Comprehension
Extrinsic semiconductors are made by doping pure or intrinsic semiconductors with suitable impurity. There are two types of dopants used in doping, Si or Ge, and using them p-type and n-type semiconductors can be obtained. A p-n junction is the basic building block of many semiconductor devices. Two important processes occur during the formation of a p-n junction: diffusion and drift. When such a junction is formed, a ’depletion layer’ is created consisting of immobile ion-cores. This is responsible for a junction potential barrier. The width of a depletion layer and the height of potential barrier changes when a junction is forward-biased or reverse-biased. A semiconductor diode is basically a p-n junction with metallic contacts provided at the ends for application of an external voltage. Using diodes, alternating voltages can be rectified.
Question: 1
Which of the following is a donor impurity atom for Ge?
Which of the following is a donor impurity atom for Ge?
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Donor impurities for semiconductors are typically from group V elements (e.g., Antimony, Arsenic). They donate electrons to the conduction band, making the material n-type. Acceptor impurities are from group III elements (e.g., Boron, Aluminium), which accept electrons from the semiconductor, making the material p-type.
Updated On: Feb 9, 2026
- Boron
- Antimony
- Aluminium
- Indium
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The Correct Option is B
Solution and Explanation
In semiconductor physics, donor impurities are elements with more valence electrons than the host semiconductor. For germanium (Ge), a Group IV element, Group V elements, possessing five valence electrons, are typically used as donor impurities.
- Boron and Aluminium are Group III elements with three valence electrons, acting as acceptor impurities for Ge.
- Antimony and Indium are Group V and III elements, respectively. Antimony, with its five valence electrons, functions as a donor impurity for Ge by donating an extra electron to Ge's conduction band.
- Indium, having three valence electrons, is an acceptor impurity for Ge.
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Question: 2
When a pentavalent atom occupies the position of an atom in the crystal lattice of Si, four of its electrons form covalent bonds with four silicon neighbours, while the fifth remains bound to the parent atom. The energy required to set this electron free is about:
When a pentavalent atom occupies the position of an atom in the crystal lattice of Si, four of its electrons form covalent bonds with four silicon neighbours, while the fifth remains bound to the parent atom. The energy required to set this electron free is about:
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In semiconductor physics, donor electrons, which are introduced by pentavalent impurities, are typically freed with low energy (around 0.05 eV in silicon). This low energy requirement is why silicon can easily be doped to control charge carrier concentration in electronic devices.
Updated On: Feb 9, 2026
- 0.5 eV
- 0.1 eV
- 0.05 eV
- 0.01 eV
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The Correct Option is C
Solution and Explanation
When a pentavalent atom (e.g., phosphorus, arsenic) replaces a silicon (Si) atom in its crystal lattice, it introduces an extra electron. Silicon atoms have four valence electrons, while pentavalent atoms have five.
- Four of the pentavalent atom's valence electrons form covalent bonds with the surrounding four silicon atoms, preserving the crystal's structure.
- The fifth valence electron remains unbound and is weakly associated with the pentavalent atom.
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Question: 3
During formation of a p-n junction:
During formation of a p-n junction:
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During the formation of a p-n junction, diffusion dominates at first, causing charge carriers to move across the junction, leading to the development of the depletion region and the formation of a built-in electric field.
Updated On: Feb 9, 2026
- a layer of negative charge on n-side and a layer of positive charge on p-side appear.
- a layer of positive charge on n-side and a layer of negative charge on p-side appear.
- the electrons on p-side of the junction move to n-side initially.
- initially diffusion current is small and drift current is large.
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The Correct Option is C
Solution and Explanation
When a p-n junction forms, charge carrier diffusion occurs. Upon initial contact between p-type and n-type materials:
1. Charge Carrier Diffusion:
* Electrons, more concentrated on the n-side, migrate to the p-side where their concentration is lower.
* Holes, more concentrated on the p-side, migrate to the n-side.
2. Depletion Region Formation:
* Electron diffusion to the p-side results in recombination with holes, leaving immobile negative charges (ionized donors) on the n-side.
* Hole diffusion to the n-side leaves immobile positive charges (ionized acceptors) on the p-side.
Consequently, the initial movement of electrons from the p-side to the n-side at the junction establishes a depletion region.
Therefore, the correct answer is (C).
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Question: 4
In reverse-biased p-n junction:
In reverse-biased p-n junction:
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In a reverse-biased p-n junction, the current is very small and primarily due to the minority carriers. The applied voltage is mostly dropped across the depletion region, leading to a small reverse current.
Updated On: Feb 9, 2026
- the drift current is of the order of few mA
- the applied voltage mostly drops across the depletion region.
- the depletion region width decreases.
- the current increases with increase in applied voltage.
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The Correct Option is B
Solution and Explanation
In a reverse-biased p-n junction, applied voltage repels electrons from the n-side and holes from the p-side away from the junction, thus widening the depletion region.
- Drift Current: The drift current in reverse bias is negligible, approximately nanoamperes (nA), not milliampere (mA). Therefore, option (A) is incorrect.
- Voltage Drop: The majority of the applied voltage in reverse bias is dissipated across the depletion region. This occurs because the electric field within this region impedes carrier flow, validating option (B) as correct.
- Depletion Region: An increasing reverse voltage leads to an expansion, not a reduction, of the depletion region width, rendering option (C) incorrect.
- Current: The reverse current remains consistently minimal and largely independent of increases in reverse voltage, making option (D) incorrect.
Consequently, option (B) is the correct choice.
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Question: 5
The output frequency of a full-wave rectifier with 50 Hz as input frequency is:
The output frequency of a full-wave rectifier with 50 Hz as input frequency is:
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In a full-wave rectifier, the output frequency is always twice the input frequency because both halves of the input signal are used in the rectification process.
Updated On: Feb 9, 2026
- 25 Hz
- 50 Hz
- 100 Hz
- 200 Hz
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The Correct Option is C
Solution and Explanation
A full-wave rectifier inverts the negative portion of the input signal. Consequently, the output frequency is twice the input frequency. Given an input frequency of 50 Hz, the output frequency is calculated as:\[f_{\text{output}} = 2 \times f_{\text{input}} = 2 \times 50 \, \text{Hz} = 100 \, \text{Hz}\]Therefore, option (C) is the correct choice.
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Question: 6
In reverse-biased p-n junction:
In reverse-biased p-n junction:
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In a reverse-biased p-n junction, the current is very small and primarily due to the minority carriers. The applied voltage is mostly dropped across the depletion region, leading to a small reverse current.
Updated On: Feb 9, 2026
- the drift current is of the order of few mA
- the applied voltage mostly drops across the depletion region.
- the depletion region width decreases.
- the current increases with increase in applied voltage.
Show Solution
The Correct Option is B
Solution and Explanation
In a reverse-biased p-n junction, the applied voltage repels electrons from the n-side and holes from the p-side, widening the depletion region.
- Drift Current: The reverse bias drift current is minuscule, measured in nanoamperes (nA), not milliamperes (mA), rendering option (A) invalid.
- Voltage Drop: The majority of the applied voltage in reverse bias is absorbed by the depletion region due to the opposing electric field, confirming option (B).
- Depletion Region: Increased reverse voltage leads to an expansion, not contraction, of the depletion region, negating option (C).
- Current: The reverse current remains nearly constant and very small, irrespective of voltage increases, making option (D) incorrect.
Therefore, option (B) is the accurate choice.
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Question: 7
The output frequency of a full-wave rectifier with 50 Hz as input frequency is:
The output frequency of a full-wave rectifier with 50 Hz as input frequency is:
Show Hint
In a full-wave rectifier, the output frequency is always twice the input frequency because both halves of the input signal are used in the rectification process.
Updated On: Feb 9, 2026
- 25 Hz
- 50 Hz
- 100 Hz
- 200 Hz
Show Solution
The Correct Option is C
Solution and Explanation
A full-wave rectifier inverts the negative portion of the input signal. Consequently, the output frequency is twice the input frequency. Given an input frequency of 50 Hz, the output frequency is calculated as:
\[
f_{\text{output}} = 2 \times f_{\text{input}} = 2 \times 50 \, \text{Hz} = 100 \, \text{Hz}
\]
Therefore, the correct option is (C).
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