LED I–V

8 MCQs9-step worked example

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

LED I–V, explained for NEET

The LED's I-V curve looks like a silicon diode's, and that resemblance is where marks are lost. Two numbers on the curve are not silicon's: the forward knee and the reverse breakdown.

NCERT Class 12 Physics, Chapter 14 (Semiconductor Electronics), page 336, states it directly — the V-I characteristic of an LED is similar to that of a Si junction diode, but the threshold voltages are much higher and slightly different for each colour, and the reverse breakdown voltages are very low, typically around 5 V.

Both deviations trace to one cause. An LED is a heavily doped p-n junction in a transparent encapsulation. Forward bias injects electrons into the p-side and holes into the n-side; near the junction these excess minority carriers recombine with majority carriers, and the released energy leaves as a photon of energy close to the band gap. Colour therefore fixes the band gap, and the band gap fixes the forward voltage at which the curve turns up. Red light comes from GaAs₀.₆P₀.₄ with a band gap near 1.9 eV, so that LED turns on near 1.9 V — not 0.7 V. Shorter-wavelength colours need more. The same band gap leaves the reverse rating low, so a supply an ordinary rectifier diode shrugs off will destroy an LED.

Read the forward region carefully too. Below threshold, current and light are both negligible. Past the knee the curve is steep, which is why an LED is driven through a series resistor or a current source rather than straight from a voltage supply. Brightness is also not monotonic in current: NCERT records that intensity rises with forward current, reaches a maximum, then falls, so LEDs are biased near the current of maximum emitting efficiency.

In NEET this surfaces as a one- or two-step item: identify the bias, compare thresholds across colours, or size the series resistor. Watch the reverse half of the axis — the LED is a forward-bias emitter, unlike the photodiode and solar cell in the neighbouring topics.

Can you answer these LED I–V 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

In an LED, the emission of light is a direct result of

Show answer and why every option is right or wrong

Answer: C. Under forward bias, electrons are injected into the p-side and holes into the n-side; these excess minority carriers recombine near the junction and the released energy is emitted as a photon of energy close to the band gap (NCERT Class 12 Physics, Chapter 14, page 336).

Why A is wrong: A is wrong because reverse bias does not inject minority carriers into either side; the LED emits only when it is forward biased, and reverse operation risks breakdown instead of emission.

Why B is wrong: B is wrong because there is no filament. LED emission is a junction recombination process, not thermal incandescence — that difference is why the LED runs at low operating voltage and needs no warm-up time.

Why D is wrong: D is wrong because it describes the reverse process, photon absorption generating carriers, which is how the photodiode and solar cell in the neighbouring topics operate. The LED runs the energy conversion the other way.

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MCQ 2Easy RecallPractice

The forward V-I characteristic of an LED has the same general shape as that of a silicon junction diode. According to NCERT, how does its threshold voltage compare?

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Answer: A. NCERT Class 12 Physics, Chapter 14, page 336 states that the LED V-I characteristic is similar to a Si junction diode's except that the threshold voltages are much higher and slightly different for each colour.

Why B is wrong: B is wrong on both counts: LED thresholds are higher than silicon's, not lower, and they vary with colour because they are set by the emitting material's band gap.

Why C is wrong: C is wrong because 0.7 V is the silicon knee. Transplanting it onto an LED is the standard error in this topic — a red LED turns up near 1.9 V, well above 0.7 V.

Why D is wrong: D is wrong because the forward current and the light output are both negligible until the applied forward voltage reaches the threshold; the curve has a definite knee.

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MCQ 3Easy RecallPractice

NCERT notes that GaAs, with a band gap of about 1.4 eV, is used for infrared LEDs, while GaAs₀.₆P₀.₄, with a band gap of about 1.9 eV, is used for red LEDs. What minimum band gap must a semiconductor have if an LED made from it is to emit visible light?

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Answer: D. The visible range runs from roughly 0.4 μm to 0.7 μm, corresponding to photon energies of about 3 eV down to about 1.8 eV, so the emitting material needs a band gap of at least about 1.8 eV (NCERT Class 12 Physics, Chapter 14, page 336).

Why A is wrong: A is wrong because a 0.7 eV photon lies deep in the infrared; such an LED would emit no visible light at all.

Why B is wrong: B is wrong because 1.1 eV is silicon's band gap, which is below the visible floor. Silicon's number is being carried over from the earlier diode topics, where it belongs.

Why C is wrong: C is wrong because 1.4 eV is exactly the GaAs figure quoted in the stem, and the stem identifies GaAs as an infrared emitter — the value sits just below the visible threshold.

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MCQ 4Direct ApplicationPractice

A red LED and a blue LED are made from materials with band gaps of about 1.9 eV and about 2.7 eV respectively. Their forward I-V characteristics are recorded on the same axes. Which curve turns upward at the higher forward voltage, and for what reason?

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Answer: B. The LED threshold is set by the band gap of the emitting material, which is why NCERT records that LED threshold voltages differ slightly from colour to colour (NCERT Class 12 Physics, Chapter 14, page 336). The 2.7 eV material therefore has the higher knee.

Why A is wrong: A is wrong twice over: longer wavelength means lower photon energy, not higher, and the red LED has the smaller band gap, so its curve turns up at the lower voltage.

Why C is wrong: C is wrong because 0.7 V is the silicon knee. NCERT states explicitly that LED threshold voltages are much higher than silicon's and are not the same for every colour.

Why D is wrong: D picks the right LED for the wrong reason. Doping level is not what sets the knee; the band gap is, and the stem supplies both band gaps precisely so that comparison can be made.

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MCQ 5Direct ApplicationPractice

A blue LED with a forward threshold voltage of 3.0 V is connected in forward bias, through a suitable series resistor, across a single 1.5 V dry cell. What is observed?

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Answer: D. Below the threshold voltage the forward current on the LED's I-V curve is negligible, so there is essentially no recombination current to produce light (NCERT Class 12 Physics, Chapter 14, page 336).

Why A is wrong: A is wrong because current capability is irrelevant when the applied forward voltage never reaches the knee — the resistor cannot raise the voltage across the LED above what the 1.5 V cell provides.

Why B is wrong: B is wrong because the sub-threshold current is negligible; the LED's I-V curve is essentially flat below the knee, so there is no dim-glow regime to observe here.

Why C is wrong: C is wrong because the emitted photon energy, and hence the colour, is fixed by the band gap of the material, not by the drive voltage. An LED does not shift colour when underdriven.

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MCQ 6Direct ApplicationPractice

An LED is connected across a 12 V DC supply with its p-side to the negative terminal and its n-side to the positive terminal, with no protective component in the circuit. What is the expected outcome?

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Answer: A. NCERT Class 12 Physics, Chapter 14, page 336 warns that the reverse breakdown voltages of LEDs are very low, typically around 5 V, so care must be taken that high reverse voltages do not appear across them. A 12 V reverse bias is well past that.

Why B is wrong: B is wrong because reverse bias withdraws carriers from the junction rather than injecting minority carriers into it; there is no recombination current to emit light, at any reverse voltage.

Why C is wrong: C is wrong because a junction blocks reverse voltage only up to its breakdown rating. For an LED that rating is a few volts, far lower than for an ordinary rectifier diode — which is exactly the point being tested.

Why D is wrong: D is wrong because voltage regulation by reverse breakdown is the Zener diode's designed role, treated in a separate topic. An LED reaching breakdown is being destroyed, not regulating.

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MCQ 7Concept TrapPractice

A student drives an LED from a variable current source and records the light output while raising the forward current from a very small value upward. Which statement matches the behaviour described in NCERT, and its consequence for choosing the operating point?

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Answer: B. NCERT Class 12 Physics, Chapter 14, page 336 records that light intensity is small at small forward current, increases to a maximum as the current rises, and then decreases on further increase — so LEDs are biased for maximum light-emitting efficiency.

Why A is wrong: A is wrong because the intensity-current relation is not monotonic; past the maximum, more current yields less light, so driving the LED as hard as possible is counterproductive as well as damaging.

Why C is wrong: C is wrong because intensity does depend on forward current — NCERT states it is small when the forward current is small. Crossing the threshold starts the emission; it does not fix the brightness.

Why D is wrong: D is wrong in the initial region: intensity rises with current up to the maximum. A monotonic fall would make the LED brightest at the threshold, which is the opposite of what is observed.

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MCQ 8CalculationPractice

A red LED of forward threshold voltage 1.8 V and a blue LED of forward threshold voltage 3.0 V are connected in series with a resistor R across a 6.0 V DC supply. The forward current through the chain is to be 1.0 × 10¹ mA. What value of R is required?

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Answer: C. Both LED drops subtract from the supply: 6.0 V − 1.8 V − 3.0 V = 1.2 V across R, so R = 1.2 V ÷ 1.0 × 10⁻² A = 1.2 × 10² Ω. Each LED's drop is set by its own band gap, which is why the two thresholds differ (NCERT Class 12 Physics, Chapter 14, page 336).

Why A is wrong: A comes from dividing the full 6.0 V supply by the current, ignoring both LED drops. Past its knee an LED holds a near-fixed forward voltage, and that voltage is not available to the resistor.

Why B is wrong: B comes from subtracting only the red LED's 1.8 V and forgetting the blue LED in series. In a series chain every forward drop must be subtracted.

Why D is wrong: D comes from subtracting only the blue LED's 3.0 V and forgetting the red one — the same omission as B, made on the other element.

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How do you solve a LED I–V question? A worked example

1
Given
A red LED emits light of wavelength λ = 6.5 × 10² nm. It is to be operated at a forward current I = 1.5 × 10¹ mA from a DC supply V_s = 5.0 V through a series resistor R. Take hc = 1.24 × 10³ eV·nm.
2
Required
(i) The approximate forward threshold voltage of this LED. (ii) The series resistance R.
3
Concept
In an LED the emitted photon energy is close to the band gap of the emitting material, and that same band gap sets the forward voltage at which the I-V curve turns up. So the emission wavelength gives the threshold voltage directly. Past the knee the forward characteristic is steep, so the operating current must be fixed by an external series resistor rather than by the LED itself.
4
Formula
E = hc/λ; V_th ≈ E/e (numerically, E expressed in eV equals V_th in volts); R = (V_s − V_th)/I.
5
Substitution
E = (1.24 × 10³ eV·nm) ÷ (6.5 × 10² nm). Then R = (5.0 V − V_th) ÷ (1.5 × 10⁻² A).
6
Calculation
E = 1.24 × 10³ / 6.5 × 10² = 1.907 eV → 1.9 eV. Hence V_th ≈ 1.9 V. Voltage across the resistor = 5.0 − 1.9 = 3.1 V. R = 3.1 ÷ 1.5 × 10⁻² = 206.7 Ω.

Note on constants: hc and the elementary charge e are physical constants carried at full precision and do not limit the significant figures. The eV-to-volt step is a definitional conversion, also exact. The measured inputs — λ (2 s.f.), V_s (2 s.f.) and I (2 s.f.) — are what set the precision, so the answer carries 2 significant figures.
7
Final answer
V_th ≈ 1.9 V and R ≈ 2.1 × 10² Ω.
8
Common trap
Substituting the silicon knee of 0.7 V for V_th gives R = 4.3 ÷ 1.5 × 10⁻² ≈ 2.9 × 10² Ω; fitting that resistor would leave the LED running well below the intended current. Omitting V_th altogether gives 5.0 ÷ 1.5 × 10⁻² ≈ 3.3 × 10² Ω, the same error in the opposite direction of magnitude. Both come from treating the LED's forward characteristic as silicon's.
9
Similar NEET-style question
A green LED with a forward threshold voltage of 2.2 V is to carry 1.0 × 10¹ mA from a 9.0 V supply. What series resistance is required? *(Answer: (9.0 − 2.2)/1.0 × 10⁻² = 6.8 × 10² Ω.)*

What to remember before solving LED I–V questions

Key Fact

Light emitting diode (LED)

In the NEET syllabus; removed from current NCERT.

A heavily doped p-n junction that emits spontaneous radiation under forward bias; excess minority carriers recombine with majority carriers near the junction, releasing photons with energy equal to or slightly less than the band gap. Light intensity rises with forward current to a maximum, then falls. The V-I characteristic is similar to a Si diode's, but threshold voltages are much higher and slightly different for each colour; reverse breakdown voltages are very low (typically around 5 V). Visible LEDs need a band gap of at least 1.8 eV.

-- NCERT Class 12 Physics (pre-2023 edition), Chapter 14, p. 488

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.

LED I–V 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

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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