LED

An LED converts electrical current directly into light. It is a p-n junction diode made from compound semiconductors where recombining electrons and holes release photons instead of heat. The color of the emitted light depends on the bandgap energy of the material, which is why a red LED and a blue LED need different forward voltages.

Core Concept

An LED is a forward-biased p-n junction made from direct-bandgap semiconductors. When electrons from the n-side recombine with holes from the p-side near the junction, the energy is released as a photon. The photon's energy equals the semiconductor's bandgap energy Eg, which determines the wavelength and therefore the color of the emitted light.

Silicon and germanium have indirect bandgaps, so recombination releases heat rather than light. LEDs use compound semiconductors. Red LEDs use GaAsP (gallium arsenide phosphide) with Eg ≈ 1.8 eV and V_F ≈ 1.8 V. Blue and white LEDs use InGaN (indium gallium nitride) with Eg ≈ 3.4 eV and V_F ≈ 3.4 V. The higher the bandgap, the shorter the wavelength and the higher the forward voltage.

The LED must always have a series current-limiting resistor to prevent thermal runaway. Unlike a Zener diode, an LED does not regulate its own current. Without a series resistor, even a small increase in supply voltage can push the current high enough to destroy the junction in milliseconds. A typical indicator LED operates at 10 to 20 mA, while high-brightness types like the Cree XP-L run at 350 mA to 3 A.

Key Equations

Photon energy and wavelength: E = h*f = h*c/lambda = Eg where h = 6.626e-34 J·s, c = 3e8 m/s.

Wavelength from bandgap: lambda (nm) = 1240 / Eg (eV)

Series resistor: Rs = (Vcc - V_F) / I_F where V_F is the LED forward voltage and I_F is the desired forward current.

Power dissipated in LED: P_LED = V_F * I_F

Power in series resistor: P_Rs = (Vcc - V_F)^2 / Rs

Given:
Supply: Vcc = 5 V
LED: green, V_F = 2.2 V, desired I_F = 15 mA = 0.015 A
Find series resistor Rs

Why this formula:
Series resistor must drop the voltage difference between supply and LED forward voltage at the required current.

Formula:
Rs = (Vcc - V_F) / I_F

Substitution:
Rs = (5 - 2.2) / 0.015
Rs = 2.8 / 0.015

Calculation:
Rs = 186.7 Ω → use 180 Ω (standard E12 value)

Actual current with 180 Ω:
I_F = (5 - 2.2) / 180 = 2.8 / 180 = 15.6 mA (acceptable)

Power in LED:
P_LED = 2.2 * 0.0156 = 34.3 mW

Power in Rs:
P_Rs = (2.8)^2 / 180 = 7.84 / 180 = 43.6 mW

Final Answer:
Rs = 180 Ω, I_F = 15.6 mA, P_LED = 34.3 mW, P_Rs = 43.6 mW

Exam Tip: The most tested LED question asks for the series resistor Rs = (Vcc - V_F) / I_F. Students often forget to subtract V_F and write Rs = Vcc / I_F, which gives a wrong (too-high) resistor and too-low current. Also remember: wavelength lambda = 1240 / Eg(eV) in nm. So a blue LED with Eg = 3.1 eV emits at 400 nm. This formula converts between energy and wavelength directly.

Key Properties

• Forward voltage V_F depends on material: Red GaAsP ≈ 1.8 V, yellow ≈ 2.1 V, green ≈ 2.2 V, blue InGaN ≈ 3.4 V, white (phosphor-coated blue) ≈ 3.2 to 3.6 V.
• Typical forward current for indicator LEDs: 10 to 20 mA. Maximum current (I_F_max) for 5 mm LEDs is typically 30 mA. Exceeding this destroys the junction.
• Luminous intensity increases with current up to a saturation point. Beyond that, the junction heats and efficiency drops (thermal droop).
• LEDs are reverse-breakdown sensitive. Peak reverse voltage (PIV) for most 5 mm indicator LEDs is only 5 V. They must not be used in AC circuits without protection.
• Efficiency (luminous efficacy) of modern white LEDs exceeds 200 lm/W, compared to 15 lm/W for incandescent bulbs. This is why LEDs replaced filament lamps.
• The emission wavelength of an LED shifts with temperature: about -0.1 nm/°C for GaN blue LEDs. This is why LED color looks slightly different when hot.
• Seven-segment displays like the common-cathode SA52-11 use eight LEDs (7 segments + decimal point) sharing a common ground rail.

Quick Revision

• LED emits light by radiative recombination in a direct-bandgap semiconductor.
• Photon wavelength: lambda = 1240 / Eg(eV) in nm.
• Series resistor: Rs = (Vcc - V_F) / I_F. Always required.
• V_F: Red ≈ 1.8 V, Green ≈ 2.2 V, Blue ≈ 3.4 V. Higher bandgap = higher V_F.
• Si and Ge have indirect bandgaps: no LED action. Use GaAsP, InGaN, GaN.
• Typical operating current: 10 to 20 mA for indicator LEDs.
• Reverse voltage rating of LEDs is low: typically 5 V. Do not apply reverse AC.
• Exam trap: Omitting V_F when computing Rs. Writing Rs = Vcc / I_F instead of Rs = (Vcc - V_F) / I_F gives a wrong answer and is the single most common LED mistake in GATE.