Chapter 14 Semiconductor Electronics NCERT Solutions Class 12 Physics

Question 14.1

In an n-type silicon, which of the following statement 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.

Analysis

n-type: Created by doping Silicon (Group 14) with Pentavalent impurity (Group 15, e.g., Arsenic, Phosphorus).
Pentavalent atoms: Have 5 valence electrons. 4 form bonds, 1 is free. Thus, they donate electrons.
Carriers: Electrons are Majority carriers. Holes are Minority carriers.

Statement (c) says: “Holes are minority carriers and pentavalent atoms are the dopants.” This is correct.

Correct Answer: (c)

Question 14.2

Which of the statements given in Exercise 14.1 is true for p-type semiconductors.

Analysis

p-type: Created by doping Silicon (Group 14) with Trivalent impurity (Group 13, e.g., Boron, Aluminium).
Trivalent atoms: Have 3 valence electrons. They create a vacancy (hole) to accept an electron.
Carriers: Holes are Majority carriers. Electrons are Minority carriers.

Looking at options from 14.1:

(d) Holes are majority carriers and trivalent atoms are the dopants.

Correct Answer: (d)

Question 14.3

Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to \((E_g)_C\), \((E_g)_{Si}\) and \((E_g)_{Ge}\). Which of the following statements is true?
(a) \((E_g)_{Si} < (E_g)_{Ge} < (E_g)_C\)
(b) \((E_g)_C < (E_g)_{Ge} > (E_g)_{Si}\)
(c) \((E_g)_C > (E_g)_{Si} > (E_g)_{Ge}\)
(d) \((E_g)_C = (E_g)_{Si} = (E_g)_{Ge}\)

Energy Band Gaps

The energy band gap depends on the strength of the atomic bonding and the size of the atom.

Carbon (Diamond): Insulator. Very strong bonds. \(E_g \approx 5.4 \text{ eV}\).
Silicon: Semiconductor. \(E_g \approx 1.1 \text{ eV}\).
Germanium: Semiconductor. Larger atom, weaker hold. \(E_g \approx 0.7 \text{ eV}\).

Order: Carbon > Silicon > Germanium.

Correct Answer: (c) \((E_g)_C > (E_g)_{Si} > (E_g)_{Ge}\)

Question 14.4

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.

Explanation

Diffusion is the movement of particles from a region of higher concentration to a region of lower concentration.

In a p-n junction, the p-side has a high concentration of holes, while the n-side has very few holes.
This concentration gradient drives the holes to diffuse from p to n.
It is not primarily due to attraction or potential difference (in fact, the potential barrier eventually stops this diffusion).

Correct Answer: (c)

Question 14.5

When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above.

Forward Bias

In forward bias, the positive terminal is connected to the p-side and negative to the n-side.

The applied external field opposes the built-in potential field.
This lowers the effective potential barrier width and height.
This allows majority charge carriers to cross the junction easily, causing current flow.

Correct Answer: (c)

Question 14.6

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.

1. Half-Wave Rectifier

[Image of Half-Wave Rectifier Waveform]

A half-wave rectifier conducts only during the positive half-cycle. It produces one output pulse for every one input cycle. The time period remains the same.

Output Frequency = Input Frequency = 50 Hz.

2. Full-Wave Rectifier

A full-wave rectifier converts both positive and negative half-cycles into positive output pulses. It produces two output pulses for every one input cycle.

Output Frequency = \(2 \times\) Input Frequency = \(2 \times 50\) = 100 Hz.

Half-wave: 50 Hz
Full-wave: 100 Hz