Free NCERT Solutions For Class 12 Physics Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits Free PDF Download

Question 1: In an n-type silicon, which of the following statements is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.

Ans:
The correct statement is (c).
In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carries. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.

Question 2: Which of the statements given in Exercise 14.1 is true for p-type semiconductor.

Ans:
The correct statement is (d).
In a p-type semiconductor, the holes are the majority carriers, while the holes are the minority carries. A p-type semiconductor is obtained when trivalent atoms, such aluminium, are doped in silicon atoms.

Ans:
The correct answer is (c).
Of the three given elements, the energy band gap of carbon is the maximum and that germanium is the least.

Question 4: In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) Free electrons in the n-region attract them.
(b) They move across the junction by the potential difference.
(c) Hole concentration in p-region is more as compared to n-region.
(d) All the above.

Ans:
The correct statement is (c).
The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

Question 5: When a forward bias is applied to a p-n junction, it
(a) Raises the potential barrier.
(b) Reduce the majority carrier current to zero.
(c) Lower the potential barrier.
(d) None of the above.

Ans:
The correct statement is (c).
When forward bias is applied to a p-n junction, it lower the values of potential barrier. In the case of forward bias, the potential barrier is opposed by the applied voltage. Hence, the potential barrier across the junction gets reduced.

Question 6: For transistor action, which of the following statements are correct.
(a) Base, emitter and collector regions should have similar size and doping concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collector junction is reverse biased.
Both the emitter junction as well as the collector junction are forward biased.

Ans:
The correct statement is (b), (c).
For a transistor action, the base region must be lightly doped so that the base region is very thin. Also, the emitter junction must be forward-biased and collector junction should be reversed-biased.

Question 7: For a transistor amplifier, the voltage gain
(a) Remains constant for all frequencies.
(b) Is high at high and low frequencies and constant in the middle frequency range.
(c) Is low at high and low frequencies and constant at mid frequencies.
(d) None of the above.

Ans:
The correct statement is (c).
The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies.

Question 8: In half-wave rectifier, what is the output frequency if the input frequency is 50Hz. What is the output frequency of a full-wave rectifier for the same input frequency

Solution 8:
Input frequency = 50Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
∴ Output frequency = 50Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
∴ Output frequency = 2 × 50 = 100 Hz

Rest questions are numerical which has been solved in the above pdf