Thermal Resistance and Temperature Rise Calculator
A conduction path in series with a surface that sheds heat by convection and radiation together. Radiation is nonlinear, so the surface temperature is relaxed to convergence rather than assumed - which matters, because at typical power-electronics temperatures radiation is often a third of the total.
Choosing something to solve for makes Component temperature an editable target and computes the chosen input from it.
Formula
| L, A | conduction path length (m) and cross-section (m^2) |
| k | thermal conductivity (W/m/K) |
| h | convection coefficient (W/m^2/K) |
| As | surface area exposed to air (m^2) |
| e, sigma | emissivity and the Stefan-Boltzmann constant |
What this model assumes, and where it stops
Assumptions
- One-dimensional conduction along a uniform path.
- Isothermal surface at a single temperature.
- The convection coefficient is constant and independent of temperature.
- The surroundings are a large black enclosure at ambient temperature.
- Steady state - no thermal capacitance, no transient.
Limitations
- The convection coefficient is the weak link. The presets are order-of-magnitude bands; real h depends on orientation, characteristic length, surface finish and whether the airflow is actually reaching the surface. Treat the answer as accurate to tens of percent, not percent.
- A single lumped resistance cannot represent spreading resistance, which often dominates when a small heat source feeds a large plate.
- The isothermal-surface assumption fails for fins and extended surfaces, where fin efficiency matters.
- Radiation assumes a clear view to ambient surroundings. Inside an enclosure, surfaces radiate to each other and the net loss is much smaller.
- No transient behaviour, so nothing about thermal time constants or pulsed loads.
- Natural-convection h itself depends on the temperature rise you are solving for, which makes the real problem implicit in a way this does not capture.
When you need a 3D field solution instead
Closed-form models like the one above hold on idealised geometry. These are the cases where they stop being good enough and a full 3D electromagnetic and thermal solution is the only way to get a trustworthy answer:
- Spreading resistance from a concentrated source into a plate, which a 1D path cannot represent at all.
- Any enclosure, where radiation exchange between surfaces and recirculating air decide the answer.
- Finned heat sinks, where fin efficiency and the boundary layer between fins set the real performance.
- Buried hot spots - the middle of a winding, the centre limb of a core - where the peak is nowhere near a surface.
- Coupled problems: loss depends on temperature, temperature depends on loss, and the two have to be solved together.
- Transient and pulsed loading, where thermal mass is doing the work.
Common questions
How much heat is lost by radiation?
In still air with a reasonably emissive surface, often 20 to 40% of the total. That is why painting or anodising a heat sink helps: bare bright aluminium has an emissivity around 0.05, anodised around 0.8. This calculator solves convection and radiation in parallel rather than ignoring radiation.
What convection coefficient should I use?
Roughly 5 to 10 W per square metre per kelvin for natural convection in still air, 25 for gentle forced air around 1 m/s, and 60 or more for stronger airflow. These are order-of-magnitude bands - real values depend on orientation, size and whether the air actually reaches the surface, so expect accuracy to tens of percent.
Why is my component hotter than this calculation suggests?
The usual causes are spreading resistance from a small heat source into a large plate, which a one-dimensional path cannot represent, and enclosure effects where surfaces radiate to each other instead of to ambient. Buried hot spots inside a winding or core are also invisible to a lumped surface model.
References
- Incropera & DeWitt - Fundamentals of Heat and Mass Transfer, 6th ed.
- Lienhard - A Heat Transfer Textbook, 5th ed. (freely available)