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NEET Electronic Devices MCQs with answers

277 questions on Electronic Devices for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. The barrier potential of a p-n junction depends on: (A) Nature of the semiconductor material (B) Amount of doping (C) Temperature Choose the most appropriate answer from the options given below:

  1. (A) and (B) only
  2. (B) and (C) only
  3. (A), (B) and (C)
  4. (A) and (C) only
Show answer and explanation

Answer: (C) (A), (B) and (C)

The built-in barrier potential depends on the band gap of the base material (e.g., ~0.7 V for Si, ~0.3 V for Ge), the doping levels of both p and n sides, and the ambient temperature (barrier potential decreases with an increase in temperature).

medium

2. A Zener diode voltage regulator maintains an output voltage of 6 V across a variable load resistor. The unregulated DC input is 12 V and the series resistor Rs is 120 Ω. If the minimum safe operating current through the Zener diode to remain in breakdown is 5 mA, what is the minimum load resistance RL,min for proper regulation?

  1. 80 Ω
  2. 120 Ω
  3. 133.3 Ω
  4. 150 Ω
Show answer and explanation

Answer: (C) 133.3 Ω

The total current supplied through Rs is Is = (Vin - Vz) / Rs = (12 - 6) / 120 = 50 mA. Since Is = Iz + IL, the maximum load current occurs when the Zener current is at its minimum: IL,max = Is - Iz,min = 50 mA - 5 mA = 45 mA. Thus, the minimum load resistance is RL,min = Vz / IL,max = 6 V / 45 mA = 133.3 Ω.

hard

3. In the circuit shown conceptually, a silicon p-n junction diode (having a knee/barrier voltage of 0.7 V) is connected in series with a 1.3 kΩ resistor and a DC supply of 4.6 V in forward bias. What is the power dissipated in the resistor?

  1. 11.7 mW
  2. 16.2 mW
  3. 3.9 mW
  4. 7.8 mW
Show answer and explanation

Answer: (A) 11.7 mW

Voltage across the resistor V_R = V_supply - V_diode = 4.6 V - 0.7 V = 3.9 V. Current I = V_R / R = 3.9 V / 1300 Ω = 3 × 10⁻³ A = 3 mA. Power dissipated in the resistor P = I²R = V_R × I = 3.9 V × 3 × 10⁻³ A = 11.7 × 10⁻³ W = 11.7 mW.

medium

4. When a p-n junction is forward biased, the depletion region:

  1. Remains unchanged
  2. Disappears permanently
  3. Becomes narrower and the barrier potential decreases
  4. Becomes wider
Show answer and explanation

Answer: (C) Becomes narrower and the barrier potential decreases

Forward bias opposes the built-in field, reducing the barrier height and the width of the depletion layer, so majority carriers cross easily.

NCERT Class 12

easy

5. A photodiode is constructed using a semiconductor with a band gap of 1.24 eV. What is the maximum wavelength of incident radiation that can be detected by this photodiode? (Take Planck's constant h = 6.63 × 10⁻³⁴ J s and c = 3 × 10⁸ m/s)

  1. 800 nm
  2. 1000 nm
  3. 1240 nm
  4. 1500 nm
Show answer and explanation

Answer: (B) 1000 nm

The maximum (threshold) wavelength is given by λ_max = hc / Eg. Converting Eg to Joules: 1.24 × 1.6 × 10⁻¹⁹ J ≈ 1.984 × 10⁻¹⁹ J. Thus, λ_max = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (1.984 × 10⁻¹⁹) ≈ 1.0 × 10⁻⁶ m = 1000 nm.

hard

6. Given below are two statements: Assertion (A): A Zener diode does not act as a voltage regulator if the unregulated input voltage falls below its breakdown voltage. Reason (R): When the reverse voltage is less than the breakdown voltage, the Zener diode operates in the reverse saturation region and behaves virtually like an open circuit.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
Show answer and explanation

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

When the input voltage is less than Vz, the reverse bias across the diode is insufficient to cause breakdown. The diode operates in the reverse saturation region where it draws negligibly small current and behaves as an open switch, so the output voltage is determined purely by the potential divider formed by Rs and RL, failing to regulate. Thus, (R) correctly explains (A).

medium

7. The Boolean expression Y = (A + B) · (A' + B') simplifies to which logic operation?

  1. AND gate
  2. XNOR gate
  3. NOR gate
  4. XOR gate
Show answer and explanation

Answer: (D) XOR gate

Expanding the expression: Y = A·A' + A·B' + B·A' + B·B'. Since A·A' = 0 and B·B' = 0, this reduces to Y = A·B' + A'·B, which is the Boolean function for an Exclusive-OR (XOR) gate.

medium

8. A Zener diode with breakdown voltage 9 V is connected through a 100 Ω series resistor to a 24 V supply (no load). The current through the series resistor is:

  1. 150 mA
  2. 240 mA
  3. 90 mA
  4. 330 mA
Show answer and explanation

Answer: (A) 150 mA

The resistor drops Vin − Vz = 15 V, so I = 15/100 A = 150 mA.

medium

9. A capacitor filter of capacitance C = 100 μF is connected across the load resistor R_L = 2 kΩ of a full-wave rectifier fed from a 50 Hz mains supply. The ripple frequency seen by the filter and the approximate time constant of discharge are:

  1. 50 Hz and 0.1 s
  2. 50 Hz and 0.2 s
  3. 100 Hz and 0.05 s
  4. 100 Hz and 0.2 s
Show answer and explanation

Answer: (D) 100 Hz and 0.2 s

For a full-wave rectifier with input frequency f = 50 Hz, the output ripple frequency is 2f = 100 Hz. The time constant for the discharge of the capacitor through the load resistor is τ = R_L × C = 2000 Ω × 100 × 10⁻⁶ F = 0.2 s.

medium

10. A Zener diode with breakdown voltage 5 V is connected through a 200 Ω series resistor to a 15 V supply (no load). The current through the series resistor is:

  1. 50 mA
  2. 100 mA
  3. 25 mA
  4. 75 mA
Show answer and explanation

Answer: (A) 50 mA

The resistor drops Vin − Vz = 10 V, so I = 10/200 A = 50 mA.

medium

11. Which of the following materials is unsuitable for fabricating a terrestrial solar cell despite having high optical absorption, primarily because of a band gap that is too small?

  1. Germanium (Eg ≈ 0.67 eV)
  2. Gallium Arsenide (Eg ≈ 1.43 eV)
  3. Silicon (Eg ≈ 1.1 eV)
  4. Cadmium Telluride (Eg ≈ 1.45 eV)
Show answer and explanation

Answer: (A) Germanium (Eg ≈ 0.67 eV)

Germanium has a small band gap of about 0.67 eV. For a small band gap, almost all radiation is absorbed near the top layer, leading to significant thermalisation losses and an open-circuit voltage that is too low to produce useful efficiency.

medium

12. Consider the following statements: Statement I: At breakdown voltage in a Zener diode, a very large change in reverse current produces an almost negligible change in the reverse voltage. Statement II: Zener breakdown occurs primarily due to the impact ionization of host atoms by rapidly accelerating minority charge carriers. In the light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because after breakdown, the reverse voltage remains virtually constant even for large changes in current. Statement II is incorrect because Zener breakdown is caused by the direct rupture of covalent bonds by a strong electric field (field ionization / quantum tunneling), whereas impact ionization is the mechanism of avalanche breakdown.

medium

13. Given below are two statements regarding light emitting diodes (LEDs): Statement I: The reverse breakdown voltage of an LED is very low, typically around 5 V. Statement II: An LED must be operated under reverse bias so that minority carriers can drift across the junction to emit photons. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect.

Statement I is correct because LEDs have low reverse breakdown voltages, usually around 5 V, so care must be taken not to reverse bias them with high voltages. Statement II is incorrect because an LED is always operated under forward bias, where injected minority carriers recombine with majority carriers near the junction, emitting energy as photons.

medium

14. Semiconductors with a band gap close to about 1.5 eV, such as Si and GaAs, are preferred for solar cells because:

  1. They emit visible light
  2. They have zero band gap
  3. They absorb a large part of the solar spectrum efficiently
  4. They are insulators
Show answer and explanation

Answer: (C) They absorb a large part of the solar spectrum efficiently

The solar spectrum peaks around 1.5 eV. Materials with band gaps near this value (Si about 1.1 eV, GaAs about 1.43 eV) absorb a large fraction of sunlight.

NCERT Class 12

easy

15. For a two-input OR gate, what are the logic states of the inputs A and B if the output Y is 0?

  1. A = 0, B = 0
  2. A = 0, B = 1
  3. A = 1, B = 0
  4. A = 1, B = 1
Show answer and explanation

Answer: (A) A = 0, B = 0

The output of an OR gate is given by Y = A + B. It yields 0 (LOW) if and only if both inputs A and B are 0 (LOW). For all other combinations, the output is 1.

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