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ArbitrLikvidat [17]
2 years ago
9

Assume a fabrication process improves the yield from 0.92 to 0.95. find the defects per area unit for each version of the techno

logy given a die area of 200 mm.
Physics
1 answer:
Irina18 [472]2 years ago
3 0

Answer:

Fabrication process that improves the yield from 0.92 to 0.95. The defects per area unit for 0.92 and 0.95 technology are 0.042 per cm^2 and 0.026 per cm^2 respectively

Explanation:

The yield is increased by a manufacturing procedure from 0.92 to 0.95 the defects will be;

Given Data

Suppose the area of the die is 2 Cm^2

The defects per unit area with a yield of 0.92 and 0.95 must be determined

Solution

Equation for yield

Yield= 1/(1+(defects × die area/2)^2

The yield equation has been rearranged

Defects= 2×(1√(yield)-1)/die area

First, we find for the technology of 0.92

Defects= 2×(1√(yield)-1)/die area

Putting the value of yield and die are which is 0.92 and 2Cm^2 respectively

Defects=2× (1√(0.92-1)/2

Defects= 0.042 per Cm^2

Now, find for the technology of 0.95

Putting the value of yield and die are which is 0.95 and 2Cm^2 respectively

Defects=2× (1√(0.95-1)/2

Defects=0.026 per Cm^2

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b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

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