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For example, I have a cicuit of single resistor R = 2k, apply with Vdc = 18V. The Power of the resistor is about 162mW. Should I pick a resistor with higher Power rating (Eg 0.25W 0.5W) or lower power rating (Eg 0.1W).

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  • \$\begingroup\$ come with a requirement. Resister wattage is determined by current handling, tell where is the resister is used/circuit etc. \$\endgroup\$
    – user19579
    Oct 31, 2018 at 10:53

3 Answers 3

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Since 0.1 < 0.162 it would be a bad idea to pick a resistor that is not rated for the actual nominal dissipation.

For high reliability, or if the environmental conditions exceed what is in the datasheet for the rated power, you should derate the part. So 0.25W or 0.5W would be better (either would be acceptable in many situations, 0.5W would likely be larger and more expensive but probably a bit more reliable).

Or you could change the circuit to use less power. For example, if the 2K is a dropping resistor for an LED, use a more efficient LED and increase the resistor. Or use a switching power supply to reduce the 18V to 5V and then the resistor might only be required to dissipate 20mW -- (5V-3V)*10mA.

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The power rating strongly depends on the removal of heat.

Very short leads, soldered to large regions of chilled copper foil, are a good approach.

Forced air cooling (some power-resistors, 5 watts and up, have metal cases with fins) also is a good approach.

If forced air cooling, is the resistor located in the airstream near the air intake where the moving air is coolest?

Can we think about using copper foil to cool resistors, whether leaded or SurfaceMount? Sure.

schematic

simulate this circuit – Schematic created using CircuitLab

Now how much are we stressing your resistor? If you have large regions of copper soldered to the resistor, using the 49 degree Centigrade per Watt numbers we computed here, you'll only have 49 * 0.16 = 7 degree Centigrade heat rise.

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You have to go for higher power resistor..

every resistor along with resistance value, power is the most important factor.

P = I^2 * R

selected resistor power > P;

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