I know that PCB with good copper area will dissipate much more heat than air surrounding the resistor. However, the datasheet for Yageo resistors is vague about whether the dissipation ratings take into account PCB or just air. I am planning to drive this one close to maximum and it got me wondering. Can I give it 3W instead of rated 1W if it is attached to copper layers on both sides with big vias? As far as the current goes, it will be 1/5 of the rated current at 3W.
PS I know the resistance will drift upwards but it is not an issue in this case (as long as not downwards). I just worry about it not blowing up.
Best Answer
The PCB is the heatsink for most of the power, so the actual maximum ambient temperature for a given power dissipation will vary greatly with the nature of the substrate and what else is mounted nearby. For example, consumer paper-base phenolics are worse than epoxy-glass which are much worse than ceramic or aluminum-core substrates.
There is also the factor of "hot spots" in the resistor construction due to current crowding caused by trimming cuts in the resistive film. Conservative designs may avoid using a resistor above 30-40% of rated power. If you've got a situation where the power is being stressed you'll need to take a lot more care.
Here is a document from Koa Speers on resistor derating. And another which uses thermal camera measurements of terminal temperatures to calculate derating. Note that they do not recommend going over 100% of rated power under any conditions. If you do, you are on your own. And this similar note.
You can also refer to this Yageo document on mounting which illustrates effects of substrates and other mounting factors:
Edit: As pointed out in comments, the Vishay link pointed to by @Cursorkeys mentions the standard conditions for measurement EN 140400 said (in that document, not the standard, which I do not have):
.. which are fairly optimistic conditions for the resistor