Heatsink calculator

Enter the power the component dissipates, its maximum junction temperature, the ambient temperature, the junction-to-case resistance from the datasheet and the resistance of the thermal pad or paste. The calculator gives the largest heatsink thermal resistance you can use, and for the heatsink you enter, the junction, case and heatsink temperatures, the safety margin and the most power it can handle.

Runs in your browser: nothing you enter is sent anywhere.

How to use it

Enter the power the part turns into heat, in watts.
Enter Tj max and θjc from the datasheet; you can also enter a lower value to keep a safety margin.
Enter the ambient temperature and the interface resistance θcs of the pad, paste or insulator.
Enter the θsa of a heatsink you are considering to check it.

How it works

Heat flows from the junction to the case, through the interface into the heatsink and into the air. The resistances add up like resistors in series:

Tj = Ta + P × (θjc + θcs + θsa)

Solving for the heatsink gives the largest allowed value:

θsa = (Tj max − Ta) ÷ P − θjc − θcs

The case temperature is Ta + P × (θcs + θsa) and the heatsink temperature Ta + P × θsa. The most power the chosen heatsink can handle is (Tj max − Ta) ÷ (θjc + θcs + θsa).

Worked example

A regulator dissipates 10 W; Tj max = 125 °C, Ta = 40 °C, θjc = 1.5 °C/W, a pad with θcs = 0.5 °C/W.

θsa ≤ (125 − 40) ÷ 10 − 1.5 − 0.5 = 6.5 °C/W
With a 5 °C/W heatsink: Tj = 40 + 10 × 7 = 110 °C, 15 °C below the limit
The case is at 95 °C, the heatsink at 90 °C; the setup can take up to 12.14 W

Tips

Heatsink ratings depend on airflow. The same heatsink has a much lower θsa with a fan than in still air; use the figure for your conditions.
Insulating pads add resistance. Take θcs from the pad or paste datasheet; at high power even a fraction of a degree per watt matters.
Use the worst case: maximum power, highest ambient temperature.

Frequently asked questions

How do I calculate the heatsink thermal resistance I need?

θsa = (Tj max − Ta) ÷ P − θjc − θcs. For 10 W, Tj max 125 °C, 40 °C ambient, θjc 1.5 °C/W and a 0.5 °C/W pad, θsa = 85 ÷ 10 − 2 = 6.5 °C/W or lower.

Is a lower or higher °C/W better?

Lower. The thermal resistance is the temperature rise per watt, so a 2 °C/W heatsink runs cooler than a 5 °C/W one with the same power.

What ambient temperature should I use?

The air temperature around the heatsink inside the enclosure on the hottest day, not the room temperature. Inside a closed box it can be well above the room temperature.

What if the required value is negative?

Then even a perfect heatsink is not enough, because θjc + θcs alone already lets the junction pass its limit. Reduce the power, use a part with lower θjc or spread the heat over several parts.

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All tools in this category: Unit converters ·Updated November 4, 2026