Voltage Regulator Heat Calculator
How much heat a linear regulator makes, how hot its junction gets, and whether it needs a heatsink — with the real current ceiling.
A linear regulator burns the difference
It does not convert voltage. Every volt it drops, at every milliamp the load draws, becomes heat inside a package about the size of a fingernail. Dropping 12 V to 5 V at half an amp is 3.5 W — more power thrown away than delivered — and a bare TO-220 would need a 217 °C rise to shed it. The datasheet says 1 A on the front page because the part is rated for the current, not for the heat, and the heat is what actually stops you.
- Efficiency is exactly Vout ÷ Vin. Nothing else. It does not improve at light load and there is no way to tune it. A 5 V rail from 24 V is 21% efficient, permanently.
- Lowering the input beats a bigger heatsink. Going from 12 V to 7 V input cuts the heat from 3.5 W to 1 W without touching the mechanics. If the supply is fixed and the gap is large, a buck converter running at 85–95% is the real answer.
- Free-air figures assume still air on an open board. Inside a sealed enclosure the ambient itself climbs, so a design that measures fine on the bench can shut down in the box on a warm day. Leave real margin, not a couple of degrees.
- The tab is usually connected to a pin. On most parts it is ground, on some it is the output. Bolting it straight to a shared chassis can short something, so fit an insulating pad — and count the extra 0.5 to 2 °C/W it adds.
- Thermal shutdown is not protection you should rely on. It exists to save the chip, not the design. The symptom is an output that works for a minute and then droops, which is a miserable fault to chase later.
How to use
- Enter the input and output voltages and the load current.
- Pick the package and whatever heatsinking you have, if any.
- Read the dissipation, the junction temperature and the verdict.
- Check the thermal limit figure — it is usually well below the datasheet current rating.
Frequently asked questions
Why does my regulator get so hot?
Because a linear regulator does not convert voltage, it burns the difference. Every volt it drops, at every milliamp the load draws, becomes heat. Dropping 12 V to 5 V at half an amp is 3.5 W dissipated in a package the size of a fingernail — more power thrown away than delivered to the load. Nothing on the front of the datasheet warns you, because the part is rated for current and it is heat that stops you.
How efficient is a linear regulator?
Exactly the output voltage divided by the input voltage, and nothing else. It does not improve at light load, it does not improve with a better part, and there is no way to tune it. A 5 V rail from 24 V is 21 percent efficient permanently. That single fact is the whole argument for a switching regulator whenever the gap is large.
Do I need a heatsink?
Work out the watts and compare against the package. A TO-220 in free air has a junction-to-ambient resistance around 62 degrees per watt, so roughly 1.5 W takes it to its limit from room temperature. Past that a heatsink stops being optional. The calculation here does it properly, including the ambient you actually have rather than the 25 degrees the datasheet assumes.
What is junction temperature and why not just measure the case?
The junction is the silicon itself, and it is always hotter than the case you can touch — that difference is the junction-to-case thermal resistance. Most parts allow 125 or 150 degrees at the junction, and thermal shutdown triggers on that, not on the case. A case reading of 80 degrees can sit alongside a junction well past its limit.
Is thermal shutdown enough protection?
It protects the chip, not your design. When it triggers, the output folds back and the symptom is a supply that works for a minute and then droops, recovers, and droops again. That is a miserable fault to chase months later, and it means the circuit is not delivering what it was meant to. Design so shutdown never happens rather than treating it as a safety net.
What is dropout voltage?
The minimum difference between input and output at which the regulator can still regulate. An ordinary 78xx needs about 2 V, so a 7805 on a 6 V supply will not hold 5 V — the output sags and follows the input. Low-dropout parts manage on a few hundred millivolts and some on tens, which is what makes running 3.3 V from a single lithium cell practical.
Should I use a switching regulator instead?
If the voltage gap is large or the current is more than a couple of hundred milliamps, usually yes. A buck converter runs 85 to 95 percent efficient regardless of the gap and stays cool. The costs are switching noise on the output, a slightly more involved circuit with an inductor, and more care over board layout. For a low-noise analogue rail from a small drop, a linear regulator is still the better part.
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