All categories

2 years ago

name the three exposure techniques in photolithography. what are the alternatives to photolithography in ic processing?

Engineering

1 answer:

zhenek [66]2 years ago

6 0

The three exposure techniques in photolithography are:

Contact
Proximity
Projection

Alternatives to photolithography in IC processing include;

X-ray
UV
Ion, and
Electron lithography

<h3>What is Photolithography?</h3>

Photolithography is a term in integrated circuit development that describes the patterned films that are formed when a beam of light falls on a substance.

This phenomenon protects the surface of sensitive materials such as glass during some operations like etching. UV and X-rays can be used for this purpose.

Learn more about photolithography here:

brainly.com/question/13650094

#SPJ11

You might be interested in

Find the current Lx in the figure

AleksandrR [38]

Explanation:

$\frac{1}{8} + \frac{1}{2} \\ 1.6 + 1.4 = 3 \\ \frac{1}{3} + \frac{1}{9} \\ 2.25 + 2 = 4.25 \: ohm$

R total = 4.25 ohm

I total = Vt/Rt

I total= 17/4.25= 4 A

Ix= 600 mA

$\frac{9}{9 + 3} \times 4 = 3\\ \frac{2}{2 + 8} \times 3 = 0.6a \\ = 0.6 \: milli \: amper$

6 0

3 years ago

PLZZ HELP

9966 [12]

Answer:

Could ask a family member to help

Explanation:

5 0

3 years ago

Read 2 more answers

A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse

d1i1m1o1n [39]

Answer:

The dude was killed around 6:30PM

Explanation:

Newton's law of cooling states:

$T = T_m + (T_0-T_m)e^{kt}$

where,

$T_0$ = initial temp

$T_m$ = temp of room

$T$ = temp after t hours

$k$ = how fast the temp is changing

$t$ = time (hours)

$T_0 = 31$ because the body was initlally 31ºC when the police found it

$T_m = 22$ because that was the room temp

$T = 30$ because the body temp drop to 30ºC after 1 hour

t = 1 because that's the time it took for the body temp to drop to 30ºC

$k=???$ we don't know k so we must solve for this

rearrange the equation to solve for k

$T = T_m + (T_0-T_m)e^{kt}$

$T - T_m= (T_0-T_m)e^{kt}$

$\frac{T - T_m}{(T_0-T_m)}= e^{kt}$

$ln(\frac{T - T_m}{T_0-T_m})=kt$

$\frac{ln(\frac{T - T_m}{T_0-T_m})}{t}=k$

plug in the numbers to solve for k

$k = \frac{ln(\frac{T - T_m}{T_0-T_m})}{t}$

$k = \frac{ln(\frac{30 - 22}{31-22})}{1}$

$k=ln(\frac{8}{9})$

Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.

$T_0 = 37$ because the body was 37ºC right after being killed

$T_m = 22$ because that was the room temp

$T = 31$ because the body temp when the police found it

$k=ln(\frac{8}{9})$ we solved this earlier

t = ??? we don't know how long it took from the time of the murder to when the police found the body

Rearrange the equation to solve for t

$T = T_m + (T_0-T_m)e^{kt}$

$T - T_m= (T_0-T_m)e^{kt}$

$\frac{T - T_m}{(T_0-T_m)}= e^{kt}$

$ln(\frac{T - T_m}{T_0-T_m})=kt$

$\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}=t$

plug in the values

$t=\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}$

$t=\frac{ln(\frac{31 - 22}{37-22})}{ln(8/9)}$

$t=\frac{ln(3/5)}{ln(8/9)}$

t ≈ 4.337 hours from the time the body was killed to when the police found it.

The police found the body at 11:00PM so subtract 4.337 from that.

11 - 4.33 = 6.66 ≈ 6:30PM

7 0

3 years ago

What happen to the clutch system when you step-on and releasing the clutch pedal?

soldi70 [24.7K]

Answer:

Step On: Your foot forces the clutch pedal down and then causes it to take up the slack. This, in turn, causes the clutch friction disk to slip, creating heat and ultimately wearing your clutch out.

Step Off: When the clutch pedal is released, the springs of the pressure plate push the slave cylinder's pushrod back, which forces the hydraulic fluid back into the master cylinder.

7 0

3 years ago

Your program will be a line editor. A line editor is an editor where all operations are performed by entering commands at the co

soldier1979 [14.2K]

Answer:

Java program given below

Explanation:

import java.util.*;

import java.io.*;

public class Lineeditor

{

private static Node head;

class Node

{

int data;

Node next;

public Node()

{data = 0; next = null;}

public Node(int x, Node n)

{data = x; next =n;}

}

public void Displaylist(Node q)

{if (q != null)

{

System.out.println(q.data);

Displaylist(q.next);

}

public void Buildlist()

{Node q = new Node(0,null);

head = q;

String oneLine;

try{BufferedReader indata = new

BufferedReader(new InputStreamReader(System.in)); // read data from terminals

System.out.println("Please enter a command or a line of text: ");

oneLine = indata.readLine(); // always need the following two lines to read data

head.data = Integer.parseInt(oneLine);

for (int i=1; i<=head.data; i++)

{System.out.println("Please enter another command or a new line of text:");

oneLine = indata.readLine();

int num = Integer.parseInt(oneLine);

Node p = new Node(num,null);

q.next = p;

q = p;}

}catch(Exception e)

{ System.out.println("Error --" + e.toString());}

}

public static void main(String[] args)

{Lineeditor mylist = new Lineeditor();

mylist.Buildlist();

mylist.Displaylist(head);

}

7 0

3 years ago