Electrical – Wheatstone Bridge – strain gauge sensor

multimeterresistorsstrain-gagewheatstone-bridge

I'm trying to optimize a Wheatstone bridge with a strain gauge sensor and I'm considering all possible wire resistances. The circuit is shown in the following figure:

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I don't understand the comment below the figure. I did not realize why we ignored Rwire1.

Best Answer

Imagine the same figure but with the voltmeter connected straight across to the terminal of R2. In this case the voltmeter sees:

Vright = I x Rwire1 + I x Rgauge + I x Rwire2

at it's right terminal.

By connecting the voltmeter directly to the terminal of Rgauge, as in the figure, the voltmeter sees:

Vright = I x Rgauge + I x Rwire2

at it's right terminal.

Consider that a small current, Ivm, does flow through the voltmeter. Then in the figure the voltmeter sees:

Vright = I x Rgauge + I x Rwire2 + Ivm x Rwire3.

Ivm may be either positive or negative but it should be in the nA or even pA range. Consequently, the voltage drop across Rwire3 should be vanishingly small.

In all equations, I is the current flowing through the gauge, the voltages are with respect to the negative terminal of the battery.

This technique is called Kelvin sensing (I think it was invented by Lord Kelvin).

Now just keep wire 2 short. If that is not possible, you could compensate for the length of wire 2 by putting the same length of wire between R3 and the battery's negative terminal.