Semiconductor Electronics: Materials, Devices And Simple CircuitsClass 12 Physics NCERT Solutions
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Q1EXERCISES
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.
Solution
The correct statement is (c) Holes are minority carriers and pentavalent atoms are the dopants.
Explanation:
- An n-type semiconductor is created by doping a pure semiconductor like silicon (which is tetravalent) with a pentavalent impurity (valency 5), such as Arsenic (As) or Phosphorous (P). These impurity atoms are called dopants.
- Each pentavalent dopant atom provides one extra electron that is free to move and contribute to conduction. Therefore, in an n-type semiconductor, electrons are the majority charge carriers.
- Holes, which are generated through thermal energy (intrinsic generation), are present in much smaller numbers compared to the electrons from the dopant atoms. Hence, holes are the minority charge carriers.
Q2EXERCISES
14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.
Solution
For p-type semiconductors, the true statement is (d) Holes are majority carriers and trivalent atoms are the dopants.
Explanation:
- A p-type semiconductor is formed by doping a pure semiconductor like silicon with a trivalent impurity (valency 3), such as Boron (B) or Aluminium (Al). These impurity atoms are called acceptors.
- Each trivalent dopant atom creates a vacancy or a 'hole' in the covalent bond structure because it has one less valence electron than silicon. This hole can accept an electron from a neighbouring atom, making the hole available for conduction.
- Therefore, in a p-type semiconductor, holes are the majority charge carriers.
- Electrons, which are generated intrinsically through thermal energy, are present in much smaller numbers and are thus the minority charge carriers.
Q3EXERCISES
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
Solution
The correct statement is (c) (E_g)_C > (E_g)_Si > (E_g)_Ge
Explanation:
Based on the text, the energy gap (E_g) between the conduction and valence bands determines whether a material is an insulator, semiconductor, or conductor. The values for Carbon, Silicon, and Germanium are given as:
- For Carbon (diamond): E_g ≈ 5.4 eV
- For Silicon: E_g ≈ 1.1 eV
- For Germanium: E_g ≈ 0.7 eV
Comparing these values, it is clear that the energy gap for Carbon is the largest, followed by Silicon, and then Germanium. This explains why Carbon is an insulator, while Silicon and Germanium are intrinsic semiconductors. Therefore, the correct relationship is (E_g)_C > (E_g)_Si > (E_g)_Ge.
Q4EXERCISES
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.
Solution
The correct statement is (c) hole concentration in p-region is more as compared to n-region.
Explanation:
During the formation of a p-n junction, two processes occur: diffusion and drift. The initial movement of charge carriers is due to diffusion.
- In a p-type semiconductor (p-region), the concentration of holes (majority carriers) is very high.
- In an n-type semiconductor (n-region), the concentration of holes (minority carriers) is very low.
- Due to this large concentration gradient across the junction, holes tend to move from the region of higher concentration (p-region) to the region of lower concentration (n-region). This process is called diffusion. Similarly, electrons diffuse from the n-region to the p-region. This movement gives rise to the diffusion current.
Q5EXERCISES
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.
Solution
The correct statement is (c) lowers the potential barrier.
Explanation:
- A p-n junction is forward biased when the p-side is connected to the positive terminal of a battery and the n-side is connected to the negative terminal.
- The direction of this applied external voltage is opposite to the direction of the built-in potential barrier (V₀) that exists across the depletion region.
- As a result, the effective barrier height is reduced to (V₀ - V), where V is the applied forward voltage.
- This lowering of the potential barrier allows a large number of majority charge carriers (holes from the p-side and electrons from the n-side) to cross the junction, resulting in a significant forward current.
Q6EXERCISES
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.
Solution
For an input frequency of 50 Hz:
-
Half-wave rectifier: The output frequency is 50 Hz. Explanation: A half-wave rectifier allows current to flow only during one half of the input AC cycle (either the positive or negative half). It blocks the other half. Therefore, the output waveform consists of one pulse for every complete cycle of the input AC voltage. The period of the output wave is the same as the period of the input wave, and hence the frequency remains the same.
-
Full-wave rectifier: The output frequency is 100 Hz. Explanation: A full-wave rectifier utilizes both halves of the input AC cycle. It inverts the negative half-cycles to make them positive. As a result, for every one complete cycle of the 50 Hz input, there are two positive pulses in the output. This means the period of the output wave is half the period of the input wave. Since frequency is the reciprocal of the period, the output frequency is double the input frequency (2 × 50 Hz = 100 Hz).