CBSE Class 12 Physics Exemplar Chapter 14: Semiconductor Electronics NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 12 PDF

This chapter delves into the fundamental concepts of Semiconductor Electronics, covering materials, devices, and simple circuits. The NCERT Solutions for Class 12 Physics Exemplar Chapter 14 provide clear explanations and step-by-step solutions to various problems, including multiple-choice questions. These solutions help students understand the behavior of semiconductors, the working of p-n junctions under different biasing conditions, and the analysis of simple electronic circuits. The chapter also touches upon the effect of temperature on conductivity and the characteristics of diodes. These solutions are designed to aid students in grasping complex topics, reinforcing their learning, and preparing effectively for their board examinations by offering a comprehensive review of the chapter's key concepts and problem-solving techniques.

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Board	CBSE
Class	Class 12
Subject	Physics Exemplar
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 14

Chapter summary

Chapter 14 of the CBSE Class 12 Physics Exemplar focuses on Semiconductor Electronics. The NCERT Solutions cover essential topics such as the conductivity of semiconductors, the behavior of p-n junctions in forward and reverse bias, and the analysis of diode circuits. It explains how temperature affects semiconductor conductivity and clarifies the potential barrier across a junction. The solutions also address the functioning of diodes in AC circuits, including rectification principles.

Learning outcomes

Understand the factors affecting semiconductor conductivity.
Analyze the behavior of p-n junctions under forward and reverse bias.
Determine the current flow in diode circuits based on biasing.
Calculate potential differences across components in AC circuits with diodes.
Explain the role of potential barriers in semiconductor devices.

Topics covered

Paper topics

Semiconductor conductivity
Temperature effect on conductivity
p-n junction
Potential barrier
Forward bias
Reverse bias
Diodes
Ideal diodes
AC supply
Capacitors in circuits
Peak voltage
RMS voltage

Important topics

p-n junction biasing (forward and reverse)
Potential barrier formation and modification
Diode behavior in AC circuits
Effect of temperature on semiconductor conductivity

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Questions and Solutions

Question 1

The conductivity of a semiconductor increases with increase in temperature, because

(a) number density of free current carries increases

(b) relaxation time increases

(d) number density of carries increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density

Solution: The conductivity of a semiconductor is determined by the number density of charge carriers and their mobility. As the temperature of a semiconductor increases, more electrons gain sufficient thermal energy to break free from their covalent bonds, thus increasing the number density of free charge carriers (both electrons and holes). While increased thermal vibrations also lead to more frequent collisions, reducing the average time between collisions (relaxation time), the increase in the number of charge carriers is the dominant factor. Therefore, the conductivity increases. The decrease in relaxation time has a lesser impact compared to the significant rise in carrier density. Hence, option (d) accurately describes this phenomenon.

Question 2

In the figure given below, $V_0$ is the potential barrier across a p-n junction, when no battery is connected across the junction. Which of the following statements correctly describes the bias conditions represented by the figures?

(a) 1 and 3 both correspond to forward bias of the junction

(b) 3 corresponds to forward bias of the junction and 1 corresponds to reverse bias of the junction

(d) 3 and 1 both correspond to reverse bias of the junction

Solution: The potential barrier across a p-n junction arises due to the diffusion of charge carriers and the formation of a depletion region. When an external voltage is applied, it can either oppose or aid this potential barrier.

Forward Bias: When the p-side of the junction is connected to the positive terminal of a battery and the n-side to the negative terminal, the applied voltage opposes the potential barrier. This reduces the width of the depletion region and lowers the potential barrier ($V_0$). Figure 1, showing a reduced potential barrier, represents forward bias.

Reverse Bias: When the p-side is connected to the negative terminal and the n-side to the positive terminal, the applied voltage aids the potential barrier. This increases the width of the depletion region and raises the potential barrier ($V_0$). Figure 3, showing an increased potential barrier, represents reverse bias.

Therefore, 1 corresponds to forward bias and 3 corresponds to reverse bias of the junction.

Question 3

In the figure given below, assuming the diodes to be ideal, which of the following statements is correct?

$-10V \stackrel{A}{ullet} \stackrel{R}{ullet} \stackrel{D_1}{\longrightarrow} D_2 \stackrel{B}{ullet}$

(a) $D_1$ is forward biased and $D_2$ is reverse biased and hence current flows from A to B

(b) $D_2$ is forward biased and $D_1$ is reverse biased and hence no current flows from B to A and vice-versa

(d) $D_1$ and $D_2$ are both reverse biased and hence no current flows from A to B and vice-versa

Solution: To determine the state of the diodes, we need to examine the voltage across their p-n junctions.

Diode $D_1$: The p-side of $D_1$ is connected to point A, which is at -10V. The n-side of $D_1$ is connected to point R. Assuming R is at a higher potential than A (which is implied by the circuit diagram showing current potentially flowing towards B), or if we consider the overall circuit context where B is likely at a higher potential than A, the voltage across $D_1$ would be such that the p-side is at a lower potential than the n-side. Specifically, if we consider the path from A to B, and assuming the resistor R is present, the voltage at the n-side of D1 is likely higher than -10V. Thus, $D_1$ is reverse biased.

Diode $D_2$: The p-side of $D_2$ is connected to the n-side of $D_1$ (point R) and the n-side of $D_2$ is connected to point B. If we assume B is at a higher potential than R (e.g., ground or positive voltage), then the p-side of $D_2$ is at a higher potential than its n-side. This condition makes $D_2$ forward biased.

Since $D_1$ is reverse biased and $D_2$ is forward biased, current can flow through $D_2$ but not through $D_1$. The question states current flows from B to A, which is consistent with $D_2$ being forward biased and allowing current from B towards A (through $D_2$), while $D_1$ blocks any current from A towards B.

Therefore, $D_2$ is forward biased and $D_1$ is reverse biased. Current flows from B to A through $D_2$.

Question 4

A 220 V AC supply is connected between points A and B (figure). What will be the potential difference $V$ across the capacitor?

$220 \text{ V AC}$

(a) 220 V

(b) 110 V

(d) $220\sqrt{2}$ V

Solution: The circuit consists of an AC supply connected to a series combination of a diode and a capacitor. The diode acts as a rectifier, allowing current to flow only during one half-cycle of the AC input.

During the positive half-cycle of the AC input voltage, the diode is forward biased and conducts. This allows current to flow and charge the capacitor. The capacitor charges up to the peak voltage of the AC supply.

During the negative half-cycle, the diode is reverse biased and does not conduct. The capacitor, once charged, holds its charge (ideally) and maintains a potential difference across it equal to the peak voltage it reached.

The given AC supply is 220 V, which is the RMS (Root Mean Square) value. The peak voltage ($V_0$) of an AC supply is related to its RMS value ($V_{rms}$) by the formula:

$V_0 = V_{rms} \times \sqrt{2}$

Substituting the given RMS voltage:

$V_0 = 220 \times \sqrt{2} \text{ V}$

Therefore, the potential difference across the capacitor will be equal to the peak voltage, which is $220\sqrt{2}$ V.

Common mistakes

Confusing forward and reverse bias conditions for diodes.
Incorrectly applying the concept of potential barrier in p-n junctions.
Misinterpreting the effect of temperature on semiconductor conductivity.
Errors in analyzing diode behavior in AC circuits.

Revision tips

Review the definitions of forward and reverse bias for p-n junctions.
Practice drawing and analyzing diode circuits with different voltage sources.
Understand the relationship between temperature and semiconductor conductivity.
Focus on the concept of potential barrier and how it changes with bias.

Practice MCQs

Q1. The conductivity of a semiconductor increases with an increase in temperature because:

Q2. In a p-n junction, when no battery is connected, V0 represents the potential barrier. Which bias conditions are represented by the figures (assuming figure 1 shows a reduced barrier and figure 3 shows an increased barrier)?

Q3. Assuming ideal diodes, in the given circuit, what is the state of diodes D1 and D2, and the direction of current flow?

Q4. If a 220 V AC supply is connected across points A and B in a circuit containing a diode and a capacitor in series, what will be the potential difference across the capacitor?

Frequently asked questions

What is the main focus of CBSE Class 12 Physics Exemplar Chapter 14?

Chapter 14 focuses on Semiconductor Electronics, covering materials, devices like diodes, and simple circuits, including their behavior under different conditions.

How does temperature affect the conductivity of a semiconductor?

The conductivity of a semiconductor increases with increasing temperature because the number density of free charge carriers increases significantly, which outweighs the decrease in relaxation time.

What is the difference between forward bias and reverse bias in a p-n junction?

In forward bias, the p-n junction's potential barrier is reduced, allowing current flow. In reverse bias, the potential barrier is increased, significantly restricting current flow.

How do ideal diodes behave in a circuit?

An ideal diode acts as a perfect conductor when forward biased (zero resistance) and a perfect insulator when reverse biased (infinite resistance).

What is the role of a diode in an AC circuit with a capacitor?

A diode in series with a capacitor connected to an AC supply acts as a half-wave rectifier, charging the capacitor to the peak voltage of the AC input.

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