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3 years ago

What is the difference between p-n junction diode and extrinsic semiconductor?

Physics

1 answer:

Anna71 [15]3 years ago

4 0

Answer:

The difference between P-N junction diode and extrinsic semiconductor is the allowable direction of flow of current

The extrinsic semiconductor can allow current to flow in both directions while the P-N junction diode permits only a unidirectional flow of current

Explanation:

A semiconductor is a substance that has an intermediate conductivity between that of conductors and non conductors

Examples of semiconductors include germanium and cadmium selenide

A semiconductor to which impurities has been added (an activity known as doping) is an extrinsic semiconductor

Based on the functioning of a semiconductor, doping result in the formation of one of two types of semiconductors including;

1) N-type semiconductor that has an extra electron and the charge carriers are electrons

2) P-type semiconductor that has one less (-e⁻) electron and holes are the charge carriers

An extrinsic semiconductor can conduct allow the flow of electricity in both ways

A P-N junction diode consists of both the P and N-type extrinsic semiconductors arranged such that current can flow in only one direction.

Therefore, the difference is that the extrinsic semiconductor can allow current flow in both directions while the P-N junction diode permits only a unidirectional flow of current.

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The space shuttle releases a satellite into a circular orbit 630 km above the Earth.

solniwko [45]

Answer:

7,539 m/s

Explanation:

Let's use this equation to find the gravitational acceleration of this space shuttle:

$\displaystyle g=\frac{GM}{r^2}$

We know that G is the gravitational constant: 6.67 * 10^(-11) Nm²/kg².

M is the mass of the planet, which is Earth in this case: 5.972 * 10^24 kg.

r is the distance from the center of Earth to the space shuttle: radius of Earth (6.3781 * 10^6 m) + distance above the Earth (630 km → 630,000 m).

Plug these values into the equation:

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg)}{[(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2}$

Remove units to make the equation easier to read.

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 )}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Multiply the numerator out.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Add the terms in the denominator.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(7008100)]^2}$

Simplify this equation.

$\displaystyle g=8.11045189 \ \frac{m}{s^2}$

The acceleration due to gravity g = 8.11045189 m/s². Now we use the equation for acceleration for an object in circular motion which contains v and r.

$\displaystyle a = \frac{v^2}{r}$

a = g, v is the velocity that the space shuttle should be moving (what we are trying to solve for), and r is the radius we had in the previous equation when solving for g.

Plug these values into the equation and solve for v.

$\displaystyle 8.11045189 \ \frac{m}{s^2} = \frac{v^2}{7008100 \ m}$

Remove units to make the equation easier to read.

$\displaystyle 8.11045189 = \frac{v^2}{7008100}$

Multiply both sides by 7,008,100.

$56838857.89=v^2$

Take the square root of both sides.

$v=7539.154985$

The shuttle should be moving at a velocity of about 7,539 m/s when it is released into the circular orbit above Earth.

5 0

3 years ago

Technician A says some compressor service procedures can be performed on the vehicle if space permits. Technician B says modern

stiv31 [10]

Answer:

Technicians A.

Explanation:

Since air compressor uses series of processes that turn incoming ambient air into a power source for tools and machinery. This means that air compressor has many different parts, and each of these parts must be maintained to ensure they function properly and optimally.

These are the basis when it comes to servicing a compressor

You need to change its oil

And clean its filters.

Inspected it's filters every three months, and have its filters replaced and connections tightened at least once every year.

To do all these can be performed on the vehicle if there is enough space just as Technician A said for the question context.

5 0

3 years ago

5.0 Points The blue color of the sky is the result of

mina [271]

The blue color of the sky is the result of scattering. The answer is letter B. it is specifically called the Rayleigh scattering. This blue color is the result of the incoming rays of the Sun into the earth’s atmosphere. It has he lowest wavelength which as we all see is the blue spectrum of light.

7 0

3 years ago

An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the

katrin [286]

<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

Mass = Volume x Density

0.03713 = 48.1 x 10⁻⁶ x Density

Density = 771.93 kg/m³

Density of the unknown liquid is 771.93 kg/m³

5 0

3 years ago

What is the peak emf generated by rotating a 940-turn, 24 cm diameter coil in the Earth’s 5·10−5 T magnetic field, given th

elena-14-01-66 [18.8K]

Answer:

The peak emf generated by the coil is 2.67 V

Explanation:

Given;

number of turns, N = 940 turns

diameter, d = 24 cm = 0.24 m

magnetic field, B = 5 x 10⁻⁵ T

time, t = 5 ms = 5 x 10⁻³ s

peak emf, V₀ = ?

V₀ = NABω

Where;

N is the number of turns

A is the area

B is the magnetic field strength

ω is the angular velocity

V₀ = NABω and ω = 2πf = 2π/t

V₀ = NAB2π/t

A = πd²/4

V₀ = N x (πd²/4) x B x (2π/t)

V₀ = 940 x (π x 0.24²/4) x 5 x 10⁻⁵ x (2π/0.005)

V₀ = 940 x 0.04524 x 5 x 10⁻⁵ x 1256.8

V₀ = 2.6723 V = 2.67 V

The peak emf generated by the coil is 2.67 V

8 0

3 years ago