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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:

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Read more about beam;

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7 0

2 years ago

Poly(cis-1,4-isoprene), or natural rubber (NR), has a tendency crystallize. The Tm of this polymer is slightly below room temper

Nonamiya [84]

Answer:

Explanation:

Crystalline melting temperature (Tm) is termed as the temperature required for a crystalline polymer to change to a fluid or glasslike crystalline spaces of a semi-crystalline polymer liquefy (expanded sub-atomic movement).

Crystallization of polymers is an interaction related with incomplete arrangement of their atomic and molecular chains. These chains crease together and structure requested districts called lamellae, which form bigger spheroidal designs named spherulites. Polymers can solidify after cooling from melting, mechanical extending, or dissolvable dissipation. Crystallization influences the optical, mechanical, and synthetic chemical properties of the polymer.

For a crystalline polymer, a required polymer chain is present in or goes along a few crystalline and amorphous zones. The crystalline zones are comprised of intermolecular & intramolecular arrangements or deliberate and thus firmly stuffed plan of atoms or chain fragments, and an absence of it brings about the development of amorphous zones.

The mechanical property boundary, for example, shear modulus expansions in the temperature of perception for polymer material framework.

The temperature reaction of direct linear polymers might be seen as partitioned into three particularly separate fragments:

1. Above Tm: The polymer stays as fluid whose consistency & viscosity would rely upon atomic molecular weight and temperature.

2. Between Tm and Tg: This area may go between close to 100% crystalline & 100% amorphous chain atomic bunches relying upon the polymer underlying consistency. The amorphous part carries on similar to supercooled fluid in this section. The generally actual conduct of the polymer in this moderate portion is similar to an elastic rubber.

3.Below Tg: The polymer material saw as glass is hard and inflexible, showing and emanating a predetermined coefficient of thermal extension. The glass is more like a crystalline strong than the fluid in personal conduct standard regarding mechanical property boundaries. In regard to the molecular atomic request, in any case, the glass all the more intently takes after the fluid. There is little contrast between the direct linear and cross-connected polymer beneath Tg.

8 0

3 years ago

Consider the base plate of an 800-W household iron with a thickness of L = 0.6 cm, base area of A =160 cm2, and thermal conducti

ratelena [41]

Given-

Power, P = 800W

Thickness, L = 0.6cm

Area, A = 160cm²

Thermal conductivity, k = 60W/mK

The heat conduction would be

$\frac{d^2T}{dx^2} + \frac{d^2T}{dy^2} + \frac{d^2T}{dz^2} + \frac{e(gen)}{k} = \frac{1}{\alpha } \frac{dT}{dt}$

Except $\frac{d^2T}{dx^2}$ all the values are 0.

Therefore,

$\frac{d^2T}{dx^2} = 0$

Thus, the boundary conditions here would be

1. $Q_x_=_0 = -kA \frac{dT (0)}{dx} = Q_o$

2. $T(L) = T_L$

4 0

4 years ago

A supply fan is operating at 30000 cfm and 4 inch of water with an efficiency of 50%. (a) Calculate the fan power at the current

serg [7]

Answer:

The pressure and power of fan is 1.77 and 11.18 Hp respectively.

Explanation:

Given:

Discharge $Q_{1} = 30000$ cfm

Pressure difference $\Delta P = 4$ inch

Efficiency $\eta = 50\%$

(A)

From the formula of fan power,

$P _{1} = \frac{Q \Delta P}{6356 \eta}$

$P_{1} = \frac{30000 \times 4}{6356 \times 0.5}$

$P_{1} = 37.76$ Hp

(B)

Fan power and pressure is given by,

We know that pressure difference is proportional to the square of discharge.

$\frac{\Delta p_{2} }{\Delta P_{1} } = (\frac{Q_{2} }{Q_{1} } ) ^{2}$

$\Delta P_{2} = (\frac{20000}{30000} ) \times 4$

$\Delta P_{2} = 1.77$

Fan power proportional to the cube of discharge.

$\frac{P_{2} }{P^{1} } = (\frac{Q_{2} }{Q_{1} } )^{3}$

$P_{2} = \ (\frac{20000}{30000} ) ^{3} \times 37.76$

$P_{2} = 11.18$ Hp

Therefore, the pressure and power of fan is 1.77 and 11.18 Hp respectively.

5 0

3 years ago

The phasor technique makes it pretty easy to combine several sinusoidal functions into a single sinusoidal expression without us

devlian [24]

Answer:

The phasor technique can't be applied directly in the following cases:

a) 45 sin(2500t – 50°) + 20 cos(1500t +20°)

b) 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)

c) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)

d) 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)

Explanation:

For a) and c), it is not possible to use the phasor technique, due this technique is only possible when the sinusoidal signals to be combined are all of the same frequency.

This is due to the vector representing a signal is showed as a fixed vector in the graph( which magnitude is equal to the amplitude of the sinusoid and his angle is the phase angle with respect to cos (ωt)), which is rotating at an angular speed equal to the angular frequency of the sinusoidal signal that represents, like a radius that shows a point rotating in a circular uniform movement.

This rotating vector represents a sinusoidal signal, in the form of a cosine (as the real part of the complex function $e^{j(wt+\alpha)}$ ), so it is not possible to combine with functions expressed as a sine, even though both have the same frequency.

If we look at the graphs of cos (ωt) and sin (ωt) we can say that the sin lags the cos in 90º, so we can say the following:

sin (ωt) = cos (ωt-90º)

This means that in order to be able to represent a sine function as a cosine, we need to rotate it 90º in the plane clockwise.

This is the reason why before doing this transformation, it is not possible to use the phasor technique for b) and d).

8 0

3 years ago