In my circuit I am using a stepper motor driver, providing me up to 3 A current per trace. Thus I calculated the required trace width with a trace calculator, which told me to use a trace width of 0.07'' (or 1.78mm). Unfortunately, the chip itself does not allow me to connect them directly onto the IC legs. These are only 0.012'' (or 0.3mm) traces.

Therefore I created this routing: enter image description here

Pin 1, 21, 24 and 26 are responsible for the high current. I used 0.012'' traces right after the pin connections, then switched to 0.016 and 0.024 and as soon as possible to 0.07''-traces (the big vias). Is that enough, or will the possible heat damage my board? How can I improve it (instead of soldering copper bars on it)?
Driver Datasheet

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    \$\begingroup\$ Your picture doesn't show pin numbers so we can only try and infer what you're referring to. If you have the opportunity and are worried about the heat, could you increase the copper weight/height from 0.5oz to 1oz? The distance that it's thin also makes it less of a concern. \$\endgroup\$ – horta Feb 24 '15 at 14:56
  • \$\begingroup\$ @horta: I will add the numbers, but can you tell me maybe how I can recalculate the copper weight to height? Thanks! \$\endgroup\$ – arc_lupus Feb 24 '15 at 14:59
  • \$\begingroup\$ @arc_lupus The online calculator you linked to will let you change the copper thickness. You need to talk to your board manfacturer to see whats possible. I've never seen a 10oz copper board but 2oz and 4oz are common enough. \$\endgroup\$ – Warren Hill Feb 24 '15 at 15:14
  • \$\begingroup\$ Peak current may be 3A but what is the RMS current. Better still, link a data sheet. \$\endgroup\$ – Andy aka Feb 24 '15 at 15:36
  • \$\begingroup\$ @Andyaka: Added \$\endgroup\$ – arc_lupus Feb 24 '15 at 15:41

Here's a little tip for you:

The thickness of trace needed can be seen as a function of the length of the trace. The longer the trace the greater the voltage drop, and thus the thicker the trace needs to be.

Online calculators all assume the trace will be a constant width the entire length. This is not the case in your (and many other) situation.

So treat your layout as two traces. One trace that is the length from the chip to the start of the wide portion, and a second that is the length of just the thicker portion of the trace.

Perform two calculations - one for each portion. You'll find the shorter portion of the trace can generally be thinner than the rest.

Of course, if you go too thin then the heat dissipated could get too high, so it's a bit of a balancing act - try and keep the thin trace as short as possible.

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    \$\begingroup\$ The calculator does not depend on the length of the trace, just the resistance is changing... \$\endgroup\$ – arc_lupus Feb 24 '15 at 22:41

At this stage, fine modifications to placements can make a difference. It can be tedious, but it's worth knowing how.

Start with the unnamed passive component, 2nd in from the bottom left. The upper track to it takes a tortuous route occupying critical space. Delete it, move the via above it down as far as you dare, then squeeze that track back in. Now you can massively straighten out the track from Pin 1.

That should let you move the capacitor from Pin 28 left a bit, maybe half its length, without changing the overall length of its routing. Which opens up quite a lot more space above the chip to improve the routing there. And so on.

You won't be able to get full width tracks all the way in, but you'll be able to get intermediate width tracks within a pad-length or so of the pad.

Takes practice and time. At some point, you decide it's done...

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The A4983 data sheet states that peak current is 2A so this means you are going to get into some trouble when driving 3A loads. Additionally it is the RMS current that is important for calculating track widths - it is the RMS current that heats the tracks.

I'd estimate, that if you were using the device at the full 2A peak current, you should be plugging in something like 1.5 amps into your trace calculator.

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  • \$\begingroup\$ Therefore my routing is already valid, after (most of it) it can already be used for 3A? \$\endgroup\$ – arc_lupus Feb 24 '15 at 15:57
  • \$\begingroup\$ Your routing might be valid but your expectations of the chip may not be. \$\endgroup\$ – Andy aka Feb 24 '15 at 15:59
  • \$\begingroup\$ If the chip is not able to drive more, that should not be a problem, I was worrying more about the traces. \$\endgroup\$ – arc_lupus Feb 24 '15 at 16:00
  • \$\begingroup\$ I'd still make the traces as wide as I could, just in case. \$\endgroup\$ – Andy aka Feb 24 '15 at 16:01

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