# LED Resistor, choosing resistor power

I want to ask, when choosing resistor for a LED, how to choose power of the resistor(for example in SMD applications).

For example:

• 1 LED 2.0V FV 20mA
• Resistor 65 Ohm
• Power supply 3.3V

With this configuration I've 20mA through resistor. Now, LED dissipate 0.04W. How I can choose correctly resistor power rating?

I should refer to LED power of 0.04W and choose a resistor of at least 1/20W? Or refer to current through line(3.3V*0.02A = 0.066W) and choose a resistor of at least 1/10W?

Should I apply a safty factor?

• Power = Current*Voltage. If you know the current and voltage drop on the resistor you know the power. – Eugene Sh. Oct 29 '15 at 17:15
• You know the current through the resistor, and you know the resistance across it. What happens if you multiply the two? And what when you multiply that by the current again? – Nick Johnson Oct 29 '15 at 17:16
• You need the power dissipated in the resistor, so the calculation is Power = (V across resistor) times (Current through resistor). The resistor doesn't care about the power dissipated in the rest of the circuit. – Peter Bennett Oct 29 '15 at 17:26

You know the current through your resistor and the value of your resistor, so you can calculate the power dissipated by the resistor (Ohm's law strikes again):

$$P = I^2 R$$

So in your case that would be 0.026 W or 26 mW.

However when choosing the power rating, I'd not consider just the normal operating point but the worst case.

So the worst case would be, your power source failing with a higher voltage and the LED failing as a short circuit, in that case the resistor is dissipating the most power.

Well typically you don't consider both happening at the same time, but you might do that calculation as well.

If the LED fails as a short the current will increase and the resistor will dissipate (Ohm's law yet again):

$$P = \frac{U^2}{R}$$