I'm trying to figure out how big of heatsink I'm going to need for my switching regulator, or if it needs one at all. The IC is LM2678-5 which is rated at ~87% efficiency for the maximum load I might have to drive with it. It's stepping down from up to 20V down to 5V (battery is rated at ~18) and the maximum current that I might have draw from it is 3A though that's highly unlikely. In the case that this happens though, how do I figure out the specs of any heatsink I might have to purchase? Is the power dissipated just 13% * voltage drop * current? That would be a whopping 5+ watts for the most extreme case. The datasheet indicates the following in terms of thermal resistance: T Package, Junction to Ambient 65 C/W(Note 5) T Package, Junction to Ambient 45 C/W (Note 6) T Package, Junction to Case 2 C/W Note 5: Junction to ambient thermal resistance (no external heat sink) for the 7 lead TO-220 package mounted vertically, with ½ inch leads in a socket, or on a PC board with minimum copper area. Note 6: Junction to ambient thermal resistance (no external heat sink) for the 7 lead TO-220 package mounted vertically, with ½ inch leads soldered to a PC board containing approximately 4 square inches of (1 oz.) copper area surrounding the leads. Would appreciate any help making sense of this. Thanks!