MICRON TECHNOLOGY, INC., Semiconductor Manufacturing Engineer

Math in the Workplace - Algebra/Geometry

MICRON
TECHNOLOGY, INC.,

Semiconductor Manufacturing Engineer - Shrink

Job Description: Responsible for all aspects of product development including design verification and circuit debug, device characterization, text methodology, yield optimization, and cost reduction.

Problem:

In the semiconductor industry, shrink refers to the reduction in die (or chip) size.

Reducing the size of the die can reduce the cost of manufacturing per die, since you get more die on each wafer.
Computers and other devices that require memory are getting smaller and smaller, yet they need more and more memory. Therefore, the memory chips need to be smaller.
The smaller the die, the smaller or shorter the circuitry. This means the electrical charges can travel faster from one point to the next. Therefore, shrinking the die can also increase its speed.

8-inch wafer

Wafer diameter = 8 inches
Area of a circle = p r²(pie x radius²)

Original chip size:

x = .5 inch
y = 1 inch

1. What is the area of the wafer?

2. What is the area of the chip?

3. Find the number of possible chips per wafer?

The process engineers tell us that we can shrink the chip by 20%...

4. What are the new x and y dimensions of the chip?

5. What is the new area of the chip?

6. The x and y dimensions are reduced by a factor of 0.8. By what percentage did the area decrease?

7. Find the number of possible chips per wafer with the new chip size?

ORIGINAL chip size: ______ good chips per wafer.
New chip size: ______ good chips per wafer.

$solution$

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Solution:

Original chip size:

x = .5 inch
y = 1 inch

1. What is the area of the wafer?

p r²= area
3.14 x 4²inches = area
3.14 x 16 = 50.24 sq inches

2. What is the area of the chip?

x ^. y = area
.5 inch x 1 inch = .5 square inch

3. Find the number of possible chips per wafer?

A ₁ $divided by$ A₂ = # of chips
50.24 $divided by$ .5 = 100.48 or 100 possible chips

The process engineers tell us that we can shrink the dimensions of the chip by 20%...

4. What are the new x and y dimensions of the chip?

100%- 20% = 80%

.5 x 80% = .4 inches 1 x 80% = .8 inches

5. What is the new area of the chip?

length ^. width = area
.4 inch x .8 inch = .32 square inches

6. The new x and y dimensions are 80% of the original. By what percentage did the area decrease?

(% of length) x (% of width) = (% of area)
80%² = .64% of original
or
new area $divided by$ original area = % of original
.32 square inches $divided by$ .5 square inches = 64%

100% - 64% = 36% decrease