An NMOS transistor with kn = 1mA/V2 and Vt = 1V is operated with VGS = 2.5V. At what value of VDS does the transistor enter the saturation region? What value of ID is obtained in saturation?

Answers

Answer 1

Answer:

a) the point at which the transistor enters the saturation region is 1.5 v

b) the value of ID is obtained in saturation is 1.125 mA

Explanation:

Given the data in the question;

for an NMOS, the condition for the saturation is;

V_DS ≥ V_GS - V_t

V_GS is 2.5 v and V_t is 1 v

so we substitute

V_DS ≥ 2.5 - 1

V_DS = 1.5 v

so the point at which the transistor enters the saturation region is 1.5 v

The drain current I_d in the saturation region;

I_d = 1/2×μₙ×Cₐₓ×W/L×( V_GS - V_t)²

= 1/2Kₙ (  V_GS - V_t)²

our Kₙ is 1 mA/V², V_GS is 2.5v and V_t is 1 v

so we substitute

I_d = 1/2(1 mA/V²)( 2.5 - 1 )²

= 1/2(1 mA/V²)( 2.25)

= 1.125 mA

therefore, the value of ID is obtained in saturation is 1.125 mA


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Technician a says that diesel engines can produce more power because air in fuel or not mix during the intake stroke. Technician be says that diesel engines produce more power because they use excess air to burn feel who is correct

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Technician be says that diesel engines produce more power because they use excess air to burn feel who is correct

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A sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It has a specific gravity of solids of 2.70. Compute its moisture content and degree of saturation before it was placed in the oven.

Answers

Answer:

a) the moisture content before it was placed in the oven is 18.18%

b) degree of saturation for soil is 72.19%

Explanation:

Given the data in the question;

Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100

so we substitute

Moisture content = [(53.3 - 45.1) / 45.1 ] × 100

= (8.2/45.1) × 100

= 18.18%

Therefore the moisture content before it was placed in the oven is 18.18%

Dry Unit Weight = dry weight / volume

Dry Unit Weight = 45.1 lb / 0.45 ft³

Dry Unit Weight = 100.22 lb/ft³

we know that;

dry unit weight = (Specific gravity × unit weight of water) / (1 + e)

we also know that; unit weight of water is 62.43 lbf/ft³

so we substitute

e = (2.70×62.43 / 100.22) - 1

e = 1.68 - 1

e = 0.68

so void ratio e = 0.68

Now we determine the degree of saturation using the equation;

degree of saturation = (Moisture content × specific gravity) / void ratio

we substitute

degree of saturation = ( 18.18% × 2.7) / 0.68

= 0.49086 / 0.68

= 0.7219 ≈ 72.19%

Therefore degree of saturation for soil is 72.19%

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Answers

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