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

Engineering

1 answer:

zhenek [66]1 year 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:

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An old refrigerator consumes 247 W of power. Assuming that the refrigerator operates for 19 hours everyday, what is the annual o

german

Answer:

The annual operating cost of the refrigerator is $102.78.

Explanation:

Power consumed by the refrigerator = 247 W = 247/1000 = 0.247 kW

Daily operation of the refrigerator = 19 hours

Annual operation of the refrigerator = 365 × 19 = 6,935 hours

Annual energy consumed = 0.247 kW × 6,935 hours = 1712.945 kWh

1 kWh of electricity cost $0.06

1712.945 kWh will cost 1712.945 × $0.06 = $102.78

Annual operating cost = $102.78

6 0

3 years ago

Read 2 more answers

One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter

lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150× $10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$ = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$ , the containers volume is V, waters mass is m.

$v_{1}$ = 150× $10^{-3}$ $m^{3}$ /1

$v_{1}$ = 0.15 $m^{3}$ / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 $m^{3}$ / Kg and 0.13 $m^{3}$ / Kg.

$T_{1}$ = 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$