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Heatsink Calculator

Required heatsink thermal resistance and resulting junction temperature

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W

For a linear regulator: (Vin − Vout) × load current.

°C
°C
°C

Use the air temperature inside the enclosure.

°C/W
°C/W

Thermal paste ~0.1–0.5, silicone pad ~0.5–1.5 °C/W.

°C/W
°C/W

TO-220 ≈ 50–65 °C/W, TO-247 ≈ 40 °C/W.

Required heatsink θsa (max)

3.67 °C/W

Total allowed θja

5.67 °C/W

Junction temp with chosen heatsink

115 °C

Case temperature

92.5 °C

Heatsink temperature

85 °C

Junction temp with no heatsink

970 °C

Verdict

Chosen heatsink is adequate (10 °C below the design limit).
  • Design junction limit is 125 °C, so the whole thermal path may add at most 5.67 °C/W at 15 W.
  • Every 1 °C/W in the path adds 15 °C at this power; a good interface material is the cheapest improvement.

Temperature at each point of the thermal path (°C)

About the Heatsink Calculator

This heatsink calculator answers the two questions every power-electronics design runs into: how good a heatsink do I need, and how hot will the part actually get? Enter the power the device dissipates, its maximum junction temperature, the ambient temperature and the thermal resistances from the datasheet, and it returns the largest sink-to-ambient thermal resistance (θsa, in °C/W) that keeps the junction within limits.

It suits anyone cooling linear regulators such as the 7805 or LM317, MOSFETs, BJTs, audio amplifier ICs or power LEDs. Enter the rating of a heatsink you are considering to see the resulting junction, case and heatsink temperatures, and compare against running the part with no heatsink at all using its junction-to-ambient rating.

The model is the standard steady-state series thermal circuit: heat flows from junction to case, through the thermal interface (pad or paste) to the heatsink, and from the heatsink to the surrounding air. A safety margin below the absolute maximum junction temperature is built in because datasheet maxima are limits, not design targets; 20–25 °C is a common margin.

With the default inputs, the required heatsink θsa (max) is 3.67 °C/W. Change any value above to recalculate instantly.

How to use the heatsink calculator

  1. 1Work out the power the device dissipates (for a regulator, voltage drop × current).
  2. 2Enter the maximum junction temperature and a safety margin.
  3. 3Enter the worst-case ambient temperature inside the enclosure.
  4. 4Copy θjc and θja from the datasheet and add θcs for your pad or paste.
  5. 5Pick a heatsink rated at or below the required θsa and enter it to check temperatures.

Formula and method

θsa(max) = (Tj,max − margin − Ta) ÷ P − θjc − θcs · Tj = Ta + P × (θjc + θcs + θsa)

Heat flow through a device and heatsink behaves like current through series resistors: each thermal resistance (in °C per watt) produces a temperature rise equal to the power times the resistance. The allowed total resistance from junction to air is the available temperature rise (design junction temperature minus ambient) divided by the power.

Subtracting the junction-to-case resistance from the datasheet and the case-to-sink resistance of the interface material leaves the maximum sink-to-ambient resistance the heatsink may have. With a specific heatsink, the junction temperature is ambient plus power times the sum of all three resistances. Without a heatsink the datasheet junction-to-ambient value θja is used directly. Results are steady-state; short pulses heat the junction less.

P
Power dissipated in the device (W)
Tj,max
Maximum junction temperature (°C)
Ta
Ambient temperature (°C)
θjc
Junction-to-case thermal resistance (°C/W)
θcs
Case-to-sink (interface) thermal resistance (°C/W)
θsa
Sink-to-ambient thermal resistance of the heatsink (°C/W)
θja
Junction-to-ambient resistance with no heatsink (°C/W)

Worked examples

15 W MOSFET in a 40 °C enclosure

Designing for 150 − 25 = 125 °C, the total budget is (125 − 40) ÷ 15 = 5.67 °C/W. Subtracting θjc 1.5 and θcs 0.5 leaves 3.67 °C/W for the heatsink. A 3 °C/W sink gives a junction of 40 + 15 × 5 = 115 °C; without one the part would try to reach 970 °C.

7805 regulator dropping 12 V to 5 V at 1 A

The regulator dissipates (12 − 5) × 1 = 7 W. With a 125 °C limit and 30 °C air, the budget is 95 ÷ 7 = 13.57 °C/W, leaving 8.07 °C/W for the heatsink after θjc and θcs. An 8 °C/W heatsink holds the junction at 124.5 °C — right at the limit.

2 W in a TO-220 with no heatsink

With no heatsink the junction reaches 25 + 2 × 62 = 149 °C — under the 150 °C maximum but well above the 125 °C design target. A small clip-on heatsink of 46.5 °C/W or better restores the margin.

Frequently asked questions

What does heatsink thermal resistance (°C/W) mean?+

It is how many degrees the heatsink rises above ambient for every watt it dissipates. A 5 °C/W heatsink carrying 10 W runs 50 °C above the air around it, so a lower number means a better heatsink.

How do I find θjc and θja?+

Both are in the thermal information table of the device datasheet. θjc (junction to case) applies when a heatsink is fitted; θja (junction to ambient) applies to the bare package standing in still air.

How much does thermal paste or a pad matter?+

The interface typically adds 0.1–0.5 °C/W with thin paste and 0.5–1.5 °C/W with silicone pads or mica insulators. At 20 W, a 1 °C/W difference is a 20 °C difference at the junction.

Why use a safety margin below the maximum junction temperature?+

Datasheet Tj max is an absolute limit, and a widely used rule of thumb is that semiconductor lifetime roughly halves for every 10 °C rise. Designing 20–25 °C below the limit covers datasheet tolerances, hot days and dusty heatsinks.

Does a fan change the heatsink rating?+

Yes. Heatsink θsa is usually specified for natural convection; forced airflow can cut it by half or more. Manufacturers publish curves of thermal resistance versus air velocity.

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