From Curuxa

This is a middle-power illumination LED (rated at 3V, 50-100mA) controlled from a digital output pin. It's designed for being powered at 5V.

Electronic circuit


Bill of materials:

  • 1x 3V, 50-100mA LED. WARNING: these are not common LEDs. Make sure yours are rated at 3V, around 0.5W
  • Perforated board of at least 5x5 holes. Individual pads, not copper strips
  • 1x 2N3904, or similar BJT NPN transistor
  • 1x 1kΩ 1/4W resistor
  • 2x 1N4007 diodes, or any other diode with same voltage drop
  • 3x Straight female headers (usually sold as 40-pin strips)
  • Some thin wire
  • Some heat shrink tubing

This module has a small circuit embedded in the wires. from one side you connect it to the Main Board and from the other side two wires go to the LED.


Layout: top view
Layout: bottom view

Printable version: Single, Double

How it works

The LED will light when the microcontroller data pin is set as a digital output at logic "1".

The LED will not light when the data pin is set at a logic "0", or when it's working as a digital or analog input.

The LED is powered through a transistor because it requires much more current than the microcontroller can source, so we can use a digital output pin from the microcontroller to saturate the transistor, which will allow the current to flow though the LED.

If you power the LED directly at 5V it would burn because it's rated at 3V. This LED requires much more current than common LEDs, so if we put a common resistor (1/4W) in series with the LED, that resistor would burn too. The solution is to put two diodes in series with the LED so there is a voltage drop on them, and the LED is actually powered at a lower voltage without using huge resistors and avoiding a huge waste of energy as heat.

If you want to use more powerful LEDs you might need to modify this circuit.




You can test LTIL-A using LTIND-A examples, they are fully compatible.

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