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?

  • \$\begingroup\$ Power = Current*Voltage. If you know the current and voltage drop on the resistor you know the power. \$\endgroup\$
    – Eugene Sh.
    Oct 29, 2015 at 17:15
  • \$\begingroup\$ 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? \$\endgroup\$ Oct 29, 2015 at 17:16
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ Oct 29, 2015 at 17:26

1 Answer 1


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):

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

With your numbers: 167 mW.

The next thing to consider is what is actually meant when a manufacturer tells you "this resistor can handle 250 mW".

This value tells you the resistor will heat up a certain amount if you put 250 mW through it. Say 60 K (I'm making that up right now). So your resistor will heat up quite substantially in case the LED fails. Is your device able to handle that safely? I don't know.

The power dissipation is often also temperature de-rated, so at elevated temperatures your resistor is only allowed to dissipate a lower power. Does this limit your application? I don't know.

  • \$\begingroup\$ I've another question that is related to resistors power. For example if I use a SMD 0R Jumper resistor, in the formula above I've P = 0 but sellers of this resistors specify a power for this resistor that go from 1/20W to 1/10W, how I can calculate in this case how much power a resistor handle if it's value is 0R? \$\endgroup\$
    – Singe
    Oct 30, 2015 at 14:19
  • \$\begingroup\$ @Singe that question was already answered here. A real 0R resistor doesn't exist right now in a package mountable on a PCB. \$\endgroup\$
    – Arsenal
    Oct 30, 2015 at 14:25

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