I'm using a small value (~60milliohm) and small-package (0402) resistor to measure current. When I probe the resistance with a (nice Agilent) DMM, the resistance is more like 100millohm though and it's messing up my current sensing. I've zeroed out the probes, I've measured across five or so different samples, and the resistance ranges around 100 +/- 15mOhm.

The resistor is placed very close to a current-sense amplifier enter image description here

Are these very susceptible to the reflow procedure? If heat up the resistor too much does the resistance increase?

  • \$\begingroup\$ How about details of the resistor rather than the sense amp? \$\endgroup\$
    – Andy aka
    Jun 2, 2014 at 18:21
  • 2
    \$\begingroup\$ What about your PCB layout? If your sensing nodes are not exactly at the resistor pads you might be picking up extra resistance due to the copper traces \$\endgroup\$
    – Kvegaoro
    Jun 2, 2014 at 18:30
  • \$\begingroup\$ Can you send a photo of it in circuit too? \$\endgroup\$
    – Will
    Jun 2, 2014 at 18:32
  • 1
    \$\begingroup\$ Are you using 4-wire sensing? Both when measuring with the multimeter and when measuring in-circuit. \$\endgroup\$
    – The Photon
    Jun 2, 2014 at 18:44
  • 3
    \$\begingroup\$ The layout is terrible, but it doesn't seem to be enough to cause the magnitude of error you are reporting. Forget about trying to measure resistance directly. Put 500mA through the traces the way it normally would travel and measure the voltage right at the pins of the chip (not on the resistor) using your Agilent DMM. Then divide the two to get the equivalent resistance. If the trace is contributing significantly you'll get lousy performance because copper has a large tempco. \$\endgroup\$ Jun 2, 2014 at 21:05

2 Answers 2


Corroborating Kvegaoro, if the layout isn't ideal, you will not get the intended 60mohm. Also consider that the land pad and soldering quality will impact your measurement.

Consider that without a true Kelvin force/sense, you will not reliably measure 60mohm. The resistance of the probe leads will be at least that much.

You can attempt to create a kelvin setup by soldering some wires right at the SenseAmp pins and force 100mA and then measure the voltage drop. Do the same for 200mA. The slope is the actual resistance.


From the picture you provided, It seems that the problem is actually you board layout rather than your competent. The problem is on the bottom tap from the sense amplifier, it should be coming straight from the resistor's pad as it in on the top one. However, in the bottom one is coming to the trace a few mils before the resistor's pad. This is introducing the resistance of the small section of the trace, which depending on you trace width, copper weight and length of that trace it could be significant compared to 60mOhm. You can look for a trace resistance calculator online to calculate the trace resistance and recalculate the expected output of the sense amplifier taking into account the trace resistance and see if the results are consistent.

  • \$\begingroup\$ Problem is, a trace calculator says the layout is not the problem. A 2mm wide trace 5 mm long with 1 ounce copper is only 1.2 mohm. \$\endgroup\$ Jun 2, 2014 at 22:24
  • \$\begingroup\$ The feedback on layout is still helpful though. Thanks. \$\endgroup\$
    – tarabyte
    Jun 3, 2014 at 0:27
  • \$\begingroup\$ @WhatRoughBeast 1.2 is 2% of 60. Depending on the intended system precision, that could still be a problem. No one has mentioned the specified precision of the 60mOhm resistor, however. \$\endgroup\$
    – RBerteig
    Feb 3, 2016 at 2:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.