Class 12 Physics Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits MCQs with answers
10 multiple-choice questions on Semiconductor Electronics: Materials, Devices and Simple Circuits, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.
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Answer: (b) a semiconductor, since its resistivity is 500 Ω m
ρ = 1/σ = 1/(2.0 × 10⁻³ S m⁻¹) = 500 Ω m, which lies in the semiconductor range 10⁻⁵ to 10⁶ Ω m; insulators start near 10¹¹ Ω m. Option (a) takes σ for ρ.
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Answer: (c) indium (In)
A p-type dopant must be trivalent (third group) so that it leaves a hole in one bond; In is trivalent. Sb, As and P are pentavalent donors and give n-type material.
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Answer: (a) 2.4 × 10¹⁹
In an intrinsic semiconductor every electron freed from a bond leaves one hole, so nₑ = nₕ = nᵢ = 2.4 × 10¹⁹ m⁻³. 5.8 × 10³⁸ is nᵢ², not a concentration.
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Answer: (c) a vacancy in a covalent bond, behaving as a particle of effective charge +q
When an electron breaks away from a bond, the vacancy it leaves has effective charge +q; bound electrons from neighbouring bonds filling it make the hole appear to move like a free positive particle. No atom or nucleus moves.
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Answer: (b) immobile ionised donors on the n-side and immobile ionised acceptors on the p-side
Electrons diffusing out of the n-region leave positive donor ions behind; holes diffusing out of the p-region leave negative acceptor ions. These fixed ions, with no free carriers, form the depletion region.
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Answer: (a) points from the n-side to the p-side and drives holes on the n-side towards the p-side
The n-side holds positive space charge and the p-side negative, so the field points from n to p. It pushes holes along the field (n to p) and electrons against it (p to n): the drift current. No net current means drift balances diffusion, not that the field is zero.
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Answer: (c) 2.6 V; the depletion region widens
Reverse bias acts in the same direction as the barrier, so the barrier becomes V₀ + V = 0.60 + 2.0 = 2.6 V and the depletion region widens. (b) is the forward-bias result V₀ − V.
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Answer: (d) the depletion region, having no free charges, has a resistance far higher than the p- and n-regions
The p- and n-sides are full of majority carriers and have low resistance; the depletion region has no free charges and so a very high resistance. In a series path the voltage drops mostly across the highest resistance. The depletion region is only about 0.1 μm thick, so (a) is wrong.
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Answer: (b) a microammeter, since the reverse current is of the order of microamperes
The reverse current is a drift of minority carriers, only a few μA, so a microammeter is used; a milliammeter suits the forward current, which is in mA.
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Answer: (b) germanium
The threshold (cut-in) voltage is about 0.2 V for germanium and about 0.7 V for silicon. Tin is a metal (Eg = 0) and diamond an insulator, so neither forms such a diode.