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Background I have a strain gauge that I need to obtain measurements from. The product page indicates that the sensor is a half bridge configuration (one active/one compensating element). The sensor has three wires: White (common), Red (active) and Black (compensating). I have never used a strain gauge before, so I may have some fundamental misconceptions about how it works.

My Understanding of the gauge is that I need to measure the resistance and use the equation: \$strain = (R - R_0)/(G*R_0)^2\$, where \$R\$ is the measured resistance, \$R_0\$ is the gauge resistance, and \$G\$ is the gauge factor. I do this for both leads (active and compensating) and subtract the two to find the actual strain.

What I have Tried Using an Agilent Data Acquisition unit, I have successfully measured the resistance of both leads as ~120\$\ \Omega \$, which is expected, and calculated a strain of essentially zero.

My Problem started when I tested the sensor by heating the it (attached to a metal plate) to 50C. I expected the strain to increase, but was surprised to see that it actually decreased. Now I am questioning my basic understanding of how this sensor works. The change was very small, but distinct from the noise (below image). Strain Over Time

Additionally, the noise I am getting from the sensor is less than ideal. I suspect some of it is caused by the furnace I am using, as there is a noticeable increase in noise when I turned it on (below image). Update The increased noise level remains even after de-energizing the furnace, so I now believe the increase is caused by the temperature. Noise Induced by Furnace

Questions

  1. Shouldn't strain increase as the temperature increases?
  2. Is this level of noise normal? How can I reduce it?
  3. Am I using the strain gauge correctly?
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  • \$\begingroup\$ What are the coefficients of thermal expansion for your plate and the strain gauge? As you heat, does the plate get bigger faster than the gauge, or vice versa? Some noise is unavoidable...how much noise can your application tolerate compared to the level of noise you have now? Can you average multiple readings? How noisy is your power supply? \$\endgroup\$ – Elliot Alderson Dec 13 '19 at 19:30
  • \$\begingroup\$ The plate is regular stainless steel and the sensor is Hoskins Alloy 875 (can't find coefficient). The gauge is designed to be used at high temperatures, and was attached to the plate by the manufacturer of the sensors so I presume they have similar coefficients. I'm using the Agilent's built in power supply, which should be fairly clean. \$\endgroup\$ – Rekamanon Dec 13 '19 at 19:47
  • \$\begingroup\$ If the plate and gauge have "similar coefficients" then there should be no stress due to heating. You haven't answered the most important questions: how much noise can you tolerate? Can you average samples? How noisy is your power supply? Only you can answer these questions. \$\endgroup\$ – Elliot Alderson Dec 13 '19 at 19:49
  • \$\begingroup\$ Ideally there would be less noise, but I can live with it as is. Yes, I plan on implementing a moving average to smooth it out. \$\endgroup\$ – Rekamanon Dec 13 '19 at 19:52
  • \$\begingroup\$ The only way your test makes sense is to show noise and temperature drift in the context of signal size. You can't know if noise and drift are too big without considering what you're trying to measure. Kudos for thoroughness, but show what you're trying to measure so we can assess your signal chain. \$\endgroup\$ – Scott Seidman Dec 16 '19 at 13:59
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schematic

simulate this circuit – Schematic created using CircuitLab

The OEM website does not show any useful information. I would call or write to them.

But my guess would be to wire the measurement instrument like this and put a plastic RF cap across the input with shielded twisted pair cable.

There is no indication on sensitivity or polarity.

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There is something wrong with the equation \$\text{strain} = (R - R_0)/(G*R_0)^2\$, where G is the gage factor, which is: $$ G= \frac{\Delta R / R}{\Delta \ell / \ell}$$

The denominator is \$ \Delta \ell / \ell\$, which is the definition of strain. Thus, when we take the denominator of your expression, it is \$ \left( \Delta R / \text{strain}\right)^2 \$, and the numerator is simply \$ \Delta R\$, and your units aren't working out. The denominator should not be squared, and the correct relationship is: $$\text{strain} = (R - R_0)/(G*R_0)$$

After you unsquare the denominator, you will see a linear relationship between \$\Delta R\$ and \$G\$. The output of your gage, however, is the result of a voltage division. Assuming the passive resistor is of the same value of the strain gage resistance under zero strain,

$$ V_{\text{out}} = V_{\text{excitation}}\frac{R + \Delta R}{2R + \Delta R}$$

assuming the active element is on the bottom, and the relationship between output and strain will be nonlinear, even if you subtract \$ V_{\text{excitation}}/2 \$. Of course, this approaches linearity if \$ R \gg \Delta R \$.

Other than that, the only way your test makes sense is to show noise and temperature drift in the context of signal size. You can't know if noise and drift are too big without considering what you're trying to measure. Kudos for thoroughness, but show what you're trying to measure so we can assess your signal chain.

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  • \$\begingroup\$ I also found the squaring of the denominator odd, but I was following an equation I found via KeySight (the new name of Agilent) keysight.com/upload/cmc_upload/All/…. I followed this advice because I am using their logger to take measurements. \$\endgroup\$ – Rekamanon Dec 16 '19 at 16:52
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    \$\begingroup\$ The equation is clearly wrong. Strain is unitless, and their equation shows the units of strain to be \$\Omega^{-1}\$ \$\endgroup\$ – Scott Seidman Dec 16 '19 at 17:23

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