What Are Light Emitting Diodes Made Of? From Semiconductor Materials to Bench Circuits

If you strip away the epoxy lens and look at the tiny die inside an LED, you are looking at a highly engineered compound semiconductor. To answer the fundamental question of what are light emitting diodes made of: they are primarily constructed from gallium, arsenic, phosphorus, nitrogen, and indium. The exact chemical recipe—such as Gallium Arsenide Phosphide (GaAsP) or Indium Gallium Nitride (InGaN)—dictates the material's bandgap. That bandgap directly determines both the color of the emitted light and the forward voltage ($V_f$) you must overcome to make it turn on.

Understanding the material science isn't just academic trivia; it is the reason a red LED turns on at 1.8V while a blue LED stubbornly refuses to glow until you push it past 3.0V. Here is how those materials translate to real-world bench design, biasing, and failure prevention.

The Material Science: Semiconductor Compounds and Forward Voltage

Silicon and germanium are great for logic gates and rectifiers, but they are indirect bandgap materials, meaning they release energy as heat rather than light. LEDs require direct bandgap compound semiconductors. By tweaking the ratios of elements in groups III and V of the periodic table, manufacturers tune the energy drop across the PN junction. When an electron falls across this gap to recombine with a hole, it releases a photon. A larger gap means a higher-energy (shorter wavelength) photon and a higher forward voltage requirement.

Symbol, Pinout, and Safe Default Part Numbers

On a schematic, the LED symbol is a standard diode triangle pointing toward a vertical cathode line, with two small arrows pointing away to indicate light emission. On the physical bench, identifying the pins is critical because reverse-biasing an LED beyond 5V will punch through the junction and destroy it.

Standard 5mm Pinout:

• Anode (+): The longer lead. Internally, it connects to the smaller post inside the epoxy dome.
• Cathode (-): The shorter lead. Internally, it connects to the larger 'anvil' base that holds the semiconductor die. The outer plastic lens also features a flat edge on the cathode side.

When stocking your lab, these are the reliable, high-availability default part numbers that offer consistent $V_f$ and luminous intensity:

• Red: Lite-On LTL-307EE ($V_f$ 2.1V, 20mA, 625nm)
• Green: Kingbright WP7113SGC ($V_f$ 2.2V, 20mA, 568nm)
• Blue: Cree C503B-BAN ($V_f$ 3.2V, 20mA, 470nm)

Biasing and Operation Regions

An LED is a non-linear device. It does not obey Ohm's law in a linear fashion. Once the applied voltage crosses the $V_f$ threshold, current increases exponentially. A voltage increase of just 0.1V can double the current, which is why you never drive an LED directly from a voltage source without current regulation.

To bias an LED for a specific job, use a series resistor for low-power indicator applications (under 30mA), or a constant-current (CC) driver IC for high-power illumination. According to Texas Instruments' LED driving fundamentals, resistors waste power as heat and offer poor regulation, but they are perfectly adequate for microcontroller status indicators.

Complete Application Circuit: 3.3V Logic Indicator

Let's design a status indicator driven directly by an ESP32 GPIO pin. The ESP32 datasheet specifies an absolute maximum GPIO current of 40mA, but the recommended safe continuous limit is 12mA per pin, with a total package limit to prevent brownouts.

Design Parameters:

• Source: ESP32 GPIO (3.3V logic high)
• LED: Blue InGaN (Cree C503B-BAN, $V_f = 3.0V$, $I_{max} = 20mA$)
• Target Current: 10mA (Safe for GPIO, plenty bright for an indicator)

Component Calculation:

1. Calculate voltage drop across the resistor: $V_R = V_{cc} - V_f = 3.3V - 3.0V = 0.3V$.
2. Calculate resistance using Ohm's Law: $R = V_R / I = 0.3V / 0.010A = 30\Omega$.
3. Select nearest standard E12 resistor value: 33$\Omega$.
4. Verify actual current: $I = 0.3V / 33\Omega = 9.09mA$ (Perfect).
5. Calculate resistor power dissipation: $P = I^2 \times R = (0.00909)^2 \times 33 = 2.7mW$. A standard 1/4W (250mW) through-hole resistor is more than sufficient.

Wiring: ESP32 GPIO Pin → 33$\Omega$ Resistor → LED Anode → LED Cathode → ESP32 GND.

Bench Scenario: When the 'Standard' Resistor Catches Fire

Theory is clean; the workbench is messy. Here is a real-world scenario that illustrates why understanding LED materials and power dissipation matters when scaling up from 5mm indicators to high-power illumination.

The Setup: A hobbyist wanted to add a bright work light to a 5V USB-powered project. They chose a 1W Cree XP-G3 White LED ($V_f$ nominal 2.9V, max current 350mA). Assuming the same resistor logic used for small LEDs would work, they calculated a series resistor for a 5V USB rail.

The Numbers:
$R = (5.0V - 2.9V) / 0.35A = 6\Omega$.
The builder didn't have a 6$\Omega$ resistor, so they used a standard 6.8$\Omega$, 1/4W resistor. The actual current was $2.1V / 6.8\Omega \approx 308mA$, which seemed fine.

The Outcome: Upon powering the circuit, the LED lit up brilliantly for about four seconds. Then, the 1/4W resistor began to smoke, the epoxy coating bubbled, and the USB port's overcurrent protection tripped, shutting down the hub.

What Went Wrong: Two critical failures occurred. First, the power dissipated by the resistor was $P = I^2 \times R = (0.308)^2 \times 6.8 = \mathbf{0.64W}$. A 1/4W (0.25W) resistor was subjected to more than double its rated power, causing thermal destruction.

Second, and more insidiously, as the high-power LED die heated up, its forward voltage dropped (a property of the InGaN material's negative temperature coefficient). As $V_f$ dropped from 2.9V to 2.7V, the voltage across the resistor increased, driving the current up past 350mA into thermal runaway.

The Fix: For any LED drawing more than 30mA, abandon the series resistor. Use a buck-converter constant-current driver like the PT4115 or a linear regulator like the LM317 configured as a current source to maintain exactly 350mA regardless of thermal $V_f$ shifts.

Failure Modes and Multimeter Testing

LEDs rarely fail randomly; they fail because of electrical or thermal abuse. Understanding how they break helps you diagnose dead boards quickly.

• Open Circuit (Bond Wire Melt): Caused by massive overcurrent. The microscopic gold wire connecting the die to the anode acts like a fuse and vaporizes. The LED looks perfectly fine but emits no light.
• Short Circuit (Junction Punch-Through): Caused by reverse voltage exceeding the 5V breakdown limit, or extreme ESD strikes. The semiconductor lattice melts into a conductive path. The circuit will function, but the LED won't light, and it will pull excess current.
• Lumen Depreciation (Phosphor Burn): Specific to white LEDs. Running the die too hot degrades the yellow YAG phosphor coating. The LED shifts to a harsh, bluish tint and dims permanently.

How to Test an LED with a Multimeter:

Don't use the continuity (beep) mode; it often doesn't supply enough voltage to overcome a blue or white LED's 3.0V bandgap. Use the Diode Test mode.

1. Set your multimeter to the diode symbol (usually shares a setting with continuity).
2. Place the red probe on the LED Anode (long leg) and the black probe on the Cathode (short leg).
3. Reading: A good red LED will display 1.600 to 2.200 (representing forward voltage) and glow faintly. A good blue/white LED might display 'OL' (Over Limit) if your meter only outputs a 2.5V test voltage, which is normal because the meter cannot overcome the 3.0V InGaN bandgap.
4. Reverse the probes (Black to Anode, Red to Cathode). The meter must read 'OL'. If it reads near 0.000 or beeps, the junction is shorted and the LED is dead.

By understanding what light emitting diodes are made of at the atomic level, you stop treating them as generic 'light bulbs' and start treating them as the sensitive, non-linear semiconductor devices they are. Respect the bandgap, calculate your thermal dissipation, and your circuits will run for decades.