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3 years ago

For a given set of input values, a NAND gate produces the opposite output as an OR gate with inverted inputs.A. True

Engineering

1 answer:

Verdich [7]3 years ago

3 0

Answer:

Explanation:

For a given set of input values, A NAND gate produces exactly the same values as an OR gate with inverted inputs.

The truth table for a NAND gate with 2 inputs is as follows:

0 0 1

0 1 1

1 0 1

1 1 0

The truth table for an OR gate, is as follows:

0 0 0

0 1 1

1 0 1

1 1 1

If we add two extra columns for inverted inputs, the truth table will be this one:

0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

1 1 0 0 0

which is the same as for the NAND gate, not the opposite, so the statement is false.

This means that the right choice is B.

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(a) Consider a germanium semiconductor at T 300 K. Calculate the thermal equilibrium electron and hole concentrations for (i) Nd

padilas [110]

Answer:

i. electron concentration, n₀ = 2 × 10¹⁵ cm⁻³, hole concentration, p₀ = 2 × 10¹¹ cm⁻³

ii. electron concentration, n₀ = 1.33 × 10¹¹ cm⁻³, hole concentration, p₀ = 3 × 10¹⁵ cm⁻³

i. electron concentration, n₀ = 2 × 10¹⁵ cm⁻³, hole concentration, p₀ = 2.205 × 10⁻³ cm⁻³

ii. electron concentration, n₀ = 1.47 × 10⁻³ cm⁻³, hole concentration, p₀ = 3 × 10¹⁵ cm⁻³

c. It means that the minority carrier contribute little to the conductivity of the semi-conductor.

Explanation:

a. For Germanium, intrinsic concentration n₁ = 2 × 10¹³ cm⁻³.

i. For the electron concentration, n₀ wit N₁ = donor concentration = 2 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 0,

n₀ = 1/2[(N₁ - N₂) +√[(N₁ - N₂)² + 4n₁²] ] since N₁ > N₂

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ - 0) +√[(2 × 10¹⁵ cm⁻³ - 0)² + 4(2 × 10¹³ cm⁻³)²] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ +√[(4 × 10³⁰ cm⁻⁶ + 16 × 10²⁶ cm⁻⁶] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ +√[(4.0016 × 10³⁰ cm⁻⁶] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ + 2.0004 × 10¹⁵ cm⁻³ ]

n₀ = 1/2[4.0004 × 10¹⁵ cm⁻³ ]

n₀ = 2.0002 × 10¹⁵ cm⁻³ ≅ 2 × 10¹⁵ cm⁻³

The hole concentration p₀ is gotten from

n₀p₀ = n₁²

p₀ = n₁²/n₀ = (2 × 10¹³ cm⁻³)²/2 × 10¹⁵ cm⁻³ = 4 × 10²⁶ cm⁻⁶/2 × 10¹⁵ cm⁻³

p₀ = 2 × 10¹¹ cm⁻³

ii. For the hole concentration, p₀ wit N₁ = donor concentration = 7 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 10¹⁶ cm⁻³,

p₀ = 1/2[(N₂ - N₁) +√[(N₂ - N₁)² + 4n₁²] ] since N₂ > N₁

p₀ = 1/2[(10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³) +√[(10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³)² + 4(2 × 10¹³ cm⁻³)²] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ +√[(9 × 10³⁰ cm⁻⁶ + 16 × 10²⁶ cm⁻⁶] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ +√[(9.0016 × 10³⁰ cm⁻⁶] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ + 3.0003 × 10¹⁵ cm⁻³ ]

p₀ = 1/2[6.0003 × 10¹⁵ cm⁻³ ]

p₀ = 3.00015 × 10¹⁵ cm⁻³ ≅ 3 × 10¹⁵ cm⁻³

Te electron concentration n₀ is gotten from

n₀p₀ = n₁²

n₀ = n₁²/p₀ = (2 × 10¹³ cm⁻³)²/3 × 10¹⁵ cm⁻³ = 4 × 10²⁶ cm⁻⁶/3 × 10¹⁵ cm⁻³

n₀ = 1.33 × 10¹¹ cm⁻³

b. For GaAs, intrinsic concentration n₁ = 2 × 10⁶ cm⁻³.

i. For the electron concentration, n₀ wit N₁ = donor concentration = 2 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 0,

n₀ = 1/2[(N₁ - N₂) +√[(N₁ - N₂)² + 4n₁²] ] since N₁ > N₂ and N₁ - N₂ = 2 × 10¹⁵ cm⁻³ >> n₁ = 2 × 10⁶ cm⁻³

n₀ = (N₁ - N₂) = 2 × 10¹⁵ cm⁻³ - 0 = 2 × 10¹⁵ cm⁻³

The hole concentration p₀ is gotten from

n₀p₀ = n₁²

p₀ = n₁²/n₀ = (2.1 × 10⁶ cm⁻³)²/2 × 10¹⁵ cm⁻³ = 4.41 × 10¹² cm⁻⁶/2 × 10¹⁵ cm⁻³

p₀ = 2.205 × 10⁻³ cm⁻³

ii. For the hole concentration, p₀ wit N₁ = donor concentration = 7 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 10¹⁶ cm⁻³,

p₀ = 1/2[(N₂ - N₁) +√[(N₂ - N₁)² + 4n₁²] ] since N₂ > N₁ and N₂ - N₁ = 10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³ = 3 × 10¹⁵ cm⁻³ >> n₁ = 2.1 × 10⁶ cm⁻³

p₀ ≅ N₂ - N₁ = 3 × 10¹⁵ cm⁻³

The electron concentration n₀ is gotten from

n₀p₀ = n₁²

n₀ = n₁²/p₀ = (2.1 × 10⁶ cm⁻³)²/3 × 10¹⁵ cm⁻³ = 4.41 × 10¹² cm⁻⁶/3 × 10¹⁵ cm⁻³

n₀ = 1.47 × 10⁻³ cm⁻³

c. It means that the minority carrier contribute little to the conductivity of the semi-conductor.

4 0

4 years ago

a circular pile, 19 m long is driven into a homogeneous sand layer. The piles width is 0.5 m. The standard penetration resistanc

Elena L [17]

Answer:

Point force (Qp) = 704 kn/m²

Explanation:

Given:

length = 19 m

Width = 0.5 m

fs = 4

Vicinity of the pile = 25

Find:

Point force (Qp)

Computation:

Point force (Qp) = fs²(l+v)

Point force (Qp) = 4²(25+19)

Point force (Qp) = 16(44)

Point force (Qp) = 704 kn/m²

5 0

3 years ago

A cylindrical specimen of a brass alloy having a length of 60 mm (2.36 in.) must elongate only 10.8 mm (0.425 in.) when a tensil

Gemiola [76]

The radius of the specimen is 60 mm

Explanation:

Given-

Length, L = 60 mm

Elongated length, l = 10.8 mm

Load, F = 50,000 N

radius, r = ?

We are supposed to calculate the radius of a cylindrical brass specimen in order to produce an elongation of 10.8 mm when a load of 50,000 N is applied. It is necessary to compute the strain corresponding to this

elongation using Equation:

ε = Δl / l₀

ε = 10.8 / 60

ε = 0.18

We know,

σ = F / A

Where A = πr²

According to the stress-strain curve of brass alloy,

σ = 440 MPa

Thus,

$sigma = 50,000 / \pi (r)^2\\\\440 X 10^6 = \frac{50,000}{3.14 X (r)^2}\\\\r = 0.06m\\r = 60mm\\\\\\$

Therefore, the radius of the specimen is 60 mm

3 0

3 years ago

Air enters an adiabatic gas turbine at 1590 oF, 40 psia and leaves at 15 psia. The turbine efficiency is 80%, and the mass flow

melamori03 [73]

Answer:

a) 158.4 HP.

b) 1235.6 °F.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up an energy balance for the turbine's inlets and outlets:

$m_{in}h_{in}=W_{out}+m_{out}h_{out}$

Whereas the mass flow is just the same, which means we have:

$W_{out}=m_{out}(h_{out}-h_{in})$

And the enthalpy and entropy of the inlet stream is obtained from steam tables:

$h_{in}=1860.7BTU/lbm\\\\s_{in}= 2.2096BTU/lbm-R$

Now, since we assume the 80% accounts for the isentropic efficiency for this adiabatic gas turbine, we assume the entropy is constant so that:

$s_{out}= 2.2096BTU/lbm-R$

Which means we can find the temperature at which this entropy is exhibited at 15 psia, which gives values of temperature of 1200 °F (s=2.1986 BTU/lbm-K) and 1400 °F (s=2.2604 BTU/lbm-K), and thus, we interpolate for s=2.2096 to obtain a temperature of 1235.6 °F.

Moreover, the enthalpy at the turbine's outlet can be also interpolated by knowing that at 1200 °F h=1639.8 BTU/lbm and at 1400 °F h=174.5 BTU/lbm, to obtain:

$h_{out}=1659.15BUT/lbm$