"Re <<- larger contact area with heat sink, helps conductivity. But isn't there some loss in conductivity from the silicon to integrated heat spreader that offests it?>>
"Yes, larger contact area is not really an advantage. Because you are effectively adding a new resister in your thermal circuit."
No, you have it backwards. The contact area between the silicon and the external world is indeed a resistor, but it does not "add" to the overall thermal resistance between the silicon and the outside world: all thermal paths, regardless of their resistance, "help." Keep adding more resistive paths, in parallel, and things get better. (Adding paths in series is a different matter, but in this scenario there is only one intermediate link between silicon and substrate: the contact area. No one is talking about adding more series links. The alternative to increased contact area is "no contact," or an air gap, which is definately much worse.)
To see this, switch from thinking of resistance and instead think of its inverse form, conductivity. (Or go ahead and compute the form 1/R = 1/ (1/R1 + 1/R2 + ...).)
If water is leaking from a bucket through a couple of holes, punching a new hole does not reduce the overall flow, even if the resistance of the hole is considerable. (The above equation is how we compute overall resistance given a lot of various holes and their associated resistances...)
In this case, the area of the contact between die and substrate/heatsink is like the size of a hole in a bucket. Making the hole larger increases the water leakage rate.
Which, for a heat sink, is what we want to do.
The heat flow, usually measured in watts/cm^2 or equvalent MKS terms, is proportional to thermal conductivity times temperature difference. And thermal conductivity is proportional to the overall cross-sectional area of a conductive path (and other things, including the bulk thermal conductivity).
--Tim May |
