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PCB/EDA/CAD / Re: High current traces meeting small component legs
« Last post by T3sl4co1l on Today at 08:27:37 am »Nope! Get out the thermal simulator, or the IRL equivalent (build and test it).
On the upside, the lateral heat spreading ability of, say, 2oz x 2 layer FR-4 is about an inch or two, so, anything within that area (if it's not stupendously intense) will tend to spread heat out into that area (i.e. a circle a couple inches across), and the power dissipated in those local areas merely contributes to overall temp rise, with a modest local rise depending on how well heat is sunk away from the heat-generating area in the first place.
You can do basic approximations of, what if we have this much current in this rectangular segment of copper, which makes this power, which flows through that segment, and etc. etc. Neither heat nor current flow in uniform directions, so the value of such approximations is quite limited, but you can still hand-wave a basic guess like to say the local area isn't going to be more than x°C above the nearby area, and that the nearby area is ballpark y°C higher due to that current flow, etc..
Tim
On the upside, the lateral heat spreading ability of, say, 2oz x 2 layer FR-4 is about an inch or two, so, anything within that area (if it's not stupendously intense) will tend to spread heat out into that area (i.e. a circle a couple inches across), and the power dissipated in those local areas merely contributes to overall temp rise, with a modest local rise depending on how well heat is sunk away from the heat-generating area in the first place.
You can do basic approximations of, what if we have this much current in this rectangular segment of copper, which makes this power, which flows through that segment, and etc. etc. Neither heat nor current flow in uniform directions, so the value of such approximations is quite limited, but you can still hand-wave a basic guess like to say the local area isn't going to be more than x°C above the nearby area, and that the nearby area is ballpark y°C higher due to that current flow, etc..
Tim