Yes, but...
Heat conduction A further advantage of BGA packages over packages with discrete leads (i.e. packages with legs) is the lower thermal resistance between the package and the PCB. This allows heat generated by the integrated circuit inside the package to flow more easily to the PCB, preventing the chip from overheating.
When the PCB is so close to the bottom of the BGA package, whatever heat is produced is radiated directly to the PCB. Assuming a fairly uniform case temperature (possibly a bad assumption) by conduction, the radiated heat out the bottom of the BGA case has to go somewhere. It can't accumulate or it would just continue to heat up until it melts. So, it heats the PCB.
I think you'll find that unless there's a hidden insulator somewhere in the package, the bottom case temperature will be fairly close to the top case temperature. If it were otherwise, the case would distort or in extreme cases, crack. I can work out the exact numbers, using the thermal resistance, if you give me the exact case style and dissipation in watts.
How much is "relatively"? Most (not all) BGA arrays have the chip mounted on the base. For example, see Fig 2 the wire bonded example at:
The heat will be coming out of the base, which will be hotter than the lid due to some thermal resistance in the case. Others have the chip mounted on the top. These are easily identified by the epoxy blob or metal cover on the bottom PCB side of the BGA. See:
for Intel's packaging handbook. Also see 14.10 section for a little on thermal performance. There's a section on thermal package stress at:
See section 4.2.1 under "Stresses generated during a thermal excursion".
True. Heat removal is not 100% efficient. Think of temperature as the voltage across a string of resistors (thermal resistance). Crank up the input power and each resistor has more voltage across it. However, the ratio of the various voltages and temperatures remains constant as long as the thermal resistances don't change. That means that fairly small thermal resistances, such as between the heat sink and the case, are not going to see much of a temperature change for increase dissipation, while large thermal resistances, such as the heat sink to the air, are going to see a large increase.
Sure. But the difference in temperature is still what's bending the board and breaking the bonds. That's what my guess(tm) was causing the Nvidia video chip failures in many laptops. The chip was literally tearing itself away from the PCB because the board was bending.
There's another problem with your analysis. If you assume that the edges of the PCB are at room temperature, or at least at case temperature, then the temperature gradient across the PCB will remain fairly constant as you increase board heating. The result is just a larger heat affected zone, and no real improvement in cooling. It would be like putting a computah inside a plastic bag (for waterproofing) and dumping it inside a bucket of cold water. The case will be very cool, but the CPU will still burn up inside.
True. If the thermal resistance between the chip and every component of the thermal circuit path were zero, and the thermal mass of the air were assumed to be infinite (a really bad assumption), then the chip, heatsink, case, and air temperature would all be the same. However, if any or all of these exhibit any thermal resistance, there will be a temperature difference across it.