Connecting strain gauge output terminals to an instrumentaiton amplifier

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.

• What is your source for Figure 2 and your description of its behavior? Dec 9 '13 at 22:53
• It is just a thought experiment. Which configuration is used in real life? I wonder if Figure2 doable or exists? Dec 9 '13 at 22:54
• You're overstating problems associated with common mode gain and instrumentation amps. I suggest you peek at a data sheet and run the numbers. Resistor mismatch and issues with mounting your strain gage will provide much realer problems to deal with! Dec 9 '13 at 23:15
• No, figure 2 is not very likely to exist, as you'd be losing many of the benefits of the instrumentation amplifier-- unless of course the box M contains an instrumentation amplifier, in which case you don't need a second instrumentation amplifier, as your signal would no longer be differential. Dec 9 '13 at 23:21
• why would one lose the benefits could u give hint? Dec 9 '13 at 23:22

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.

• do u mean most of the offset comes from the noise with unbalanced line impedances instead of common-mode voltage? did i understand u well enough? so u just hook up the terminals to the amplifier inputs without using any other circuitry for handling the common-mode voltage? Dec 9 '13 at 23:17
• Do you use any sense techniques on your excitation voltage? Dec 9 '13 at 23:19
• re u asking to me or to andy aka? Dec 9 '13 at 23:24
• @ScottSeidman We use mainly current fed bridges and that means excitation sense wires aren't needed. Even if they were we calibrate out errors when the gauges get put on the machines and wired up. Dec 10 '13 at 9:12
• @user16307 exactly - the gauges get wired directly to the instrument amplifiers. The common mode voltages are largely rejected by the CMRR of the amps but even if they weren't the likely common mode voltage is 50Hz AC and this is easily dealt with. Maybe you can give an example of how sensitive your application is? Dec 10 '13 at 9:15

There are two ways of handling this. If the bridge is accessible, very often a potentiometer is used to replace one of the static resistors to balance the circuit at the desired zero.

If the offset can't be removed at the bridge, the next safest is to remove it AFTER the instrumentation amplifier with a simple difference amp. If you try to remove it before the inamp, you lose many of the benefits of using an instrumentation amplifier. Of course, if the offset is big enough to saturate then in-amp, you must do something more dramatic.

Further, amps designed for remote bridges often have specialized "sense" circuitry used in conjuction with a feedback circuit to keep excitation voltages (and grounds) rock solid.

http://www.futek.com/files/PDf/Manuals_and_Technical_Documents/Transducer%20Interfacing%20handbook%20by%20AD.PDF is a great resource to peek through.

• but if one wants to remove the offset caused by the common mode voltage after the inamp then one needs to measure it. how to measure that? Dec 9 '13 at 23:10
• what do you mean by offset? is it the "common-mode gain * common-mode voltage" = vcm*Acm? do u mean this as offset? Dec 9 '13 at 23:12
• First, offsets due to common mode voltage are likely to be MUCH SMALLER than offsets in your bridge. To deal with these very real offsets, and not the nearly zero common mode rejection ration problems, especially with something like a strain gage, you must have a zeroing and calibration procedure. Dec 9 '13 at 23:13
• To learn what I mean about offset, try taking 1% off the value of one of your resistors and watch what happens. Common mode gain is NOT the problem with Wheatstone bridges. Dec 9 '13 at 23:16
• thnx for the book. i ll look into it. Dec 9 '13 at 23:23

Yes, it's quite possible to drive the strain gauge such that the midpoint is driven to "close to zero". In practice, you'd drive the top and bottom of the bridge in such a way as to get the desired excitation and the average of the two voltages at the output is zero. You'd still go directly into an instrumentation amplifier type arrangement (or perhaps a single-ended amp, but an in-map might be lower noise) with the bridge legs.

This technique is used in some instrumentation. Below is an example of what I am talking about.

Rather than driving one bridge leg to zero as shown, the two legs could be buffered and averaged (two equal value resistors between them) and that junction driven to zero, which is quite similar to the situation you describe. Then an in-amp would not have to deal with the common mode voltage, and most of the noise in driving that input to zero would appear as common mode signal to the in-amp.

• Could provide a citation for this figure, please? Feb 8 '14 at 2:17