← Blocks

Select an appropriate resistor for an LED

Driving Loads

Overview

Calculate the series resistor value for an LED using:

R=VsuppVfILEDR = \frac{V_\text{supp} - V_\text{f}}{I_\text{LED}}

You need three values from your design:

Circuit description

LED series resistor circuit

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

Gotchas

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

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

Vsupp=12VVf=2.5VILED=10mAR=122.5V0.01A=950ΩP=I2R=0.012×950=95mWV_\text{supp} = 12 \text{V} \\[0.1cm] V_\text{f} = 2.5V \\[0.1cm] I_\text{LED} = 10 \text{mA} \\[0.5cm] R = \frac{12 - 2.5 \text{V} }{0.01 \text{A}} = 950 \Omega \\[0.5cm] P = I^2 R = 0.01^2 \times 950 = 95\text{mW}

For this example, 95 mW exceeds the rating of a 0402 (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.