Semiconductor Electronics: Materials, Devices and Simple Circuits

Correct Option: (c) Holes are minority carriers and pentavalent atoms are the dopants.

Justification:
In an n-type semiconductor, silicon (tetravalent) is doped with pentavalent atoms (donors) such as As, Sb, or P. Each dopant atom donates one extra electron to the conduction band. Therefore:
- Electrons are the majority carriers.
- Holes are the minority carriers.
- The dopants are pentavalent atoms.

Option (a) is wrong because it states trivalent atoms are dopants (that would give p-type). Option (b) is wrong because electrons are majority carriers, not minority. Option (d) is wrong because holes are not majority carriers in n-type.

Hence, option (c) is correct.

Correct Option: (d) Holes are majority carriers and trivalent atoms are the dopants.

Justification:
In a p-type semiconductor, silicon (tetravalent) is doped with trivalent atoms (acceptors) such as B, Al, or In. Each dopant atom creates a hole (electron vacancy) in the valence band. Therefore:
- Holes are the majority carriers.
- Electrons are the minority carriers.
- The dopants are trivalent atoms.

This matches statement (d) from Exercise 14.1, which is the correct answer for p-type semiconductors.

Correct Option: (c) (E_g)_C > (E_g)_{Si} > (E_g)_{Ge}

Given/Known values:
- $(E_g)_C \approx 5.4\,\text{eV}$ (Carbon/Diamond — insulator)
- $(E_g)_{Si} \approx 1.1\,\text{eV}$ (Silicon — semiconductor)
- $(E_g)_{Ge} \approx 0.7\,\text{eV}$ (Germanium — semiconductor)

Concept: As we go down Group 14 of the periodic table (C → Si → Ge), the atomic size increases, the outermost electrons are farther from the nucleus, and the interatomic bonding becomes weaker. This results in a decrease in the energy band gap.

Conclusion:
( E_g)_C > (E_g)_{Si} > (E_g)_{Ge}

Hence, option (c) is correct.

Correct Option: (c) hole concentration in p-region is more as compared to n-region.

Justification:
Diffusion is the movement of charge carriers from a region of higher concentration to a region of lower concentration. In a p-n junction:
- The p-region has a very high concentration of holes.
- The n-region has a very low concentration of holes.

Due to this concentration gradient, holes diffuse from the p-region to the n-region. This is purely a diffusion process driven by the concentration difference, not by electric attraction or an externally applied potential difference.

Option (a) is incorrect because free electrons do not attract holes across the junction in an unbiased state. Option (b) is incorrect because in an unbiased junction there is no externally applied potential difference.

Hence, option (c) is correct.

Correct Option: (c) lowers the potential barrier.

Justification:
In forward bias, the positive terminal of the battery is connected to the p-side and the negative terminal to the n-side. The applied voltage opposes the built-in potential barrier at the junction.

- The external voltage pushes majority carriers (holes from p-side and electrons from n-side) towards the junction.
- This reduces (lowers) the depletion layer width and consequently lowers the potential barrier.
- As a result, a significant majority carrier current flows across the junction.

Option (a) is wrong — raising the barrier occurs in reverse bias. Option (b) is wrong — majority carrier current is not reduced to zero; in fact it increases substantially.

Hence, option (c) is correct.

Given:
- Input frequency, $f_{in} = 50\,\text{Hz}$

Case 1: Half-Wave Rectifier

Concept: In a half-wave rectifier, only one half-cycle (either positive or negative) of the AC input is allowed to pass through the diode. The other half-cycle is blocked. Therefore, for every one complete cycle of input, there is one pulse of output.

$$f_{output} = f_{input} = 50\,\text{Hz}$$

Output frequency of half-wave rectifier = 50 Hz

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Case 2: Full-Wave Rectifier

Concept: In a full-wave rectifier, both half-cycles of the AC input are utilised (both are converted to the same polarity). Therefore, for every one complete cycle of input, there are two pulses of output.

$$f_{output} = 2 \times f_{input} = 2 \times 50 = 100\,\text{Hz}$$

Output frequency of full-wave rectifier = 100 Hz