Question

Which of the following is a donor impurity atom for Ge?

• A. Boron
• B. Antimony
• C. Aluminium
• D. Indium
• A. 0.5 eV
• B. 0.1 eV
• C. 0.05 eV
• D. 0.01 eV
• A. a layer of negative charge on n-side and a layer of positive charge on p-side appear.
• B. a layer of positive charge on n-side and a layer of negative charge on p-side appear.
• C. the electrons on p-side of the junction move to n-side initially.
• D. initially diffusion current is small and drift current is large.
• A. the drift current is of the order of few mA
• B. the applied voltage mostly drops across the depletion region.
• C. the depletion region width decreases.
• D. the current increases with increase in applied voltage.
• A. 25 Hz
• B. 50 Hz
• C. 100 Hz
• D. 200 Hz
• A. the drift current is of the order of few mA
• B. the applied voltage mostly drops across the depletion region.
• C. the depletion region width decreases.
• D. the current increases with increase in applied voltage.
• A. 25 Hz
• B. 50 Hz
• C. 100 Hz
• D. 200 Hz

Answer 1

The Correct Option is B

In semiconductor physics, donor impurities are elements that have more valence electrons than the host semiconductor. For germanium (Ge), which is a group IV element, donor impurities are typically from group V elements, which have five valence electrons. - Boron and Aluminium are group III elements, which have three valence electrons and are acceptor impurities for Ge. - Antimony and Indium are group V and III elements, respectively. Antimony is a donor impurity for Ge because it has five valence electrons, which means it can donate an extra electron to the conduction band of Ge. - Indium is an acceptor impurity for Ge, as it has three valence electrons. Thus, the correct donor impurity atom for Ge is Antimony.

Answer 2

In this scenario, a pentavalent atom (such as phosphorus or arsenic) is substituted into the silicon (Si) crystal lattice. The pentavalent atom has five valence electrons, while silicon has only four valence electrons. - Four of the pentavalent atom's electrons form covalent bonds with four silicon atoms surrounding it, maintaining the regular bonding structure. - The fifth electron, which doesn't participate in bonding, is loosely bound to the pentavalent atom. This fifth electron is known as the donor electron, and the energy required to free this electron from the parent atom is referred to as the ionization energy of the donor. The typical energy required to release this electron in silicon is approximately 0.05 eV. Thus, the energy required to set the electron free is about 0.05 eV.

Answer 3

When a p-n junction is formed, the diffusion of electrons and holes occurs. Initially, when the p-type and n-type materials come into contact: 1. Diffusion of charge carriers: - Electrons from the n-side (which have a higher concentration) diffuse to the p-side (where they are in lower concentration). - Similarly, holes from the p-side diffuse to the n-side. 2. Formation of a depletion region: - As electrons move from the n-side to the p-side, they recombine with holes, leaving behind a layer of negative charge (due to the immobile ionized donors) on the n-side. - Similarly, holes moving to the n-side leave behind a layer of positive charge (due to immobile ionized acceptors) on the p-side. Thus, initially, the electrons on the p-side of the junction move to the n-side, causing the formation of a depletion region. Therefore, the correct answer is (C).

Answer 4

In a reverse-biased p-n junction, the applied voltage causes the electrons in the n-side to move away from the junction, and holes from the p-side to do the same, leading to the expansion of the depletion region. - Drift Current: The drift current in reverse bias is extremely small, on the order of nanoamperes (nA), not milliampere (mA), making option (A) incorrect. - Voltage Drop: In reverse bias, most of the applied voltage is dropped across the depletion region, as the electric field in this region opposes the flow of carriers, which makes option (B) correct. - Depletion Region: As the reverse voltage increases, the depletion region width increases, not decreases, making option (C) incorrect. - Current: The reverse current is almost constant and very small, regardless of the increase in reverse voltage, making option (D) incorrect. Thus, the correct answer is (B).

Answer 5

A full-wave rectifier inverts the negative half of the input signal, and hence the output frequency is double the input frequency. If the input frequency is 50 Hz, the output frequency will be: \[ f_{\text{output}} = 2 \times f_{\text{input}} = 2 \times 50 \, \text{Hz} = 100 \, \text{Hz} \] Thus, the correct answer is (C).

Answer 6

In a reverse-biased p-n junction, the applied voltage causes the electrons in the n-side to move away from the junction, and holes from the p-side to do the same, leading to the expansion of the depletion region. - Drift Current: The drift current in reverse bias is extremely small, on the order of nanoamperes (nA), not milliampere (mA), making option (A) incorrect. - Voltage Drop: In reverse bias, most of the applied voltage is dropped across the depletion region, as the electric field in this region opposes the flow of carriers, which makes option (B) correct. - Depletion Region: As the reverse voltage increases, the depletion region width increases, not decreases, making option (C) incorrect. - Current: The reverse current is almost constant and very small, regardless of the increase in reverse voltage, making option (D) incorrect. Thus, the correct answer is (B).

Answer 7

A full-wave rectifier inverts the negative half of the input signal, and hence the output frequency is double the input frequency. If the input frequency is 50 Hz, the output frequency will be: \[ f_{\text{output}} = 2 \times f_{\text{input}} = 2 \times 50 \, \text{Hz} = 100 \, \text{Hz} \] Thus, the correct answer is (C).