PCB Design
How to calculate junction temperature
To work out how hot a part runs, treat heat like current. Watts flow through thermal resistance, θ, in °C per watt, and give a temperature rise, just as amps through ohms give volts. The junction inside the chip sits at T = Tambient + P × θtotal, where θtotal is every resistance between the silicon and the air.
Thermal Ohm's law
Example: a linear regulator turns 12 V into 5 V at 250 mA. It drops the other 7 V across itself:
The heat starts in the silicon, the junction, crosses θJC (junction to case, 5 °C/W here) to the tab, then the copper takes it into the air. On a bare pad, about 70 °C/W:
Most chips are rated to 125 °C. This one would shut down or slowly cook.
Live experiment
Pour copper under the regulator
A 5 V regulator at 250 mA turns every volt above 5 into heat, through θJC of 5 °C/W and the copper under its tab to a 25 °C room. On a bare pad from 12 V the junction reaches 156 °C; a square inch of copper brings it to 86 °C.
“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.
Series and parallel paths
Every rule you know for resistors works for heat. Paths one after another, junction to case to board to air, add. Paths side by side combine in parallel, and help.
Pour a square inch of copper around the tab, about 30 °C/W, and the junction falls to 25 + 1.75 × 35 ≈ 86 °C. Add thermal vias into a plane on the other side, a second 40 °C/W path alongside 40 °C/W of top copper, and the two combine to 20 °C/W: about 69 °C. Or attack the cause: feed the regulator 9 V instead of 12, and it drops only 1 W.
Choosing a heatsink
Turn it round to find what you need:
Subtract θJC and the thermal pad or paste, θCS, and the rest is the most the heatsink may have. Aim well under the rated limit: a common rule of thumb says every 10 °C hotter roughly halves a part's life, so designers often target 100 °C or less. And use the real ambient: inside a closed box it may be 40 °C or more, not 25.
Where you'll meet it
Every power part's datasheet lists its θ values. A datasheet's θJA is measured on a standard test board, which yours probably isn't, so treat it as a guide. LED bulbs sit on aluminium boards because metal is a very low θ, and phone chargers are laid out around their hottest parts.
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.
“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.
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Practice questions
Three quick questions from the lessons. Pick an answer to check it.
Show the answer
Answer: Add up to 80 °C/W: the heat must cross both, and the junction gets hotter. In a row the heat crosses one, then the other, so they add: 25 + 1.75 × (5 + 80) = 174 °C.
- Why not “Cool it best of all”? In series both resistances are in the way, so it's the worst arrangement, not the best.
- Why not “Make 20 °C/W”? 20 °C/W is what the two paths give side by side. One after the other, the heat must cross both, so they add.
- Why not “Make 40 °C/W”? 40 °C/W would mean the second path had no resistance at all. In series their resistances add.
Show the answer
Answer: Over 150 °C: watts are small, but °C/W can be large. 25 + 1.75 × (5 + 70) = 156 °C. A couple of watts through a small pad is enough to cook a chip.
- Why not “About 50 °C”? About 50 °C would need a total of only about 14 °C/W. A bare pad is 70 °C/W.
- Why not “About 27 °C: 1.75 W is tiny”? Small watts can still make a big rise when the thermal resistance is large: 1.75 W × 75 °C/W is 131 °C.
- Why not “Exactly 125 °C, its rating”? The rating is a limit, not where it settles. 25 + 1.75 × (5 + 70) works out at about 156 °C.
Show the answer
Answer: 0.7 W. (12 − 4.8) × 0.1 = 0.72 W, as heat.
- Why not “Nothing”? A linear regulator always turns the spare volts times the current into heat.
- Why not “7 W”? 7 W would need 1 A. At 100 mA the regulator burns about 0.7 W.
- Why not “0.07 W”? That's a tenth of the real figure: (12 − 4.8) V × 0.1 A = 0.72 W.
“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.
Questions people ask
How do you calculate junction temperature?
T_junction = T_ambient + P × θ_total, where P is the power in watts and θ_total is the thermal resistance from junction to air in °C/W. A part dissipating 1.75 W with 75 °C/W in total, in a 25 °C room, reaches 25 + 1.75 × 75 ≈ 156 °C.
What does °C/W mean?
Degrees of temperature rise per watt of heat flowing. A heatsink rated 10 °C/W runs 10 °C above the air for every watt it sheds. Lower is better: it means heat escapes more easily.
How do I choose a heatsink?
Work out the total allowed: (T_max − T_ambient) ÷ P. Subtract the part's θJC and the interface's θCS, and what's left is the most the heatsink may have. For 2 W, 100 °C max, 25 °C air and θJC of 5 °C/W, that leaves under 32.5 °C/W for interface and heatsink together.
How much does PCB copper cool a part?
A lot. A bare minimum pad for a small power package might be around 70 °C/W to the air; a square inch of copper poured around it, about 30 °C/W. Thermal vias into a plane on the other side add a second path in parallel.
Related topics
- How does a linear voltage regulator work?A zener sets the voltage and a pass transistor carries the current: the circuit inside every 7805 and LDO.
- PCB trace width for currentCount the squares: how a trace's width sets its ohms, its voltage drop and its heat, and how wide to draw it.
- Electrical power explained: P = V × IP = V × I: how fast energy turns into heat, and why a resistor has a power rating.
- What is a ground plane on a PCB?A whole layer of copper for ground: milliohms for steady current, about a nanohenry for fast current.