JEE Physics Semiconductor Electronics MCQs Set 17

Practice MCQs for JEE Physics Semiconductor Electronics

Review structured MCQ sets for JEE Physics Semiconductor Electronics. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Access Semiconductor Electronics Questions and Solutions

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: A diode D is connected in series with an external resistance R = 100 ohm and an emf of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit (diode forward biased) will be:
(a) 20 mA
(b) 35 mA
(c) 30 mA
(d) 40 mA
Show Answer & Explanation

Answer: (c) 30 mA

Explanation:
1. \( I=\frac{3.5-0.5}{100}=0.03\ \text{A}=30\ \text{mA} \)

Teacher's Note:
1. In forward bias the diode drops its barrier potential and the rest falls across the resistor.

Question 2: Each of two NOT gates takes one of the inputs A and B (outputs \( Y_1=\bar A \) and \( Y_2=\bar B \)). These outputs are fed to a NOR gate whose output is Y. Which logic gate is represented by this combination?
(a) AND
(b) NOR
(c) OR
(d) NAND
Show Answer & Explanation

Answer: (a) AND

Explanation:
1. \( Y=\overline{\bar A+\bar B}=A\cdot B \) (De Morgan), which is the AND operation.

Teacher's Note:
1. A NOR gate with inverted inputs behaves as an AND gate.

Question 3: The barrier of a p-n junction depends on: (I) type of semiconductor material, (II) amount of doping, (III) temperature. Which one of the following is correct?
(a) I and II only
(b) II only
(c) II and III only
(d) I, II and III
Show Answer & Explanation

Answer: (d) I, II and III

Explanation:
1. The barrier height is set by the doping concentrations, the band gap of the material and the thermal generation of carriers, so all three affect it.

Teacher's Note:
1. The contact potential depends on material, doping and temperature.

Question 4: The output of a NAND gate is divided into two in parallel and fed to the two inputs of another NAND gate (the second NAND gate has its inputs tied together). The resulting gate is a:
(a) AND gate
(b) NOR gate
(c) OR gate
(d) NOT gate
Show Answer & Explanation

Answer: (a) AND gate

Explanation:
1. First gate: \( X=\overline{A\cdot B} \). The second gate with both inputs X gives \( \bar X \) (a NOT gate).
2. \( C=\overline{\overline{AB}}=AB \): an AND gate.

Teacher's Note:
1. A NAND gate with joined inputs is an inverter, so NAND followed by NOT is AND.

Question 5: One way in which the operation of an n-p-n transistor differs from that of a p-n-p transistor is that:
(a) The emitter junction injects minority carriers into the base region of a p-n-p
(b) The emitter injects holes into the base of the p-n-p and electrons into the base region of the n-p-n
(c) The emitter junction injects holes into the base of an n-p-n
(d) The emitter junction is reverse biased in n-p-n
Show Answer & Explanation

Answer: (b) The emitter injects holes into the base of the p-n-p and electrons into the base region of the n-p-n

Explanation:
1. In an n-p-n transistor the n-type emitter sends electrons into the p-type base; in a p-n-p transistor the p-type emitter sends holes into the n-type base.
2. The emitter junction is forward biased in both.

Teacher's Note:
1. The majority carriers of the emitter are the ones injected into the base.

Question 6: Transfer characteristics (output voltage versus input voltage) of a base-biased transistor in CE configuration have three regions: region I (flat, high output: cut-off), region II (steeply falling: active) and region III (flat, low output: saturation). For using the transistor as a switch, it is used in:
(a) Region III only
(b) Both region I and region III
(c) Region II
(d) Region I only
Show Answer & Explanation

Answer: (b) Both region I and region III

Explanation:
1. A switch needs two distinct states: OFF (cut-off, high output) and ON (saturation, low output).
2. So the transistor is operated in regions I and III; region II is used for amplification.

Teacher's Note:
1. Switching uses cut-off and saturation, amplification uses the active region.

Question 7: The input resistance of a silicon transistor is 100 ohm. The base current is changed by 40 microampere, which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 kilo-ohm. The voltage gain of the amplifier is:
(a) 2000
(b) 3000
(c) 4000
(d) 1000
Show Answer & Explanation

Answer: (a) 2000

Explanation:
1. \( \beta=\frac{\Delta I_C}{\Delta I_B}=\frac{2\ \text{mA}}{40\ \mu\text{A}}=50 \).
2. \( A_V=\beta\frac{R_L}{R_i}=50\times\frac{4000}{100}=2000 \)

Teacher's Note:
1. Voltage gain of a CE amplifier is current gain times the ratio of load to input resistance.

Question 8: A Zener diode having breakdown voltage equal to 15 V is used in a voltage regulator circuit. The 20 V source is connected through a 250 ohm series resistor to the Zener diode in parallel with a 1 kilo-ohm load. The current through the diode is:
(a) 5 mA
(b) 10 mA
(c) 15 mA
(d) 20 mA
Show Answer & Explanation

Answer: (a) 5 mA

Explanation:
1. Current through the series resistor \( =\frac{20-15}{250}=20\ \text{mA} \).
2. Load current \( =\frac{15}{1000}=15\ \text{mA} \).
3. Zener current \( =20-15=5 \) mA.

Teacher's Note:
1. The series current splits between the load and the Zener diode.

Question 9: Pure Si at 500 K has equal numbers of electron and hole concentrations of \( 1.5\times10^{16}\ \text{m}^{-3} \). Doping by indium increases \( n_h \) to \( 4.5\times10^{22}\ \text{m}^{-3} \). The doped semiconductor is:
(a) p-type with electron concentration \( n_e=5\times10^9\ \text{m}^{-3} \)
(b) n-type with electron concentration \( n_e=5\times10^{22}\ \text{m}^{-3} \)
(c) p-type with electron concentration \( n_e=2.5\times10^{10}\ \text{m}^{-3} \)
(d) n-type with electron concentration \( n_e=2.5\times10^{23}\ \text{m}^{-3} \)
Show Answer & Explanation

Answer: (a) p-type with electron concentration \( n_e=5\times10^9\ \text{m}^{-3} \)

Explanation:
1. Indium is trivalent, so the doped crystal is p-type.
2. \( n_en_h=n_i^2\Rightarrow n_e=\frac{(1.5\times10^{16})^2}{4.5\times10^{22}}=5\times10^9\ \text{m}^{-3} \)

Teacher's Note:
1. Mass action law: the product of electron and hole concentrations stays equal to n_i squared.

Question 10: A common emitter amplifier has a voltage gain of 50, an input impedance of 100 ohm and an output impedance of 200 ohm. The power gain of the amplifier is:
(a) 500
(b) 1000
(c) 1250
(d) 50
Show Answer & Explanation

Answer: (c) 1250

Explanation:
1. Power gain = voltage gain times current gain.
2. Current gain \( =A_V\frac{R_{in}}{R_{out}}=50\times\frac{100}{200}=25 \), so \( \text{power gain}=50\times25=1250 \)

Teacher's Note:
1. Current gain is the voltage gain times the ratio of input to output impedance.

Question 11: For a common base circuit, if \( \frac{I_C}{I_E}=0.98 \), then the current gain for the common emitter circuit will be:
(a) 49
(b) 98
(c) 4.9
(d) 25.5
Show Answer & Explanation

Answer: (a) 49

Explanation:
1. \( \beta=\frac\alpha{1-\alpha}=\frac{0.98}{0.02}=49 \)

Teacher's Note:
1. Beta equals alpha over one minus alpha.

Question 12: The correct relation between \( \alpha \) and \( \beta \) for a transistor is:
(a) \( \beta=\frac{1-\alpha}{\alpha} \)
(b) \( \beta=\frac\alpha{1-\alpha} \)
(c) \( \alpha=\frac{\beta-1}\beta \)
(d) \( \alpha\beta=1 \)
Show Answer & Explanation

Answer: (b) \( \beta=\frac\alpha{1-\alpha} \)

Explanation:
1. \( \alpha=\frac{I_C}{I_E},\ \beta=\frac{I_C}{I_B},\ I_E=I_B+I_C\Rightarrow\beta=\frac{\alpha}{1-\alpha} \)

Teacher's Note:
1. Use the emitter current being the sum of base and collector currents.

Question 13: For a common emitter configuration, if alpha and beta have their usual meanings, which of the following relations between them are incorrect? (A) \( \frac1\alpha=\frac1\beta+1 \); (B) \( \alpha=\frac{\beta}{1-\beta} \); (C) \( \alpha=\frac{\beta}{1+\beta} \); (D) \( \alpha=\frac{\beta^2}{1+\beta^2} \)
(a) (A) and (C)
(b) (B) and (D)
(c) Only (B)
(d) (C) and (D)
Show Answer & Explanation

Answer: (b) (B) and (D)

Explanation:
1. \( \alpha=\frac\beta{1+\beta} \) and \( \frac1\alpha=\frac{1+\beta}{\beta}=1+\frac1\beta \) are correct, so (A) and (C) are right.
2. (B) and (D) are incorrect.

Teacher's Note:
1. Start from beta = alpha/(1 - alpha) and invert it.

Question 14: A truth table has outputs: A = 0, B = 0 gives Y = 0; A = 0, B = 1 gives Y = 1; A = 1, B = 0 gives Y = 1; A = 1, B = 1 gives Y = 1. It represents:
(a) OR gate
(b) NAND gate
(c) AND gate
(d) NOR gate
Show Answer & Explanation

Answer: (a) OR gate

Explanation:
1. The output is 1 whenever at least one input is 1: the OR function.

Teacher's Note:
1. OR gives zero only when both inputs are zero.

Question 15: In an unbiased n-p junction, electrons diffuse from the n-region to the p-region because:
(a) Holes in the p-region attract them
(b) Electrons travel across the junction due to potential difference
(c) The electron concentration in the n-region is more as compared to that in the p-region
(d) Only electrons move from n to p region and not the vice versa
Show Answer & Explanation

Answer: (c) The electron concentration in the n-region is more as compared to that in the p-region

Explanation:
1. Diffusion is driven by a concentration gradient, not by an electric field; the n-side has far more free electrons.

Teacher's Note:
1. Diffusion is motion from high to low concentration.

Question 16: In the dc voltage regulator circuit, a series resistor \( R_s \) carries the current from the input voltage \( V_i \); the load current is \( I_L \) and the Zener diode current is \( nI_L \). The load voltage is \( V_L \). The value of \( R_s \) is:
(a) \( \frac{V_i-V_L}{nI_L} \)
(b) \( \frac{V_i+V_L}{nI_L} \)
(c) \( \frac{(V_i-V_L)(n+1)}{I_L} \)
(d) \( \frac{V_i-V_L}{(n+1)I_L} \)
Show Answer & Explanation

Answer: (d) \( \frac{V_i-V_L}{(n+1)I_L} \)

Explanation:
1. Current through \( R_s \) \( =I_L+nI_L=(n+1)I_L \).
2. \( R_s=\frac{V_i-V_L}{(n+1)I_L} \)

Teacher's Note:
1. The series resistor carries the sum of the load and Zener currents.

Question 17: The circuit consists of an ac source \( v_m\sin\omega t \) in series with a resistor, with a diode connected across the output terminals (in parallel with the output), its cathode towards the bottom terminal, so that it conducts when the upper terminal is positive. The output of the given circuit:
(a) Would be zero at all times
(b) Would be like a half wave rectifier with positive cycles in the output
(c) Would be like a half wave rectifier with negative cycles in the output
(d) Would be like that of a full wave rectifier
Show Answer & Explanation

Answer: (c) Would be like a half wave rectifier with negative cycles in the output

Explanation:
1. During the positive half cycle the diode conducts and shorts the output: output is about zero.
2. During the negative half cycle the diode is reverse biased, so the output follows the source: only negative cycles appear.

Teacher's Note:
1. A diode in parallel with the output removes the half cycle in which it conducts.

Question 18: A doped semiconductor has impurity levels 30 meV below the conduction band. In a thermal collision, an amount kT of energy is given to the extra electron loosely bound to the impurity ion, and the electron is just able to jump into the conduction band. The temperature T is (nearly):
(a) 300 K
(b) 348 K
(c) 400 K
(d) 500 K
Show Answer & Explanation

Answer: (b) 348 K

Explanation:
1. \( kT=30\ \text{meV}\Rightarrow T=\frac{0.030}{8.62\times10^{-5}}\approx348\ \text{K} \)

Teacher's Note:
1. Use k = 8.62 x 10^-5 eV per kelvin.

Question 19: The energy of a photon of sodium light (\( \lambda=589 \) nm) equals the band gap of a semiconducting material. The value of E/kT at 300 K is (nearly):
(a) 41
(b) 81
(c) 163
(d) 326
Show Answer & Explanation

Answer: (b) 81

Explanation:
1. \( E=\frac{1240}{589}=2.105\ \text{eV} \); \( kT=8.62\times10^{-5}\times300=0.0259\ \text{eV} \).
2. \( \frac{E}{kT}\approx81 \)

Teacher's Note:
1. Convert the wavelength to electron-volts first.

Question 20: The band gap in germanium is \( \Delta E=0.68 \) eV. Assuming that the number of hole-electron pairs is proportional to \( e^{-\Delta E/2kT} \), the percentage increase in the number of charge carriers in pure germanium as the temperature is increased from 300 K to 320 K is (nearly):
(a) 57
(b) 128
(c) 228
(d) 275
Show Answer & Explanation

Answer: (b) 128

Explanation:
1. Exponent change \( =\frac{0.68}{2\times8.62\times10^{-5}}\left(\frac1{300}-\frac1{320}\right)=0.822 \).
2. \( \frac{n_2}{n_1}=e^{0.822}=2.275 \), an increase of about 128%.

Teacher's Note:
1. The ratio of carrier numbers is an exponential of the inverse temperature difference.

Question 21: When the base current in a transistor is changed from 30 microampere to 80 microampere, the collector current changes from 1.0 mA to 3.5 mA. The current gain beta is:
(a) 25
(b) 50
(c) 75
(d) 100
Show Answer & Explanation

Answer: (b) 50

Explanation:
1. \( \beta=\frac{\Delta I_C}{\Delta I_B}=\frac{2.5\ \text{mA}}{50\ \mu\text{A}}=50 \)

Teacher's Note:
1. AC current gain uses the changes in the currents.

Question 22: A load resistor of 2 kilo-ohm is connected in the collector branch of an amplifier circuit using a transistor in common-emitter mode. The current gain is \( \beta=50 \). If the input current is changed by 50 microampere, the output voltage changes by:
(a) 2.5 V
(b) 5 V
(c) 10 V
(d) 50 V
Show Answer & Explanation

Answer: (b) 5 V

Explanation:
1. \( \Delta I_C=\beta\Delta I_B=50\times50\ \mu\text{A}=2.5\ \text{mA} \)
2. \( \Delta V=\Delta I_CR_L=2.5\times10^{-3}\times2000=5\ \text{V} \)

Teacher's Note:
1. Output voltage change equals the collector current change times the load resistance.

Question 23: The drift current in a p-n junction is 20.0 microampere. The number of electrons crossing a cross-section per second in the depletion region is (in units of \( 10^{12} \)):
(a) 12.5
(b) 62.5
(c) 125
(d) 250
Show Answer & Explanation

Answer: (c) 125

Explanation:
1. \( n=\frac Ie=\frac{20\times10^{-6}}{1.6\times10^{-19}}=1.25\times10^{14}=125\times10^{12} \)

Teacher's Note:
1. Number per second is the current divided by the electronic charge.

Free study material for Physics

Semiconductor Electronics Objective Questions & Solutions for JEE Physics

Download Multiple Choice Questions: Semiconductor Electronics (JEE Physics)

Review structured objective questions for JEE Physics Semiconductor Electronics. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

Concept Clarification for Semiconductor Electronics

Cross-reference your completed choices with comprehensive NCERT solutions for JEE Physics to ensure absolute clarity across all sub-topics in this chapter.

Additional Study Resources for JEE Physics

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest JEE Physics Semiconductor Electronics MCQs Set 17?

You can get most exhaustive JEE Physics Semiconductor Electronics MCQs Set 17 for free on StudiesToday.com. These MCQs for JEE Physics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Physics JEE material?

Yes, our JEE Physics Semiconductor Electronics MCQs Set 17 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Physics MCQs help in scoring full marks in JEE exams?

By solving our JEE Physics Semiconductor Electronics MCQs Set 17, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.

Do you provide answers and explanations for JEE Physics Semiconductor Electronics MCQs Set 17?

Yes, Physics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Physics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Physics Semiconductor Electronics MCQs Set 17 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.