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

Bridge — the series

3 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · applied
  2. 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.

    part 2 · applied
  3. 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.

    part 3 · applied

All series