Can i get away with a pcb heatsink for a triac?

so, i have a bunch of to-220 L4006L5TP triacs laying around and i have to design an AC phase control pcb, reading the datasheet i saw that the Rθ(J-A) is 50°C/W, the Rθ(J-C) is 3.3°C/W and the voltage drop is 1.6V, since the max current of the design is 1A i have to dissipate 1.6W to the pcb itself. according to this question i need a heatsink with thermal resistance less than 36.7°C/W to 16.7°C/W to keep the triac in the 80°C to 60°C range considering that the air inside the case is 45°C is it possible to achieve this with only using the pcb as heatsink with horizontal mounted triac and some CPU thermal paste? if yes, how?

tl;dr: is it possible to use the pcb as a heatsink and achieve less than 36.7°C/W using some thermal paste?if yes, how?

3 Answers

According to Fig. 5 & Fig. 11 in the datasheet the power dissipation will be about 0.95 W.

You will need about 2 sq inch. This means top and bottom. There should be thermal vias where the device thermal pad is solder to the PCB pad.

I would use the TO-263AB because it will be easier to solder the thermal pad to the PCB.

If you use the TO-220 you still need a thermal pad on the PCB to solder it to. Screw it down before soldering. Use solder paste if you have some.

In the app note below, section 3.1.1 Example: Calculating the Required Board Size to Hit a Target θJA is very similar to your project.

This example is for a 0.94 W device with a θJC of 7.3°C W. It required 2.23 sq. in. for a target θJA of 40°c. Using the same calculations with a θJC of 1.9°C/W the area was reduced to 1.91 sq. in.

So 2 sq. in. sounds like a safe bet with about 25°C margin.

At 1A you will be fine with a minimal amount of heatsink area- perhaps 1 square inch should be more than enough.

The actual triac dissipation will be maybe 1W. The forward voltage drops with current, see the curve in datasheet.

To confirm, check out the below curve for a TO-252 SMT package, happens to be for a voltage regulator, but applicable:

Standard thickness copper foil (1.4 mils, 35 microns) has 70 degree Centigrade per watt edge-edge of thermal resistance. For any size square of foil.

This, with 8 squares around the Triac's mounting stud, will have 70/8 = 9 degree Centrigrade of heat rise. If you surround that with 8 more squares (each being 3X larger than the mounting stud, you'll have another 9 degree C rise.

simulate this circuit – Schematic created using CircuitLab

• so the entire pcb has 70°C/W and i can divide this into lesser amounts? Nov 18 '18 at 22:12
• Imagine heat flowing OUT from the center. Each of those squares, small or large, presents 70 degree Centigrade per watt, of thermal resistance. If your triac dumps 1 watt onto the PCB, and the outer edges of the larger squares are at 25C, the inner-most square where the triac is bolted will be 9 +9 degree C warmer. Nov 18 '18 at 22:40
• shouldn't it be lower as the area of the copper increases? Nov 18 '18 at 23:45
• @ Lucas More area only helps if radiation cooling or air cooling is available. Otherwise the # squares in the path will set the thermal resistance. Nov 19 '18 at 4:00
• If you are designing a satellite, where the air escaped during launch, and you are not encouraged to "radiate" to nearby boxes and raise their own temperature, then "conducted heat flows" are your only way to get rid of heat. At some point, the metal BOLTS mounting the PCBs to your chassis have become a key path for the heat, along with the hollow metal spacers. In that case, the PCB copper must extent under the bolt heads. Nov 22 '18 at 23:53