# Help in Thermal PAD and glue calculation (equivalence) in shared heatsink

Hello if I have 15 chips to dissipate 1W each (thus 15W in total), by math calculation, which of the next three possibilities is better:

1) using a unique shared heatsink @1.7K/W with 0.5mm thermal pad @1.5W/mK

2) using a unique shared heatsink @1.7K/W with 1.5mm thermal pad @8W/mK

3) using 15 heatsinks @25.6K/W each with 0.1mm thick thermal glue @0.9W/mK

By feeling the three performances appear pretty similar: one shared heatsink @1.7K/W seems slight better than the sum of the 15 heatsinks, but the thermal glue is slight better than the 8W/mK thermal pad.

Can we prove the three cases by exact calculation and formulas using MKS metric system please?

I need the thermal resistance model in term of K/W of each of the three mentioned cases (heatsink plus heat transfer material).

I think that calculation are independent from the surface area as it is just the sum of the top of each chip package and it is the same in all the three cases. In any case I give you: area of the top of the chip is 60mm^2 each.

Ambient temperature is fixed. No fans.

This is a real case. Thanks!

(by now don't consider compression ot the pad: I know it is not fully linear with the conducibility: if you have time considering an extra case of a 30% of reduction of the thickness of the pad due to the screws...)

In a thermal system, the path of most resistance (least conductivity) will be the biggest factor in how much heat will move through the system.

If you have plenty of cooling power, your limitation is going be getting it to the heatsinks.

1.5W/mK*0.0005m=0.00075W/K this is lower than the other pad:

8W/mK*0.0015m=0.012W/K

0.9W/mK*0.0001m=0.00009W/K which is the lowest conductivity (highest resistivity)

Regardless of the heatsink on go with the highest conductivity pad which would be the 8W/mK pad.

• The way you're using the formulae, the thicker a 0.01W/mK piece of insulator the better the heat sink. I'm not unfamiliar with the same way of thinking in a rush. W/mK actually comes from (W*m)/(m2*K). It's a very mean number. – Asmyldof Jan 17 '18 at 8:28
• What about using the 15 separate heatsinks? I need the thermal resistance model in term of K/W of each of the three mentioned cases (heatsink plus heat transfer material). – Gianluca G Jan 17 '18 at 8:40
• I think the last sentence should be: "So if you're putting the same heatsink on, go with the highest conductivity pad which would be the 8W/mK pad." Sorry to be pedantic, but it really jarred when I first read it. – DiBosco Jan 17 '18 at 8:49
• I don't think the PAD with 8W/mK is the best using the same heatsink.--------------RPad1= THICKNESS / ( AREA * K) ------------ Area = 0,01 m * 0,006 m = 0,00006 m2 ---------------- K= 1,5 W/mK------------ Thickness= 0,0005m------------- RPad1= 0,0005 / (0,00006 * 1,5)= 5,6 C/W ---------------- K= 8 W/mK-------------- Thickness= 0,0015m--------------- RPad2= 0,0015 / (0,00006 * 8)= 3,12 C/W ---------------- K= 0,9 W/m*K Thickness= 0,0001m Rglue= 0,0001 / (0,00006 * 0,9)= 1,85 C/W – Gianluca G Jan 17 '18 at 9:29
• Nobody can help to clarify if any error? – Gianluca G Jan 18 '18 at 8:59