Photodiode

8 MCQs9-step worked example

Source: NCERT Electronic DevicesOfficial key: NTA-verifiedLast updated: 23 Sep 2026

Photodiode, explained for NEET

A photodiode is a p-n junction with a transparent window, and the common slip is to bias it the way you would bias an LED. It is operated in reverse bias.

Light on the window generates electron-hole pairs, but only if the photon energy clears the band gap: hν > E_g. The junction is built so that this happens in or near the depletion region, where the junction field pulls the pair apart before it recombines — electrons to the n-side, holes to the p-side. That separated charge adds to the reverse current, and the photocurrent tracks the intensity of the incident light (NCERT Class 12 Physics, Chapter 14, page 336).

Why reverse and not forward? Reverse current is carried by minority carriers. Illumination adds equal excess electrons and holes, Δn = Δp. In an n-region the majority density is already huge, so Δn/n is tiny, while the minority density is small, so Δp/p is large. The photo-effect shows up as a measurable fractional change in the minority-carrier-dominated reverse current — that is the regime you can read an intensity off.

Separate the neighbours in this unit in one line each: an LED is forward biased and emits; a solar cell uses the same photogeneration but runs with no external bias, to deliver power. The photodiode's job is detection — it is a photodetector for optical signals.

On the characteristics, an illuminated photodiode sits in the third quadrant — reverse voltage, reverse current — one nearly flat curve per intensity, stacked so that brighter light means larger reverse current.

Watch out: the cutoff is one-sided. A photon below E_g makes no pair however intense the beam, so a long-wavelength source can be invisible to a wide-gap diode. NEET touches this topic lightly — under half a question a year on average — but at medium negative-marking risk, so fix the bias direction and the hν > E_g test and move on.

Can you answer these Photodiode MCQs?

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

A photodiode used to measure the intensity of incident light is normally operated under which bias condition?

Show answer and why every option is right or wrong

Answer: C. C is correct. NCERT Class 12 Physics, Chapter 14, page 336 states that a photodiode is operated in reverse bias, because the change in current with light intensity is easier to observe there.

Why A is wrong: A is wrong because forward bias is how an LED is driven to emit; under forward bias the small photogenerated contribution is swamped by the large majority-carrier current.

Why B is wrong: B is wrong because unbiased operation describes the solar cell, which is built to deliver power rather than to register an intensity reading.

Why D is wrong: D is wrong because breakdown operation describes the Zener diode; a photodiode is kept below breakdown so that its reverse current reports the illumination and not an avalanche.

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MCQ 2Easy RecallPractice

Illumination of a photodiode generates electron-hole pairs only when the incident photon energy hν satisfies which condition, where E_g is the band gap of the semiconductor?

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Answer: A. A is correct. NCERT Class 12 Physics, Chapter 14, page 336 gives the generation condition as photon energy greater than the band gap — the absorbed photon lifts an electron across the gap and leaves a hole behind.

Why B is wrong: B is wrong because a photon below the band gap cannot promote an electron across it, so it generates no pair regardless of beam intensity.

Why C is wrong: C is wrong because equality is only the threshold; photons above it are absorbed too, with the surplus appearing as carrier kinetic energy.

Why D is wrong: D is wrong because nothing in the generation condition requires twice the gap — that would exclude photons the stated condition admits.

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MCQ 3Easy RecallPractice

A photodiode is fabricated so that photogeneration of electron-hole pairs takes place at which location in the device?

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Answer: D. D is correct. NCERT Class 12 Physics, Chapter 14, page 336 notes the diode is made so that generation occurs in or near the depletion region, where the junction field separates the electron and the hole before they recombine.

Why A is wrong: A is wrong because a pair created deep in the p-region sees no junction field, and the minority electron largely recombines before it can reach the junction.

Why B is wrong: B is wrong for the mirror reason on the other side: with no junction field to sweep it out, the pair recombines instead of contributing to the photocurrent.

Why C is wrong: C is wrong because the contacts only collect carriers; there is no junction field there and it is not where the pairs are generated.

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MCQ 4Direct ApplicationPractice

A photodiode is fabricated from a semiconductor of band gap 2.8 eV. Can it detect radiation of wavelength 6.0 × 10³ nm? Choose the statement with both the correct verdict and the correct reason. (Take hc = 1.24 × 10³ eV nm.)

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Answer: B. B is correct. E = hc/λ = (1.24 × 10³ eV nm)/(6.0 × 10³ nm) ≈ 0.21 eV, which fails the hν > E_g condition stated in NCERT Class 12 Physics, Chapter 14, page 336, so no pairs are generated.

Why A is wrong: A is wrong because penetration depth is irrelevant here: a sub-gap photon is not absorbed into a pair at any depth.

Why C is wrong: C is wrong because reverse bias only collects carriers that already exist; it cannot supply the energy a sub-gap photon lacks.

Why D is wrong: D is wrong in its reason even though it reaches the right verdict: photodiodes are routinely used in the infrared, and the disqualifier here is the photon energy against the band gap, not the spectral region.

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MCQ 5Direct ApplicationPractice

A photodiode is made from a semiconductor of band gap 1.24 eV. What is the longest wavelength it can detect? (Take hc = 1.24 × 10³ eV nm.)

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Answer: D. D is correct. The threshold sits at hν = E_g, so λ_max = hc/E_g = (1.24 × 10³ eV nm)/(1.24 eV) = 1.0 × 10³ nm, i.e. 1.0 μm; anything longer carries less than the band-gap energy.

Why A is wrong: A is wrong by a factor of ten — it comes from dropping a power of ten while evaluating hc/E_g.

Why B is wrong: B is wrong because 5.0 × 10² nm is a mid-visible wavelength assumed rather than computed; hc/E_g fixes this diode's threshold at 1.0 × 10³ nm.

Why C is wrong: C is wrong because it carries the band-gap number in eV straight across into nanometres — a unit slip, not a calculation.

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MCQ 6CalculationPractice

Two photodiodes P and Q are made from semiconductors of band gap 1.4 eV and 2.5 eV respectively. Both are reverse biased and illuminated by monochromatic light of wavelength 7.0 × 10² nm. Which of them registers a current above its dark current? (Take hc = 1.24 × 10³ eV nm.)

Show answer and why every option is right or wrong

Answer: A. A is correct. The photon energy is E = (1.24 × 10³ eV nm)/(7.0 × 10² nm) ≈ 1.8 eV, which clears P's 1.4 eV gap but falls short of Q's 2.5 eV gap, so only P generates pairs (condition from NCERT Class 12 Physics, Chapter 14, page 336).

Why B is wrong: B is wrong because it inverts the test: a wider band gap demands more photon energy, not less, so the 2.5 eV diode is precisely the one this photon cannot excite.

Why C is wrong: C is wrong because it assumes any illumination yields a photocurrent; at 1.8 eV the photon is below Q's gap and generates nothing in it.

Why D is wrong: D is wrong because 1.8 eV does exceed 1.4 eV, so P responds; only Q is blind at this wavelength.

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MCQ 7Concept TrapPractice

Reverse bias is preferred when a photodiode is used to read light intensity. Which statement gives the reason?

Show answer and why every option is right or wrong

Answer: B. B is correct. NCERT Class 12 Physics, Chapter 14, page 336 argues from Δn = Δp: in an n-region the majority density already dwarfs Δn while the minority density does not, so the photo-effect is easier to measure as a fractional change in the minority-carrier-dominated reverse current.

Why A is wrong: A is wrong because sub-gap photons are not absorbed into pairs however wide the depletion region is; widening it does not relax the hν > E_g condition.

Why C is wrong: C is wrong because the band gap is a material property and is not shifted by the applied bias — and shifting the threshold is not what the reverse-bias argument is about.

Why D is wrong: D is wrong because the forward current does change slightly under illumination; the point is that the change is a negligible fraction of a large current, not that it is absent.

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MCQ 8Direct ApplicationPractice

The I-V characteristics of a photodiode are recorded at four illumination intensities I₁ < I₂ < I₃ < I₄. Which statement describes the family of curves correctly?

Show answer and why every option is right or wrong

Answer: C. C is correct. The illuminated photodiode is characterised in the reverse-bias region, where each curve is nearly independent of the reverse voltage while the reverse current scales with intensity, so the brightest illumination gives the largest current (NCERT Class 12 Physics, Chapter 14, page 336).

Why A is wrong: A is wrong because it describes a forward-biased junction; the photodiode's working characteristic is taken under reverse bias.

Why B is wrong: B is wrong because it is the intensity, not the reverse voltage, that fixes the current level here — that near-independence of voltage is exactly what makes the reading an intensity measurement.

Why D is wrong: D is wrong because it reverses the trend: more photons mean more pairs swept out by the junction field, so the reverse current grows with intensity.

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How do you solve a Photodiode question? A worked example

1
Given.
Band gap E_g = 2.0 eV. Incident wavelength λ = 5.0 × 10² nm. Constant hc = 1.24 × 10³ eV nm (taken as exact for this calculation). Diode reverse biased.
2
Required.
(a) Whether electron-hole pairs are generated at this wavelength. (b) The threshold wavelength λ_max.
3
Concept.
Photogeneration requires the photon energy to clear the band gap, hν > E_g. The pair is created in or near the depletion region, where the junction field separates it and the carriers add to the reverse current (NCERT Class 12 Physics, Chapter 14, page 336).
4
Formula.
Photon energy E = hc/λ. Threshold at hν = E_g, so λ_max = hc/E_g.
5
Substitution.
E = (1.24 × 10³ eV nm)/(5.0 × 10² nm). λ_max = (1.24 × 10³ eV nm)/(2.0 eV).
6
Calculation.
E = 2.48 eV. λ_max = 6.2 × 10² nm. The constant hc is a given exact value and does not contribute to the significant-figure count; the two data values (2.0 eV and 5.0 × 10² nm) each carry two significant figures, so both answers are quoted to two.
7
Final answer.
(a) 2.5 eV exceeds the 2.0 eV band gap, so pairs are generated and the reverse current rises above the dark current — a photocurrent flows. (b) λ_max = 6.2 × 10² nm.
8
Common trap.
Reading the wavelength comparison the wrong way round. A longer wavelength means a smaller photon energy, so any source beyond 6.2 × 10² nm stays undetected no matter how bright the lamp is made. The companion slip is assuming forward bias would give a bigger signal — it gives a larger current, but a far smaller fractional response.
9
Similar NEET-style question.
A photodiode of band gap 1.5 eV is illuminated in turn by 6.0 × 10² nm and 9.0 × 10² nm light of the same intensity. In which case does the reverse current rise above the dark value? (Photon energies 2.07 eV and 1.38 eV; only the 6.0 × 10² nm source clears the gap.)

What to remember before solving Photodiode questions

Key Fact

Photodiode

In the NEET syllabus; removed from current NCERT.

A special purpose p-n junction diode with a transparent window, operated under reverse bias. Photons with hν > E_g generate electron-hole pairs in or near the depletion region; the junction field separates them (electrons to the n-side, holes to the p-side), giving an emf. The photocurrent is proportional to the incident light intensity. It is used in reverse bias because the fractional change in the minority-carrier-dominated reverse current with light intensity is much larger, so easier to measure, than the fractional change in forward current.

-- NCERT Class 12 Physics (pre-2023 edition), Chapter 14, p. 487

Key Fact

Optoelectronic junction devices

In the NEET syllabus; removed from current NCERT.

Semiconductor diodes in which carriers are generated by photons (photo-excitation): (i) photodiodes, used for detecting optical signals (photodetectors); (ii) light emitting diodes (LED), which convert electrical energy into light; (iii) photovoltaic devices, which convert optical radiation into electricity (solar cells).

-- NCERT Class 12 Physics (pre-2023 edition), Chapter 14, p. 486

More in Electronic Devices: 3 exam traps and mistakes · 3 formulas · 3 question patterns from its other lessons.

Photodiode questions from past NEET papers

No question in our NEET 2020–2025 set targets this topic directly.

All 17 past-paper questions from Electronic Devices →

Sources

NCERT refs: Class 12 Physics Chapter 14, p.336

Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.

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