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Diode as rectifier and LED characteristics: NEET MCQs with answers

51 questions on this topic of Electronic Devices (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. For an LED to emit radiation in the visible region of the electromagnetic spectrum (wavelength roughly 400 nm to 700 nm), the semiconductor material used must have an energy band gap (E_g) approximately:

  1. Less than 0.7 eV
  2. Between 1.0 eV and 1.5 eV
  3. At least 1.8 eV
  4. Greater than 5.0 eV
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Answer: (C) At least 1.8 eV

The maximum wavelength for visible light is approximately 700 nm, corresponding to the red threshold. Using E_g = hc / λ ≈ 1240 / 700 ≈ 1.8 eV, the band gap must be at least 1.8 eV for emissions in the visible spectrum.

medium

2. In a bridge rectifier circuit, an alternating voltage v = V_m sin(ωt) is applied. At any instant during conduction, how many diodes are in the forward-biased state and what is the peak inverse voltage (PIV) across each reverse-biased diode (assuming ideal diodes)?

  1. 1 diode conducts, PIV = 2V_m
  2. 2 diodes conduct, PIV = V_m
  3. 2 diodes conduct, PIV = 2V_m
  4. 4 diodes conduct, PIV = V_m
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Answer: (B) 2 diodes conduct, PIV = V_m

In a bridge rectifier, pairs of diagonally opposite diodes conduct during each half-cycle, meaning two diodes conduct simultaneously. The maximum reverse voltage appearing across each non-conducting diode is equal to the peak secondary voltage V_m.

hard

3. In a full-wave bridge rectifier, each of the four diodes has an internal forward resistance of 20 Ω and the load resistance is 960 Ω. The rectification efficiency of this circuit is:

  1. 40.6%
  2. 78.0%
  3. 81.2%
  4. 84.5%
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Answer: (B) 78.0%

In a bridge rectifier, current passes through two conducting diodes simultaneously, so the total diode resistance in series is 2r_d = 40 Ω. The efficiency is η = [8/π²] × [R_L / (2r_d + R_L)] = 0.812 × [960 / (40 + 960)] = 0.812 × 0.96 ≈ 0.7795 or 78.0%.

medium

4. Given below are two statements: Statement I: In a bridge rectifier, the Peak Inverse Voltage (PIV) across any non-conducting diode is equal to the peak secondary voltage V_m. Statement II: In a centre-tapped full-wave rectifier, the Peak Inverse Voltage (PIV) across each diode is 2V_m. In 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
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Answer: (A) Both Statement I and Statement II are correct

In a bridge rectifier, two diodes conduct while the other two are in reverse bias, experiencing a maximum reverse voltage of V_m. In a centre-tapped rectifier, the non-conducting diode faces the total voltage of both halves of the secondary, giving PIV = 2V_m. Thus, both statements are correct.

easy

5. An LED has a forward operating voltage drop of 2.0 V and requires a continuous forward current of 15 mA to glow with normal brightness. To operate it safely from a 5.0 V DC power supply, the value of the series current-limiting resistor should be:

  1. 200 Ω
  2. 133 Ω
  3. 333 Ω
  4. 467 Ω
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Answer: (A) 200 Ω

The voltage drop across the series resistor is V_R = V_supply - V_LED = 5.0 V - 2.0 V = 3.0 V. The required resistance is R = V_R / I = 3.0 / (15 × 10⁻³) = 200 Ω.

hard

6. In a full-wave rectifier circuit without filter, the AC input is given by v = V₀ sin(ωt). The root-mean-square value of the AC ripple component present in the output voltage is:

  1. V₀ √(1 - 2/π)
  2. V₀ (1/√2 - 2/π)
  3. V₀ / (2√2)
  4. V₀ √(1/2 - 4/π²)
Show answer and explanation

Answer: (D) V₀ √(1/2 - 4/π²)

The total RMS output voltage is V_rms = V₀ / √2, and the DC component is V_dc = 2V₀ / π. The RMS value of the AC ripple component is V'_rms = √[V_rms² - V_dc²] = √[(V₀ / √2)² - (2V₀ / π)²] = V₀ √(1/2 - 4/π²).

hard

7. A bridge full-wave rectifier uses four identical diodes, each with a dynamic forward resistance r_d = 10 Ω. It delivers power to a load resistor R_L = 480 Ω. What percentage of the total secondary power delivered by the transformer is converted into useful DC power?

  1. 78.0%
  2. 81.2%
  3. 40.6%
  4. 39.0%
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Answer: (A) 78.0%

In a bridge rectifier, two diodes conduct simultaneously in series during each half cycle, so total forward resistance is 2r_d = 20 Ω. Rectification efficiency η = [0.812 / (1 + 2r_d / R_L)] = 0.812 / (1 + 20/480) = 0.812 / (1 + 0.0417) ≈ 0.812 / 1.0417 ≈ 78.0%.

medium

8. A green LED having a forward turn-on voltage of 2.4 V is powered by a 9.0 V battery in series with a protective current-limiting resistor R. If the desired operating current through the LED is 30 mA, the required resistance R and the power dissipated in R are respectively:

  1. 300 Ω and 0.270 W
  2. 220 Ω and 0.198 W
  3. 220 Ω and 0.072 W
  4. 150 Ω and 0.135 W
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Answer: (B) 220 Ω and 0.198 W

The voltage drop across the resistor is V_R = V_supply - V_LED = 9.0 V - 2.4 V = 6.6 V. Resistance R = V_R / I = 6.6 V / (30 × 10⁻³ A) = 220 Ω. Power dissipated in the resistor is P_R = I²R = V_R × I = 6.6 V × 0.030 A = 0.198 W.

medium

9. An LED made of gallium arsenide phosphide (GaAsP) glows with an energy band gap of 2.0 eV. What is the minimum potential difference required across its terminals to make it conduct and emit light, and why is its reverse breakdown voltage usually low (~5 V)?

  1. About 0.7 V; because of thick depletion layer
  2. About 2.0 V; because of heavy doping on both sides of the junction
  3. About 1.1 V; because light absorption destroys the barrier
  4. About 0.3 V; because mobility of electrons is exceptionally high
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Answer: (B) About 2.0 V; because of heavy doping on both sides of the junction

The threshold or operating forward voltage of an LED is roughly equal to its band gap potential, V ≈ E_g / e ≈ 2.0 V. Because LEDs are heavily doped to ensure efficient recombination, the depletion region is very thin, which leads to a low reverse breakdown voltage (typically ~5 V).

medium

10. In a full-wave rectifier using an AC input of frequency 50 Hz, the output ripple frequency is:

  1. 100 Hz
  2. 50 Hz
  3. 25 Hz
  4. 200 Hz
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Answer: (A) 100 Hz

A full-wave rectifier uses both half-cycles, so the output has two pulses per input cycle, giving a ripple frequency of 2 × 50 = 100 Hz.

NCERT Class 12

easy

11. In a full-wave rectifier without a filter, an AC sinusoidal voltage of peak value 30 V is applied across the secondary circuit. If the diodes are assumed to be ideal and the load resistor is 1.0 kΩ, what is the average (DC) current flowing through the load resistor?

  1. 27.0 mA
  2. 9.55 mA
  3. 19.1 mA
  4. 13.5 mA
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Answer: (C) 19.1 mA

For a full-wave rectifier, the average (DC) output voltage is given by V_dc = 2V_m / π. Here, V_m = 30 V, so V_dc = (2 × 30) / π ≈ 60 / 3.1416 ≈ 19.1 V. Therefore, the average load current is I_dc = V_dc / R_L = 19.1 V / 1000 Ω = 19.1 mA.

medium

12. Match the semiconductor optoelectronic device in List I with its essential operational feature in List II: List I: P. Light Emitting Diode Q. Photodiode R. Solar Cell S. Zener Diode List II: 1. Operates in reverse breakdown to regulate voltage 2. Generates emf under no external bias via photovoltaic effect 3. Operates under reverse bias to detect optical signals 4. Recombination of minority injected carriers under forward bias

  1. P-4, Q-3, R-2, S-1
  2. P-3, Q-4, R-2, S-1
  3. P-4, Q-1, R-2, S-3
  4. P-2, Q-3, R-4, S-1
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Answer: (A) P-4, Q-3, R-2, S-1

An LED operates under forward bias where injected minority carriers recombine with majority carriers to emit photons (P-4). A photodiode is reverse biased to detect optical signals (Q-3). A solar cell generates emf without external bias (R-2). A Zener diode operates in reverse breakdown region (S-1).

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