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I've read before that each 1 mm\$^2\$ of copper can handle up to 1 A. Is that true? Is there an easier formula than the one used on online calculators that requires temperature and such?

To give more details, I have 3 main points in my circuit where current is going to be at most 500 mA which are now unrelated (to make 1.5 A in total). My tracks' width is 0.02 in which is 0.5 mm. My width is going to be most likely 1 mm (or maybe less, depending on the etching equality which I cannot guarantee) so with my calculations I'll have around 0.5 A with no factor of safety.

My questions are:

  1. Are my calculations correct? Is there a better way to do them?
  2. Any tips or suggestions?
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  • \$\begingroup\$ What is the copper thickness? \$\endgroup\$ – Leon Heller Jun 11 '15 at 2:10
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    \$\begingroup\$ There's more to it than track thickness. How much temperature rise can you tolerate? What about voltage drop? Add another half dozen parameters if there are any controlled impedance tracks. \$\endgroup\$ – Matt Young Jun 11 '15 at 2:21
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Here's a VERY simplistic analysis:

An inch of a 20 mil wide trace of half-ounce copper will have a resistance of about 0.05 Ohm, and a surface area of 0.02 inch. If you ran an amp through it, you'd have 0.05 Volts, so 0.05 Watts, 50 mW. At a constant current, power would be proportional to length, and so would surface area.

For comparison, an old one Watt axial-leaded carbon composition resistor is about 0.6 inch long and 0.23 diameter. Call that 0.6 x 0.75 in surface area, or less than half a square inch. The circuit board will conduct heat away from your trace more effectively than air would from the resistor.

This really doesn't seem easier than an on-line calculator. This one seems to work fairly simply: http://www.4pcb.com/trace-width-calculator.html I put in your requirement of 0.5 Amp, assumed a standard 0.5 ounce copper, and allowed 20C rise (more than the default of 10). Assuming also you start at 25 C room temperature, you get a 15 mil wide trace, whether your trace length is one inch or ten.

You can order boards with heavier copper (1 and 2 ounce per sq ft are not unusual), and the resistance will go down appropriately.

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  • \$\begingroup\$ okay see now i'm more confused than before, here is the test i did in mm current 0.5A, thickness 0.1 mm, 10C rise, 25C room temp results in required width of 0.0404 mm (mine is 0.5), with the same paramaters i tested to see the max and it was like 3 A, ama i doing something wrong here or does really all the width i need is 0.04 mm ! @William Watson also cause 15 mil is like 0.381 mm which means it would still work but i want to know why we had different results, thanks and ill accept it once i get response \$\endgroup\$ – AndrewxXx Jun 11 '15 at 6:03
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    \$\begingroup\$ A copper layer 0.1 mm think is VERY thick. A "one ounce copper" layer is only 0.0347 mm thick (34.7 micrometers). For digital circuits, we usually only have "half ounce" copper (so only 17 um). With more power, we might use one ounce, or in an extreme case, two ounce. That's why the discrepancy: 0.1 mm of copper is thick, so traces of copper that thick don't need to be as wide. \$\endgroup\$ – William Watson Jun 12 '15 at 17:59
  • \$\begingroup\$ well i honestly duno how thick it is as they vendor said around 0.1mm but i don't have the right tool to test it, i spent like 4 days now routing that motherboard on my own to get a 0.04in width on all tracks which is 1.016mm, using that calculator it showed that all i need is only 0.116 width (with 0.0347 mm thickness and 0.5A), so that mean i'm on the safe side with 9.xx safty factor ? \$\endgroup\$ – AndrewxXx Jun 13 '15 at 4:12
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    \$\begingroup\$ It sounds like you're likely well on the safe side, unless your board vendor has a very unusual process, with much thinner copper layers than industry standard. My guess is that the 0.1 mm answer was intended to be just saying that the copper will be "thin enough so that you don't have to worry about it affecting the thickness of the board mechanically," rather than an attempt to give you a hard number for calculation of resistance or trace width. \$\endgroup\$ – William Watson Jun 15 '15 at 20:11
  • \$\begingroup\$ yup and true, thanks anyway i've reached tracks of 0.06 in which should be more than enough :D thanks for your help \$\endgroup\$ – AndrewxXx Jun 16 '15 at 13:30

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