| Electronics > Projects, Designs, and Technical Stuff |
| Using PCB as heatsink! |
| (1/3) > >> |
| ali_asadzadeh:
Hi, I want to know if there is a calculator for calculating thermal design for PCB, In one of my designs I have added some Copper fills to act as a heat-sink, so the question is where can I find some tool for calculations? I'm using LM1117 in SOT-223 as the Vcore regulator for my Spartan 6 device (XC6SLX9) so, I want to make sure everything is fine. |
| tszaboo:
555: You can dissipate something like 5 W in a 50 x 50 mm PCB if it is 4 layer with high copper on all layers. |
| thinkfat:
--- Quote from: NANDBlog on January 21, 2020, 01:43:39 pm ---555: You can dissipate something like 5 W in a 50 x 50 mm PCB if it is 4 layer with high copper on all layers. --- End quote --- With ample thermal vias, of course. |
| MagicSmoker:
--- Quote from: ali_asadzadeh on January 21, 2020, 01:40:48 pm ---SOT-223 --- End quote --- The best you can realistically hope for with standard pcb construction (2-4 layers of 35um Cu) is a thermal resistance of 40C/W. Allotting a bigger area to the heatsink doesn't help because the lateral thermal resistance of the thin copper is too high to make use of it; anything beyond 15mm per side as a heatsink for an SOT-223 component will likely end up as wasted area, in fact. You can get marginally lower thermal resistance with a larger SMT package (e.g. - D2Pak) but if you have to allot, say, 50mm per side of board area just to dissipate 5W then you really need to rethink your approach (ie - a PTH component with a small aluminum heatsink or a, perhaps, a buck converter [pre-]regulator instead of a linear one). |
| schmitt trigger:
What MagicSmoker said. A common 1 oz copper is only 0.0347 mm thick(*), so it will not conduct heat readily. I have used as a rule of thumb, for 1 oz copper, I've used 1 sq-in or roughly 25 x 25 mm, as the largest practical copper area to be used as a heatsink. (*) Source https://pcbprime.com/pcb-tips/copper/ |
| Navigation |
| Message Index |
| Next page |