Sorry for asking you this, but I could find no answer in the documentation of this calibrator.
Maybe someone can explain it to me...
We got a Fluke 5700 as a replacement for the old Datron 4808 for a while, and now we have massive (out of tolerance) problems calibrating some multimeters in the lower ohms ranges, that were calibrated one year before using the Datron 4808. Normally nearly all used to be "in tolerance" after one year.
The only way to get the correct reading on several Yokogawa 7551 in 100-Ohms-range:
EX SNS on calibrator: ON 2-wire-comp on calibrator: ON NULL/Relative on instument = set on calibrator "0" and ON
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Is this the correct way to calibrate an instrument in 2-wire-Ohms mode and how does the 2-wire-compensation of the 5700 works?The 4808 calibrator I used before only had "2-w" and "4-w" modes (REM SENSE on and off).
The nominal values for 2-w were higher than the nominals for 4-w (about 0.5 Ohms, what is the resistance of the Pomoma cable used in calibration of the calibrator plus calibrator internal wiring to "Rx").
This is plausible to me.
This is how I understood 2-w/4-w on
Datron 4808:
5700: (where is the 2-wire-comp-device, how does it work?)

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My
speculation: 2-wire-comp-device, how 2-wire-comp
could work:

The 5700 uses the sense-wiring from instrument to calibrator sense to measure voltage drop by instuments measurement current I and connects a voltage source with same voltage in series but in reversed polarity into the current path... (sorry for my confused writing...)
In other words: the resulting voltage drop along the current path equals zero, so voltage at the instruments input is equal voltage over Rx.
Anybody agree or disagree?