3 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.

1.10.4 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 200 mm²

Short answer

1.10.1 yield_1=0.959 and yield_2=0.909.

1.10.2. Cost per die of first wafer =0.148

Cost per die of second wafer =0.165.

1.10.3. New area per die=1.912 cm^2 and yield_1=0.957

New area per.die=2.85 cm^2 and yield_2=0.905.

1.10.4 Defects=0.042 per cm^2 and defects=0.026 per cm^2

Step-by-step solution

Determine the formula.

Write a formula for Die area

$Diearea=\frac{Waferarea}{Diesperwafer}=\frac{{\mathrm{\pi r}}^2}{Diesperwafer}$ ……(1)

Write a formula for Total area

${\mathrm{\pi r}}^2$ ……(2)

Write a formula for Radius

$\frac{Diameter}{2}$

.…(3)

Write a formula for Cost per die

$\cos tperdie=\frac{\cos tperwafer}{Diesperwafer}$ ……(4)

Write a formula for yield

$yield=\frac{1}{{\left(1+\left(defects\times diearea/2\right)\right)}^2}$ .…(5)

Write a formula for defects

$defects=\frac{\left(2*{\displaystyle \frac{1}{sqrtyield}}-1\right)}{diearea}$ ' .…(6)

Find the yield for both wafers.

Given :defects=0.020

$Diearea=\frac{3.14{\left({\displaystyle \frac{15}{2}}\right)}^2}{84}=2.1$

role="math" localid="1650932782104" $$\begin{gathered} yiel{d}_1=\frac{1}{{\left(1+0.020\times 2.1/2\right)}^2} \\ =0.959 \end{gathered}$$

Wafer 2:Now yield of 2^nd wafer is calculated similarly.

Radius=20/2=10cm

Total area=$$\begin{gathered} \mathrm{\pi }\times {10}^2 \\ =3.14\times 100 \\ =314.159 \end{gathered}$$

Area per die=314.159/100=3.14

role="math" localid="1650932903756" $$\begin{gathered} yiel{d}_2=\frac{1}{{\left(1+0.031\times 3.14/2\right)}^2} \\ =0.909 \end{gathered}$$

Calculate the per-die cost for each wafers

1.10.2

Wafer 1:

Cost_1=$\frac{12}{84\times 0.959}=0.148$

Wafer 2:

$Cost\_2=\frac{15}{100\times 0.909}=0.165$

find the die area and yield if increased 

Wafer 1:

There is a 10% increase in the number of dies

10% of 84=8.4

New number of dies =84+8.4=92.4

There is a 15% increase in the defects per cm^2

15% of 0.020=0.003

New defects per area=0.020+0.003=0.023

New area per die=176.71/92.4 =1.912 cm²

$yiel{d}_1=\frac{1}{{\left(1+0.023\times {\displaystyle \frac{1.912}{2}}\right)}^2}=0.957$

Wafer 2:

There is a 10% increase in the number of dies

10% of 100=10

New number of dies 100+10=110

There is a 15% increase in the defects per cm^2

15% of 0.031=0.0046

New defects per area=0.031+0.00465=0.0356 defects per cm²

New area per die=314.159/110=2.85 cm²

$yiel{d}_2=\frac{1}{{\left(1+{\displaystyle \frac{0.0356\times 2.85}{2}}\right)}^2}=0.905$

Find defects per unit area

1.10.4

Assuming a die area of 2 cm^2

For yields of 0.92 and 0.95, we must determine the faults per unit

area.Rearranging the yield equation,

defects=2*(1/sqrt(yield-1)/diearea

For 0.92 technology

defects=2*(1/sqrt(0.92)-1)/2=0.042 per cm²

For 0.95 technology

defects=2*(1/sqrt(0.95)-1)/2=0.026 per cm²