JEE Physics Semiconductor Electronics MCQs Set 08

Practice 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.

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Multiple Choice Questions

Question 1: A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength:
(a) 4000 Å
(b) 6000 Å
(c) 4000 nm
(d) 6000 nm
Show Answer & Explanation

Answer: (a) 4000 Å

Explanation:
1. The cut-off wavelength is \( 12400/2.5 \approx 4960 \) Å.
2. Only wavelengths shorter than this can be detected, so 4000 Å works.

Teacher's Note:
1. Wavelength must be less than or equal to the cut-off.

Question 2: The symbolic representations of four logic gates are given: (i) a gate with a flat back, rounded front and a small circle at the output; (ii) a triangle with no circle; (iii) a flat back, rounded front with no circle; (iv) a curved back with a pointed front. The symbols for OR, NOT and NAND gates are respectively:
(a) (i), (iii), (iv)
(b) (iii), (iv), (ii)
(c) (iv), (i), (iii)
(d) (iv), (ii), (i)
Show Answer & Explanation

Answer: (d) (iv), (ii), (i)

Explanation:
1. The curved-back pointed shape (iv) is OR.
2. The triangle (ii) is NOT.
3. The AND shape with a small circle (i) is NAND.

Teacher's Note:
1. The small circle means inversion.
2. Flat back with round front is AND; with a circle it is NAND.

Question 3: Which one of the following statements is false?
(a) The resistance of an intrinsic semiconductor decreases with an increase of temperature
(b) Pure Si doped with trivalent impurities gives a p-type semiconductor
(c) Majority carriers in an n-type semiconductor are holes
(d) Minority carriers in a p-type semiconductor are electrons
Show Answer & Explanation

Answer: (c) Majority carriers in an n-type semiconductor are holes

Explanation:
1. In an n-type semiconductor, electrons are the majority carriers.
2. So the statement that holes are majority carriers is false.

Teacher's Note:
1. Check the other three: they are all true.

Question 4: Inputs A and B go to an OR gate and the OR output and input C go to an AND gate with output Y. To get Y = 1, which input is correct?
(a) A = 1, B = 1, C = 0
(b) A = 0, B = 1, C = 0
(c) A = 1, B = 0, C = 0
(d) A = 1, B = 0, C = 1
Show Answer & Explanation

Answer: (d) A = 1, B = 0, C = 1

Explanation:
1. \( Y = (A + B) \cdot C \).
2. Only option (d) has C = 1 together with A + B = 1.

Teacher's Note:
1. Y can never be 1 if C = 0.

Question 5: The band energy gap of a light emitting diode (LED) is:
(a) 1.1 eV
(b) 1.3 eV
(c) 1.5 eV
(d) 1.8 eV
Show Answer & Explanation

Answer: (d) 1.8 eV

Explanation:
1. Visible light LEDs need photon energies above about 1.8 eV.
2. Of the given options, 1.8 eV is the one in the visible range.

Teacher's Note:
1. Red light is about 1.8 eV and blue about 3 eV.

Question 6: In a common emitter transistor, the current gain is 100 and the base current is 2 mA. If the load resistance is 0.25 \( \Omega \), the voltage across the load resistance is:
(a) 5 V
(b) 5 mV
(c) 50 mV
(d) 500 mV
Show Answer & Explanation

Answer: (c) 50 mV

Explanation:
1. \( I_C = \beta I_B = 100 \times 2 = 200 \) mA.
2. \( V = I_C R_L = 0.2 \times 0.25 = 0.05 \) V = 50 mV.

Teacher's Note:
1. Convert mA to A before multiplying.

Question 7: A Zener diode with a breakdown voltage of 15 V is used in a voltage regulator. A 20 V supply feeds a 250 \( \Omega \) series resistor, and the Zener diode is in parallel with a 1 k\( \Omega \) load. The current through the Zener diode is:
(a) 5 mA
(b) 10 mA
(c) 15 mA
(d) 20 mA
Show Answer & Explanation

Answer: (a) 5 mA

Explanation:
1. Voltage across the series resistor = 20 − 15 = 5 V, so the total current = 5/250 = 20 mA.
2. Load current = 15/1000 = 15 mA.
3. Zener current = 20 − 15 = 5 mA.

Teacher's Note:
1. Total current splits between the Zener and the load.

Question 8: Pure Si at 500 K has equal numbers of electrons and holes, each \( 1.5 \times 10^{16} \) m\( ^{-3} \). Doping by indium increases \( n_h \) to \( 4.5 \times 10^{22} \) m\( ^{-3} \). The doped semiconductor is:
(a) p-type with \( n_e = 5 \times 10^{9} \) m\( ^{-3} \)
(b) n-type with \( n_e = 5 \times 10^{22} \) m\( ^{-3} \)
(c) p-type with \( n_e = 2.5 \times 10^{10} \) m\( ^{-3} \)
(d) n-type with \( n_e = 2.5 \times 10^{23} \) m\( ^{-3} \)
Show Answer & Explanation

Answer: (a) p-type with \( n_e = 5 \times 10^{9} \) m\( ^{-3} \)

Explanation:
1. Indium is trivalent, so the semiconductor is p-type.
2. \( n_e = n_i^2/n_h = (1.5 \times 10^{16})^2/(4.5 \times 10^{22}) = 5 \times 10^9 \) m\( ^{-3} \).

Teacher's Note:
1. Use the mass action law \( n_e n_h = n_i^2 \).

Question 9: In forward biasing of a p-n junction:
(a) the positive terminal of the battery is connected to the p-side and the depletion region becomes thick
(b) the positive terminal is connected to the n-side and the depletion region becomes thin
(c) the positive terminal is connected to the n-side and the depletion region becomes thick
(d) the positive terminal is connected to the p-side and the depletion region becomes thin
Show Answer & Explanation

Answer: (d) the positive terminal is connected to the p-side and the depletion region becomes thin

Explanation:
1. Forward bias means p to positive and n to negative.
2. It opposes the barrier field, so the depletion layer becomes thin.

Teacher's Note:
1. Remember: forward bias = thin depletion layer.

Question 10: If a small amount of antimony is added to a germanium crystal:
(a) it becomes a p-type semiconductor
(b) the antimony becomes an acceptor atom
(c) there will be more free electrons than holes in the semiconductor
(d) its resistance is increased
Show Answer & Explanation

Answer: (c) there will be more free electrons than holes in the semiconductor

Explanation:
1. Antimony is pentavalent and acts as a donor.
2. The crystal becomes n-type with more electrons than holes and a lower resistance.

Teacher's Note:
1. Group 15 impurity gives n-type.

Question 11: Four logic gates are represented by (i) an OR shape, (ii) an AND shape, (iii) a triangle with a small circle, (iv) an AND shape with a small circle. Which are for AND, NAND and NOT gates, respectively?
(a) (ii), (iii) and (iv)
(b) (iii), (ii) and (i)
(c) (iii), (i) and (iv)
(d) (ii), (iv) and (iii)
Show Answer & Explanation

Answer: (d) (ii), (iv) and (iii)

Explanation:
1. AND is (ii).
2. NAND is the AND shape with a circle, (iv).
3. NOT is the triangle with a circle, (iii).

Teacher's Note:
1. The circle means complement.

Question 12: In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 k\( \Omega \) is 2 V. The base resistance is 1 k\( \Omega \) and the current amplification is 100. The input signal voltage is:
(a) 1 mV
(b) 10 mV
(c) 0.1 V
(d) 1.0 V
Show Answer & Explanation

Answer: (b) 10 mV

Explanation:
1. \( I_C = 2/2000 = 1 \) mA, so \( I_B = 10 \ \mu \)A.
2. \( V_{in} = I_B R_B = 10 \times 10^{-6} \times 1000 = 10 \) mV.

Teacher's Note:
1. Work backwards from the output to the input.

Question 13: Transfer characteristics (output voltage versus input voltage) of a base-biased transistor in CE configuration show three regions: I (cut-off, output high and flat), II (active, steeply falling) and III (saturation, flat low). To use the transistor as a switch, it is used:
(a) in region II
(b) in region I
(c) in region III
(d) in both regions I and III
Show Answer & Explanation

Answer: (d) in both regions I and III

Explanation:
1. A switch has only two states, off and on.
2. These correspond to cut-off (region I) and saturation (region III).

Teacher's Note:
1. Region II (active) is for amplifier use.

Question 14: The input resistance of a silicon transistor is 100 \( \Omega \). The base current changes by 40 \( \mu \)A, causing a collector current change of 2 mA. Used as a CE amplifier with a load of 4 k\( \Omega \), the voltage gain is:
(a) 4000
(b) 1000
(c) 2000
(d) 3000
Show Answer & Explanation

Answer: (c) 2000

Explanation:
1. \( \beta = 2\text{ mA}/40\ \mu\text{A} = 50 \).
2. \( A_V = \beta R_L/R_{in} = 50 \times 4000/100 = 2000 \).

Teacher's Note:
1. Compute beta, then voltage gain.

Question 15: For a common emitter transistor, the input current is 5 \( \mu \)A and \( \beta = 100 \). The circuit operates with a load resistance of 10 k\( \Omega \). The voltage across the load is:
(a) 5 V
(b) 10 V
(c) 12.5 V
(d) 7.5 V
Show Answer & Explanation

Answer: (a) 5 V

Explanation:
1. \( I_C = 100 \times 5 \ \mu\text{A} = 0.5 \) mA.
2. \( V = 0.5 \text{ mA} \times 10 \text{ k}\Omega = 5 \) V.

Teacher's Note:
1. Voltage across load is Ic times RL.

Question 16: A common emitter amplifier has a transistor with transconductance 0.03 mho and current gain 25 and a given load. If the transistor is replaced by another with transconductance 0.02 mho and current gain 20 (same load), and the original voltage gain is G, the new voltage gain is:
(a) \( \dfrac{5}{4}G \)
(b) \( \dfrac{2}{3}G \)
(c) 1.5 G
(d) \( \dfrac{1}{3}G \)
Show Answer & Explanation

Answer: (b) \( \dfrac{2}{3}G \)

Explanation:
1. Voltage gain = \( g_m R_L \).
2. The ratio is \( 0.02/0.03 = 2/3 \), so the new gain is \( \frac{2}{3}G \).

Teacher's Note:
1. Transconductance times load resistance gives voltage gain.

Question 17: In an n-type semiconductor, which statement is true?
(a) Holes are majority carriers and trivalent atoms are dopants
(b) Electrons are majority carriers and trivalent atoms are dopants
(c) Electrons are minority carriers and pentavalent atoms are dopants
(d) Holes are minority carriers and pentavalent atoms are dopants
Show Answer & Explanation

Answer: (d) Holes are minority carriers and pentavalent atoms are dopants

Explanation:
1. n-type material is made with pentavalent dopants (donors).
2. Electrons are majority carriers and holes are minority carriers.

Teacher's Note:
1. Match carrier type with dopant valency.

Question 18: The mobility of electrons in Ge is 0.4 m\( ^2 \)/V s and the electronic charge is \( 1.6 \times 10^{-19} \) C. How many donor atoms per m\( ^3 \) are needed for a semiconductor of conductivity 500 mho/m?
(a) \( 8 \times 10^{21} \)
(b) \( 8 \times 10^{15} \)
(c) \( 5 \times 10^{21} \)
(d) \( 8 \times 10^{16} \)
Show Answer & Explanation

Answer: (a) \( 8 \times 10^{21} \)

Explanation:
1. \( \sigma = n e \mu \).
2. \( n = 500/(1.6 \times 10^{-19} \times 0.4) = 7.8 \times 10^{21} \approx 8 \times 10^{21} \) m\( ^{-3} \).

Teacher's Note:
1. Holes are neglected for heavily doped material.

Question 19: The barrier potential of a p-n junction depends on: (a) type of semiconductor material, (b) amount of doping, (c) temperature. Which is correct?
(a) (a) and (b) only
(b) (b) only
(c) (b) and (c) only
(d) (a), (b) and (c)
Show Answer & Explanation

Answer: (d) (a), (b) and (c)

Explanation:
1. The barrier potential \( V_B = (kT/e)\ln(N_AN_D/n_i^2) \).
2. It depends on the material through \( n_i \), on the doping and on temperature.

Teacher's Note:
1. All three factors appear in the formula.

Question 20: Which statement about a transistor is not correct?
(a) the emitter is highly doped
(b) the base is lightly doped
(c) the base is thin in size
(d) the collector is comparable with the emitter in doping
Show Answer & Explanation

Answer: (d) the collector is comparable with the emitter in doping

Explanation:
1. The emitter is heavily doped and the collector only moderately doped.
2. The collector is not doped like the emitter, so the statement is incorrect.

Teacher's Note:
1. Doping order: emitter > collector > base.

Question 21: NOT gates are connected to the inputs A and B separately and their outputs are fed to an OR gate. The combination represents:
(a) NAND
(b) AND
(c) NOR
(d) OR
Show Answer & Explanation

Answer: (a) NAND

Explanation:
1. \( Y = \bar{A} + \bar{B} = \overline{AB} \) by De Morgan's theorem.
2. This is the NAND gate.

Teacher's Note:
1. Bubbled-input OR is the same as NAND.

Question 22: A diode D is connected in series to an external resistance R = 100 \( \Omega \) and a battery of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be:
(a) 35 mA
(b) 30 mA
(c) 40 mA
(d) 20 mA
Show Answer & Explanation

Answer: (b) 30 mA

Explanation:
1. The diode is forward biased and drops 0.5 V.
2. \( I = (3.5 - 0.5)/100 = 30 \) mA.

Teacher's Note:
1. Subtract the diode drop before applying Ohm's law.

Question 23: The input signal to a CE amplifier with voltage gain 150 is \( V_i = 2\cos(15t + \pi/3) \). The corresponding output signal will be:
(a) \( 300\cos(15t + 4\pi/3) \)
(b) \( 300\cos(15t + \pi/3) \)
(c) \( 75\cos(15t + 2\pi/3) \)
(d) \( 2\cos(15t + 5\pi/6) \)
Show Answer & Explanation

Answer: (a) \( 300\cos(15t + 4\pi/3) \)

Explanation:
1. Amplitude = 150 × 2 = 300.
2. A CE amplifier inverts the signal: a phase shift of \( \pi \) gives \( \pi/3 + \pi = 4\pi/3 \).

Teacher's Note:
1. CE amplifier output is 180 degrees out of phase.

Question 24: A Zener diode is connected across a 12 V source so that 6 V drops across it. The power drawn by the Zener diode is 2.4 mW. What is the maximum value of the series resistance in the circuit?
(a) 5 k\( \Omega \)
(b) 15 k\( \Omega \)
(c) 12 k\( \Omega \)
(d) 6 k\( \Omega \)
Show Answer & Explanation

Answer: (b) 15 k\( \Omega \)

Explanation:
1. \( I_Z = P/V_Z = 2.4 \text{ mW}/6 \text{ V} = 0.4 \) mA.
2. \( R = (12 - 6)/0.4 \text{ mA} = 15 \) k\( \Omega \).

Teacher's Note:
1. Find the Zener current from its power.

Question 25: A 20 V battery drives current through a 2 k\( \Omega \) series resistor to a 5 V Zener diode connected in parallel with a 1 k\( \Omega \) load. The current through the Zener diode is:
(a) 5 mA
(b) 7.5 mA
(c) 2.5 mA
(d) 12.5 mA
Show Answer & Explanation

Answer: (c) 2.5 mA

Explanation:
1. Current in the series resistor = (20 − 5)/2 k\( \Omega \) = 7.5 mA.
2. Load current = 5/1 k\( \Omega \) = 5 mA.
3. Zener current = 7.5 − 5 = 2.5 mA.

Teacher's Note:
1. Use KCL after fixing the voltage at \( V_Z \).

Question 26: In a transistor the current gain is 100. If the load resistance is 3 k\( \Omega \) and the input resistance is 1 k\( \Omega \), the voltage gain is:
(a) 100
(b) 150
(c) 200
(d) 300
Show Answer & Explanation

Answer: (d) 300

Explanation:
1. \( A_V = \beta R_L/R_{in} = 100 \times 3/1 = 300 \).

Teacher's Note:
1. Voltage gain is beta times the resistance ratio.

Question 27:
Assertion (A): A transistor amplifier in common emitter configuration has a low input impedance.
Reason (R): The base-to-emitter region is forward biased.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A forward biased junction has low resistance.
2. So the input impedance is low.

Teacher's Note:
1. Reverse biased junctions have high resistance.

Question 28:
Assertion (A): In a common emitter transistor amplifier, the input current is much less than the output current.
Reason (R): The common emitter transistor amplifier has a very high input impedance.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. Because \( I_C = \beta I_B \), the input current is small, so A is true.
2. The input impedance is low, not high, so R is false.

Teacher's Note:
1. Low input impedance and high output impedance.

Question 29:
Assertion (A): In common base configuration, the current gain of the transistor is less than unity.
Reason (R): The collector terminal is reverse biased for amplification.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( \alpha = I_C/I_E < 1 \), because the base current is lost.
2. Reverse biased collector is true, but it is not the reason.

Teacher's Note:
1. The gain is below 1 due to recombination in the base.

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