Semiconductor Electronics: Materials, Devices and Simple Circuits Test 34

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 34

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Weekly Quiz Competition

Question 1
1 / -0

The curve represents the $$I-V$$ characteristics of which optoelectronic device?

A
Solar Cell

B
Photo-diode

C
Light Emitting Diode (LED)

D
None of these

Solution

1.The span of the solar cell $$I-V$$ characteristic curve ranges from short circuit current $$I_{sc}$$ at zero output volts, to zero current at te full open circuit voltage$$V_{oc}$$
2.With the solar cell open circuited that is not connected to any load, the current will be at its maximum(zero) and the voltage across the cell is at its maximum known as open circuit voltage $$(V_{oc})$$
3.At the other extreme when the solar cell is short circuited that is the positive and negative loads connected together, the current flowing out of the cell reaches its maximum known as short circuit current$$I_{sc}$$
Question 2
1 / -0

What is the reason to operate the photodiodes in reverse bias?

A
fractional change due to the photo-effects on the majority carrier dominated forward bias current is more easily measurable than the fractional change in the reverse bias current.

B
reverse bias is more efficient than forward bias.

C
fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current.

D
reverse bias configuration result in larger amount of current (in $$\mu A$$)

Solution

1.In reverse biased $$p-n$$ junction, the width of depletion region increases as we increases the applied reverse bias voltage across the diode.
2.So,by applying a larger voltage, more of the incident photons are converted to electric current, or the efficiency increases.
3.On the other hands when we apply forward bias in the $$p-n$$ junction, the width of the depletion region reduces, so, only a small portion of the incident photons gets converted to electric current.
so, this means that reverse bias is more efficient than forward bias that's the reason to operate the photo diodes in reverse bias.
Question 3
1 / -0

In , excess minority carriers recombine with majority carriers near the p-n junction.

A
Photodiode

B
Light Emitting Diode

C
Solar Cells

D
Zener diode

Solution

A light-emitting diode (LED) is a semiconductor device that emits visible light when an electric current passes through it
A P-N junction can convert absorbed light energy into electric current. Here the reverse happens that is the P-N junction emits light when electrical energy is applied to it. The charge carriers recombine in a forward-biased P-N junction as the electrons cross from the N-region and recombine with the holes existing in the P-region. Free electrons are in the conduction band of energy levels, while holes are in the valence energy band. A part of the energy gets dissipated in order to recombine the electrons and the holes and this energy is emitted in the form of heat and light.
Question 4
1 / -0

The thickness of depletion region is of the order of

A
$$0.1 \mu m$$

B
$$1 \mu m$$

C
$$0.1 mm$$

D
$$1 mm$$

Solution

Depletion region is a region near the p-n junction where flow of charge carriers (free electrons and holes) is reduced over a given period and finally results in zero charge carriers.
The width of depletion region which is generally 1 micro meter, depends on the amount of impurities added to the semiconductor. Impurities are the atoms (pentavalent and trivalent atoms) added to the semiconductor to improve its conductivity.
Question 5
1 / -0

When an external voltage ($$V$$) is applied across the diode in reverse bias with a barrier potential of $$V_o$$, the effect barrier height under reverse bias is

A
$$V_o + V$$

B
$$V_o - V$$

C
$$V - V_o$$

D
None of these

Solution

When a reverse bias voltage (V) is applied externally across the diode, and the barrier potential is $$V_{o}$$. The potential difference adds to the barrier potential. The effective barrier potential increases to $$(V_{o}+V)$$.
Question 6
1 / -0

Across the depletion region, there is a barrier potential which

A
prevents the movement of electron from the n region into the p region

B
prevents the movement of electron from the p region into the n region

C
prevents the movement of holes from the n region into the p region

D
prevents the movement of holes from the p region into the n region

Solution

The potential difference due to negative immobile ions on p-side of the junction and positive immobile ions on n-side of the junction is called potential barrier. It prevents the movement of electrons from n to p side of junction and holes from p to n side of the junction.
Question 7
1 / -0

In the figure, the numbers 1, 2, 3 represent barrier potential ($$V_o$$) of a p-n junction under forward bias

A
1 - high voltage battery 2 - without battery 3 - low voltage bettery

B
1 - high voltage battery, 2 - low voltage battery, 3 - without battery

C
1 - without battery, 2 - low voltage battery, 3 - high voltage battery

D
1 - low voltage battery 2 - without battery 3 - high voltage bettery
Solution
If a forward bias voltage is applied to a p-n junction diode with a barrier potential $$V_{o}$$, the effective potential reduces to $$(V_{o}-V)$$. More is the forward bias voltage, more the potential reduces.
So, in
Barrier potential remains same, so the battery is not connected externally.
Low voltage battery is connected externally.
High voltage battery is connected externally.
Question 8
1 / -0

For creating a $$p-n$$ junction,

A
we take one slab of p-type semiconductor and physically join it to another n-type semiconductor

B
continuous contact at the atomic level is necessary.

C
two macroscopic smooth flat slabs can be used

D
None of these

Solution

When we take two independent p and n-type semiconductor bars joined end to end, they cannot form a p-n junction by the physical contact. A continuous contact at the atomic level is required to make a p-n junction.
This is because even when most smoothly polished surface when pressed against each other, have a very small area of contact. Moreover, even the cleanest surface may have some contamination like minute dust particles which do not allow two surfaces to come in atomic contact.
Question 9
1 / -0

Due to the concentration gradient across p-, and n- sides, the motion of charge carries gives rise to_______ current across the junction

A
drift

B
diffusion

C
both diffusion and drift

D
None of these

Solution

Diffusion of charge carriers across the junction in p-n junction diode occurs due to the difference in concentration of charge carriers on both sides (i.e. p and n side) of the junction.
For example, when a p-n junction is formed, there is more number of electrons on the n-side than on p-side whereas more number of holes on p-side than on n-side. As a result, electrons diffuse from n to p side whereas holes diffuse from p to n side.
Question 10
1 / -0

The direction of the applied voltage ($$V$$) is opposite to the built-in potential $$V_o$$ . The effective barrier height under forward bias is

A
($$V_o + V$$).

B
($$V_o V$$).

C
($$V V_o$$).

D
($$V_o \times V$$).

Solution

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 34

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 34

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Question 1

Solar Cell

Photo-diode

Light Emitting Diode (LED)

None of these

Solution

Question 2

fractional change due to the photo-effects on the majority carrier dominated forward bias current is more easily measurable than the fractional change in the reverse bias current.

reverse bias is more efficient than forward bias.

fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current.

reverse bias configuration result in larger amount of current (in $$\mu A$$)

Solution

Question 3

Photodiode

Light Emitting Diode

Solar Cells

Zener diode

Solution

Question 4

$$0.1 \mu m$$

$$1 \mu m$$

$$0.1 mm$$

$$1 mm$$

Solution

Question 5

$$V_o + V$$

$$V_o - V$$

$$V - V_o$$

None of these

Solution

Question 6

prevents the movement of electron from the n region into the p region

prevents the movement of electron from the p region into the n region

prevents the movement of holes from the n region into the p region

prevents the movement of holes from the p region into the n region

Solution

Question 7

1 - high voltage battery 2 - without battery 3 - low voltage bettery

1 - high voltage battery, 2 - low voltage battery, 3 - without battery

1 - without battery, 2 - low voltage battery, 3 - high voltage battery

1 - low voltage battery 2 - without battery 3 - high voltage bettery

Solution

Question 8

we take one slab of p-type semiconductor and physically join it to another n-type semiconductor

continuous contact at the atomic level is necessary.

two macroscopic smooth flat slabs can be used

None of these

Solution

Question 9

drift

diffusion

both diffusion and drift

None of these

Solution

Question 10

($$V_o + V$$).

($$V_o V$$).

($$V V_o$$).

($$V_o \times V$$).

Solution

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