Concept

Kelvin connection — where it appears

A four-terminal measurement in which current is forced through one pair of leads and voltage read through another, so the leads' resistance is not in the answer. It does not remove the leads but moves their error out of the current path, where it is first order, and into the voltage path, where the voltmeter's own input impedance divides it.

Named by 4 essays across 2 fields — each of them below, with the objects they name alongside it.

Measuring with 50 mΩ of lead in each wire. computed by solving, not by drawing at 61 resistances, twice each. The two-wire arrangement measures the leads too, so its error is 2×50 mΩ over whatever is being measured: one per cent at 10 Ω, and 10000% at 1 mΩ. The four-wire arrangement senses on a separate pair that carries almost no current, and its error stays under 1.0e-2% across the whole range.

Two terminals measure the leads as well

Fifty milliohms in each lead makes a two-wire measurement one per cent high at ten ohms, ten per cent high at one ohm, and a hundred per cent high at a tenth. Not approximately — the reading is the resistance plus the leads, and below about ten ohms most of what is being reported is the wire between the instrument and the thing.

instruments · Four-terminal $A tenth of a per cent of lead is 999 µε of strain that is not there. computed by solving, not by drawing. A 350 Ω quarter bridge at zero load, against the resistance in each lead. With both leads in the changing arm the output is 4.99500 mV at 350 mΩ — an apparent fractional change of 0.1998%, which at a gauge factor of 2 is 999 microstrain. With one lead in that arm and one in the arm beside it the output is zero to the last bit, at every lead resistance drawn. The lower panel is what the second arrangement costs: the sensitivity falls as 1/(1 + Rlead/R), which is 0.100% at the same lead and is a calibration constant rather than a drift.$

The leads that are in the bridge

A strain gauge on the end of two long wires cannot be told from a strain gauge under load: both leads in the changing arm is bit for bit the same netlist as a quarter bridge whose fractional change is larger by twice the lead over the gauge, which at 350 milliohms on a 350 ohm gauge is 999 microstrain that is not there. Moving one of those leads into the arm beside it leaves the output at exactly zero for every lead resistance drawn, and turns a twenty-kelvin drift of 76.8 microstrain into 0.077. That factor is two over the strain being read — 999 at a thousand microstrain and 9990 at two hundred — and it costs 0.100 per cent of sensitivity.

applied · Bridge

Four wires against a 10 MΩ voltmeter. computed by solving, not by drawing at 81 resistances, twice each, with a voltmeter of 10 MΩ and 50 mΩ in every lead. The four-wire error is not zero: it is the voltmeter's own divider, −(R + 2R_lead)/(R + 2R_lead + R_m), which grows with the resistance being measured rather than shrinking. The two-wire error is that same quantity plus the leads, so it passes through zero at 1000 Ω — where the reading is right to 1.8e-12 while the four-wire reading is 0.0100% low — and above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance.

The voltmeter four wires do not remove

A four-terminal measurement is described everywhere as removing the leads from the answer. It moves them. What is left is the voltmeter's own input resistance, and it grows with the resistance being measured rather than shrinking: with a ten-megohm voltmeter and fifty milliohms of lead, the four-wire reading is 0.0100 per cent low at a kilohm, where the two-wire reading is exactly right — 1.8 × 10⁻¹² — because its lead error and its loading error cancel. Above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance.

instruments · Four-terminal

Two leads nobody counts cost 1995 parts per million, and two more leads remove all of it. computed by solving, not by drawing. The error in the reported strain against the resistance in each of the two excitation leads, at 1000 µε on a quarter bridge of 350 Ω excited at 10 V. With four wires the instrument takes the excitation to be the supply's voltage, so every reading is scaled by R/(R + 2r): 1995 parts per million at 0.35 Ω, 54029 at 10. Two more leads brought back from the bridge's own terminals, carrying only the instrument's input current, leave 3.5e-4 parts per million. Exciting with a current instead of a voltage does the same thing with no extra leads at all, because the lead resistance is in series with a source that does not care.

The two leads nobody counts

The leads that are in the bridge moved a lead out of the changing arm and turned a 999-microstrain error into nothing. The two leads carrying the excitation are still there, and they scale every reading: 0.35 ohms each on a 350-ohm bridge is 1,995 parts per million, exactly −2r/(R + 2r), the same on a full bridge as on a quarter one, and drifting 0.153 microstrain over twenty kelvin — twice what the three-wire fix left behind. Two more wires brought back from the bridge's own terminals leave 0.00035 parts per million. So does exciting the bridge with a current, which needs no extra wires at all.

applied · Bridge

Named alongside it

The objects these essays reach for when they reach for this one.

Four-terminal sensing Lead resistance Excitation Loading Low resistance measurement Measurement error Strain gauge Temperature coefficient Wheatstone bridge Desensitivity Design tradeoff Input impedance

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