WBCHSE Class 12 Physics Semiconductor Electronics Short Question And Answers

Semiconductors And Electronics Short Question And Answers

Question 1. Draw the voltage-current characteristics of a p-n junction and compare them with the characteristics of a resistor.
Answer:

Two characteristics are shown. Comparing them, we can say:

The ordinary resistor obeys Ohm’s law and hence its characteristic curve is linear, but the p-n junction does not obey Ohm’slaw.  Current can flow in any direction through an ordinary resistor, with the reverse bias of a p-n junction, the magnitude of the current is negligible

Question 2. In which kind of biasing of a p-n junction, do the net holes flow from the n -n-region to the p -p-region?

In the case of reverse biasing, the net flow of holes is from n -region to p -region. Because, in this case, the majority carrier holes in the p -p-region cannot enter the n -n-region, but the minority carrier holes can move easily from the n -n-region to the p -p-region

Question 3. Explain why the resistance of a semiconductor decreases with an increase in temperature.
Answer:

With the increase in temperature, a greater number of bonds inside the semiconductor are broken. Hence, a large number of electrons come out from those bonds. As a result, the number of charge carriers increases and consequently, the resistance decreases

Question 4. If the emitter and base of a transistor have the same doping concentration, how will the base current and the j collector current be affected?
Answer:

The base current will increase and the collector current j will decrease. Most of the majority carriers coming from the emitter will be neutralized in the base by electron-hole recombine nation. This results in a large base current. Only a few majority carriers will reach the collector, resulting in a small collector current.

WBBSE Class 12 Semiconductor Electronics Short Q&A

Question 5. How will you test whether a transistor is damaged or not
Answer:

The resistance of the forward-biased emitter-base junction is low while that of the reverse-biased collector-base junction A damaged transistor, when checked by an AVO  meter will show low resistance for both junctions

Question 6.  Can two separate p-n junction diodes placed back to back be used to form a p-n-p transistor?
Answer:

No. When two p-n junction diodes are placed back to back, the n -n-region will form the base. This region will be of considerable thickness and doping will also be very high. For a transistor, the thickness of the base should be very small and as the doping. So two p-n junction diodes placed back to back cannot form a p-n-p transistor.

Question 7. In half-wave rectification, what is the output frequency if the input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency?
Answer:

For a half-wave rectifier, the output frequency is equal to the input frequency. Output frequency = 50 Hz A full-wave rectifier rectifies both halves of the input AC.

∴  Output frequency = 2 × input frequency = 2 × 50 = 100 H

Question 8. What is the effect of doping on the depletion layer of a p-n junction?
Answer:

With the increase in the rate of doping, the thickness of the depletion layer generated at the junction decreases. Even if one part (p -part, say) of a p-n junction is heavily doped, the thickness of the depletion layer on that side (p -side) becomes less than that of the other side

Question 9. Why does the conductivity of a pure, semiconductor increase with temperature rise?
Answer:

The temperature of a pure semiconductor is increased, and some electrons acquire sufficient kinetic energy to break the bond and come out of the valance band. These elec¬ trons act as charge carriers inside the crystal. So, the conductivity of a pure semiconductor increases with rise in temperature

Short Questions on p-n Junction Diodes

Question 10. In which kind of biasing of a p-n junction, do the net holes flow from the n -n-region to the p-region?
Answer:

In the case of reverse biasing, the net flow of holes is from n -region to p -region. Because, in this case, the majority carrier holes in the p -p-region cannot enter the n -n-region, but the minority carrier holes can move easily from the n -n-region to the p -p-region

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Question 11. Explain why the resistance of a semiconductor decreases with an increase in temperature.
Answer:

With the increase in temperature, a greater number of bonds inside the semiconductor are broken. Hence, a large number of electrons come out from those bonds. As a result, the number of charge carriers increases and consequently, the resistance decreases

Question 12. To which factor is die pulsating nature of the output due?
Answer:

In the process of full wave rectification, the output current will be the same for each half cycle of the alternating current. As the Input voltage is sinusoidal, the BO pulsating nature of the output voltage occurs.

Question 13. In a transistor, the emitter-base junction is always forward-biased, while the collector-base junction is reverse-biased. Why?
Answer:

This bias system is the condition of the transistor to be active. As the emitter-base is forward biased, the input resistance i.e., the resistance of the emitter-base junction becomes very small. Again, as the collector-base junction is reverse biased, the output resistance i.e., the resistance of the emitter-collector becomes very high. The circuit with low input resistance and high output resistance acts as the best amplifier.

Question 14. Why does a photodiode function in reverse bias?
Answer:

In a reverse-biased photodiode, the electric current flows through it from negative to positive potential, which is similar to the current in an electric cell

Practice Short Questions on Logic Gates

Question 15. How is a light-emitting diode fabricated?
Answer:

In an LED, the upper layer of p-type semiconductors is deposited by diffusion on the n-type layer of the semiconductor. The metalized contacts are provided for applying the forward bias voltage to the p-n junction diode from the battery through a resistance that controls the brightness of light emitted

Question 16. In the following diagram, is the junction diode forward-biased or reverse-biased?

Answer:

In the given diagram, the voltage at the p-side is less than the voltage at the n-side of the diode. So, the diode is reverse-biased.

Question 17. 

1. In the following diagram, which bulb out of B₁ and, B₂ will glow and why?
2. Explain briefly the three processes due to which generation of emf takes place in a solar cell.

Answer: 

1. Bulb B₁, will glow as diode D₁ is forward-biased. Bulb B₂ will not glow as diode D₂ is reverse-biased.

Question 18. Three processes due to which generation of emf takes place in a solar cell are:

1. Generation of electron-hole pair* due to light, close to the junction
2. Separation of electrons and holes due to the electric field of the depletion region. Electrons are swept to the n-side and holes to the p -the sides.
3. The electrons reaching the n-side are collected by the front contact and holes reaching the side are collected by the back contact. Thus p -side becomes positive and the n-side becomes negative giving rise to photovoltage

Conceptual Short Questions on Intrinsic and Extrinsic Semiconductors

Question 19. In the following diagram’ S’ is a semiconductor. Would you increase or decrease the value of R to keep the reading of the ammeter A constant when S is heated Give a reason for your answer

We will increase the value of R. When semiconductor S U is heated, Its resistance decreases with a temperature rise. As the semiconductor S is connected in series, the net resistance of the circuit also decreases. So, by Increasing the value of R we can keep the resistance of the circuit constant and lienee the current in the circuit Le. The reading of ammeter A can be kept constant

Question 20. Name the junction diode whose 1-V characteristics are drawn below

Answer:

The junction diode is a solar cell

Real-Life Scenarios Involving Semiconductor Devices Questions

Question 20. HOW is a light-emitting diode fabricated? Briefly, the state is working.
Answer:

In an LED, the upper layer of p-type semiconductors is deposited by diffusion on the n-type layer of the semiconductor. The metalized contacts are provided for applying the forward bias voltage to the p-n junction diode from the battery through a resistance that controls the brightness of light emitted.