Thermal resistance behaves like a series circuit
Heat flowing out of a semiconductor passes through a chain of stages, each with its own resistance measured in degrees Celsius per watt. From the silicon junction to the package case is θjc, fixed by the package. From the case to the heat sink is θcs, set by the interface material and the mounting pressure. From the sink to the surrounding air is θsa, which is what you buy when you buy a heat sink. They add in series, exactly like resistors, and each one turns watts into a temperature rise across itself.
So the junction temperature is Tambient + P × (θjc + θcs + θsa). Rearranged for design, the sink you need is (Tj_target − Tambient) / P − θjc − θcs. A part dissipating 5 W with a 125 °C junction limit, a 25 °C margin so a 100 °C target, in 25 °C air, with θjc of 1.5 and θcs of 0.5, has a 75 °C budget: 75 / 5 is 15 °C/W total, minus 2 for the package and the interface, leaving 13 °C/W for the sink. That is a modest clip-on part.
When the answer comes out negative
Push the power to 40 W in the same situation and the total allowed is 75 / 40, which is 1.875 °C/W, and the package plus interface already consume 2. The subtraction gives a negative number, which is not a very good heat sink — it is the arithmetic telling you that even a sink with zero thermal resistance, bolted to a block of ice-cold metal, would leave the junction above target. This calculator says so explicitly rather than printing a small negative figure that looks like an answer.
The fixes are all upstream. Dissipate less: a switching regulator instead of a linear one removes most of the heat rather than moving it. Split the load across two devices so each handles half. Choose a package with lower θjc, which usually means a larger tab or an exposed pad. Improve the interface — thermal grease is often three times better than an insulating pad, though it gives up the isolation. Lower the ambient with ventilation. Or check whether the junction limit really is 125 °C and whether your margin is larger than it needs to be.
The interface is the term people underestimate
| Interface | Rough θcs | Notes |
|---|---|---|
| Direct metal-to-metal, dry | 0.5-2 °C/W | Surface roughness leaves air gaps; air is an insulator |
| Thermal grease, thin and even | 0.1-0.5 °C/W | Best plain option; too much is worse than too little |
| Silicone insulating pad | 0.5-1.5 °C/W | Buys electrical isolation, costs thermal performance |
| Mica plus grease | 0.3-0.8 °C/W | Isolation with better conduction, fiddly to assemble |
On a low-power design the interface is a rounding error. On a design where the whole budget is 2 °C/W, choosing a pad over grease can consume half of it. Mounting pressure matters as much as the material: a loose screw leaves a gap that no compound bridges, and an over-torqued one can crack a package tab.
Measuring instead of trusting
You cannot put a probe on a junction. What you can measure is the case, or the sink close to the device, and work back: Tjunction ≈ Tcase + P × θjc. That requires knowing the power, which for a linear regulator is (Vin − Vout) × Iout and for a switching device is harder and usually estimated from efficiency. Run the circuit at full load in the enclosure it will live in, with the lid on, until the temperature stops climbing — which can take twenty minutes — then measure.
Two things reliably make reality worse than the calculation. Heat sink ratings assume natural convection with fins vertical and air free to move; mounted horizontally in a closed box, the same sink can be half as effective. And ambient inside an enclosure is not room temperature; it is room temperature plus whatever everything else in the box is contributing. Both push the same direction, so build in margin and confirm it on the bench.
The power figure that feeds this page usually comes from a regulator choice — see the power supply rail calculator for working out what each rail wastes, and the PCB trace width calculator for the copper carrying it.
Questions people ask
What does °C/W actually mean?
Degrees of temperature rise per watt of heat passing through. A 10 °C/W heat sink dissipating 3 W sits 30 degrees above the air around it. The unit makes thermal design tractable because the stages add: package, interface and sink in series give a total that multiplies by the power to give the total rise. Lower is better, and the numbers span a huge range — a bare small-outline package might be 200 °C/W while a large extruded sink with a fan is under 0.5.
How do I know how much power my part is dissipating?
For a linear regulator it is the voltage across it times the current through it, plus a small quiescent term: a 12 V input, 5 V output regulator at 500 mA is dissipating 3.5 W, which is more than it delivers. For a MOSFET switching a load it is conduction loss, I²·Rds(on) times the duty cycle, plus switching losses that grow with frequency. For an amplifier it depends on the signal. What it is not, in any of these cases, is the power delivered to the load — only the part that turns into heat inside the device matters, and getting that number right matters more than any refinement in the thermal calculation.
Why should I design below the maximum junction temperature?
Because the maximum is an absolute limit, not an operating target, and everything around it is uncertain. Ambient varies, the power figure is an estimate, the interface degrades as grease dries and screws relax, and semiconductor life falls steeply with junction temperature — a common rule of thumb is that lifetime roughly halves for every 10 °C. Designing 20 to 30 °C below the limit costs a slightly larger sink and buys a part that survives a hot day and several years of service.
Is a bigger heat sink always better?
It is better thermally and it stops helping at a point worth knowing about. Once θsa is small compared with θjc and θcs, extra sink barely moves the junction temperature, because the bottleneck has moved inside the package and the interface. If the sink is 0.5 °C/W and the package is 1.5, doubling the sink saves a quarter of a degree per watt. Beyond that the return is in airflow, in a better interface, or in a different device. Size and mounting also have to be practical — a large sink poorly attached is worse than a small one bolted tight with grease.
Does a fan change the calculation?
Yes, and substantially: forced air can improve a sink's thermal resistance by a factor of two to four compared with natural convection. It also changes the failure mode. A passive design that runs hot is stable and predictable; a forced-air design that depends on a fan fails abruptly when the fan stops or the filter clogs. If you use forced air, use the manufacturer's rating at your actual airflow rather than the natural convection figure, and think about what happens at zero airflow — either the design survives it or something has to shut down.