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:

enter image description here

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


1 Answer 1


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.

  • \$\begingroup\$ Thanks for the answer. Why do I need to use a Wheatstone bridge to measure small changes in resistance? I could put a voltmeter in parallel with the strain gauge and read the value of the voltmeter. I do not understand why the websites say that the Wheatstone bridge provides greater precision. \$\endgroup\$ Mar 16, 2019 at 16:37
  • 2
    \$\begingroup\$ @CarmenGonzález in a word the answer is "gain". If you just connect a voltmeter across the strain gauge, the minimum voltage will be something like 1.2V and the maximum will be something like 1.3V. These are purely made-up values to illustrate my point. You will have a difficult time trying to get a high-resolution view of the strain gauge value. With the Wheatstone bridge you can balance the voltage range to something like -0.05V to +0.05V, so you can greatly increase the gain on the voltmeter and get a high-resolution view of the strain gauge value. \$\endgroup\$
    – Mr. Snrub
    Mar 16, 2019 at 19:23
  • \$\begingroup\$ @Mr.Snrub But why does the Wheatstone Bridge provide better precision? And, from the resistance differences, how can I know the deformation of the sample? \$\endgroup\$ Mar 17, 2019 at 0:26
  • 1
    \$\begingroup\$ @CarmenGonzález As described above, I would say that the Wheatstone Bridge doesn't really "provide better precision", it enables a higher precision measurement because you can apply a much higher gain to the output of a Wheatstone Bridge than you can to a regular measurement of the strain gauge. (But you could also do the same trick with an instrumentation amplifier, just apply a large DC offset to center the signal around zero before applying the gain.) \$\endgroup\$
    – Mr. Snrub
    Mar 17, 2019 at 0:47
  • \$\begingroup\$ @CarmenGonzález Regarding your second question I'm no expert on strain gauges, but I'm sure there is a published relationship of resistance versus applied strain. (It is probably dependent on the axis of the strain as well.) \$\endgroup\$
    – Mr. Snrub
    Mar 17, 2019 at 0:48

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