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What is the math behind traces and clearance calculations? I am designing a PCB, which will carry 12V and 6A, what should be trace width and trace clearance?

Similarly, what should it be for 12V 3A, and 5V 3A. Is there a general rule of thumb, using which we can decide trace width and clearance?

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That's actually two separate questions. It's the circuit voltages that determine the clearance requirements, while the current levels determine the width (and thickness) requirements.

Trace Width

Dealing with the latter first, it's the width and thickness of a copper trace on a PCB that determine its cross-sectional area, in the same way that the diameter does for an ordinary wire. The cross-sectional area determines its resistance per unit length, at which point, it's up to you to decide two things:

  • How much voltage drop (ΔV = I × R) can you tolerate from one end of the trace to the other?

  • How much heating of the trace (Power = I2R) can you tolerate?

One or the other of these will be the limiting factor for each trace.

For example, you might have "1 oz." copper on your PCB. This is shorthand notation for "1 ounce of copper per square foot", which translates to a thickness of 1.38 mils, or 0.035 mm. A trace that's 10 mils (0.254 mm) wide, then, has a cross-sectional area of 13.8 mil2 which is roughly equivalent to an AWG38 wire. It will have a resistance of about 0.75 Ω/ft. and the current capacity is on the order of 10s of mA.

To handle higher currents, you might select "2 oz." copper (0.070 mm thick) and use traces that are, say, 100 mils (2.54 mm) wide. This gives you a cross-sectional area of 276 mil2 which is roughly equivalent to an AWG24 wire.

Note that since the traces on a PCB are very flat and wide, they're actually much better at getting rid of heat to the enviroment than the equivalent circuilar wire is. So as far as I2R losses are concerned, you can put a lot more current through a PCB trace — but you still need to pay attention to the temperature rise and the associated thermal management.

Clearance

The required spacing between conductors is determined by the voltage difference between them and the amount of leakage current you can tolerate. Leakage current is primarily associated with surface contamination of the PCB (e.g., residual flux, as well as accumulated dust, moisture, etc.).

One guideline comes from safety testing services such as UL, which requires a creepage distance of 5mm per kilovolt for circuits that are supposed to be "isolated" from each other (material group I, pollution degree 2 from UL840).

Obviously, this guideline gives very small values for low voltages (0.05 mm or 0.002 in. at 10 V), so the limiting factor actually becomes the line/space widths that your PCB fab house is capable of.

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  • \$\begingroup\$ Side note: Sometimes you'll see a board that has some of the high-current traces built up with solder. This is less effective than you might think. The resistivity of solder is about ten times that of copper, which means that you would need to build up the solder to ten times the thickness of the copper (0.35 mm on top of 1 oz. copper, for the full width of the trace) just to cut the resistance in half. \$\endgroup\$ – Dave Tweed Jan 17 '17 at 20:41
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Besides Dave's excellent answer, you might want to look at IPC-2152 standard, which defines the "Standard for Determining Current-Carrying Capacity in Printed Board Design".

The sole industry standard for determining appropriate internal and external conductor sizes on printed boards as a function of the current carrying capacity required and the acceptable conductor temperature rise. This document provides guidance on how thermal conductivity, vias, copper planes, power dissipation and printed board material and thickness all factor into the relationship between current, conductor size, and temperature. 97 pages. Released August 2009.

It can be aquired here.

But there are handy calculators available like the Saturn PCB toolkit

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