power problem (newbie here!)

[Acetronics2]

I don't understand your last sentence; what is : 10° C/W?

Hi,

It's the thermal resistance of the dissipator to use ... it's like Resistance's "Ohms" ... but for heat dissipation value.

That means it will heat at a rate of 10°C per dissipated Watt ... in your design, the temp of dissipator will be ( at blind sight ! ) ~ 25°C + (2 x 10°C) = 45°C

Alain