Kind of amazing that they would make a mistake like this, something that was obvious to random people on the forum.
While the die size determines the heat transfer at the silicon junction, the size and more importantly type of package is also important as the thermal energy must pass through the package as well and the lower the thermal conductivity the higher the core die temp is going to reach.
Yeah, I was just pointing out the size of the packages is not the major determining factor. A "chip scale" flip chip BGA package would probably be tiny but still be able to remove a lot of heat (from what I understand)
It is strange. It would be helpful to have more information from them on why they decided to forgo the heat pad. Maybe they think the plastic will be able to dissipate the heat.
The thermal conductivity of plastic is about 0.2-0.5 W/mK, compared to 200 for aluminum.
But, remember the 'm' stands for the thickness of the material, not the surface area. If the package material is very thin, it can still conduct a decent amount of heat.
If you have a surface area of 1cm (0.0001m2), thermal conductivity of 0.5 W/mK, 50C temperature difference, and 0.25mm thick layer of plastic, you should be able to move
-(0.5W/mK) * 0.0001m2 * (50K / 0.0025m)
=-0.5W/K*0.0001m*(20000K/m)
=-0.5W/K*0.0001*20000K
=-0.5W*0.0001*20000
=-1W.
So, just one watt of cooling using those parameters. Maybe 2 watts if you count both sides.
It's also possible that they could be using a type of plastic with a higher thermal conductivity as well.
And, it's also possible my math could be completely off. The units did cancel properly, but I could be doing that totally wrong
EDIT: A 12-inch wafer is 0.775mm thick, and a lqfp44 package is 1.2mm thick (not counting the pins). So the thickness of the packaging should be 0.21mm on either side. But there could be other materials in there as well. If so, that should increase the amount of heat able to be emitted package a great deal.
Anyway, we'll find out in a few days what's actually possible.