royyang said:

from china ,my english is not good.

i want to make a heat sink use Aluminium for a SCR, the aluminium thermal conductivity is about 200W/m.k, if the size of the piece of aluminium is 100mm x 100mm x1mm( lenght x width x thickness), can i get the result as below:

thermal resistance= thickness/ thermal conductivity .area=0.001/200 x 0.01 x 0.01=0.05 C/W

but this result seems too small, anything i make wrong?

Thanks!

That actually isn't true as 100 mm is 0.1 m, not 0.01 m. The thermal resistance of a 100 mm x 100 mm x 1 mm sheet of aluminum is 0.0005 deg C/W. That number may look too good to be true as a good commercial heat sink may have a thermal resistance of 0.15 deg C/W, but this is because the thermal efficiency of the fins is less than 100%. The thermal resistance of an entire heatsink is as so:

R_hs = 1/(h * [area_of_HS_base + {number_of_fins * efficiency_of_fin * area_of_fin}])

where h = heat transfer coefficient of the fin.

That takes a decent amount of calculations to figure out (

see here) and there really isn't enough space nor a good way to enter equations into the forum here (such as LaTex) so I won't do them here. However, that site has the equations you need to calculate how good of a heatsink your designs might be, so plug some numbers into it and see what you get.