I want to use a shunt resistor to measure the current flowing though a section of my circuit. I am deciding between using an off the shelf shunt resistor or simply drawing a wide track on my PCB and calculating the resistance (and hence voltage drop) of that section.

What are the pros and cons or each method, and from your experience which method is best?. For my application the accuracy is not highly important (avg of 10A flowing and I need to detect current to an accuracy of 1A).

From what i can see the advantage of using an off the shelf shunt resistor is the precision and accuracy of the resistance, however it is more costly and the requirement of soldering will add uncertainty to the resistance.

Drawing a section of copper track on my PCB seems cheaper and eliminates the uncertainly brought by the solder joint.

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    \$\begingroup\$ I will be producing approximately 30 PCBs. Is it necessary to measure each PCB or is a calculation based on the track dimension sufficient? @PlasmaHH \$\endgroup\$ Feb 9, 2016 at 13:55
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    \$\begingroup\$ @Marla a thin track will not really be possiable considering I will be pushing an average of 10A across the track. However at 10A the voltage drop accross even a small resistance will be possiable to measure using my equipment \$\endgroup\$ Feb 9, 2016 at 13:57
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    \$\begingroup\$ @user3095420: That is something you have to tell; depending on the quality of your fab house there can be a significant error in the resistance, which may or may not be higher than the error you can tolerate (whatever that is). Should not be hard to find a 1% 1W 10mΩ resistor below 30¢ \$\endgroup\$
    – PlasmaHH
    Feb 9, 2016 at 14:01
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    \$\begingroup\$ Keep in mind that copper has a rather large temperature coefficient, too. An actual resistor will be more accurate over a wider temperature range. \$\endgroup\$
    – Dave Tweed
    Feb 9, 2016 at 14:04
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    \$\begingroup\$ I think we're having a slight confusion-of-perspective here. what I think @Marla meant is that you'll need a narrow track compared to the width of track you would normally use for carrying a 10A load. The thinner section in the trace would psrform as your shunt resistor & no, I don't think that mass-produced pcbs would have >10% variance in trace conductivity, so checking a few from every batch should work. \$\endgroup\$ Feb 9, 2016 at 14:08

1 Answer 1


The three largest errors for using an etched copper resistor are the copper thickness, etching tolerance on the copper width, and the temperature coefficient of the resistivity.

The fab house ought to be able to give you tolerances for the first two.

If the board contains plated thru holes, there will be a build up of copper on your shunt track, that will not be as well controlled in thickness as the original copper. It will help both track thinness and thickness tolerance if you can get the shunt area masked before hole plating.

At 10A, you will need a wide track, so the effect of lateral tolerances will be small.

The resistivity tempco of copper is about 0.4% per degree. That's a 10% change in 25 degrees. That means a copper current shunt will give you an 'indication' rather than a 'measurement'. You could measure the temperature of the board and correct for it, but that sounds a lot more trouble then specifying a low tempco resistor in the first place.

If you can tolerate the uncertainty in resistance from all these terms, then an on-board copper track is a cheap and cheerful way to get what you want.

If you want higher accuracy and go for a soldered component, consider using a '4 terminal' resistor. That will reduce uncertainties around track and solder conductivity, and geometry details.

  • \$\begingroup\$ Having the board area masked before plating would probably make a real shunt resistor look cheap, at a quick guess. \$\endgroup\$
    – Ecnerwal
    Feb 9, 2016 at 14:25
  • \$\begingroup\$ great summary. Based on all the contributions so far It looks like purchasing a shunt is both cheaper and more accurate. \$\endgroup\$ Feb 9, 2016 at 14:31

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