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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.
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 in silicon atoms.
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
The correct statement is (c).
Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least.
The energy band gap of these elements are related as: (Eg)C > (Eg)Si > (Eg)Ge
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.
The correct statement is (c).
The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.
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.
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.
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.
Input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
∴Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
∴Output frequency = 2 × 50 = 100 Hz
Which of the statements given in the previous question is true for p-type semiconductos.
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 in silicon atoms.
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Energy band gap of the given photodiode, Eg = 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10−9 m
The energy of a signal is given by the relation:
E = 
Where,
h = Planck’s constant
= 6.626 × 10−34 Js
c = Speed of light
= 3 × 108 m/s
E 
= 3.313 × 10−20 J
But 1.6 × 10−19 J = 1 eV
∴E = 3.313 × 10−20 J
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.
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