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Photodiode and solar cell: NEET MCQs with answers

50 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. Which of the following is NOT an essential criterion for the selection of a semiconductor material to fabricate a solar cell?

  1. High direct band gap greater than 3.0 eV
  2. Band gap between 1.0 eV and 1.8 eV
  3. High electrical conductivity
  4. High optical absorption coefficient (≈ 10⁴ cm⁻¹)
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Answer: (A) High direct band gap greater than 3.0 eV

The essential criteria for solar cell materials include a suitable band gap (1.0 to 1.8 eV), high optical absorption, electrical conductivity, and availability of the raw material. A material with a band gap greater than 3.0 eV absorbs very little of the solar spectrum, making it completely unsuitable.

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2. Consider the following two statements regarding a photodiode: Statement I: The reverse current in a photodiode remains constant when the intensity of incident light of a fixed frequency (hν > Eg) is increased. Statement II: The reverse saturation current in dark conditions is due to thermally generated minority carriers. Choose the correct option:

  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: (C) Statement I is incorrect but Statement II is correct.

Statement I is incorrect because the reverse current directly increases linearly with an increase in incident light intensity as more electron-hole pairs are created per second. Statement II is correct because the dark current under reverse bias arises from thermally generated minority charge carriers.

hard

3. A photodiode with a quantum efficiency of 100% detects optical signals of wavelength 600 nm. If an optical power of 3.3 mW falls on the diode, the photo-generated current produced is approximately: (Take h = 6.6 × 10⁻³⁴ J·s, c = 3.0 × 10⁸ m/s, e = 1.6 × 10⁻¹⁹ C)

  1. 0.8 mA
  2. 3.3 mA
  3. 4.8 mA
  4. 1.6 mA
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Answer: (D) 1.6 mA

Energy of one photon E = hc/λ = (6.6 × 10⁻³⁴ × 3.0 × 10⁸) / (600 × 10⁻⁹) = 3.3 × 10⁻¹⁹ J. Number of incident photons per second N = P/E = (3.3 × 10⁻³) / (3.3 × 10⁻¹⁹) = 1.0 × 10¹⁶ s⁻¹. With 100% quantum efficiency, photo-current I = N × e = 1.0 × 10¹⁶ × 1.6 × 10⁻¹⁹ A = 1.6 mA.

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4. A photodiode has a band gap of 1.5 eV. It is exposed to four different light beams of wavelengths λ₁ = 700 nm, λ₂ = 800 nm, λ₃ = 900 nm, and λ₄ = 1000 nm. Which of these wavelengths can be successfully detected by the photodiode? (Take hc ≈ 1240 eV·nm)

  1. λ₁ only
  2. All four wavelengths
  3. λ₁, λ₂ and λ₃ only
  4. λ₁ and λ₂ only
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Answer: (D) λ₁ and λ₂ only

For detection, the energy of the incident photon must be greater than or equal to the band gap (hc/λ ≥ Eg), which implies λ ≤ hc/Eg = 1240 / 1.5 ≈ 826.7 nm. Among the given wavelengths, only 700 nm and 800 nm are less than 826.7 nm.

easy

5. For the fabrication of an efficient terrestrial solar cell, which range of energy band gap (Eg) of the semiconductor material is considered most optimal, considering the solar spectrum reaching Earth?

  1. 0.2 eV to 0.6 eV
  2. 1.0 eV to 1.8 eV
  3. 2.8 eV to 3.5 eV
  4. 4.0 eV to 5.2 eV
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Answer: (B) 1.0 eV to 1.8 eV

The solar radiation spectrum reaching the surface of the Earth peaks around photon energies of 1.5 eV. Therefore, semiconductors with band gap energies close to 1.5 eV (typically between 1.0 eV and 1.8 eV, such as Si with 1.1 eV and GaAs with 1.43 eV) maximize solar energy conversion efficiency.

hard

6. In a solar cell, the short-circuit current (Isc) is directly proportional to the incident light intensity, whereas the open-circuit voltage (Voc) varies:

  1. Inversely with light intensity
  2. Linearly with light intensity
  3. Logarithmically with light intensity
  4. As the square root of light intensity
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Answer: (C) Logarithmically with light intensity

The open-circuit voltage Voc is related to photocurrent I_ph by Voc = (k_B T / e) ln(I_ph / I₀ + 1). Because I_ph is proportional to light intensity, Voc increases logarithmically with light intensity and saturates at high intensities.

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7. In an experiment with a reverse-biased photodiode, monochromatic light of wavelength 620 nm produces a photocurrent of 10 μA when the light source is placed at a distance of 1.0 m. If the same light source is moved to a distance of 2.0 m, what will be the new photocurrent?

  1. 5.0 μA
  2. 2.5 μA
  3. 10.0 μA
  4. 20.0 μA
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Answer: (B) 2.5 μA

Intensity from a point source follows the inverse-square law: I_light ∝ 1/r². Moving the source from 1.0 m to 2.0 m reduces the intensity by a factor of (2.0/1.0)² = 4. Since photocurrent is directly proportional to intensity, I_photo = 10 μA / 4 = 2.5 μA.

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8. GaAs has an energy band gap of 1.43 eV and Si has an energy band gap of 1.1 eV. Why does GaAs produce a higher open-circuit voltage (Voc) than Si when used in solar cells under standard solar illumination?

  1. GaAs has a higher dielectric constant than Si
  2. GaAs absorbs photons only in the ultraviolet region
  3. GaAs has a larger band gap, allowing a higher built-in potential barrier
  4. GaAs has higher thermal conductivity than Si
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Answer: (C) GaAs has a larger band gap, allowing a higher built-in potential barrier

The open-circuit voltage Voc is fundamentally limited by the band gap of the material (Voc < Eg / e). Since GaAs has a larger band gap (1.43 eV) than Si (1.1 eV), its built-in potential is higher, resulting in a substantially larger open-circuit voltage.

medium

9. When a p-n junction photodiode is in reverse bias with a steady reverse saturation current I₀ in the dark, illumination by light of frequency ν (where hν > Eg) increases the total reverse current to I. What is the photocurrent generated solely by light?

  1. I + I₀
  2. √(I × I₀)
  3. I / I₀
  4. I - I₀
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Answer: (D) I - I₀

Under illumination, the total current in reverse bias is given by I = I_sc - I₀ or effectively I_total = I_ph + I_dark in magnitude. Hence, the photocurrent produced by illumination alone is the difference between total current and dark current: I_ph = I - I₀.

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10. A photodiode made from a semiconductor with Eg = 2.0 eV is illuminated by four monochromatic laser sources: L₁: 550 nm L₂: 650 nm L₃: 700 nm L₄: 450 nm Which of these lasers will produce a detectable photocurrent? (Take hc ≈ 1240 eV nm)

  1. L₄ only
  2. L₂ and L₃ only
  3. L₁ and L₄ only
  4. L₁, L₂, and L₄ only
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Answer: (C) L₁ and L₄ only

The threshold wavelength for detection is λ_threshold = 1240 / Eg = 1240 / 2.0 = 620 nm. A photon can create an electron-hole pair only if its wavelength is less than or equal to 620 nm. Among the given lasers, L₁ (550 nm) and L₄ (450 nm) have λ ≤ 620 nm, so only they can produce a photocurrent.

easy

11. A photodiode circuit has a reverse bias voltage supply of 10 V and a series load resistor of 10 kΩ. In the dark, the current is negligible. When illuminated, the voltage drop across the load resistor is found to be 2.0 V. What is the actual reverse voltage appearing across the photodiode?

  1. 2.0 V
  2. 8.0 V
  3. 6.0 V
  4. 10.0 V
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Answer: (B) 8.0 V

By Kirchhoff's Voltage Law in the series loop, V_supply = V_diode + V_load. Therefore, V_diode = V_supply - V_load = 10 V - 2.0 V = 8.0 V.

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12. Why is a photodiode preferred to be operated under reverse bias rather than forward bias to measure light intensity?

  1. The fractional change in minority carrier current is much larger than that in majority carrier current.
  2. The reverse bias reduces the barrier height, allowing easier flow of photo-generated carriers.
  3. In forward bias, electron-hole pairs recombine before they can be separated by the electric field.
  4. Reverse bias completely prevents thermal generation of electron-hole pairs.
Show answer and explanation

Answer: (A) The fractional change in minority carrier current is much larger than that in majority carrier current.

Under illumination, the absolute increase in carrier density (Δn = Δp) is the same for both carriers. However, because the initial minority carrier density is extremely small compared to majority carrier density, the fractional increase (Δn/n or Δp/p) is significantly higher for minority carriers, which dominate reverse saturation current.

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