Extremely simple, low-defect lattice. Wonder what that would do in terms of acoustic phonon velocity. 😀
Glad we got the OP sorted out. I do encourage looking around for different insulator materials, though. Bill Waslo had a thread on Al2O3 (Alumina) insulators that performed quite well. Far, far more important than the marginal differences in heat sink paste.
The flux density at the bottom of the semiconductor package is the most severe, and this is where the thermal conductivity of graphene will serve the best use.
By bonding the semi to a copper heat spreader, you can increase the area of the thermal transfer by roughly double the thickness of the copper spreader. If for example, you have a 250 mil square die (IRF450) on a copper tab (60 mils), the effective area of transfer is 370 mils by 370 mils. That area will carry the bulk of the flux, and standard insulating goop will give a large thermal drop.
If you use graphene, and a 250 mil thick copper tab at least an inch square, you basically create a spread of 500 plus 370, or 870 by 870 for area. The 870 by 870 area can now use the insulating medium, and the thermal drop will scale as the area change.
So, the thermal rise will be (.37x.37)/(.87x.87), or .180 of the initial rise.
If for example, the IR die is pushing 100 watts and the die temperature increased 100 degrees with the simple insulator, using the graphene and a .25 thick copper spreader will drop the temp rise to 18 degrees.
John
By bonding the semi to a copper heat spreader, you can increase the area of the thermal transfer by roughly double the thickness of the copper spreader. If for example, you have a 250 mil square die (IRF450) on a copper tab (60 mils), the effective area of transfer is 370 mils by 370 mils. That area will carry the bulk of the flux, and standard insulating goop will give a large thermal drop.
If you use graphene, and a 250 mil thick copper tab at least an inch square, you basically create a spread of 500 plus 370, or 870 by 870 for area. The 870 by 870 area can now use the insulating medium, and the thermal drop will scale as the area change.
So, the thermal rise will be (.37x.37)/(.87x.87), or .180 of the initial rise.
If for example, the IR die is pushing 100 watts and the die temperature increased 100 degrees with the simple insulator, using the graphene and a .25 thick copper spreader will drop the temp rise to 18 degrees.
John
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