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The Keithley 2450 sourcemeter has two ways you can do a 4-probe measurement: in the front there are four banana connection ports labeled SENSE HI, SENSE LO,FORCE HI, and FORCE LO. In the back, there are four tri-ax connections with similar labels.

My confusion is that co-ax and tri-ax cables usually carry the signal line in their inner conductor and the return line in the outer conductor, as I understand it, so I would think these cables would carry the FORCE/SENSE HI and LO, and you would only need two triax cables to do a four probe measurement. What am I not understanding?

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The triaxial cables are for low-current, high-precision measurements in the nanoamp range. From the 2450 datasheet:

Current specs showing a "triax-only" footnote for the lowest current values

The real-panel diagram shows how the conductors are connected. The second (middle) conductor is not merely another ground:

Rear panel shot of the 2450 showing the triax connections

Cinch Connectivity has an app note that describes what's going on. The goal is to prevent leakage current through the cable insulation. This is done by applying a buffered copy of the force voltage to the middle conductors. This ensures that there's (almost) no voltage across the insulator surrounding the center conductors, which carry the real force and sense signals. Here's a diagram from the app note:

Force-measure block diagram showing triaxial guard conductors

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  • \$\begingroup\$ Thank you for all this information. It makes sense to hold them at the same voltage so that there’s less leakage current. Why do they use this buffer method to produce a near identical voltage, rather than just connecting the inner conductor and the shield to the same voltage source? \$\endgroup\$
    – Roy
    Commented Feb 6 at 14:50
  • \$\begingroup\$ @Roy If there's a grounded outer conductor (for EM shielding) and a voltage on an inner conductor, then there's going to be leakage no matter what. The idea is to make sure that none of that leakage comes from the force or sense lines. If a small current leaks from the buffered voltage source, it doesn't matter, because that voltage isn't used in the measurement. \$\endgroup\$
    – Adam Haun
    Commented Feb 6 at 17:31

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