Solutions- Class-11-Science-Physics-Chapter-14-Semiconductors-Maharashtra Board

Semiconductors

Maharashtra Board-Class-11-Science-Physics-Chapter-14

Notes Part-1

Question 1. Choose the correct option.

(i) Electric conduction through a semiconductor is due to:

(A) electrons

(B) holes

(C) none of these

(D) both electrons and holes

Answer :

(D) both electrons and holes

(ii) The energy levels of holes are:

(A) in the valence band

(B) in the conduction band

(C) in the band gap but close to valence band

(D) in the band gap but close to conduction band

Answer :

(C) in the band gap but close to valence band

(iii) Current through a reverse biased p-n junction, increases abruptly at:

(A) breakdown voltage

(B) 0.0 V

(C) 0.3V

(D) 0.7V

Answer :

(A) breakdown voltage

(iv) A reverse biased diode, is equivalent to:

(A) an off switch

(B) an on switch

(C) a low resistance

(D) none of the above

Answer :

(A) an off switch

(v) The potential barrier in p-n diode is due to:

(A) depletion of positive charges near the junction

(B) accumulation of positive charges near the junction

(C) depletion of negative charges near the junction,

(D) accumulation of positive and negative charges near the junction

Answer :

(D) accumulation of positive and negative charges near the junction

Question 2. Answer the following questions.

(i) What is the importance of energy gap in a semiconductor?

Answer :

Importance of energy gap in a semiconductor :

  • A semiconductor, like Si or Ge, has a 1 eV energy gap(EG) between its valence and conduction bands, making it a perfect insulator. The valence band is fully occupied near 0K, while the conduction band is empty.
  • At room temperature, only a few valence electrons gain thermal energy above Eg, the conduction band, leaving empty energy. These electrons travel into the valence band, where they form holes. In the presence of an external electric field, free electrons in the conduction band and holes in the valence band act as charge carriers, resulting in restricted electrical conductivity. The substance is no longer an insulator.

(ii) Which element would you use as an impurity to make germanium an n-type semiconductor?

Answer :

Germanium crystal should be doped with a pentavalent impurity such as phosphorus (P), arsenic (As) or antimony (Sb). These atoms with one electron extra can donate it to the semiconductor to make it n-type.

(iii) What causes a larger current through a p-n junction diode when forward biased?

Answer :

Forward biasing in a pn-junction diode reduces the potential barrier, allowing charge carriers to diffuse across the junction. This narrowed depletion region results in low resistance, resulting in a large forward current across the junction.

(iv) On which factors does the electrical conductivity of a pure semiconductor depend at a given temperature?

Answer :

Importance of the depletion region in a pn-junction diode :

The region of the junction between a p-type region and an n-type region within a single semiconducting crystal is depleted of free (or mobile) charge carriers. This is called the depletion region.

  • The depletion region in a diode acts as a potential barrier, allowing diffusion of majority charge carriers across the junction.
  • Under forward bias, the external voltage lowers the barrier height, allowing charge carriers to diffuse into the other region, causing the diode to conduct.
  • Under reverse bias, the external voltage adds to the barrier potential, increasing the barrier height, stopping diffusion and causing the diode to enter a non-conducting state.
  • This allows for control of the diode's conduction state like a switch, with forward bias acting like a closed switch and reverse bias acting like an open switch.

(v) Why is the conductivity of a n-type semiconductor greater than that of p-type semiconductor even when both of these have same level of doping?

Answer :

  • In a p-type semiconductor, holes are majority charge carriers.
  • Holes behave as positive charges. When a p-type semiconductor is connected to a battery, holes are attracted towards the negative terminal of the battery.
  • Holes are transported, when there is movement of electrons.
  • The drift speed of these electrons in a p-type semiconductor is less than that in an n-type semiconductor. Also the mobility of the holes is less than that of the electrons.
  • The electrical conductivity depends on the mobility of charge carriers. Hence, the conductivity of an n-type semiconductor is greater than that of a p-type semiconductor even when both of these have the same level of doping.

Question 3. Answer in detail.

(i) Explain how solids are classified on the basis of band theory of solids.

Answer :

Classification of solids on the basis of the band theory of solids :

Solids are classified as conductors, semiconductors, and insulators based on their electrical characteristics. Band theory explains the variation in electrical conductivities.

  • Metals have an overlap between the conduction and valence bands, allowing for large numbers of electrons and energy levels. Electrons can shift into higher energy states, absorb energy from the field, and participate in electrical conduction when an external electric field is present.
  • Semiconductors have a small forbidden energy gap (EG) of about 1 eV, allowing some electrons near the top of the valence band to absorb thermal energy and participate in electrical conduction when an electric field is applied.
  • Insulators have a wide energy gap (EG ≥ 3 eV), requiring less than 3 eV for electron transfer from the valence band to the conduction band.

(ii) Distinguish between intrinsic semiconductors and extrinsic semiconductors.

Answer :

Intrinsic semiconductors

Extrinsic semiconductors.

A pure, perfect semiconductor with an ideal crystal structure is called an intrinsic semiconductor. An intrinsic semiconductor doped with a small and controlled amount of a specific dopant, i.e., impurity atoms, is called an extrinsic semiconductor.
Close to 0 K, each tetravalent semiconductor atom has a complete octet of valence electrons by forming four covalent bonds with its nearest neighbours. Close to 0 K, each substitutional impurity atom in a tetravalent semiconductor has an incomplete bond, with the hole or the surplus electron

weakly bound to the dopant atom.

Above 0 K, pairs of electrons and holes, in the conduction band (CB) and the valence band (VB), respectively are created by breaking of covalent bonds due to thermal energy. Above 0 K, acceptor atoms create holes in the VB without adding electrons to the CB; donor atoms create electrons in the CB without adding holes to the VB.

(iii) Explain the importance of the depletion region in a p-n junction diode.

Answer :

Importance of the depletion region in a pn-junction diode :

The region of the junction between a p-type region and an n-type region within a single semiconducting crystal is depleted of free (or mobile) charge carriers. This is called the depletion region.

  • The depletion region in a diode acts as a potential barrier, allowing diffusion of majority charge carriers across the junction.
  • Under forward bias, the external voltage lowers the barrier height, allowing charge carriers to diffuse into the other region, causing the diode to conduct.
  • Under reverse bias, the external voltage adds to the barrier potential, increasing the barrier height, stopping diffusion and causing the diode to enter a non-conducting state.
  • This allows for control of the diode's conduction state like a switch, with forward bias acting like a closed switch and reverse bias acting like an open switch.

(iv) Explain the I-V characteristic of a forward biased junction diode.

Answer :

I-V characteristic of a forward biased diode : Initially, the current is very low and then there is a sudden rise in the current. The point at which current rises sharply is shown as the ‘knee’ point on the I-V characteristic curve. The corresponding voltage is called the ‘knee voltage’. It is about 0.7 V for silicon and 0.3 V for germanium.

(v) Discuss the effect of external voltage on the width of depletion region of a p-n junction.

Answer :

  • In an unbiased pn-junction, the n-side has a higher electric potential than the p-side. An external voltage can be applied to control the width of the depletion region and barrier height.

  • Under forward bias[Fig.(a)], the applied voltage opposes the barrier, allowing electrons and holes to diffuse into the p-side and n-side. Some electrons and holes recombine with uncompensated donor and acceptor ions, reducing the width of the depletion region.
  • Under reverse bias [Fig.(b)], the applied voltage increases the barrier potential. The reverse bias draws the majority carriers out of each region and the depletion region widens.

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