Diodes

LED

A diode that gives off light when current flows forwards through it. The colour depends on the semiconductor material, and so does the forward voltage: about 1.8 V for red and about 3 V for blue or white.

Think of it as: A one-way valve that glows when water rushes through it.

Pins

A (anode, +)
: The longer leg. Goes towards the positive supply.
K (cathode, −)
: The shorter leg, next to the flat edge of the rim. On SMD LEDs there is a mark (often green) on this end.

Through-hole LEDs are 3 mm or 5 mm domes; SMD LEDs are tiny chips (0603, 0805) with a mark on the cathode end.

What to look for inside

  • A red LED. The voltage across it slowly rises to about 1.95 V and falls again.
  • Below about 1.6 V the barrier in the middle (dark band) holds the carriers apart: no current, no light.
  • Past the knee, electrons and holes rush into the middle and meet. Each meeting gives off one small packet of light (a photon): the red wiggles.
  • More current means more meetings per second, so a brighter LED.

The numbers that matter

SpecTypical
Forward voltage V_F
Use it to work out the series resistor.
1.8 V red · 2.1 V green · 3 V blue/white
Forward current I_F
Modern LEDs are bright at 2 to 5 mA. 20 mA is usually the maximum.
2 to 20 mA
Reverse voltage
LEDs are poor at blocking. More than about 5 V backwards can kill them.
≈ 5 V
Luminous intensity
How bright, measured straight ahead.
mcd

Where you’ll use it

  • Indicator lights: power on, status, errors.
  • Driven straight from a GPIO pin through a resistor, or through a transistor for more current.
  • Lighting: white LED strips, torches, displays.

Common mistakes

  • No series resistor. The current is limited only by luck, and the LED burns out.
  • Fitting it backwards. It simply stays dark (and can be damaged by more than 5 V).
  • Running LEDs in parallel off one resistor. They never share evenly; give each its own resistor.

Lessons that use it

Go deeper

R=Us−UFIFR = \tfrac{U_s - U_F}{I_F}

Series resistor: e.g. (5 V − 1.8 V) ÷ 10 mA = 320 Ω, use 330 Ω.

Ephoton=hcλ≈q UFE_{photon} = \tfrac{hc}{\lambda} \approx q\,U_F

Shorter wavelength (blue) means more energy per photon, which is why blue LEDs need a higher voltage.

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