NCERT Solutions For Chapter 14: Semiconductor Electronics

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NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits are provided in the article below. A semiconductor is a material whose resistivity is between a conductor such as metallic copper and an insulator such as glass. Its resistivity falls as the temperature rises which is completely opposite to any metal.

Class 12 Physics Chapter 14 Semiconductor Electronics belong to Unit 9 - Electronic Devices which has a weightage of 7 marks in the CBSE Board examinations. Class 12 Physics Semiconductor Electronics NCERT Solutions covers the concepts of intrinsic and extrinsic semiconductors, P-n Junction, and Rectifiers

Download PDF: NCERT Solutions for Class 12 Physics Chapter 14

NCERT Solutions for Class 12 Physics Chapter 14

Class 12 Physics Chapter 14 – Topics Covered

Semiconductors have resistivity or conductivity in between that of metals and insulators.

ρ ~ 10^-5. 10⁶ Ωm, σ ~ 10⁺⁵ .10^-6 Sm^-1

Types of Semiconductors: There are two types of semiconductors – Elements Semiconductors and Compound Semiconductors.

(i) Elements Semiconductors are available in natural form, e.g. germanium and silicon.
(ii) Compound Semiconductors are made by compounding the metals, e.g. InP, CdS, polyaniline, GaAs, CdSe, anthracene, etc.

On the basis of purity, semiconductors are classified as intrinsic semiconductors and extrinsic semiconductors.
Intrinsic Semiconductors are pure semiconductor that does not have any significant dopant species present

n_e = n_h = n_i

where, n_e and n_h are the number densities of electrons and holes respectively and n_i is the intrinsic carrier concentration.

Extrinsic Semiconductors are pure semiconductors that are doped with an impurity.

Extrinsic semiconductors are classified into two types: p-type semiconductors and n-type semiconductors.

Formation of Depletion Region in p-n junction: During the formation of a p-n junction, due to the concentration gradient across the p and n sides, the holes diffuse from the p-side to the n-side and electrons diffuse from the n-side to the p-side.

Formation of Depletion Region in p-n junction

Formation of Depletion Region in p-n junction

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CBSE CLASS XII Related Questions

1.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))
View Solution
2.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.
View Solution
3.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:
- attract with a force \( \frac{F}{2} \)
- repel with a force \( \frac{F}{2} \)
- repel with a force \( F \)
- attract with a force \( F \)
View Solution
4.
Write any two features of nuclear forces.
View Solution
5.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.
View Solution
6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:
- \( \frac{q}{4 \pi \epsilon_0 l^2} \) pointing along MA
- \( \frac{q}{\pi \epsilon_0 l^2} \) pointing along AM
- \( \frac{q}{2 \pi \epsilon_0 l^2} \) pointing along AM
- Zero
View Solution

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NCERT Solutions For Chapter 14: Semiconductor Electronics

NCERT Solutions for Class 12 Physics Chapter 14

Class 12 Physics Chapter 14 – Topics Covered

CBSE CLASS XII Related Questions

1.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

2.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

3.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

4.
Write any two features of nuclear forces.

5.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

Similar Physics Concepts

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NCERT Solutions For Chapter 14: Semiconductor Electronics

NCERT Solutions for Class 12 Physics Chapter 14

Class 12 Physics Chapter 14 – Topics Covered

CBSE CLASS XII Related Questions

1.Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

2. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

3.Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

4. Write any two features of nuclear forces.

5.Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

6.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

2.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

3.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

4.
Write any two features of nuclear forces.

5.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be: