# Am I misunderstanding steady state power disspation calculations?

I am looking at the Littlefuse 5KP series. It says on the datasheet: I am trying to understand how they arrive at $$\P_D\$$ = 8.0W.

Given thermal resistances of 40°C/W or 8.0°C/W, and 100°C temperature difference, should $$\P_D\$$ not be either 2.5W or 12.5W?

Oddly enough for the Littlefuse SMBJ series, they use the thermal resistance junction to lead to calculate $$\P_D\$$. (The equivalent of 12.5W above.)

Where does the discrepancy come from? Is it due to the difference of leaded vs SMD components? Inconsistency in the datasheet? Or am I misunderstanding something?

Also: What is the point of specifying $$\P_D\$$ "on an infinite heat sink" for a round component with leads? Where is the heat sink supposed to go? Why is this seemingly purely theoretical value useful at all?

There is probably a thermal limit for a temperature gradient due to coefficients of expansion of conductor and epoxy insulator from a 64'C rise from leads to junction in an infinite heatsink that limit this specification. This is their safe limit for mechanical stress.

The key words are "infinite heatsink" so thermal resistant Rjl junction to lead applies. This could be an oil tank in theory. But in practical terms a design could provide a chassis clamp around the body that will add thermal resistance and thus Pmax derating.

This heatsink Rca would be added to the $$\R_{θJL}=\$$ 8'C/W then *8W=64'C rise plus an ambient spec range must limit this rating.

Although the Tmax is 175'C absolute max, this would affect life expectancy, so a part rated for reasonable MTBF will provide a rated Pmax. for reliable operation.

Thermal resistance junction-lead $$\R_{th,JL}\$$ is the dominating heat path. Actually $$\R_{th,JL}\$$ is in parallel to the thermal resistance junction-ambient $$\R_{th,JA}\$$, so the relevant thermal resistance might even be 6.67 K/W.