JEE Physics Semiconductor Electronics MCQs Set 09

Physics Objective Questions and Answers: 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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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: An npn transistor in CE configuration has a load of 800 \( \Omega \) in the collector with a drop of 0.8 V across it. The current amplification factor is 0.96 and the input resistance is 192 \( \Omega \). The voltage gain and the power gain respectively are:
(a) 4, 3.84
(b) 3.69, 3.84
(c) 4, 4
(d) 4, 3.69
Show Answer & Explanation

Answer: (a) 4, 3.84

Explanation:
1. Using \( A_V = \alpha R_L/R_{in} = 0.96 \times 800/192 = 4 \).
2. Power gain = \( \alpha \times A_V = 0.96 \times 4 = 3.84 \).

Teacher's Note:
1. Power gain is the product of voltage and current gains.

Question 2: In a CE transistor amplifier the audio signal voltage across the collector resistance of 2 k\( \Omega \) is 4 V. The current amplification factor is 100 and the base resistance is 1 k\( \Omega \). The input signal voltage is:
(a) 30 mV
(b) 15 mV
(c) 10 mV
(d) 20 mV
Show Answer & Explanation

Answer: (d) 20 mV

Explanation:
1. \( I_C = 4/2000 = 2 \) mA, so \( I_B = 20 \ \mu \)A.
2. \( V_{in} = 20 \times 10^{-6} \times 1000 = 20 \) mV.

Teacher's Note:
1. Find the base current first.

Question 3: In a P-N-P transistor in CE mode, \( I_C = 2 \) mA, \( I_B = 5 \ \mu \)A and \( R_B = 500 \ \Omega \). The transconductance (in A/V) is:
(a) 0.8
(b) 0.7
(c) 0.6
(d) 0.5
Show Answer & Explanation

Answer: (a) 0.8

Explanation:
1. \( V_{BE} = I_B R_B = 5 \times 10^{-6} \times 500 = 2.5 \) mV.
2. \( g_m = I_C/V_{BE} = 2 \times 10^{-3}/2.5 \times 10^{-3} = 0.8 \) A/V.

Teacher's Note:
1. Transconductance is output current per unit input voltage.

Question 4: The resistivity of an n-type germanium crystal is 2 \( \Omega \) m. The velocity of electrons per unit electric field is 0.4 (SI units). The number density of charge carriers is:
(a) \( 6.8 \times 10^{18} \)/m\( ^3 \)
(b) \( 6.3 \times 10^{18} \)/m\( ^3 \)
(c) \( 5.2 \times 10^{16} \)/m\( ^3 \)
(d) \( 7.8 \times 10^{18} \)/m\( ^3 \)
Show Answer & Explanation

Answer: (d) \( 7.8 \times 10^{18} \)/m\( ^3 \)

Explanation:
1. \( \sigma = 1/\rho = 0.5 \) S/m and \( \sigma = n e \mu \).
2. \( n = 0.5/(1.6 \times 10^{-19} \times 0.4) = 7.8 \times 10^{18} \) m\( ^{-3} \).

Teacher's Note:
1. Mobility is drift velocity per unit field.

Question 5: Inputs A and B go to an OR gate. B also goes through a NOT gate whose output and input C go to a second OR gate. The two OR outputs go to an AND gate with output Y. Y equals:
(a) \( (\bar{A}+B)\cdot(\bar{B}+C) \)
(b) \( (A+B)\cdot(\bar{B}+C) \)
(c) \( (A+\bar{B})\cdot(B+\bar{C}) \)
(d) \( (\bar{A}+B)\cdot(\bar{B}+\bar{C}) \)
Show Answer & Explanation

Answer: (b) \( (A+B)\cdot(\bar{B}+C) \)

Explanation:
1. The first OR gives A + B.
2. The second OR gives \( \bar{B} + C \).
3. The AND gate multiplies them.

Teacher's Note:
1. Write the output of each gate stage by stage.

Question 6: The resistivity of a pure semiconductor is 0.5 \( \Omega \) m. The electron and hole mobilities are 0.39 and 0.19 m\( ^2 \)/V s. The intrinsic carrier concentration is:
(a) \( 2.16 \times 10^{19} \)/m\( ^3 \)
(b) \( 4.32 \times 10^{19} \)/m\( ^3 \)
(c) \( 10^{20} \)/m\( ^3 \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( 2.16 \times 10^{19} \)/m\( ^3 \)

Explanation:
1. \( \sigma = 1/0.5 = 2 \) S/m and \( \sigma = n_i e(\mu_e + \mu_h) \).
2. \( n_i = 2/(1.6 \times 10^{-19} \times 0.58) = 2.16 \times 10^{19} \) m\( ^{-3} \).

Teacher's Note:
1. Add both mobilities for an intrinsic semiconductor.

Question 7: A NOR gate is followed by a NOT gate formed by a NAND gate with joined inputs, and then by another NOT gate. The network is equivalent to:
(a) OR gate
(b) NOR gate
(c) NOT gate
(d) AND gate
Show Answer & Explanation

Answer: (b) NOR gate

Explanation:
1. Two NOT gates cancel each other.
2. Only the first NOR gate remains.

Teacher's Note:
1. Count inverters after the NOR gate.

Question 8: Which of the following represents a forward-biased diode? The diode arrow points from the left towards the resistor R on the right in each case.
(a) left end at −4 V, right end of R at −3 V
(b) left end at −2 V, right end of R at +2 V
(c) left end at 3 V, right end of R at 5 V
(d) left end at 0 V, right end of R at −2 V
Show Answer & Explanation

Answer: (d) left end at 0 V, right end of R at −2 V

Explanation:
1. Forward bias needs the anode (arrow side) at a higher potential than the cathode side.
2. Only 0 V to −2 V satisfies this.

Teacher's Note:
1. Current flows from higher to lower potential in the arrow direction.

Question 9: An NPN transistor in CE configuration has the base connected to a 5 V supply through \( R_1 = 8.6 \) k\( \Omega \) and the collector connected to a 10 V supply through \( R_2 = 100 \ \Omega \); the emitter is grounded. With \( V_{BE} = 0.7 \) V and \( \beta = 100 \), \( V_{CE} \) is:
(a) 5 V
(b) 8 V
(c) 3 V
(d) 4 V
Show Answer & Explanation

Answer: (a) 5 V

Explanation:
1. \( I_B = (5 - 0.7)/8.6 \text{ k}\Omega = 0.5 \) mA, so \( I_C = 50 \) mA.
2. \( V_{CE} = 10 - 50 \text{ mA} \times 100 \ \Omega = 5 \) V.

Teacher's Note:
1. Find the base current, then the collector current.

Question 10: In a combination of logic gates, A is passed through a NOT gate into an AND gate, with B as the other input; B is passed through a second NOT gate into another AND gate with A as the other input. The outputs of both AND gates enter an OR gate. Y in terms of A and B is:
(a) \( \overline{A\cdot B} \)
(b) \( A \cdot \bar{B} + \bar{A} \cdot B \)
(c) \( \overline{A\cdot\bar{B} + A\cdot B} \)
(d) \( \overline{A+B} \)
Show Answer & Explanation

Answer: (b) \( A \cdot \bar{B} + \bar{A} \cdot B \)

Explanation:
1. The first AND gives \( \bar{A}B \) and the second gives \( A\bar{B} \).
2. The OR gives \( A\bar{B} + \bar{A}B \), the XOR function.

Teacher's Note:
1. This is the standard XOR construction.

Question 11: In a transistor circuit, the input voltage \( V_i = 20 \) V is applied through \( R_B = 500 \) k\( \Omega \) to the base, \( V_{BE} = 0 \) and \( V_{CE} = 0 \). The collector resistor is 4 k\( \Omega \) from a 20 V supply. The values of \( I_B, I_C \) and \( \beta \) are:
(a) \( I_B = 40 \ \mu \)A, \( I_C = 10 \) mA, \( \beta = 250 \)
(b) \( I_B = 25 \ \mu \)A, \( I_C = 5 \) mA, \( \beta = 200 \)
(c) \( I_B = 20 \ \mu \)A, \( I_C = 5 \) mA, \( \beta = 250 \)
(d) \( I_B = 40 \ \mu \)A, \( I_C = 5 \) mA, \( \beta = 125 \)
Show Answer & Explanation

Answer: (d) \( I_B = 40 \ \mu \)A, \( I_C = 5 \) mA, \( \beta = 125 \)

Explanation:
1. \( I_B = 20/500 \text{ k}\Omega = 40 \ \mu \)A.
2. \( I_C = 20/4 \text{ k}\Omega = 5 \) mA.
3. \( \beta = 5 \text{ mA}/40 \ \mu\text{A} = 125 \).

Teacher's Note:
1. Since \( V_{CE} = 0 \), the full supply drops across \( R_C \).

Question 12: In a p-n junction diode, a change in temperature due to heating:
(a) affects only the reverse resistance
(b) affects only the forward resistance
(c) does not affect the resistance of the p-n junction
(d) affects the overall V-I characteristics of the p-n junction
Show Answer & Explanation

Answer: (d) affects the overall V-I characteristics of the p-n junction

Explanation:
1. Temperature changes the carrier density and barrier potential.
2. So both forward and reverse parts of the V-I curve change.

Teacher's Note:
1. Reverse saturation current roughly doubles for every 10 degrees rise.

Question 13: A Zener diode of voltage 12 V works with a load resistance \( R_L = 1 \) k\( \Omega \) and a series resistance \( R_S = 100 \ \Omega \). If the current through the Zener diode is 18 mA, the applied external voltage is:
(a) 15 V
(b) 20 V
(c) 25 V
(d) 10 V
Show Answer & Explanation

Answer: (a) 15 V

Explanation:
1. Load current = 12/1000 = 12 mA.
2. Total current = 18 + 12 = 30 mA.
3. Supply = 12 + 30 mA × 100 \( \Omega \) = 15 V.

Teacher's Note:
1. Add the load and Zener currents to get the series current.

Question 14: An intrinsic semiconductor is doped with boron to a density of \( 1.5 \times 10^{18} \) atoms/cm\( ^3 \) and with phosphorus to \( 10^{18} \) atoms/cm\( ^3 \). The type of semiconductor and its charge carrier density are:
(a) p-type, \( 0.5 \times 10^{18} \) cm\( ^{-3} \)
(b) n-type, \( 0.5 \times 10^{18} \) cm\( ^{-3} \)
(c) p-type, \( 2.5 \times 10^{18} \) cm\( ^{-3} \)
(d) n-type, \( 2.8 \times 10^{18} \) cm\( ^{-3} \)
Show Answer & Explanation

Answer: (a) p-type, \( 0.5 \times 10^{18} \) cm\( ^{-3} \)

Explanation:
1. Boron is an acceptor and phosphorus a donor, and they compensate each other.
2. Net acceptor density = 1.5 − 1.0 = 0.5 \( \times 10^{18} \), so the material is p-type.

Teacher's Note:
1. Compensation: subtract the smaller dopant density.

Question 15: What is the minimum number of NAND gates required to obtain the Boolean expression AB + CD?
(a) 3
(b) 4
(c) 5
(d) 6
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. \( AB + CD = \overline{\overline{AB}\cdot\overline{CD}} \).
2. Use one NAND for AB, one for CD and a third to combine the two outputs.

Teacher's Note:
1. Apply De Morgan's theorem twice.

Question 16: A half wave rectifier has input signal \( V = 220\sin\omega t \) and a load resistance of 2 k\( \Omega \). The output dc current is about:
(a) 140 mA
(b) 35 mA
(c) 49 mA
(d) 70 mA
Show Answer & Explanation

Answer: (b) 35 mA

Explanation:
1. \( I_{dc} = I_m/\pi = V_m/(\pi R_L) = 220/(\pi \times 2000) \approx 35 \) mA.

Teacher's Note:
1. Half wave: dc value is peak divided by pi.

Question 17: In a CE circuit, the base is connected to a 10 V supply through 1 M\( \Omega \) and the collector to a 10 V supply through 4 k\( \Omega \); the emitter is grounded. With \( V_{BE} = 0.7 \) V and \( \beta = 100 \), \( V_{CE} \) is:
(a) 6.28 V
(b) 7.00 V
(c) 6.5 V
(d) 6.00 V
Show Answer & Explanation

Answer: (a) 6.28 V

Explanation:
1. \( I_B = (10 - 0.7)/10^6 = 9.3 \ \mu \)A, \( I_C = 0.93 \) mA.
2. \( V_{CE} = 10 - 0.93 \times 4 = 6.28 \) V.

Teacher's Note:
1. Base current first, then collector drop.

Question 18: An electric field is applied to a semiconductor. Let the carrier density be n and the average drift speed v. If the temperature is increased:
(a) both n and v will increase
(b) n will increase but v will decrease
(c) v will increase but n will decrease
(d) both n and v will decrease
Show Answer & Explanation

Answer: (b) n will increase but v will decrease

Explanation:
1. Heating generates more carriers, so n increases.
2. Greater lattice vibrations increase scattering, so mobility and drift speed v decrease.

Teacher's Note:
1. Carrier density and mobility change in opposite ways.

Question 19: What is the conductivity of pure silicon at 300 K if the electron-hole pairs per cm\( ^3 \) are \( 1.072 \times 10^{10} \), \( \mu_e = 1350 \) cm\( ^2 \)/V s and \( \mu_h = 480 \) cm\( ^2 \)/V s?
(a) \( 3.14 \times 10^{-6} \) mho/cm
(b) \( 3 \times 10^{6} \) mho/cm
(c) \( 10^{-6} \) mho/cm
(d) \( 10^{6} \) mho/cm
Show Answer & Explanation

Answer: (a) \( 3.14 \times 10^{-6} \) mho/cm

Explanation:
1. \( \sigma = n_i e(\mu_e + \mu_h) \).
2. \( = 1.072 \times 10^{10} \times 1.6 \times 10^{-19} \times 1830 = 3.14 \times 10^{-6} \) mho/cm.

Teacher's Note:
1. Use consistent units of cm.

Question 20: The mobility of electrons in n-type Ge is 5000 cm\( ^2 \)/V s and the conductivity is 5 mho/cm. If the effect of holes is negligible, the impurity concentration is:
(a) \( 6.25 \times 10^{15} \)/cm\( ^3 \)
(b) \( 9.25 \times 10^{14} \)/cm\( ^3 \)
(c) \( 6 \times 10^{13} \)/cm\( ^3 \)
(d) \( 9 \times 10^{13} \)/cm\( ^3 \)
Show Answer & Explanation

Answer: (a) \( 6.25 \times 10^{15} \)/cm\( ^3 \)

Explanation:
1. \( n = \sigma/(e\mu) = 5/(1.6 \times 10^{-19} \times 5000) = 6.25 \times 10^{15} \) cm\( ^{-3} \).

Teacher's Note:
1. Use \( \sigma = ne\mu \).

Question 21: If the two ends of a p-n junction are joined by a wire:
(a) there will not be a steady current in the circuit
(b) there will be a steady current from the n-side to the p-side
(c) there will be a steady current from the p-side to the n-side
(d) there may or may not be a current depending upon the resistance of the connecting wire
Show Answer & Explanation

Answer: (a) there will not be a steady current in the circuit

Explanation:
1. The built-in potential is exactly balanced by the contact potentials at the wire ends.
2. So no net emf drives a current.

Teacher's Note:
1. An energy source cannot appear from nothing.

Question 22: A hole diffuses from the p-side to the n-side in a p-n junction. This means that:
(a) a bond is broken on the n-side and the electron freed jumps to the conduction band
(b) a conduction electron on the p-side jumps to a broken bond to complete it
(c) a bond is broken on the n-side and the freed electron jumps to a broken bond on the p-side to complete it
(d) a bond is broken on the p-side and the freed electron jumps to a broken bond on the n-side to complete it
Show Answer & Explanation

Answer: (c) a bond is broken on the n-side and the freed electron jumps to a broken bond on the p-side to complete it

Explanation:
1. A hole moving to the n side means an electron from a bond on the n side moves into a vacancy on the p side.
2. So an n-side bond is broken and a p-side bond is completed.

Teacher's Note:
1. A hole moves opposite to the electrons that fill it.

Question 23: To reduce the ripples in a rectifier circuit with a capacitor filter: (a) \( R_L \) should be increased; (b) input frequency should be decreased; (c) input frequency should be increased; (d) capacitors with high capacitance should be used.
(a) Only a
(b) Only a and b
(c) a, b and d
(d) a, c and d
Show Answer & Explanation

Answer: (d) a, c and d

Explanation:
1. Ripple depends on the time constant \( R_L C \) and on the frequency.
2. A larger \( R_L \), a larger C and a higher frequency all reduce ripple.

Teacher's Note:
1. Make the capacitor discharge slowly between peaks.

Question 24: The energy of radiation emitted by an LED is:
(a) greater than the band gap of the semiconductor used
(b) always less than the band gap
(c) always equal to the band gap
(d) equal to or less than the band gap of the semiconductor used
Show Answer & Explanation

Answer: (d) equal to or less than the band gap of the semiconductor used

Explanation:
1. Photon energy comes from recombination across the gap and is approximately equal to \( E_g \).
2. It may be a little less when recombination happens through impurity levels.

Teacher's Note:
1. LED colour depends on the gap.

Question 25: In a common emitter transistor circuit, the base current is 40 \( \mu \)A. The base resistor is 245 k\( \Omega \), \( V_{CC} = 10 \) V and the collector resistor is 2 k\( \Omega \). Then \( V_{BE} \) is:
(a) 2 V
(b) 0.2 V
(c) 0.8 V
(d) zero
Show Answer & Explanation

Answer: (b) 0.2 V

Explanation:
1. Voltage across the base resistor = \( 40 \ \mu\text{A} \times 245 \text{ k}\Omega = 9.8 \) V.
2. \( V_{BE} = 10 - 9.8 = 0.2 \) V.

Teacher's Note:
1. Apply KVL to the base loop.

Question 26: Two similar blocks of intrinsic silicon are doped uniformly to obtain equal dopant densities \( N_D = N_A \) per m\( ^3 \). The n-type and p-type samples are connected in parallel and a current is passed through them. Then:
(a) the same current passes through each sample
(b) the current through the n-type sample is more
(c) the current through the p-type sample is more
(d) the current through the p-type sample is twice that through the n-type sample
Show Answer & Explanation

Answer: (b) the current through the n-type sample is more

Explanation:
1. Conductivity \( = n e \mu \) for the majority carrier.
2. With equal densities, the higher electron mobility makes the n-type conductivity larger, so it carries more current.

Teacher's Note:
1. Mobility of electrons exceeds that of holes.

Question 27:
Assertion (A): A transistor amplifier operates in the active region.
Reason (R): In the active region, the transistor characteristic is linear.
(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 active region lies between cut-off and saturation.
2. The characteristic is linear there, so signals are amplified without distortion.

Teacher's Note:
1. Switch: cut-off and saturation. Amplifier: active region.

Question 28:
Assertion (A): A photodiode and a photovoltaic cell are based on the same principle.
Reason (R): Both use the same method of operation.
(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. Both are based on photo-generation of carriers in a junction.
2. A photodiode operates in reverse bias while a solar cell works without bias, so R is false.

Teacher's Note:
1. Principle same; method different.

Question 29:
Assertion (A): A transistor can be used as a switch.
Reason (R): Both linear and non-linear voltage bias dependence occurs in it.
(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. A transistor can be driven between cut-off and saturation.
2. Linear and non-linear regions both exist, but the second does not explain A.

Teacher's Note:
1. Switching uses the non-linear regions.

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Multiple Choice Questions (MCQs) for JEE Physics Semiconductor Electronics

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