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Heatsink and junction temperature calculator

Heat flows out of a chip like current through resistors. Each step from the silicon to the air has a thermal resistance, θ, in degrees per watt. Type the power, or a linear regulator's voltages and current, and the θ values from the datasheet to get the junction temperature and the heatsink it needs.

What feeds the regulator.

A 7805 gives 5 V.

250m = 250 mA.

From the datasheet. About 5 for a TO-220.

The thermal pad or paste. Leave empty for none.

A bare pad is about 70, a square inch of copper about 30.

Inside a closed box, often 40 or more.

125 °C for most chips.

Junction temperature, Tj = Ta + P × (θJC + θCS + θSA)

86 °C
Power, P
1.75 W
Heat path, θ in total
35 °C/W
Rise above the room
61 °C
Room to spare
39 °C
Most θSA allowed
52.1 °C/W
Most power on this path
2.86 W

The formula

Watts through °C/W give a temperature rise, just as amps through ohms give volts:

Tj = Ta + P × (θJC + θCS + θSA)

θJC is junction to case, from the datasheet. θCS is the thermal pad or paste between the case and the heatsink. θSA is the heatsink, or the board's copper, to the air. In a row, they add. A linear regulator turns every volt it drops into heat:

P = (Vin − Vout) × I

A worked example

World 35's regulator turns 12 V into 5 V at 250 mA:

P = (12 − 5) × 0.25 = 1.75 W

Its θJC is 5 °C/W. On a bare pad, about 70 °C/W to the air, the junction reaches 25 + 1.75 × (5 + 70) ≈ 156 °C: well over a typical 125 °C limit. Pour a square inch of copper around its tab, about 30 °C/W:

Tj = 25 + 1.75 × (5 + 30) ≈ 86 °C

To stay under 125 °C, the copper or heatsink can have at most (125 − 25) ÷ 1.75 − 5 ≈ 52 °C/W.

Two ways out

Two heat paths side by side share the heat like resistors in parallel. Copper on top at 40 °C/W and thermal vias into a plane below at 40 °C/W make 40 × 40 ÷ (40 + 40) = 20 °C/W together, so type 20 for θSA: the junction drops to about 69 °C. Or attack the cause: feed the regulator 9 V instead of 12, and it burns only 1 W.

Free lesson

Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“How hot does it get?” is lesson 1 of 6 in Gerber Gulf, PCB Design World 3. Free with an account. Lessons unlock in order, so you’ll do the 14 before it first.

Or start from the very first lesson, no account needed

The ideas behind it

Questions people ask

How do I calculate junction temperature?

Tj = Ta + P × θ, where Ta is the air temperature, P the power in watts and θ the thermal resistance from junction to air in °C/W. In a row, the resistances add: θJC + θCS + θSA. 1.75 W through 5 + 30 °C/W in a 25 °C room gives 25 + 1.75 × 35 ≈ 86 °C.

What heatsink do I need?

Work out the total allowed, (Tj max − Ta) ÷ P, then take off θJC and the pad's θCS. What's left is the most the heatsink may have. For 1.75 W, a 125 °C limit and 25 °C air: 100 ÷ 1.75 − 5 ≈ 52 °C/W. Lower is better, and aiming for 100 °C leaves room for a hot day.

How much heat does a linear regulator make?

Every volt it drops, times the current: P = (Vin − Vout) × I. A 5 V regulator fed 12 V at 250 mA burns 7 × 0.25 = 1.75 W. Fed 9 V it burns only 1 W, which is why a lower input, or a buck converter first, runs much cooler.

Is 125 °C a safe junction temperature?

It's the usual maximum, not a target. Chips wear out faster the hotter they run, and a common rule of thumb says every 10 °C halves the life. Designers aim for 100 °C or less.

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