JEE Physics Semiconductor Electronics MCQs Set 26

Download JEE MCQs for JEE Physics: Semiconductor Electronics

Access targeted multiple-choice questions for Semiconductor Electronics designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Chapter-wise Objective Questions: Semiconductor Electronics

Access the complete set of multiple-choice questions for Semiconductor Electronics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: The inputs A and B each pass through a NOT gate, and the two outputs are fed to a NAND gate with output Y. The logic gate equivalent to this circuit is
(a) NOR
(b) NAND
(c) AND
(d) OR
Show Answer & Explanation

Answer: (d) OR

Explanation:
1. \( Y=\overline{\bar A\cdot\bar B} \).
2. By De Morgan's law \( Y=A+B \), which is an OR gate.

Teacher's Note:
1. Identify the device region (forward or reverse bias, Zener breakdown, saturation) before applying Ohm's law.
2. Check the sign of the current and that the answer respects the device limits.

Question 2: In a common emitter circuit, \( V_{CC}=5 \) V, \( \beta_{DC}=200 \), \( R_B=100\ \text k\Omega \), \( R_C=1\ \text k\Omega \) and \( V_{BE}=1.0 \) V. The base supply \( V_{BB} \) can vary from 0 to 5.0 V. The minimum base current and the input voltage at which the transistor will go to saturation are, respectively
(a) \( 25\ \mu\text A\text{ and }2.8\ \text V \)
(b) \( 20\ \mu\text A\text{ and }3.5\ \text V \)
(c) \( 25\ \mu\text A\text{ and }3.5\ \text V \)
(d) \( 20\ \mu\text A\text{ and }2.8\ \text V \)
Show Answer & Explanation

Answer: (c) \( 25\ \mu\text A\text{ and }3.5\ \text V \)

Explanation:
1. Saturation: \( I_C=\dfrac{5}{1\text{ k}\Omega}=5 \) mA, so \( I_B=\dfrac{5\text{ mA}}{200}=25\ \mu \)A.
2. \( V_{BB}=I_BR_B+V_{BE}=25\times10^{-6}\times10^5+1.0=3.5 \) V.

Teacher's Note:
1. Identify the device region (forward or reverse bias, Zener breakdown, saturation) before applying Ohm's law.
2. Check the sign of the current and that the answer respects the device limits.

Question 3: A silicon diode (threshold 0.7 V) is connected forward biased in series with a 200 \(\Omega\) resistor, an ammeter and a 3 V battery. The reading of the ammeter is
(a) 13.5 mA
(b) zero
(c) 11.5 mA
(d) 15 mA
Show Answer & Explanation

Answer: (c) 11.5 mA

Explanation:
1. The diode drops 0.7 V when conducting.
2. \( I=\dfrac{3-0.7}{200}=11.5 \) mA.

Teacher's Note:
1. Identify the device region (forward or reverse bias, Zener breakdown, saturation) before applying Ohm's law.
2. Check the sign of the current and that the answer respects the device limits.

Question 4: Which of the following diode connections is forward biased? (The diode's p-side is at the left and the n-side leads through a resistor to the right.) (i) left end at +2 V, right end at -2 V; (ii) left end at -3 V, right end at -3 V; (iii) left end at 2 V, right end at 4 V; (iv) left end at -2 V, right end at +2 V.
(a) (ii) only
(b) (i) only
(c) (iii) only
(d) (iv) only
Show Answer & Explanation

Answer: (b) (i) only

Explanation:
1. A diode is forward biased when the p-side is at a higher potential than the n-side.
2. Only (i) has the p-side (+2 V) above the n-side (-2 V); (ii) has no bias, (iii) and (iv) are reverse biased.

Teacher's Note:
1. Identify the device region (forward or reverse bias, Zener breakdown, saturation) before applying Ohm's law.
2. Check the sign of the current and that the answer respects the device limits.

Question 5: The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap (in eV) for the semiconductor is
(a) 0.9
(b) 0.5
(c) 0.7
(d) 1.1
Show Answer & Explanation

Answer: (b) 0.5

Explanation:
1. The threshold photon energy equals the band gap.
2. \( E_g=\dfrac{1240}{2480}=0.5 \) eV.

Teacher's Note:
1. Identify the device region (forward or reverse bias, Zener breakdown, saturation) before applying Ohm's law.
2. Check the sign of the current and that the answer respects the device limits.

Question 6: Holes are charge carriers in
(a) intrinsic semiconductors and p-type semiconductors
(b) ionic solids and metals
(c) metals only
(d) ionic solids only
Show Answer & Explanation

Answer: (a) intrinsic semiconductors and p-type semiconductors

Explanation:
1. In an intrinsic semiconductor each broken bond gives an electron-hole pair, and in p-type material holes are the majority carriers.
2. Metals conduct by electrons and ionic solids by ions.

Teacher's Note:
1. Identify the device region (forward or reverse bias, Zener breakdown, saturation) before applying Ohm's law.
2. Check the sign of the current and that the answer respects the device limits.

Question 7: In the forward bias arrangement of a p-n junction rectifier, the p end is connected to the
(a) positive terminal of the battery, and the conventional current is from p to n
(b) negative terminal of the battery, and the current is from n to p
(c) positive terminal of the battery, and the current is from n to p
(d) negative terminal of the battery, and the current is from p to n
Show Answer & Explanation

Answer: (a) positive terminal of the battery, and the conventional current is from p to n

Explanation:
1. Forward bias needs the p-side at the higher potential.
2. Conventional current then flows through the junction from the p-side to the n-side.

Teacher's Note:
1. Identify the device region (forward or reverse bias, Zener breakdown, saturation) before applying Ohm's law.
2. Check the sign of the current and that the answer respects the device limits.

Question 8: The plate resistance of a triode is \( 3\times10^3\ \Omega \) and its mutual conductance is \( 1.5\times10^{-3} \) A/V. The amplification factor of the triode is
(a) \( 5\times10^{-5} \)
(b) 45
(c) \( 2\times10^5 \)
(d) 4.5
Show Answer & Explanation

Answer: (d) 4.5

Explanation:
1. \( \mu=r_p\times g_m \).
2. \( \mu=3\times10^3\times1.5\times10^{-3}=4.5 \).

Teacher's Note:
1. Identify the device region (forward or reverse bias, Zener breakdown, saturation) before applying Ohm's law.
2. Check the sign of the current and that the answer respects the device limits.

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Semiconductor Electronics Objective Questions & Solutions for JEE Physics

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FAQs

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