# Temperature raise inside a case

I'm using a D2PACK triac model ACST830-8G datasheet

According to datasheet (over 1cm2 cooper) it supports 2A, considering Tamb = 43°

TABLE

GRAPHICS

I want to use this triac inside a case and without heat sink.

Inside a small case, can I assume this triac will handle 2A at 43 °?

I worried about this point because, as I understand, it will not be in "free air convection" as the second image says.

• Are there any other significant heat-generating components in the case? Why do you think the temperature inside the case will not exceed 43 °C? Is the case ventilated - and can it be placed in forced air flow? Is there any chance of using a metal case and a TO-220 triac which is thermally connected to the case? – Andrew Morton Jan 24 '17 at 19:50

Then no, you can't assume $2\:\textrm{A}$. Imagine that you place this inside a thermally sealed environment (perfectly so.) Then no heat energy would ever be lost and the temperature would rise forever. Reality will be between these cases, obviously. But if your circumstance is worse than "free air convection" then the temperature rise of the die will be higher and you must derate on the basis of what you know about the differences. If you can't do a finite element analysis, plus convection model, based upon a detailed physical model of your circumstances, then the next best here would be for you to make a test and find out what happens in the exact circumstance that you make and use. But if worse than "free air" you definitely will need to lower that figure.

Free air convection assumes that there is enough air around the part (or in the room) that the ambient temperature of the air would not be affected by the heat from the part.

If the part is in an enclosure then you introduce a thermal barrier between the air in the room and the enclosure and heat the enclosure air up. So interpreting the graph, the air in your enclosure will probably be more than 43C while running at 2A because of the added thermal resistance. The temperature range of the part is 125C (which you probably wouldn't the part or other parts in you enclosure to reach but it will still operate).

One thing you will want to avoid is thermal runaway where the heat of the part increases the internal resistance and leads to more heat being generated by the part until you have a meltdown. This will be determined by the thermal resistance of the enclosure. The a good way to determine what the internal temperature would be to experimentally verify it with the design in the enclosure or find the max power dissipation of the design and put a dummy load in the enclosure and measure the temperature.

You'll get some heat movement even thru plastic. Copper foil, the standard 1 ounce/foot^2, has thermal resistance 70 degree C per watt, per square of foil..any size square....the heat flowing from one edge to the opposite edge of the square. Not perpendicular to the foil, but with the layer of foil. The plastic epoxy has approximately 100X that thermal resistance, but some heat does flow.

Suppose we have a plastic blob, 1,000x the thickness (which is 1.4 mils) of standard copper foil....thus 1.4 inches. The copper would have 70/1,000 or 0.07 degree C. The plastic would be 100X thermal resistive, or 7 degree Centigrade per watt, if the plastic blob is a cube, 1.4 inches on a side.