Semiconductor Electronics Important Questions

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Very Short Answer Questions [1 Marks Questions]

Ques. Why are semiconductors used in electronics?

Ans. Because we can regulate the flow of electrons in this material, such as with a regulating current, semiconductors are employed in many electrical circuits. Semiconductors are also employed for a variety of other purposes. In actuality, a solar cell is made up of semiconductors that are light-sensitive.

Ques. What are Some devices that use semiconductors?

Ans. List of common semiconductor devices

• DIAC
• Diode (rectifier diode)
• Gunn diode
• IMPATT diode
• Laser diode
• Light-emitting diode (LED)
• Photocell
• Phototransistor

Ques. Is silicon a semiconductor?

Ans. Because of its stable structure, silicon, a common element, is utilised as the raw material for semiconductors. Silicon purification needs a lot of energy.

Ques. Define the current amplification factor in a common–emitter mode of the transistor.

Ans. At constant collector-emitter junction voltage, the current amplification factor is the ratio of a minor change in collector current to a little difference in base current.

Ques. Why is a semiconductor damaged by a strong current?

Ans. When a big current passes through a semiconductor, it generates a lot of heat, which causes the covalent bonds in the semiconductor to break. As a result, the semiconductor is harmed.

Ques. Is diode a semiconductor?

Ans. A diode is a semiconductor device that functions as a current one-way switch. It permits current to flow freely in one direction while drastically restricting current flow in the other.

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Short Answer Questions [2 Marks Questions]

Ques. What is a semiconductor in basic electronics?

Ans. A semiconductor is a type of crystalline substance that is halfway between a conductor and an insulator in terms of electrical conductivity. Semiconductors are used in the production of diodes, transistors, and integrated circuits, among other electronic devices.

Ques. Explain briefly how a photodiode operates.

Ans. A photodiode is a special purpose p-n junction diode with a transparent window to allow light to pass through it. It works with a reverse bias. The absorption of photons causes electron-hole pairs to form when the photodiode is irradiated with light (photons) with energy (hv) larger than the semiconductor's energy gap (Eg). The diode is made in such a way that the creation of e-h pairs occurs in or near the depletion area.

Ques. Name the p-n junction diode which emits spontaneous radiation when forward biased. How do we choose the semiconductor to be used in these diodes, if the emitted radiation is to be in the visible region?

Ans. The "light-emitting diode," or LED, is a p-n junction diode that generates spontaneous radiation when forward biassed.

The visible tight ranges between 400 nm and 700 nm, with corresponding energy of 2.8 eV to 1.8 eV. As a result, the energy gap of the semiconductor utilised in LEDs should be 1.8 eV in order for the emitted radiation to be visible. Gallium arsenide with phosphorus doping and gallium phosphide are two such semiconductors.

Ques. Give reasons for the following:

1. High reverse voltage does not appear across an LED.
2. Sunlight is not always required for the working of a solar cell.

Ans.

1. It's because LEDs have a very low reverse breakdown voltage, about 5 V.
2. Because any light with photon energy greater than the bandgap energy may be used by solar cells.

Ques. Zener diode has a saturation current of 20A and reverses breakdown voltage of 100V whereas the corresponding values are 40μA and 40V. Find the current through the circuit.

Ans. Here, Z₁ is forward biased whereas Z₂ is reverse biased.

Clearly, Z₁ acts as a conductor and reverse saturation current will move from Z₂.

Thus,

RZ₂ = 40/(40×10⁻⁶)

RZ₂ = 10⁶ Ω

Now, 50V will appear across Z₂.

So, Current

I = 50/10⁶

I = 50×10⁻⁶ A

Also check:

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Short Answer Questions [3 Marks Questions]

Ques.

1. Explain with the help of a diagram how the depletion region and potential barrier are formed in a junction diode.
2. If a small voltage is applied to a p-n junction diode, how will the barrier potential be affected when it is (i) forward biased and (ii) reverse biased? (CBSE AI2015)

Ans. (i) Because the n-type has more electrons and the p-type has more holes when a p-n junction is formed, the electrons from the n-side diffuse into the p-side & the holes from the p-side diffuse into the n-side.

Potential Barrier

A potential difference between the two regions is established by the buildup of electric charges of opposing polarities in the two regions across the junction. This is referred to as the potential or junction barrier. The potential barrier that has formed across the junction prevents charge carriers from moving from p to n and vice versa. There is a zone on each side of the intersection where mobile charges have depleted and only immobile charges remain. The depletion layer or zone is the area around the junction that is devoid of any mobile charge carriers.

(ii) (a) In forwarding bias the potential barrier decreases.

(b) In reverse bias the potential barrier increases.

Ques. Explain, with the help of a circuit diagram, the working of a photo-diode. Write briefly how it is used to detect optical signals. (CBSE Delhi 2013)

Ans. The absorption of photons causes electron-hole pairs to form when the photodiode is irradiated with light (photons) with energy (hv) larger than the energy gap (Eg) of the semiconductor. The diode is made in such a way that the creation of e-h pairs occurs in or near the depletion area.

Electrons and holes are separated before they recombine due to the electric field of the junction. The electric field is directed in such a way that electrons reach the n-side while holes reach the p-side. An emf is created when electrons are captured on the n-side and holes are gathered on the p-side. Current flows when an external load is attached.

The photocurrent's magnitude is determined by the amount of incident light that strikes it. This facilitates the detection of optical signals.

Photocurrent effect

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Long Answer Questions [5 Marks Questions]

Ques. With what considerations in view, a photodiode is fabricated? State its working with the help of a suitable diagram. Even though the current in the forward bias is known to be more than in the reverse bias, the photodiode works in reverse bias. What is the reason? (CBSE Delhi 2015)

Ans. It has a transparent window that allows light to pass through to the diode. It is constructed in such a way that the creation of e-h pairs occurs in or near the diode's depletion zone. The absorption of photons causes electron-hole pairs to form when the photodiode is irradiated with light (photons) with energy (hw) larger than the semiconductor's energy gap (Eg). The diode is made in such a way that the creation of e-h pairs occurs in or near the depletion area. Electrons and holes are separated before they recombine due to the electric field of the junction.

The electric field is directed in such a way that electrons reach the n-side while holes reach the p-side. An emf is created when electrons are captured on the n-side and holes are gathered on the p-side. Current flows when an external load is attached. The photocurrent's magnitude is determined by the intensity of incoming light (photocurrent is proportional to incident light intensity). The diagram is exactly how it appears. When a reverse bias is used, it is simpler to see how the current changes as the light intensity changes. As a result, even though the current in the forward bias is higher, the photodiode is employed in reverse bias mode.

Potential Barrier

Ques. State briefly the processes involved in the formation of the p-n junction explaining clearly how the depletion region is formed. (CBSE Delhi 2014)

Ans. As we all know, an n-type semi-conductor has more electrons than a hole, whereas a p-type semi-conductor has more holes than an electron. The holes diffuse from the p-side to the n-side & the electrons diffuse from the n-side to the p-side due to the difference in charge carrier concentration in the two areas of the p-n junction.

An ionised donor is left on the n-side when an electron diffuses from n to p side. Because it is bounded by the surrounding atoms, the ionised donor (+ve charge) is immobile. As a result, on the n-side of the connection, a layer of positive charge develops. On the p-side, a layer of negative charge develops in a similar way.

As a result, on either side of the junction, a space-charge zone with immobile ions and no charge carrier is produced, known as a depletion layer or depletion region.

Potential Barrier

The imaginary battery, which seems to be linked across the junction with its +ve end on the n-type & the -ve end on the p-type, is the potential barrier.

Ques. Using the necessary circuit diagrams, show how the V – l characteristics of a p-n junction are obtained in

1. Forward biasing
2. Reverse biasing

How are these characteristics made use of in rectification? (CBSE Delhi 2014)

Ans. (a) A p-n junction diode under forwarding bias - The positive terminal is attached to the p-side, while the negative terminal is connected to the n-side. Applied voltage decreases as it passes through the depletion area.

The electron in the n-region goes towards the p-n junction, while the holes in the p-region move towards it. As the depletion layer's breadth diminishes, it provides less resistance. The majority of carriers are diffused throughout the junction. The forward current is the result of forwarding bias.

(b) A p-n junction diode under reverse bias - The battery's positive terminal is attached to the n-side, while the negative terminal is attached to the p-side. The possibility of a barrier is supported by reverse bias. As a result, the barrier height rises, as does the width of the depletion zone. There is no transmission across the junction due to the preponderance of carriers. After being boosted by the high reverse bias voltage, a few minority carriers pass the junction. This creates a reverse current, which is a current that flows in the opposite direction.

For V-l curves

V - I Curve

As a half-wave rectifier, a p-n junction diode is employed. Its work is based on the fact that when the p-n junction is forward biassed, the resistance is low, and when it is reverse biassed, the resistance is high. In rectification, the diode's properties are exploited.

Ques. Write the two processes that take place in the formation of a p-n junction. Explain the formation of the depletion region and barrier potential in a p-n junction. (CBSE Delhi 2017)

Ans. Drift and diffusion

The n-type has more electrons than the p-type, whereas the p-type has more holes. The electrons from the n-type diffuse into the p-region and the holes from the p-type diffuse into the n-region when a p-n junction forms. Near the connection, these dispersing electrons and holes merge. Each combination removes an electron and a hole from the equation. The n-region near the junction becomes positively charged as electrons are lost, and the p-region near the junction becomes negatively charged as holes are lost.

A potential difference between the two regions is established by the buildup of electric charges of opposing polarities in the two regions across the junction. This is referred to as the potential or junction barrier. The potential barrier that has formed across the junction prevents charge carriers from moving from p to n and vice versa. As a result, an area forms on each side of the junction where mobile charges have depleted and only immobile charges remain. The depletion layer or zone is the area around the junction that is devoid of any mobile charge carriers.

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