Help With Thermal Modeling? - Page 1
Forum section: electronics / projects

author: fourfathom
author_rank: Super Contributor
timestamp: August 25, 2023, 09:11:35 pm
content: I am trying to design a Ham Radio device that contains six or eight 1W Class-D or similar amplifiers, plus controller, etc., with a total worst-case system power dissipation of around 20 Watts. I want to put this in an extruded aluminum box (dimensions about 200 X 90 X 50mm or perhaps larger). Can anyone give me a rule of thumb for calculating internal temperature rise? I would like to avoid internal fans, so probably the bottom surface of the case wouldn't help much with the dissipation. There will be some convection internally above the circuit board, but possibly not much.I'm not a mechanical engineer by any means, and while I have looked at some equations for this type of analysis I haven't motivated myself to make the effort to figure them out -- I've got too many other tasks at the moment. Is there a simple rule of thumb I can use to tell me if this enclosure size is even practical?BTW, I am a One Man Band on this, and while the finished product may be sold at a very marginal profit, I'm doing this for fun, and to help out some amateur (and professional) scientists who use this stuff to study the ionosphere. I've got a similar design currently being used, but that one is modular, in a 1RU (Rack Unit) enclosure, with the individual amplifier boards mounted for vertical air-flow:https://turnislandsystems.com/meet-the-beaconblaster-6/https://turnislandsystems.com/evolution-of-the-beacon-blaster-6/I am trying to shrink that one down to a single board, for easy assembly and lower cost. The 1RU design stays quite cool.Any advice will be greatly appreciated!

author: IanB
author_rank: Super Contributor
timestamp: August 25, 2023, 09:31:53 pm
content: Do you have the planned box or a similar box available?The most reliable approach would be to put a 20 W heat source inside the box and leave it running for a while as a test, while measuring the internal and surface temperatures. For the heat source you could use a 21 W automotive bulb or a 20 W mains incandescent bulb.It is theoretically possible to do calculations, but experiment and experience are always better options.

author: Tomorokoshi
author_rank: Super Contributor
timestamp: August 25, 2023, 09:37:36 pm
content: Here is a very coarse comparison. I worked on a device where the box had similar dimensions, with a large aluminum heatsink with multiple fins. The box and the heatsink were thermally connected using an aluminum block. The module was mounted in open air on a DIN rain in an enclosed 12" x 12" x 8" steel cabinet. Around 22W input power was dissipated in stud-mounted parts in the aluminum block, with internal ambient at around 50C, and no fans. The maximum module temperature reached around 160C. The box had a lot of holes, however, no power was dissipated in the box, so it behaved as an additional heatsink.Therefore the assembly was characterized at about (160C - 50C) / 22W = 5C/W.Without convection to open air, conduction to the environment, etc. things get hot.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 25, 2023, 09:46:27 pm
content: The ballpark figure I stuck in my head is 150 in^2 C/W. So, one square inch is 150 C/W or about the dissipation of a TO-92. Well -- one square inch of sheet is two faces, or 75 C/W, really, which sounds about right as heatsinks go.Actual figure is of course dependent on convection (ducting, chimney effect), temp rise (it's roughly quadratic with temp; this figure is calibrated towards the low end IIRC), and other effects like radiation (which is negligible at low levels, but rises to a few to 10% at maximum electronics operating temperatures).Now, if you aren't getting conduction inside that box,double the figurebecause you have convection insideandout and they act in series.So you can see it's a big deal getting thermal contact to the enclosure.There's no good way to bolt something down inside an extruded enclosure, but you might opt for a diecast box, and put a thermal pad under the PCB, or bolt devices to the bottom or sides.Even as poor as ABS is, if you have an ABS enclosure but some heat spreading inside (heavy aluminum or copper foil, adhesive backed), and thermal pads from PCB to that, you can probably still do better than the extrusion with only static air inside.You could also put a fan or two inside the extrusion, which is weird; sealed up I mean, no air holes, just circulating what's inside -- but that will indeed greatly reduce the convection coefficient from still air to whatever velocity it's getting. Which helps doubly between components/heatsinks themselves, and the enclosure. (You might want a temp sensor and cutout to protect when the fan eventually stops. Even without exposure to ambient dust, the bearings will only last so long.)Tim

author: fourfathom
author_rank: Super Contributor
timestamp: August 26, 2023, 04:13:46 am
content: Thank you all, this is very helpful.IanB, I don't have a candidate box, nor my measurement tools at present. When I get back home I plan to take a smaller box that I *do* have, and put some power resistors in it, then measure the air and surface temperatures. At least then I can check my assumptions.Tomorokoshi, yes, this is what I am afraid of.Tim, that equation looks like it needs a reciprocal somwehere. Otherwise as I read it, the temperature drops as the surface area is reduced. Or am I reading it wrong? Here, please check my math, using that 150 deg C / W value as I assume it works:My proposed enclosure has about 70 sq in of surface (not inclusing faceplates).Cut that in half because the internal circuit board blocks any airflow from whatever internal convection from reaching the bottom of the enclosure (the heat-generating parts are on the top side of the PCB).Cut that in half again, because of your comment about poor internal airflow.This gives me 17.5 sq in effective surface. 150 deg C per W / 17.5 sq in = 8.57 deg c delta case temperature.With my 20W (max) dissipation that gives a case temp rise of 8.57 * 20 = 171 deg C.This would obviously be an unworkable case design, but is my math correct? Yes, I know, approximations, factors, etc., but is it in the general ballpark? Or do I misunderstand your equation?If I eventually get close to a suitable mech design, I might try adding vent cutouts to the PCB, to encourage some airflow underneath the PCB. That might give me a few % improvement...Thank you again, this is one of the reasons I hang out at EEVBlog rather than StackExchange!

author: IanB
author_rank: Super Contributor
timestamp: August 26, 2023, 04:46:31 am
content: A basic heat transfer equation is this:Q = h A (T1 - T2)Where Q = heat flow (W), h is a heat transfer coefficient (W/m2/K) and (T1 - T2) is the temperature difference (K).You can do this in US customary units also (Btu/h, Btu/h/ft2/F, and F), but if you start with W it is easier to work in SI units.For a flat surface in the open air with no fans, h is a natural convection film coefficient, which varies a lot depending on the temperature difference, the angle of the surface (horizontal or vertical), and the circulation space around the case.Here is one reference for typical values of h (you can find others):https://www.researchgate.net/post/How_much_heat_transfer_coefficient_of_airIf you assume a nominal value of 10 W/m2/K, and say 70 sq in is about 0.045 m2, and a heat transfer rate of 20 W, then you would have this:20 = 10 * 0.045 * (T1 - T2)From this you can find that T1 - T2 is approximately 44 K. This would be the temperature difference between the surrounding air and the metal enclosure.You could say the same for the temperature difference between the metal enclosure and the air inside it, so this would give a total difference of about 88 K.If the room air is at 20°C, then the air inside the box might be at 110°C.Now the circuit components themselves are seeing a surrounding air temperature of 110°C, so their heat sinks and internal temperatures will be much hotter than that.Where this all becomes approximate is that h = 10 W/m2/K is only a guess. It could be as low as 5, it could be as high as 25 or more. This is where you really need tests to narrow things down.It's a really good idea to fix the hot components directly to the enclosure with heat sink compound to maximize the cooling potential. It can also help to put fins on the outside of the enclosure.Another reference:https://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 26, 2023, 07:09:47 am
content: Quote from: fourfathom on August 26, 2023, 04:13:46 amIt does have a reciprocal somewhere -- watts is on the bottom.So if watts goes up, C goes up, at constant area. Or area goes up, watts goes up, for constant temp rise.QuoteTo be a bit more precise, that would be internal (e.g. component) temperature.At such temperatures, convection will be rather strong, and the PCB's conduction and the other airspace won't be negligible. Probably it would be 130-150C, not 170. But still, pretty toasty.Whereas if you can conduction cool everything to the enclosure, you incur some 10s of C for those materials, and then just the one convective drop.If the enclosure is reasonably conductive (or it's clamshell but you can get heat into both pieces, say by splitting your board in halves and using them in parallel), the full 70 in^2 would give 2.14 C/W or 43C rise, toasty but not disastrous.Tim

author: fourfathom
author_rank: Super Contributor
timestamp: August 27, 2023, 05:45:43 am
content: Ian, Tim, thanks for the additional info. I am looking forward to(and dreading) my upcoming experiments as I try to figure out the best enclosure for the new design.-Paul