Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits
Exercises
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.
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Answer
The correct statement is (c).
In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped with silicon atoms.
14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.
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Answer
The correct statement is (d).
In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped with silicon atoms.
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 $\left(E _{\mathrm{g}}\right) _{\mathrm{C}},\left(E _{\mathrm{g}}\right) _{\mathrm{Si}}$ เคคเคฅเคพ $\left(E _{\mathrm{g}}\right) _{\mathrm{Ge}}$ Which of the following statements is true?
(a) $\left(E _{g}\right) _{\mathrm{Si}}<\left(E _{g}\right) _{\mathrm{Ge}}<\left(E _{g}\right) _{\mathrm{C}}$
(b) $\left(E _{g}\right) _{\mathrm{C}}<\left(E _{g}\right) _{\mathrm{Ge}}>\left(E _{g}\right) _{\mathrm{Si}}$
(c) $\left(E _{g}\right) _{\mathrm{C}}>\left(E _{g}\right) _{\mathrm{Si}}>\left(E _{g}\right) _{\mathrm{Ge}}$
(d) $\left(E _{g}\right) _{\mathrm{C}}=\left(E _{g}\right) _{\mathrm{Si}}=\left(E _{g}\right) _{\mathrm{Ge}}$
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Answer
To determine the correct relationship between the energy band gaps of carbon (C), silicon (Si), and germanium (Ge), we need to consider the known values of their energy band gaps:
- The energy band gap for carbon (in its diamond form) is approximately $ E_g \approx 5.5 , \text{eV} $.
- The energy band gap for silicon is approximately $ E_g \approx 1.1 , \text{eV} $.
- The energy band gap for germanium is approximately $ E_g \approx 0.66 , \text{eV} $.
Based on these values, we can summarize the relationships as follows:
$ E_g \text{ (C)} > E_g \text{ (Si)} > E_g \text{ (Ge)} $
This means that carbon has the largest band gap, followed by silicon, and germanium has the smallest band gap.
Now, let’s evaluate the options provided:
(a) $ E_g \text{ (Si)} < E_g \text{ (Ge)} < E_g \text{ (C)} $ - False (because $ E_g \text{ (Ge)} < E_g \text{ (Si)} $)
(b) $ E_g \text{ (C)} < E_g \text{ (Ge)} > E_g \text{ (Si)} $ - False (because $ E_g \text{ (C)} > E_g \text{ (Ge)} $)
(c) $ E_g \text{ (C)} > E_g \text{ (Si)} > E_g \text{ (Ge)} $ - True (this is consistent with the known values)
(d) $ E_g \text{ (C)} = E_g \text{ (Si)} = E_g \text{ (Ge)} $ - False (they are not equal)
Thus, the correct statement is:
(c) $ E_g \text{ (C)} > E_g \text{ (Si)} > E_g \text{ (Ge)} $.
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.
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Answer
In an unbiased p-n junction, holes diffuse from the p-region to the n-region primarily due to the concentration gradient. Let’s analyze the options:
(a) Free electrons in the n-region attract them.
- This statement is partially true. While it is true that free electrons in the n-region can attract holes, the primary reason for the diffusion of holes is the concentration gradient rather than the attraction by electrons.
(b) They move across the junction by the potential difference.
- This statement is misleading in the context of an unbiased junction. In an unbiased p-n junction, there is no external potential difference applied. The movement of holes is primarily due to diffusion rather than a potential difference.
(c) Hole concentration in the p-region is more as compared to the n-region.
- This statement is true. Holes will diffuse from the region of higher concentration (p-region) to the region of lower concentration (n-region) due to the concentration gradient.
(d) All the above.
- Since not all the statements are true, particularly (b), this option cannot be correct.
Given this analysis, the most accurate answer is:
(c) Hole concentration in the p-region is more as compared to n-region.
This is the primary reason for the diffusion of holes from the p-region to the n-region in an unbiased p-n junction.
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.
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Answer
The correct statement is (c).
When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced.
14.6 In half-wave rectification, what is the output frequency if the input frequency is $50 \mathrm{~Hz}$. What is the output frequency of a full-wave rectifier for the same input frequency.
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Answer
Input frequency $=50 \mathrm{~Hz}$
For a half-wave rectifier, the output frequency is equal to the input frequency.
$\therefore$ Output frequency $=50 \mathrm{~Hz}$
For a full-wave rectifier, the output frequency is twice the input frequency.
$\therefore$ Output frequency $=2 \times 50=100 \mathrm{~Hz}$
BETA
