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

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

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

ne  = nh = ni 

where, ne and nh are the number densities of electrons and holes respectively and ni 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.
    A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.


      • 2.
        The radius of a nucleus of mass number 125 is:

          • 6.0 fm
          • 30 fm
          • 72 fm
          • 150 fm

        • 3.
          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: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))


            • 4.
              Differentiate between inductive reactance, capacitive reactance and impedance of an ac circuit.
              An ideal inductor and an ideal capacitor are connected in series across an ac voltage. Plot a graph showing variation of net reactance of the circuit with frequency of the applied ac voltage.


                • 5.
                  The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:

                    • \( 0.5 \times 10^3 \, \text{m}^{-1} \)
                    • \( 6.0 \times 10^2 \, \text{m}^{-1} \)
                    • \( 7.5 \times 10^2 \, \text{m}^{-1} \)
                    • \( 1.5 \times 10^3 \, \text{m}^{-1} \)

                  • 6.
                    Write any two features of nuclear forces.

                      CBSE CLASS XII Previous Year Papers

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