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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.

enter image description here

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    \$\begingroup\$ Vishay have an interesting app-note on this: link When I tried to decrease the surface temperature of a 2512 I found that massive thermal pours and vias only gave a modest improvement over the normal tracking I started with. \$\endgroup\$
    – Cursorkeys
    Commented May 10, 2021 at 13:29
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    \$\begingroup\$ The actually resistive material in an SMD resistor is only a thin layer on top of the ceramic substrate. The thermal resistance is mainly limited by the substrate interconnect, so probably it makes no big difference. \$\endgroup\$
    – tobalt
    Commented May 10, 2021 at 13:39
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    \$\begingroup\$ Consider that vias and large copper pours do increase the thermal load - but if no airflow, it will heat up to nearly the same temperature, reduced by increased radiative area effects only, and with a time delay due to the added mass. If you can put a small fan there, problem solved. \$\endgroup\$
    – rdtsc
    Commented May 10, 2021 at 14:39
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    \$\begingroup\$ @rdtsc If it's a still a closed system then a fan won't help either. And if it's an open system then the increased area should allow more heat loss through convection. I think loss by radiation is going to be a small component as the delta T isn't very large? \$\endgroup\$
    – Cursorkeys
    Commented May 10, 2021 at 15:53
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    \$\begingroup\$ Even if the board were in a sealed box, a fan would couple more thermal energy into the air and thus reduce the resistor temperature (the whole box warms up slightly however.) Could also try an insulating thermal pad and heat-pipe, etc. \$\endgroup\$
    – rdtsc
    Commented May 10, 2021 at 16:53

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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.

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You can also refer to this Yageo document on mounting which illustrates effects of substrates and other mounting factors:

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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):

2.3.3: FR4 base material 100 mm x 65 mm x 1.4 mm, 35 μm Cu-layer, pad/circuit path 2.0 mm width.

.. which are fairly optimistic conditions for the resistor

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  • \$\begingroup\$ Sounds contradictory to the vishay appnote posted by Cursorkeys, no ? Their conclusion is basically, the only meaningful to make resistors dump heat faster is by widening the solder connects. The thermal resistance of the PCB is anyway lower compared to intra-part resistances. \$\endgroup\$
    – tobalt
    Commented May 10, 2021 at 14:03

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