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babunello [35]
3 years ago
14

Draw a 3-D physical structure of an NMOS transistor. Label four terminals: body, drain, gate, and source. And also label silicon

oxide thickness tox, channel length L and channel width W.

Engineering
1 answer:
Vanyuwa [196]3 years ago
4 0

Answer:

Answer is attached.

Explanation:

A NMOS is a n-channel MOSFET or Metal Oxide

Semiconductor Field Effect Transistor. This type

of transistor might be an enhancement or

depletion type nMOS transistor designed using

layers of Metal-oxide, Silicon-oxide and Silicon

fabricated on a substrate.

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Explanation:

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3 years ago
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A three-phase line has a impedance of 0.4+j2.7 per phase. The line feeds 2 balanced three-phase loads that are connected in para
mamaluj [8]

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c)  54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

V_p=\frac{3810.5}{\sqrt{3} }=2200 V

The complex power S is given as:

S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA

where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA

The line current I is given as:

I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)}  =100\angle -36.87^o\ A

The phase voltage at the sending end is:

V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV

The magnitude of the line voltage at the source end of the line (V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V

b) The Total real and reactive power loss in the line is:

S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

5 0
4 years ago
thermodynamics A nuclear power plant based on the Rankine cycle operates with a boiling-water reactor to develop net cycle power
IrinaK [193]

Answer:

(a) the percent thermal efficiency is 27.94%

(b) the temperature of the cooling water exiting the condenser is 31.118°C

Explanation:

3 0
3 years ago
A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the
Olegator [25]

Answer:

119.35 mm

Explanation:

Given:

Inside diameter, d = 100 mm

Tensile load, P = 400 kN

Stress = 120 MPa

let the outside diameter be 'D'

Now,

Stress is given as:

stress = Load × Area

also,

Area of hollow pipe = \frac{\pi}{4}(D^2-d^2)

or

Area of hollow pipe = \frac{\pi}{4}(D^2-100^2)

thus,

400 × 10³ N = 120 × \frac{\pi}{4}(D^2-100^2)

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or

D = 119.35 mm

7 0
3 years ago
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A particle travels along a straight line with a velocity v = (12 – 3t2) m/s. When t = 1 s, the particle is located 10 m to the l
arlik [135]

Answer:

The displacement from t = 0 to t = 10 s,  is -880 m

Distance is 912 m

Explanation:

v = (12 - 3t^2) m/s = ds/dt.  .  . . . . . . . .  A

integrate above equation we get

s = 12t - t^3 + C

from information given in the question  we have

t = 1 s, s = -10 m

so distance s will be

-10 = 12 - 1 + C,

C = -21

s(t) = 12t - t^3 - 21

we know that acceleration is given as

a(t) = dv/dt = -6t  

[FROM EQUATION A]

Acceleration at  t = 4 s, a(4) = -24 m/s^2

for the displacement from t = 0 to t = 10 s,

s(10) - s(0) = (12*10 - 10^3 - 21) - (-21) = -880 m

the distance the particle travels during this time period:

let v = 0,

3t^2 = 12

t = 2 s

Distance = [s(2) - s(0)] + [s(2) - s(10)] = [1\times 2 - 2^3] + [(12\times 2 - 2^3) - (12\times 10 - 10^3)] = 912 m

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