# Will the equation result in higher junction temperature if adding a heat sink?

The formula below demonstrates the relationship between power absorbed and thermal resistance. I feel confused.

Won't adding more heatsink result in higher junction temperature according to the formula below?

Since junction temperature is the product of power absorbed times thermal resistance of the system plus ambient temperature(air), adding more heatsink only increases the thermal resistance of the system, since thermal resistance is adds up like a series resistance in resistor.

I'm considering the case with only junction thermal resistance and casing thermal resistance as compared to the one with added thermal resistance due to a heatsink.

The case-to-ambient thermal resistance is in parallel with the case to heatsink + heatsink-to-ambient thermal resistance.

Normally the latter is so small (when a substantial heatsink is added) that we can ignore the former entirely. Also the heatsink specification probably includes some allowance for the expected package.

Eg. IRF530

• Junction to ambient: 62°C/W
• Case to heatsink: 0.5°C/W
• Junction to case 1.7°C/W

(calculated case to ambient 59.8°C/W)

Ohmite RA-T2X-38E heatsink to ambient: 3.9°C/W

Equation B in your exercise can be generalized. Ta is normally ambient, but it can be any known temperature. But, you must always find the total thermal resistance to a known temperature.