Connecting strain gauge output terminals to an instrumentaiton amplifier

gaugeinstrumentation-amplifierstrain-gage

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Before posing the question the figures are about as follows:

In Figure 1 a strain gauge's outputs v1 and v2 is connected to an instrumentation amplifier’s inputs. For simplicity buffering part is not drawn and I'm assuming the signals are balanced with equal line impedance. Let’s say I measure v1 and v2 with respect to ground such as: v2=0.390mV and v1=0.400mV. So that v1-v2=0.005mV. If the inamp is not ideal and obtaining common-mode voltage vcm = (v1+v2)/2= 0.395mV will affect the output voltage since every non-ideal amplifier has common mode gain.

In Figure 2 the same outputs from the same strain gauge goes into a mysterious circuit called M which converts v1 and v2 to v1’ and v2’ such as v1’= (v1-v2)/v and v2’=-(v1-v2)/2. So that v1’= 0.0025mV and v2'=-0.0025mV so that again v1-v2=0.005mV. But in this case obtaining common-mode voltage vcm = (v1+v2)/2= 0mV will not affect the output voltage since it is zero.

Now my question is: Which way is used in real life? If exists what is the mysterious circuit M in my Figure 2 which gets rid of the the common-mode voltage? I’m asking this question because we are using instrumentation amplifiers often for strain gauges.

Best Answer

I design strain gauge amplifiers and use a fair amount of gain and my observations are that imbalances in the bridge resistors are the biggest source of error. Having said that I go for amps that have good CMRR because the gauges are wired up to 10m distant and can pick up a fair amount iof noise. I don't use anything in between the bridge and IA.

If you're using quarter bridge active devices, as per your diagram I use constant current feed for excitation because it's twice as linear ie fewer theoretical errors.

I also use auto balance mechanisms on the REF pin of the IA fed from a DAC.