Select an appropriate resistor for an LED

Overview

Calculate the series resistor value for an LED using:

R = \frac{V_\text{supp} - V_\text{F}}{I_\text{F}}

Where;

$V_\text{supp}$ -> supply rail (GLV or logic voltage)
$V_\text{F}$ -> LED forward voltage (datasheet)
$I_\text{F}$ -> target forward current (datasheet; 1–5 mA is typical for SMD indicators)

Circuit description

LED series resistor circuit

A single resistor R is placed in series between the supply rail and the LED anode. Current $I_\text{LED}$ flows through R and the LED to ground. R drops the difference between $V_\text{supp}$ and $V_\text{f}$ , setting the current through the LED.

Gotchas

The resistor drops the excess supply voltage, dissipating it as heat. At higher supply voltages, the power dissipated in the resistor $P_R$ becomes significant - don’t assume a 0402 is fine.

For example, driving an LED from 12 V. Note a higher current for this example:

V_\text{supp} = 12 \text{V} \\[0.1cm] V_\text{F} = 2.5V \\[0.1cm] I_\text{F} = 10 \text{mA} \\[0.5cm] R = \frac{12 - 2.5 \text{V} }{0.01 \text{A}} = 950 \Omega \\[0.5cm] \text{Select }R=1k\Omega \text{, nearest common value} \\[0.5cm] P_R = V^2 / R = (12-2.5)^2 \div 1000 = 90\text{mW}

For this example, the 90 mW exceeds the rating of a 0402 package (62 mW) and sits right at the limit of a 0603 (100 mW). Use 0805 (125 mW rated) or larger for any 12V GLV-supplied indicator.