Optical sources: LEDs and lasers

LEDs and laser diodes: bandgap and wavelength, spontaneous vs stimulated emission, output power, threshold, mode spacing, bandwidth and temperature effects.

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Why it matters

Every fibre-optic link, optical sensor, barcode scanner, laser rangefinder and spectrophotometer starts with a light source. Whether you choose an LED or a laser diode decides how much power reaches the fibre, how far and how fast you can send data, and how much the source costs and drifts with temperature. Instrumentation engineers must size drive currents, predict output power and wavelength, and understand why the spectrum of the source limits bandwidth.

Key ideas

Light from a p–n junction. In a forward-biased junction of a direct-bandgap semiconductor (GaAs, AlGaAs, InGaAsP), injected electrons and holes recombine and release photons of energy close to the bandgap: E_g ≈ hν = hc/λ, so λ(µm) ≈ 1.24/E_g(eV). Silicon and germanium have indirect bandgaps and are very poor emitters. The material and alloy composition therefore fix the wavelength: GaAs/AlGaAs for 0.8–0.9 µm, InGaAsP for 1.3 µm and 1.55 µm, the low-loss windows of silica fibre.

LEDs (spontaneous emission). Photons are emitted at random times, phases and directions, so LED light is incoherent, has a wide spectral width (about 30–60 nm at 850 nm, 50–150 nm at 1.3 µm) and a broad (Lambertian) radiation pattern. Surface emitters and edge emitters are the two structures; edge emitters couple more power into small fibres.

  • Internal quantum efficiency η_int is the fraction of recombinations that produce photons. Internal optical power P_int = η_int·(hc/λ)·(I/q).
  • Much of the light is trapped by total internal reflection at the chip surface, so the external efficiency is only a few per cent, and coupling into a fibre loses more.
  • Output power rises almost linearly with current — convenient for analog intensity modulation.
  • Modulation bandwidth is limited by the carrier lifetime τ: optical 3 dB bandwidth f_3dB = 1/(2πτ), typically tens to a few hundred MHz.

Laser diodes (stimulated emission). A laser needs (1) population inversion, obtained by heavy current injection into a thin active layer, (2) stimulated emission, in which a passing photon triggers an identical photon (same frequency, phase, direction and polarisation), and (3) optical feedback from a resonant cavity — in a Fabry–Perot laser, the cleaved crystal facets act as mirrors.

  • Below the threshold current I_th the device behaves like a weak LED; above it gain exceeds the cavity losses and output rises steeply: P = η_d·(hc/λq)·(I − I_th), where η_d is the differential (external) quantum efficiency. The slope dP/dI in W/A is the slope efficiency.
  • A Fabry–Perot cavity supports longitudinal modes spaced by Δλ = λ²/(2nL), so its spectrum shows several lines within a few-nm envelope. Distributed-feedback (DFB) lasers use a grating to select a single mode with linewidth well below 1 nm, which is needed for long-haul 1.55 µm links and WDM. VCSELs emit from the surface and are cheap for short 850 nm links.
  • Light is narrow in spectrum and beam, so far more power can be launched into a single-mode fibre, and direct modulation reaches several GHz.
  • Temperature sensitivity: I_th rises roughly as exp(T/T₀) (T₀ is the characteristic temperature), and the wavelength shifts with temperature, so laser transmitters use a monitor photodiode with automatic power control and often a thermoelectric cooler.

LED versus laser — summary. LEDs: cheap, robust, long life, simple drive, little temperature sensitivity, wide spectrum (so more chromatic dispersion), low coupled power, low bandwidth — used with multimode fibre over short distances. Lasers: narrow spectrum, high coupled power, high bandwidth, but costlier, threshold and temperature control needed, eye-safety precautions.

Formulas

λ = hc/E_g, λ(µm) ≈ 1.24/E_g(eV) h = 6.626 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s, E_g bandgap energy (J or eV), λ wavelength (m or µm).

ν = c/λ ν optical frequency (Hz).

P_int = η_int·(hc/λ)·(I/q) P_int internal LED optical power (W), η_int internal quantum efficiency, I drive current (A), q = 1.602 × 10⁻¹⁹ C.

f_3dB = 1/(2πτ) LED optical 3 dB modulation bandwidth (Hz); τ minority-carrier lifetime (s).

P = η_d·(hc/λq)·(I − I_th) for I > I_th P laser output power (W), η_d differential quantum efficiency, I_th threshold current (A). hc/(λq) in volts equals 1.24/λ(µm).

Δλ = λ²/(2nL) Δλ longitudinal mode spacing (m), n refractive index of the active region, L cavity length (m).

Worked examples

Example 1 (standard). A GaAs LED (E_g = 1.43 eV) has internal quantum efficiency 0.6 and is driven at 50 mA. Find the emission wavelength and internal optical power.

  1. λ = 1.24/E_g = 1.24/1.43 = 0.867 µm (867 nm).
  2. Photon energy in volts: hc/(λq) = 1.24/0.867 = 1.43 V.
  3. P_int = η_int × (hc/λq) × I = 0.6 × 1.43 V × 0.050 A = 0.0429 W. Answer: λ ≈ 867 nm, P_int ≈ 42.9 mW (the power leaving the chip is much smaller because of internal reflection).

Example 2 (GATE level). A 1.31 µm Fabry–Perot laser has I_th = 20 mA, differential quantum efficiency 0.4, cavity length 300 µm and refractive index 3.5. Find the output power at 50 mA and the longitudinal mode spacing.

  1. hc/(λq) = 1.24/1.31 = 0.947 V.
  2. P = η_d × 0.947 V × (I − I_th) = 0.4 × 0.947 × 0.030 = 0.01136 W = 11.4 mW.
  3. Δλ = λ²/(2nL) = (1.31 × 10⁻⁶)²/(2 × 3.5 × 300 × 10⁻⁶) = 1.716 × 10⁻¹²/2.1 × 10⁻³ = 8.17 × 10⁻¹⁰ m. Answer: P ≈ 11.4 mW; Δλ ≈ 0.82 nm.

Example 3 (LED bandwidth). An LED has a carrier lifetime of 5 ns. Find its optical 3 dB bandwidth.

  1. f_3dB = 1/(2πτ) = 1/(2π × 5 × 10⁻⁹).
  2. = 3.18 × 10⁷ Hz. Answer: about 31.8 MHz.

Common mistakes

  • Using the drive current alone for laser power; only the current above threshold produces stimulated output.
  • Forgetting to convert eV to joules, or using λ in nm with the 1.24 rule (it gives µm).
  • Claiming laser light is attenuated less in fibre because it is coherent. Fibre loss depends on wavelength, not coherence; lasers win on coupled power, spectral width (less dispersion) and speed.
  • Treating silicon as an LED material. Indirect bandgap makes it inefficient.
  • Confusing optical and electrical 3 dB bandwidth of an LED; the electrical one is lower (f_3dB,elec = f_3dB,opt/√2 for a single-pole response).

For GATE IN

  • Wavelength from bandgap and vice versa; photon energy and frequency.
  • LED internal/external power and quantum efficiency; laser output above threshold and slope efficiency.
  • Mode spacing of Fabry–Perot lasers.
  • Conceptual comparisons: spontaneous vs stimulated emission, LED vs laser spectral width and modulation speed.

Quick check

  1. What wavelength does a 0.8 eV bandgap emit?
  2. A laser has slope efficiency 0.3 W/A and I_th = 15 mA. What is the output at 35 mA?
  3. Name the three requirements for laser action.
  4. Why is silicon not used for LEDs? Answers: 1. 1.55 µm. 2. 6 mW. 3. Population inversion, stimulated emission, optical feedback (resonant cavity). 4. It has an indirect bandgap, so radiative recombination is rare.

Try answering each one aloud before you open it.

  1. 1.What is an LED and how does it work as an optical source?Concept

    An LED, or Light Emitting Diode, is a semiconductor device that emits light when an electric current passes through it. It works as an optical source by converting electrical energy into light energy through the process of electroluminescence. When electrons recombine with holes within the semiconductor material, photons are emitted, producing light. LEDs are widely used due to their efficiency, long lifespan, and ability to emit light in various colors.

  2. 2.Explain the basic principle of operation of a laser as an optical source.Concept

    A laser needs three things: population inversion (more carriers in the upper state than the lower, obtained in a laser diode by heavy forward current), stimulated emission (an incoming photon triggers an identical photon with the same frequency, phase and direction) and optical feedback from a resonant cavity, such as the cleaved facets of a Fabry–Perot diode. Above the threshold current, round-trip gain exceeds cavity loss and output rises steeply and linearly with current. The result is narrow-spectrum, directional, coherent light that couples efficiently into single-mode fibre.

  3. 3.What are the main differences between LEDs and lasers as optical sources?Concept

    An LED relies on spontaneous emission: incoherent light, spectral width of tens of nm, a broad Lambertian beam, no threshold and a nearly linear P–I curve. A laser diode relies on stimulated emission in a cavity: narrow linewidth (sub-nm for DFB), a narrow beam, a threshold current and much higher coupled power and modulation bandwidth (GHz versus about 100 MHz). LEDs are cheaper, longer-lived and less temperature-sensitive, so they suit short multimode links and sensors; lasers suit long-haul, high-rate and single-mode systems.

  4. 4.Why are lasers preferred over LEDs in long-distance optical communication systems?Application

    A laser launches far more power into a fibre, especially single-mode fibre, because its beam is narrow, so the loss budget allows longer spans. Its narrow spectral width greatly reduces chromatic dispersion, the pulse spreading that limits bit rate × distance with broad-spectrum LEDs. It can also be modulated directly at several GHz. Fibre attenuation itself is set by wavelength, not by coherence, so the advantage is in coupled power, dispersion and speed.

  5. 5.How does temperature affect the performance of LEDs and lasers?Application

    Rising temperature shrinks the bandgap, shifting emission to longer wavelengths, and increases non-radiative recombination, so LED output falls gradually. In a laser diode the threshold current rises roughly as exp(T/T₀), so at a fixed drive current the output can drop sharply, and the wavelength drifts (and can mode-hop in Fabry–Perot lasers). Laser transmitters therefore use a back-facet monitor photodiode for automatic power control and often a thermoelectric cooler to hold temperature and wavelength.

  6. 6.Calculate the wavelength of light emitted by an LED with a bandgap energy of 2 eV.Numerical

    To calculate the wavelength (λ) of light emitted by an LED, use the formula: λ = hc / E, where h is Planck's constant (6.626 × 10^-34 J·s), c is the speed of light (3 × 10^8 m/s), and E is the energy in joules. First, convert the bandgap energy from electron volts to joules: 2 eV = 2 × 1.602 × 10^-19 J = 3.204 × 10^-19 J. Then, λ = (6.626 × 10^-34 J·s × 3 × 10^8 m/s) / 3.204 × 10^-19 J = 6.2 × 10^-7 m or 620 nm.

  7. 7.A laser emits light at a wavelength of 1550 nm. Calculate the frequency of this light.Numerical

    To calculate the frequency (f) of light, use the formula: f = c / λ, where c is the speed of light (3 × 10^8 m/s) and λ is the wavelength. First, convert the wavelength from nanometers to meters: 1550 nm = 1550 × 10^-9 m. Then, f = 3 × 10^8 m/s / 1550 × 10^-9 m = 1.935 × 10^14 Hz.

  8. 8.What are the safety considerations when working with lasers?Application

    When working with lasers, safety considerations include avoiding direct eye exposure to the laser beam, as it can cause serious eye damage. It's important to use appropriate laser safety goggles that match the laser's wavelength. Additionally, ensure that the laser is operated in a controlled environment to prevent accidental exposure. Proper signage and training are also essential to ensure that all personnel are aware of the potential hazards and safety protocols.

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