Circuit Response
every model has an edge, and the figure carries it
Concept

Lead resistance — where it appears

The resistance of the wires between an instrument and what it measures, which a two-terminal reading includes and a four-terminal one does not. Fifty milliohms in each wire is most of a one-ohm reading, which is the whole reason a four-terminal measurement exists.

Named by 7 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 The millivolts in the wire, which are nobody's signal. computed by solving, not by drawing at 145 frequencies. A 10 mm run of one-ounce copper carries the return of a 100 mA load and the reference of a 10 mV sensor. Its 5.00 mΩ and 10.0 nH put 500.0 µV in series with the sensor at low frequency — 5.00% of the reading — rising a decade per decade above 79.6 kHz until at 1.59 MHz the error is the whole signal. The full solve and the interfering current times the shared impedance agree to 4.4e-7. The slider is the length of the shared run: both the resistance and the inductance are proportional to it, so every point on the curve moves down together and the corner stays at 79.6 kHz.

The millivolts in the wire

Ten millimetres of one-ounce copper is five milliohms and ten nanohenries, and if a hundred-milliamp load and a ten-millivolt sensor both return through it, half a millivolt of somebody else's current is added to the reading — five per cent of it, before anything has been amplified. Above 79.6 kilohertz the error rises a decade per decade with no ceiling, and shortening the shared run moves the whole curve down and the corner not at all.

instruments · Common-impedance The same millivolts, subtracted — and what four resistors leave behind. computed by solving, not by drawing at 145 frequencies. A 10 mm run of one-ounce copper carries the return of a 100 mA load and the reference of a 10 mV sensor. Its 5.00 mΩ and 10.0 nH put 500.0 µV in series with the sensor at low frequency — 5.00% of the reading — rising a decade per decade above 79.6 kHz until at 1.59 MHz the error is the whole signal. The full solve and the interfering current times the shared impedance agree to 4.4e-7. The slider is the length of the shared run: both the resistance and the inductance are proportional to it, so every point on the curve moves down together and the corner stays at 79.6 kHz. The second curve is the same measurement made differentially: a unity-gain difference amplifier across the sensor, its four resistors carrying the worst-case skew 0.1% allows. It divides the error by 500.5 — 53.99 dB against a closed-form (1 + G)/(4·tolerance) of 53.98 dB — at every frequency, because four resistors have no frequency in them. What is left is 998.9 nV, which is not zero: the amplifier subtracts the interference and the resistors decide how much of it survives.

The rejection four resistors decide

Measuring a ten-millivolt sensor against a ground that somebody else's hundred milliamps is also using puts five hundred microvolts of their current into the reading. Subtracting the two ends of that conductor with a difference amplifier removes it — by 53.99 decibels with one-tenth-per-cent resistors, against a closed form of 53.98, which is a factor of five hundred and not a removal. The amplifier has nothing to do with it: the number is one plus the gain over four times the resistor tolerance.

instruments · Common-mode rejection A 350 Ω bridge, with ideal leads. The straight line is the expression every textbook gives, Vδ/4; the curve is the solve. They part company at half the fractional change — 0.498% at 1.0% — so the tangent is worth one per cent only up to 2.020%. Driving the bridge from a current source instead halves the departure at every point and moves that edge to 4.040%.

The bridge that is linear near one point

The expression for a Wheatstone bridge's output is V delta over four, and the solve says it is V delta over two times two plus delta — low by half the fractional change, exactly, at every change tested. So the tangent is worth one per cent only up to a two per cent change and a tenth of a per cent only up to two tenths. Driving the bridge from a current source instead halves the departure at every point and doubles both edges, which is a change of one component and no change at all to the four resistors being measured.

applied · Bridge 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 sensingKelvin connectionLoadingExcitationStrain gaugeWheatstone bridgeCommon-impedanceLow resistance measurementMeasurement errorModel rangeReturn currentTemperature coefficient

All concepts