Circuits that do a job, and the range they do it over

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.

19 min read 7 figures The schematic is a label Measured, not tabulated Every model has an edge

Assumes: The bridge that is linear near one point · Two terminals measure the leads as well

The bridge that is linear near one point measured a Wheatstone bridge’s second-order term, found it to be exactly half the fractional change, and then put resistance in the leads and watched the range collapse. At three and a half ohms in each lead the fractional change at which the standard expression is one per cent out fell from 2.020 per cent to 0.020 — a hundredfold — and the essay reported that a voltage-driven bridge there has no usable range at all.

That number is right and the reading of it is not, and the difference is the whole of this essay. The collapse is a scale factor, which every strain instrument calibrates out on the bench before it measures anything. What the leads actually cost is something else, it does not calibrate out, and which of the two a bridge pays depends not on how large the leads are but on where they are.

What is being solved

Four resistors, an excitation, and a difference read between the two midpoints — the same netlist what a network answers is built on, solved by nodal analysis with no expression substituted anywhere. One arm carries the gauge at $R(1+\delta)$ ; the others are at $R$ .

The leads are two more resistances and the question is which arms they land in.

A gauge at the far end of a cable has one wire carrying excitation to it and one carrying excitation away. If both of those wires return to the same node of the bridge — the two-wire connection — both lead resistances are in series with the gauge, and the arm is $R(1+\delta) + 2R_\text{lead}$ .

If a third wire is run, so that the excitation arrives on one wire, leaves on another, and the bridge’s midpoint is sensed on a third that carries no current — the three-wire connection — then one lead resistance is in the gauge’s arm and the other is in the arm beside it. Each of the two left-hand arms is larger by $R_\text{lead}$ and the gauge’s arm is larger by that and by $R\delta$ .

Those are two different netlists made of the same components, and the second is not a smaller version of the first.

The third wire is worth being exact about, because it is the one component in this circuit that does nothing. It runs from the gauge’s junction with the excitation return to the bridge’s midpoint, and the bridge’s midpoint is where a high-impedance amplifier reads. It therefore carries no current, it drops no voltage, and it is not an element of the netlist at all — its whole function is to move where the node is. The gauge’s arm and the arm beside it each get one current-carrying lead, and what the third wire buys is that the node between them sits at the gauge rather than at the instrument.

That is why the arrangement cannot be described as a smaller lead resistance. Nothing about the copper has changed. What has changed is which arm of the matrix each piece of it is stamped into.

The two-wire case needs no new element

$R(1+\delta) + 2R_\text{lead}$ is $R(1 + \delta + 2R_\text{lead}/R)$ , so the two-wire arrangement is the same netlist as a quarter bridge with no leads at all whose fractional change is larger by $2R_\text{lead}/R$ . That is an identity rather than an approximation, and the figures below check it by stamping the two-wire bridge as five elements and comparing the solve against the shifted one: they agree to the last bit at every lead resistance and every strain drawn.

Which means the instrument cannot separate them, and this is not a limitation of a particular instrument. The lead resistance and the strain enter the matrix in the same place, so no measurement made at the bridge’s own terminals contains the information needed to tell them apart — the same argument two terminals measure the leads makes about an ohmmeter, arriving here as a strain rather than as a resistance.

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. — Fig. 1 The two arrangements at zero load. With both leads in the changing arm the output is 4.99500 mV at 350 mΩ of lead, which the bridge’s own tangent reads as a fractional change of 0.1998 per cent — 999 microstrain at a gauge factor of two. With one lead in each of the two left-hand arms the output is zero to the last bit at every lead resistance drawn. The lower panel is the sensitivity, which falls as one over one plus the lead over the gauge in both cases.

A tenth of a per cent of the gauge’s own resistance in each lead — 350 milliohms of copper, which on the rung below’s own scale is a metre of thin instrument wire there and back — reads as a full-scale strain on a gauge rated for a thousand microstrain. Three and a half ohms reads as 9901 microstrain and ten ohms as 27,778. These are not corrections; they are larger than the quantity being measured.

The reproduction, which is a zero

The three-wire connection is the one worth checking first, because its claim is an exact cancellation and an exact cancellation is the kind of claim a solve can refuse.

At zero strain the two left-hand arms are $R + R_\text{lead}$ each, so their divider is one half whatever $R_\text{lead}$ is, and the right-hand divider is one half because both its arms are $R$ . The output is a difference of two halves. The solve returns 0.0000 volts — not a small number, the number zero — at 50 milliohms, 100, 350, one ohm, three and a half, and ten.

That is the calibration this essay’s later numbers are quoted against, in the way one step computed twice insists on: a bridge arrangement whose answer must be exactly right, so that the arrangements which are not can be quoted against something rather than against each other.

Separating the scale factor from the curvature

Now the correction to the rung below.

A bridge with leads in it has two departures from a straight line and they are different objects. One is the departure from $V\delta/4$ , the expression written for a bridge with no leads, and it contains the sensitivity the leads have taken away. The other is the departure from this bridge’s own tangent at balance, which is what a calibration establishes and what a nonlinearity means.

Fig. 2 Three and a half ohms in each lead, three-wire. At a fractional change of 1.0 per cent the departure from Vδ/4 is 1.466 per cent and the departure from the bridge’s own tangent is 0.481 — against 0.488 for a bridge with no leads at all. The gap between the two curves is 0.9901 per cent and it is the same at every change. The slider is the lead resistance.

The gap is constant because it is a sensitivity: the three-wire bridge’s slope at balance is 2.4752474 volts per unit of fractional change against 2.5 with ideal leads, which is $1/(1 + R_\text{lead}/R)$ to six digits and is 0.9901 per cent low. Subtract it and what is left is the curvature, and the curvature is not merely similar to the lead-free case — it is slightly better. At a fractional change of twenty per cent the departure from the tangent is 9.0090 per cent with three and a half ohms of lead against 9.0909 with none, because the second-order term is $\delta/2$ divided by the same $1 + R_\text{lead}/R$ .

That slope is worth one more sentence, because it is the whole of what a three-wire lead does and it is a single expression that the solve returns rather than is given. Both left-hand arms grow by $R_\text{lead}$ , so the divider’s denominator is $2(R + R_\text{lead}) + R\delta$ where it was $2R + R\delta$ , and the numerator’s $\delta$ term is untouched. The output is therefore $V R \delta / 2(2R + 2R_\text{lead} + R\delta)$ : a slope divided by $1 + R_\text{lead}/R$ , and a second-order term divided by the same thing. The measured slopes are 2.4975024 at 350 milliohms of lead, 2.4928774 at one ohm and 2.4752474 at three and a half, against 2.4999999 with none — which is 0.09991, 0.2849 and 0.9901 per cent, and $1/(1 + R_\text{lead}/R)$ gives the same three numbers.

One expression, one denominator, and both of the things the leads do come out of it. That is why the two curves in the figure above are parallel rather than merely close, and it is why the ratio between them can be quoted as a single number at every fractional change instead of as a curve.

A 350 Ω bridge, with 3.50 Ω in each lead. The straight line is the expression every textbook gives, Vδ/4; the curve is the solve. They part company at half the fractional change — 1.478% at 1.0% — so the tangent is worth one per cent only up to 0.020%. Driving the bridge from a current source instead halves the departure at every point and moves that edge to 2.040%. — Fig. 3 The same bridge drawn the way the rung below drew it, against Vδ/4: 1.478 per cent out at a fractional change of one per cent, with the one per cent edge at 0.020 per cent and the current-driven edge at 2.040. Every number here is correct; what it is measuring is that the tangent belongs to a different bridge.

So the sentence to carry from the rung below is narrower than it was written. Lead resistance in a three-wire bridge takes sensitivity and nothing else. A gain error of a per cent is a calibration constant, and a strain instrument calibrated with a shunt resistor across one arm measures it in the same operation that establishes the span.

350 mΩ of lead is a scale factor, not a nonlinearity. computed by solving, not by drawing. A three-wire quarter bridge with 350 mΩ in each lead. The upper curve is the departure from the bridge's own tangent at balance — 0.485% at 1.0% — and it is the same shape and very nearly the same size as the departure with no leads at all. The lower curve is the departure from Vδ/4, the expression that has no term for a lead: 0.584% at the same change. The gap between them is 0.0999% and it is the same at every change, because it is the sensitivity the leads cost and a sensitivity is calibrated out. — Fig. 4 Ten times less lead, and the two curves have nearly converged: 0.584 per cent against Vδ/4 and 0.485 against the bridge’s own tangent at a fractional change of one per cent, with 0.0999 per cent of sensitivity between them. The curvature is the same curve at both lead resistances.

What does not calibrate out

Copper’s resistance rises with temperature, and it does so by a number that is not a fitted parameter: $\alpha = 1/(234.5 + 25)$ per kelvin from 25 °C, which is 0.3854 per cent per kelvin. A lead is a piece of copper, so a lead resistance is a temperature measurement that the bridge cannot distinguish from the one it was installed to make — unless it is in an arm where the change cancels.

Twenty kelvin on the leads: 76.8 µε one way, 0.077 the other. computed by solving, not by drawing. Copper's resistance rises by 0.3854% per kelvin, so 20 K adds 27.0 mΩ to a 350 mΩ lead. Read against a real strain of 1000 µε, that shift is 76.8 µε of apparent strain with both leads in the changing arm and 0.077 µε with one lead in each — a factor of 999. The two are not the same quantity made smaller: the first is an offset that does not know what the strain is, and the second is a span error proportional to it. — Fig. 5 Twenty kelvin on the leads, read against a real strain of 1000 microstrain. With both leads in the changing arm the reading moves by 76.8 microstrain; with one lead in each of the two left-hand arms it moves by 0.077 — a factor of 999. At three and a half ohms of lead the two are 753 microstrain and 0.761, a factor of 990. The two curves are parallel because both drifts are proportional to the lead; what separates them is that only one of them is proportional to the strain.

The two drifts are not the same quantity made smaller, and the figure’s most useful feature is their opposite signs. The two-wire drift is an offset: twenty kelvin adds 27.0 milliohms to a 350 milliohm lead, which adds to the arm exactly as strain does, and it does so whether the gauge is loaded or not. The three-wire drift is a span error: the two left-hand arms grow together, the balance is untouched, and all that changes is the denominator that sets the slope — which is why it is negative and why it is proportional to the reading rather than added to it.

An offset of 77 microstrain on a structure at rest looks exactly like a load. A span error of 0.0077 per cent on a full-scale reading is invisible beside the gauge factor’s own tolerance. That is the whole case for the third wire, and it is a factor of a thousand rather than a factor of two.

The factor is almost independent of the lead and it is not independent of the strain, and that is the most useful thing in the figure. Across the whole sweep it is 999.0 at 350 milliohms, 997.2 at one ohm and 990.1 at three and a half — so no lead length makes the two arrangements converge. But change the strain being read and it moves in proportion: at 200 microstrain the two arrangements differ by 9990, at 1000 by 999, and at 10,000 by 99.9.

The reason is in the two mechanisms and it is worth stating as an expression, because it is the essay’s one general result. The two-wire drift is an offset of $2\Delta R_\text{lead}/R$ that does not contain the strain at all; the three-wire drift is a span error of $\Delta R_\text{lead}\,\delta/R$ that is proportional to it. Their ratio is therefore $2/\delta$ , which the solve returns to a part in a thousand at every combination drawn.

So the third wire’s advantage is largest exactly where a strain gauge is hardest to use, and it shrinks as the reading grows. A gauge worked near its full thousand microstrain gets a factor of a thousand from it. One reading a hundred microstrain — a stiff member, a load cell well below rating, a thermal expansion — gets ten thousand, and cannot be built any other way.

It is also worth reading the numbers as a temperature range rather than as a drift, because that is how a specification is written. A gauge rated at a thousand microstrain full scale, wired with two wires and 350 milliohms of lead, gives up one per cent of full scale for every 2.6 kelvin the cable moves. Wired with three, it gives up one per cent for every 2,600. The cable is the same cable.

Measuring with 500 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 2×500 mΩ over whatever is being measured: one per cent at 100 Ω, and 100000% 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. — Fig. 6 The same argument on a bare resistance rather than a bridge, from the instruments field. Five hundred milliohms in each wire makes a two-wire measurement one per cent out at 100 Ω and a hundred thousand per cent out at a milliohm, while a four-wire measurement stays inside 0.01 per cent across the whole range — because the sense pair carries almost no current and therefore drops almost nothing.

The bridge’s third wire and the ohmmeter’s fourth are the same idea and they are not the same arrangement, which is worth being precise about. Four-wire sensing removes the lead drop by measuring somewhere the current is not. The three-wire bridge does not remove it at all — the lead is still carrying excitation and still dropping volts — it puts an equal drop in the arm next door so that the difference the bridge reads does not contain it. The first is an avoidance and the second is a cancellation, and a cancellation is only as good as the match, which is the section below.

The half bridge, which the leads treat differently again

The rung below established that a half bridge — two gauges in opposite directions, which any bending member provides — removes the second-order term exactly rather than approximately. That result survives the leads intact.

The leads take sensitivity from a half bridge and nothing else. computed by solving, not by drawing. Two arrangements of the same four resistors, each measured against its own tangent at balance so that the sensitivity the leads cost is out of the comparison. The quarter bridge departs by 4.757% at a tenth, at every lead resistance drawn; the half bridge departs by 2.6e-10, which is the floor of the arithmetic. Both lose the same 0.100% of sensitivity to 350 mΩ of lead, so the choice between them is not about the leads at all — it is about the second gauge. — Fig. 7 Both arrangements measured against their own tangents, so that the sensitivity the leads cost is out of the comparison. The quarter bridge departs by 4.757 per cent at a tenth, at every lead resistance drawn; the half bridge departs by 2.6 × 10⁻¹⁰, which is the floor of the arithmetic. Both lose exactly the same 0.100 per cent of sensitivity to 350 milliohms of lead.

Two things in that figure are worth separating. The half bridge is exactly linear with leads in it, which is not obvious: putting a resistance in each of the two arms that are changing in opposite directions leaves their sum at $2R + 2R_\text{lead}$ , and a sum with no $\delta$ in it is a denominator with nothing to expand. And the two arrangements lose the same sensitivity to the same copper, so the choice between a quarter and a half bridge is not a choice about leads at all — it is a choice about whether there is a second gauge and somewhere to put it.

The excitation, which halves it

One more component decides how much of the lead resistance is paid for. Driving the bridge from a current source instead of a voltage source halves the desensitisation exactly: three-wire leads of 350 milliohms cost 0.09991 per cent of sensitivity driven by ten volts and 0.04998 per cent driven by twenty milliamperes, and at three and a half ohms the two are 0.9901 and 0.4975 per cent.

The reason is the same one the rung below gave for the nonlinearity. A voltage source lets the total current fall as the bridge’s total resistance rises, so the leads cost once through the division and once through the current; a current source holds the total fixed and only the division changes. It is the same change of one component that doubled the linear range, buying the same factor of two on a different quantity — the pattern the reading a data sheet does not take collects, where the instrument’s own arrangement is a larger term than the part’s specification.

What it does not say

It does not say the third wire makes lead resistance disappear. It makes the matched part of it disappear. The cancellation above is exact because both leads are the same resistance and the same temperature, and two wires in one cable at the same length are very nearly that — but a joint, a connector, or one wire running past something warm is not. What survives is the mismatch, and it enters exactly as the two-wire case does: a millidegree of difference between the leads is a millidegree of apparent strain in the same proportion. The three wires convert a common quantity into a differential one, which is the same trade what matching does about temperature prices for a transistor pair.

It does not say the sense wire is free. It carries no current only if what it is connected to draws none, and an instrumentation amplifier’s input bias current through several ohms of lead is a real offset — small here, and the reason the voltmeter four wires do not remove exists.

And it does not say the numbers transfer to a full bridge. All four of a full bridge’s arms are at the far end of the cable, so the leads carry the excitation into the whole bridge rather than into one arm of it, and the desensitisation is then a property of the excitation and not of the arrangement — which is exactly why a full bridge is the one that gets sense wires of its own.

What this opens

The obvious next question is the mismatch, and it has a shape this collection already knows. The three-wire cancellation converts a common lead resistance into a differential one, so what reaches the reading is the difference between two wires rather than either of them — which is the same conversion, with the same failure mode, that the millivolts in the wire measures for a return path. The number that would settle it is the apparent strain per milliohm of mismatch, and a stamped netlist with one arm a milliohm larger than the other returns it: 1.4286 microstrain per milliohm, at 350 milliohms of lead and at three and a half ohms alike, to six digits. A mismatch enters as half a two-wire lead does, and — this is the useful half — it does not know how long the cable is. That is a specification on the cable’s uniformity rather than on its resistance, and it is not a number cable is sold by. A ten-milliohm difference, which is one crimp, is 14.3 microstrain and puts the three-wire connection back where the two-wire one was at a twentieth of the lead.

The second is the excitation’s own leads, which this netlist does not have. The bridge above is driven at its own top node; a real one is driven through the same cable, and a voltage source at the far end of two more wires is a voltage source with a series resistance — which desensitises without cancelling and is the reason a six-wire connection exists at all. That is a divider with a load on it, which is what the divider, and the thing it does not know about is for, and the machinery to solve it is one more pair of elements.

The number worth carrying

Both leads in the changing arm: 999 microstrain of apparent load at 350 milliohms, drifting 76.8 microstrain over twenty kelvin. One lead in each of the two left-hand arms: zero apparent load at every lead resistance, drifting 0.077 microstrain over the same twenty kelvin, at a cost of 0.100 per cent of span. The ratio between the two drifts is $2/\delta$ — a thousand at full scale on a thousand-microstrain gauge and ten thousand at a tenth of it — and it is bought with one wire.

The habit that goes with it is about what a departure is being measured from. Every model has an edge, and $V\delta/4$ ’s edge is not a fractional change at all — it is a schematic that has no lead in it. Measuring the real bridge against that expression mixes a calibration constant with a curvature and reports their sum as a range, which is how a hundredfold collapse turns out to be a per cent of gain. The departure worth quoting is always the one from the instrument’s own tangent, and finding that tangent is a solve rather than a formula.