I have several questions about the current sensing through a resistance. The resistances are 2-SMD package and I will use the 4-wire kelvin method for the layout.(see image) Specifically, there are three resistors, one of 1 mOhm and another two of 10 mOhm.the layout would be approximately like this: enter image description here The PCB is two layers with components only in the top layer and the ground plane in the bottom layer. Due to restrictions in the layout, I can not find a way to route the sense tracks through the upper layer since they cut different power planes, so I will have to use vias and to route the sense tracks in the bottom layer. Due to my limited knowledge of electronics, I am presented with the following questions:

  1. Each sense track will route using two vias (top to bottom layer and then bottom to top layer). As long as the resistance of each sense tracks are the same, should the use of vias not affect the accuracy of the measurement, right?
  2. Route the sense tracks in the bottom layer (where is located the ground plane), would it affect the accuracy of the measurement?
  3. Should I increase the trace width to minimize the effects of the EMI?


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    \$\begingroup\$ Excellent layout, excellent innovation. \$\endgroup\$ Commented Nov 26, 2018 at 17:42

1 Answer 1


If your sense circuit is well designed (high input impedance say > 1K- and good common mode rejection) it should be quite insensitive to resistance in the sense lines, so matching is not required and the resistance of the vias and traces is of little importance. Of course it would be better in general to use parts with four connections but your method probably is pretty good.

Cutting the ground plane may have other effects, especially if high current traces cross the cut, but sense lines should not be affected.

Trace width is of little importance if your sense circuit is well designed, but you may wish to keep them close together (specifically minimize the loop area of the differential input) to minimize susceptibility. Usually the circuit is quite low impedance due to the shunt resistor source so it's not really that critical.


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