JEE Physics Semiconductor Electronics MCQs Set 10

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Chapter-wise Objective Questions: Semiconductor Electronics

View or download the dedicated Semiconductor Electronics MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Physics evaluations.

Multiple Choice Questions

Question 1: The intrinsic carrier density in germanium at 300 K is \( 2.5 \times 10^{13} \) per cm\( ^3 \). If the electron density in an n-type crystal at 300 K is \( 0.5 \times 10^{17} \) per cm\( ^3 \), the hole density per cm\( ^3 \) is about:
(a) \( 2.5 \times 10^{13} \)
(b) \( 5 \times 10^{6} \)
(c) \( 1.25 \times 10^{10} \)
(d) \( 0.2 \times 10^{4} \)
Show Answer & Explanation

Answer: (c) \( 1.25 \times 10^{10} \)

Explanation:
1. \( n_h = n_i^2/n_e = (2.5 \times 10^{13})^2/(0.5 \times 10^{17}) = 1.25 \times 10^{10} \) per cm\( ^3 \).

Teacher's Note:
1. Use the mass action law.

Question 2: In a half wave rectifier, the peak value of the sinusoidal input signal is 10 V. The dc component of the output is:
(a) \( 10/\sqrt{2} \) V
(b) \( 10/\pi \) V
(c) 10 V
(d) \( 20/\pi \) V
Show Answer & Explanation

Answer: (b) \( 10/\pi \) V

Explanation:
1. For half wave rectification \( V_{dc} = V_m/\pi \).
2. So \( V_{dc} = 10/\pi \) V.

Teacher's Note:
1. Full wave gives \( 2V_m/\pi \).

Question 3: Assuming the diodes to be of silicon with zero forward resistance, two diodes are connected in antiparallel across a 2 k\( \Omega \) resistor in series with a 20 V battery. The current I is:
(a) 0
(b) 9.65 mA
(c) 10 mA
(d) 10.36 mA
Show Answer & Explanation

Answer: (b) 9.65 mA

Explanation:
1. One of the two antiparallel diodes is always forward biased and drops 0.7 V.
2. \( I = (20 - 0.7)/2000 = 9.65 \) mA.

Teacher's Note:
1. Silicon drop is 0.7 V.

Question 4: An npn transistor is biased to work as an amplifier. Which statement is false?
(a) The electrons go from the base region to the collector region
(b) The electrons go from the collector region to the base region
(c) The electrons go from the emitter region to the base region
(d) The holes go from the base region to the emitter region
Show Answer & Explanation

Answer: (b) The electrons go from the collector region to the base region

Explanation:
1. In an npn transistor electrons flow emitter to base to collector.
2. The collector never sends electrons to the base, so (b) is false.

Teacher's Note:
1. Check the direction of carrier flow.

Question 5: Three diodes, each with forward resistance 50 \( \Omega \) and infinite backward resistance, are connected in three parallel branches. Branch 1 has a forward diode with a 150 \( \Omega \) resistor, branch 2 a forward diode with a 50 \( \Omega \) resistor, branch 3 a reverse-connected diode with a 50 \( \Omega \) resistor. This combination is connected in series with a 100 \( \Omega \) resistor and a 6 V battery. The current through the 100 \( \Omega \) resistor is:
(a) 0
(b) 36 mA
(c) 43 mA
(d) 50 mA
Show Answer & Explanation

Answer: (b) 36 mA

Explanation:
1. Branch 1 = 50 + 150 = 200 \( \Omega \) and branch 2 = 50 + 50 = 100 \( \Omega \); branch 3 does not conduct.
2. Parallel = \( 200 \times 100/300 = 66.7 \ \Omega \).
3. Total R = 166.7 \( \Omega \), so \( I = 6/166.7 = 36 \) mA.

Teacher's Note:
1. A reverse-biased ideal diode behaves as an open circuit.

Question 6: Inputs A and B are each passed through a NOT gate, and the outputs \( \bar{A} \) and \( \bar{B} \) feed an OR gate with output Y. The circuit is equivalent to:
(a) OR
(b) NAND
(c) AND
(d) NOR
Show Answer & Explanation

Answer: (b) NAND

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

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

Question 7: A truth table has outputs: (A, B) = (1, 1) gives 0; (0, 1) gives 1; (1, 0) gives 1; (0, 0) gives 1. It belongs to the:
(a) NOT gate
(b) OR gate
(c) XOR gate
(d) NAND gate
Show Answer & Explanation

Answer: (d) NAND gate

Explanation:
1. The output is 0 only when both inputs are 1.
2. That is the NAND gate.

Teacher's Note:
1. NAND is the complement of AND.

Question 8: Two NOT gates are connected at the two inputs of a NAND gate. This combination will behave like a:
(a) NAND gate
(b) AND gate
(c) OR gate
(d) NOR gate
Show Answer & Explanation

Answer: (c) OR gate

Explanation:
1. \( Y = \overline{\bar{A}\cdot\bar{B}} = A + B \).
2. This is an OR gate.

Teacher's Note:
1. By De Morgan's law.

Question 9: A truth table has outputs: (A, B) = (0, 0) gives 0; (0, 1) gives 1; (1, 0) gives 1; (1, 1) gives 0. It belongs to the:
(a) OR gate
(b) XOR gate
(c) AND gate
(d) NAND gate
Show Answer & Explanation

Answer: (b) XOR gate

Explanation:
1. The output is 1 only when the inputs are different.
2. That is the XOR gate.

Teacher's Note:
1. XOR is the 'odd' detector.

Question 10: 'The output is LOW if and only if all the inputs are HIGH.' Indicate the logic gate for which this statement is true:
(a) AND
(b) OR
(c) NOR
(d) NAND
Show Answer & Explanation

Answer: (d) NAND

Explanation:
1. NAND gives 0 only for all-high inputs.
2. At every other input combination its output is 1.

Teacher's Note:
1. Check the truth table of NAND.

Question 11: In a logic circuit, inputs W and X feed one OR gate and inputs W and Y feed another OR gate. The outputs of both OR gates feed an AND gate. The output is:
(a) \( W\cdot(X+Y) \)
(b) \( X\cdot(X\cdot Y) \)
(c) \( W + (X+Y) \)
(d) \( W + (X\cdot Y) \)
Show Answer & Explanation

Answer: (d) \( W + (X\cdot Y) \)

Explanation:
1. Output = \( (W+X)(W+Y) = W + WY + XW + XY = W + XY \).
2. Using distributive law of Boolean algebra.

Teacher's Note:
1. \( (W+X)(W+Y) = W + XY \).

Question 12: Inputs A and B go to an OR gate and also to a NAND gate. The outputs of the OR gate and the NAND gate are fed to an AND gate. This configuration is equivalent to:
(a) NAND
(b) OR
(c) XOR
(d) NOR
Show Answer & Explanation

Answer: (c) XOR

Explanation:
1. \( Y = (A+B)\cdot\overline{AB} = A\bar{B} + \bar{A}B \).
2. This is the XOR gate.

Teacher's Note:
1. Check with the truth table for all four inputs.

Question 13: A pure semiconductor S in series with a variable resistor R and a constant voltage source V is heated. To keep the ammeter reading constant, the value of R should be:
(a) decreased, because the semiconductor resistance increases
(b) increased, because the semiconductor resistance decreases
(c) kept the same
(d) first increased and then decreased
Show Answer & Explanation

Answer: (b) increased, because the semiconductor resistance decreases

Explanation:
1. Heating decreases the resistance of the semiconductor.
2. To keep the current constant, the total resistance must stay the same, so R must be increased.

Teacher's Note:
1. The resistance of a semiconductor falls on heating.

Question 14: Pure silicon with \( 5 \times 10^{28} \) atoms/m\( ^3 \) is doped with 1 ppm of pentavalent arsenic. Given \( n_i = 1.5 \times 10^{16} \) m\( ^{-3} \), the number of holes per m\( ^3 \) is:
(a) \( 5 \times 10^{22} \)
(b) \( 4.5 \times 10^{9} \)
(c) \( 1.5 \times 10^{16} \)
(d) \( 3 \times 10^{10} \)
Show Answer & Explanation

Answer: (b) \( 4.5 \times 10^{9} \)

Explanation:
1. Donor density = \( 5 \times 10^{28} \times 10^{-6} = 5 \times 10^{22} \) m\( ^{-3} \), so \( n_e \approx 5 \times 10^{22} \).
2. \( n_h = n_i^2/n_e = 2.25 \times 10^{32}/5 \times 10^{22} = 4.5 \times 10^{9} \) m\( ^{-3} \).

Teacher's Note:
1. Electron density is nearly equal to the donor density.

Question 15: The contributions of electrons and holes to the current of a semiconductor are in the ratio 3 : 1 and the ratio of their drift velocities is 5 : 2. The ratio of electron to hole concentration is:
(a) 5/6
(b) 6/5
(c) 3/2
(d) 2/5
Show Answer & Explanation

Answer: (b) 6/5

Explanation:
1. \( I \propto n v \).
2. \( (n_e v_e)/(n_h v_h) = 3 \) gives \( n_e/n_h = 3 \times (2/5) = 6/5 \).

Teacher's Note:
1. Current is proportional to carrier density times drift velocity.

Question 16: What minimum kinetic energy must a hole have to diffuse from the p-side to the n-side if the junction (barrier potential 0.2 V) is forward biased at 0.1 V?
(a) 0.2 eV
(b) 0.1 eV
(c) 0.3 eV
(d) 0.4 eV
Show Answer & Explanation

Answer: (b) 0.1 eV

Explanation:
1. Forward bias reduces the barrier to 0.2 − 0.1 = 0.1 V.
2. The hole needs a minimum KE of 0.1 eV.

Teacher's Note:
1. Unbiased needs 0.2 eV, reverse biased at 0.1 V needs 0.3 eV.

Question 17: A silicon p-n junction in forward bias with a series resistance has a knee voltage of 0.75 V and carries 10 mA when connected to a 2.75 V battery. The value of the series resistance is:
(a) 100 \( \Omega \)
(b) 200 \( \Omega \)
(c) 275 \( \Omega \)
(d) 75 \( \Omega \)
Show Answer & Explanation

Answer: (b) 200 \( \Omega \)

Explanation:
1. Voltage across the resistor = 2.75 − 0.75 = 2.0 V.
2. \( R = 2/0.01 = 200 \ \Omega \).

Teacher's Note:
1. Subtract the knee voltage first.

Question 18: A silicon diode (drop 0.7 V, forward resistance 2 \( \Omega \)) and a 2 \( \Omega \) resistor are connected in series with a 1.7 V battery. The current is:
(a) 0.25 A
(b) 0.5 A
(c) 1 A
(d) 0.4 A
Show Answer & Explanation

Answer: (a) 0.25 A

Explanation:
1. Effective emf = 1.7 − 0.7 = 1.0 V.
2. Total resistance = 2 + 2 = 4 \( \Omega \), so \( I = 0.25 \) A.

Teacher's Note:
1. Treat the diode as a 0.7 V cell with 2 ohm resistance.

Question 19: A 12 V battery is connected to a 5 k\( \Omega \) resistor through a Ge diode and a Si diode connected in antiparallel (Ge forward, Si reverse). Ge starts conducting at 0.3 V. The output voltage \( V_0 \) across the resistor and the current I are:
(a) \( V_0 = 11.7 \) V, \( I = 2.34 \) mA
(b) \( V_0 = 11.3 \) V, \( I = 2.26 \) mA
(c) \( V_0 = 12 \) V, \( I = 2.4 \) mA
(d) \( V_0 = 11.7 \) V, \( I = 2.26 \) mA
Show Answer & Explanation

Answer: (a) \( V_0 = 11.7 \) V, \( I = 2.34 \) mA

Explanation:
1. Only the Ge diode conducts and drops 0.3 V.
2. \( V_0 = 12 - 0.3 = 11.7 \) V and \( I = 11.7/5 \text{ k}\Omega = 2.34 \) mA.

Teacher's Note:
1. In antiparallel, only the forward diode conducts.

Question 20: A Zener diode has a breakdown voltage of 9.1 V and a maximum power dissipation of 364 mW. The maximum current it can handle is:
(a) 20 mA
(b) 40 mA
(c) 60 mA
(d) 80 mA
Show Answer & Explanation

Answer: (b) 40 mA

Explanation:
1. \( I_{max} = P/V_Z = 0.364/9.1 = 0.04 \) A = 40 mA.

Teacher's Note:
1. Power equals voltage times current.

Question 21: A GaAs semiconductor has an energy gap of 1.43 eV. The maximum wavelength emitted when an electron and hole recombine is about:
(a) 8671 Å
(b) 6200 Å
(c) 4350 Å
(d) 12400 Å
Show Answer & Explanation

Answer: (a) 8671 Å

Explanation:
1. \( \lambda = 12400/1.43 = 8671 \) Å.

Teacher's Note:
1. Use \( \lambda = hc/E_g \); this is infrared.

Question 22: In a common emitter amplifier with current gain 50 and emitter current 6.6 mA, the collector current is approximately:
(a) 6.47 mA
(b) 6.6 mA
(c) 0.13 mA
(d) 3.3 mA
Show Answer & Explanation

Answer: (a) 6.47 mA

Explanation:
1. \( I_E = I_B + I_C = I_B(1+\beta) \), so \( I_B = 6.6/51 = 0.13 \) mA.
2. \( I_C = 50 \times 0.13 = 6.47 \) mA.

Teacher's Note:
1. Emitter current includes the base current.

Question 23: In an npn transistor circuit the collector current is 10 mA. If 95% of the electrons emitted reach the collector, the base current is about:
(a) 0.53 mA
(b) 1 mA
(c) 0.95 mA
(d) 5 mA
Show Answer & Explanation

Answer: (a) 0.53 mA

Explanation:
1. \( I_E = 10/0.95 = 10.53 \) mA.
2. \( I_B = I_E - I_C = 0.53 \) mA.

Teacher's Note:
1. Base current is the part of the emitter current that does not reach the collector.

Question 24: In an NPN transistor, \( 10^{10} \) electrons enter the emitter in \( 10^{-6} \) s and 2% of them recombine in the base. The current gains \( \alpha \) and \( \beta \) are:
(a) \( \alpha = 0.98, \beta = 49 \)
(b) \( \alpha = 0.98, \beta = 50 \)
(c) \( \alpha = 0.02, \beta = 49 \)
(d) \( \alpha = 0.49, \beta = 98 \)
Show Answer & Explanation

Answer: (a) \( \alpha = 0.98, \beta = 49 \)

Explanation:
1. 98% of the carriers reach the collector, so \( \alpha = 0.98 \).
2. \( \beta = \alpha/(1-\alpha) = 0.98/0.02 = 49 \).

Teacher's Note:
1. Alpha is the fraction reaching the collector.

Question 25: For a CE amplifier the current gain is 69. If the emitter current is 7 mA, the base current and collector current are respectively:
(a) 0.1 mA and 6.9 mA
(b) 0.07 mA and 6.93 mA
(c) 0.5 mA and 6.5 mA
(d) 0.1 mA and 7 mA
Show Answer & Explanation

Answer: (a) 0.1 mA and 6.9 mA

Explanation:
1. \( I_B = I_E/(1+\beta) = 7/70 = 0.1 \) mA.
2. \( I_C = 7 - 0.1 = 6.9 \) mA.

Teacher's Note:
1. Use \( I_E = (1+\beta) I_B \).

Question 26: A common emitter amplifier has a voltage gain of 50, an input impedance of 200 \( \Omega \) and an output impedance of 400 \( \Omega \). The power gain is:
(a) 1250
(b) 2500
(c) 625
(d) 100
Show Answer & Explanation

Answer: (a) 1250

Explanation:
1. Current gain = \( A_V \times R_{in}/R_{out} = 50 \times 200/400 = 25 \).
2. Power gain = \( 50 \times 25 = 1250 \).

Teacher's Note:
1. Power gain = voltage gain times current gain.

Question 27:
Assertion (A): Metals are better conductors than semiconductors.
Reason (R): The valence band and conduction band have a large energy gap in the case of metals.
(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. Metals have free electrons and are better conductors.
2. In metals the bands overlap, so there is no gap. R is false.

Teacher's Note:
1. Metals have no forbidden gap.

Question 28:
Assertion (A): Photodiode always works under reverse bias.
Reason (R): It is easier to detect current changes in reverse bias.
(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. The reverse dark current is very small, so the change due to light is easily detected.
2. R explains A.

Teacher's Note:
1. Fractional change in current is large in reverse bias.

Question 29:
Assertion (A): GaAs can be used in making infrared LEDs.
Reason (R): The band gap of GaAs lies between 1.8 eV and 3 eV.
(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. The gap of GaAs is about 1.4 eV, which corresponds to infrared.
2. So A is true and R is false.

Teacher's Note:
1. Visible LEDs need 1.8 to 3 eV.

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