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NCERT Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits
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Semiconductor Electronics: Materials, Devices and Simple Circuits
Chapter: 14
| Part – II |
EXERCISE
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.
Ans: (c) Holes are minority carriers and pentavalent atoms are the dopants.
2. Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
Ans: (d) Holes are majority carriers and trivalent atoms are the dopants.
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 (Eg)C , (Eg )Si and (Eg )Ge. Which of the following statements is true?
(a) (Eg)Si < (Eg)Ge < (Eg)C
(b) (Eg)C < (Eg)Ge > (Eg)Si
(c) (Eg)C > (Eg)Si > (Eg)Ge
(d) (Eg)C = (Eg)Si = (Eg)Ge
Ans: (c) (Eg)C > (Eg)Si > (Eg)Ge
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.
Ans: (c) Hole concentration in p-region is more as compared to n-region.
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.
Ans: (c) Lowers the potential barrier.
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.
Ans: Here, In half-wave rectification, if the input frequency is 50 Hz, then the output frequency will also be 50 Hz:
In full-wave rectification, the output frequency is twice the input frequency.
Therefore, for an input frequency of 50 Hz, the output frequency will be:
2 × 50
= 100 Hz.
