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NemiM [27]
3 years ago
8

Assume a 15 cm diameter wafer has a cost of 12, contains 84 dies, and has0.020 defects /cm2Assume a 20 cm diameter wafer has a c

ost of 15, contains 100 dies, and has0.031 defects/cm 21. 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_____________.
Computers and Technology
1 answer:
Vitek1552 [10]3 years ago
6 0

Answer:

1. yield_1=0.959 and yield_2=0.909

2. Cost_1=0.148 and Cost_2=0.165

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

4. defects=0.042 per cm^2 and defects=0.026 per cm^2

Explanation:

1. Find the yield for both wafers.

yield= 1/(1+(defects per unit area*dies per unit area/2))^2

Wafer 1:

Radius=Diameter/2=15/2=7.5 cm

Total Area=pi*r^2=pi(7.5)^2=176.71 cm^2

Area per die= 176.71/84=2.1 cm^2

yield_1= 1/(1+(0.020*2.1/2))^2

yield_1=1/1.04244=0.959

Wafer 2:

Radius=Diameter/2=20/2=10 cm

Total Area=pi*r^2=pi(10)^2=314.159 cm^2

Area per die= 314.159/100=3.14 cm^2

yield_2= 1/(1+(0.031*3.14/2))^2

yield_2=1/1.0997=0.909

2. Find the cost per die for both wafers.

Cost per die= cost per wafer/Dies per wafer*yield

Wafer 1:

Cost_1=12/84*0.959=0.148

Wafer 2:

Cost_2=15/100*0.909=0.165

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.

Wafer 1:

There is a 10% increase in the number of dies

10% of 84 =8.4

New number of dies=84.4+8=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 defects per cm^2

New area per die= 176.71/92.4=1.912 cm^2

yield_1= 1/(1+(0.023*1.912/2))^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 cm^2

New area per die= 314.159/110=2.85 cm^2

yield_2= 1/(1+(0.0356*2.85/2))^2=0.905

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

Assuming a die area of 2cm^2

We have to find the defects per unit area for a yield of 0.92 and 0.95

Rearranging the yield equation,

yield= 1/(1+(defects*die area/2))^2

defects=2*(1/sqrt(yield) - 1)/die area

For 0.92 technology

defects=2*(1/sqrt(0.92) - 1)/2

defects=0.042 per cm^2

For 0.95 technology

defects=2*(1/sqrt(0.95) - 1)/2

defects=0.026 per cm^2

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<u />

========= Shape.java  ===========

//import the Scanner class

import java.util.Scanner;

public class Shape{

   //required fields

  private String shape;

   private double area;

   private double perimeter;

   //default constructor

  public Shape(){

       this.shape = "unknown";

       this.area = 0.0;

       this.perimeter = 0.0;

   }

   //constructor with one parameter

   public Shape(String shape){

       this.setShape(shape);

       this.area = 0.0;

       this.perimeter = 0.0;

   }

   //accessors and mutators

  public void setShape(String shape){

       this.shape = shape;

   }

  public String getShape(){

       return this.shape;

   }

   public double getPerimeter(){

       return this.perimeter;

   }

  public double getArea(){

       return this.area;

   }

  public void setPerimeter(Scanner scr){

       if(this.getShape().equals("circle")){

           System.out.println("Enter the radius of the circle");

           double radius = scr.nextDouble();

           this.perimeter = 2 * 3.142 * radius;

       }

       else if(this.getShape().equals("rectangle")){

           System.out.println("Enter the width");

           double width = scr.nextDouble();

           System.out.println("Enter the height");

           double height = scr.nextDouble();

           this.perimeter = 2 * (width + height);

       }

       else if(this.getShape().equals("square")){

           System.out.println("Enter the height or width");

           double height = scr.nextDouble();

           this.perimeter = 4 * height;

       }

       else if(this.getShape().equals("unknown")){

           System.out.println("You must define a shape first before calculating perimeter");

           this.perimeter = 0.0;

       }

       else {

           System.out.println("You must define a shape first before calculating perimeter");

           this.perimeter = 0.0;

       }

   }

   public void setArea(Scanner scr){

       if(this.getShape().equals("circle")){

           System.out.println("Enter the radius of the circle");

           double radius = scr.nextDouble();

           this.area = 3.142 * radius * radius;

       }

       else if(this.getShape().equals("rectangle")){

           System.out.println("Enter the width");

           double width = scr.nextDouble();

           System.out.println("Enter the height");

           double height = scr.nextDouble();

           this.area = width * height;

       }

       else if(this.getShape().equals("square")){

           System.out.println("Enter the height or width");

           double height = scr.nextDouble();

           this.area = height * height;

       }

       else if(this.getShape().equals("unknown")){

           System.out.println("You must define a shape first before calculating area");

           this.area = 0.0;

       }

       else {

           System.out.println("You must define a shape first before calculating area");

           this.area = 0.0;

       }

   }

   //Own methods

   //1. Method to show the properties of a shape

   public void showProperties(){

       System.out.println();

       System.out.println("The properties of the shape are");

       System.out.println("Shape : " + this.getShape());

       System.out.println("Perimeter : " + this.getPerimeter());

       System.out.println("Area : " + this.getArea());

   

   }

   //2. Method to find and show the difference between the area and perimeter of a shape

   public void getDifference(){

       double diff = this.getArea() - this.getPerimeter();

       System.out.println();

       System.out.println("The difference is " + diff);

   }

}

========= ShapeTest.java  ===========

import java.util.Scanner;

public class ShapeTest {

   public static void main(String [] args){

       Scanner scanner = new Scanner(System.in);

       // create an unknown shape

       Shape shape_unknown = new Shape();

       //get the shape

       System.out.println("The shape is " + shape_unknown.getShape());

       //set the area

       shape_unknown.setArea(scanner);

       //get the area

       System.out.println("The area is " + shape_unknown.getArea());

       //set the perimeter

       shape_unknown.setPerimeter(scanner);

       //get the perimeter

       System.out.println("The perimeter is " + shape_unknown.getPerimeter());

       // create another shape - circle

       Shape shape_circle = new Shape("circle");

       //set the area

       shape_circle.setArea(scanner);

       //get the area

       System.out.println("The area is " + shape_circle.getArea());

       //set the perimeter

       shape_circle.setPerimeter(scanner);

       //get the area

       System.out.println("The perimeter is " + shape_circle.getArea());

       //get the properties

       shape_circle.showProperties();

       //get the difference between area and perimeter

       shape_circle.getDifference();

   }

}

<h2>Sample output:</h2>

The shape is unknown

You must define a shape first before calculating area

The area is 0.0

You must define a shape first before calculating perimeter

The perimeter is 0.0

Enter the radius of the circle

>> 12

The area is 452.448

Enter the radius of the circle

>> 12

The perimeter is 452.448

The properties of the shape are

Shape : circle

Perimeter : 75.408

Area : 452.448

The difference is 377.03999999999996

<h2>Explanation:</h2>

The code above is written in Java. It contains comments explaining important parts of the code. Please go through the code for more explanations. For better formatting, the sample output together with the code files have also been attached to this response.

Download java
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> java </span>
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> java </span>
57aa40cd91dceaa5454e65fc5d209315.png
5 0
3 years ago
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