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To: Mani1 who wrote (77899)	4/22/2002 1:22:38 AM
From: tcmay	Read Replies (3) \| Respond to of 275872

"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

To: Mani1 who wrote (77899)	4/23/2002 6:40:34 AM
From: hmaly	Read Replies (1) \| Respond to of 275872

Mani RE...Yes, larger contact area is not really an advantage. Because you are effectively adding a new resister in your thermal circuit.<<<<<<<<<<<< Good explaination and I understand your anology. However barring AMD or Intel putting a HS directly on the die (which would be impossible to meet financial and differing conditions), what else could be done?

To: Mani1 who wrote (77899)	4/23/2002 7:13:41 AM
From: Bill Jackson	Read Replies (2) \| Respond to of 275872

Mani, >>Yes, larger contact area is not really an advantage. Because you are effectively adding a new resister in your thermal circuit.>> I think you are replacing the initial gradient from die to the traditional heat sink through thermal compound with a lower thermal resistance to the header(copper?). This gradient will be less steep, less delta T at the interface, and thus the spreader will be able to remove more heat than the heat sink to die method. Of course you then need a heat sink to the spreader. Since this spreader is made very flat, like the heat sink, and it is also a lot stronger than the die you will be able to squeeze the thermal paste into a much thinner layer by adding force that might break the die if used on a naked die. Thus we have another low thermal path. These two replace the fragile direct to die heat sink and is a lower thermal R path than the direct to die one. One assumes that the spreader/heatsink are matched to make sure this works well. Ever see and alphas? 2 bolts to attach the HS to the header. The two bolts and tight fitting holes prevent torque transfer when right tool is used to tighten them. Bill