I followed ThePhoton's suggestion in the comments.

Here is how to solve this:

First take offset by tying inputs together and to the GND and measure the output voltage Vout. In this case -39.3478mV as shown below:

So this should be subtracted from any output voltage after any measurement. And when I do that gain looks 800 constant:

*I suppose if a strain_gage is mounted and fixed, an offset measurement can be taken before any moving measurement(without the procedure above). Voffset is the output voltage when the inputs are tied to GND.

Imagine when mounted the offset stress of the gage is X amount(Newton) and the voltage is Vxo = V1+Voffset . If the stress is varied ΔX amount there will be increase in output voltage as ΔV and the output will be V2=(V1+Voffset)+ΔV.
Now V2-V1 = ΔV. So the amplifier's Voffset is included in the first offset already.

Neil hit things on the head. I'll try and write it up a little.

The paper you are looking at is a little aged and the terms they use and the way they use them may be a little unfamiliar. First off, take a look at the wikipedia page on the topic: Common-mode rejection ratio. There you will see the following equation:

$$V_o=A_d\left(V_{\left(+\right)} - V_{\left(-\right)}\right)+\frac{1}{2}A_{cm}\left(V_{\left(+\right)} + V_{\left(-\right)}\right)$$

Let's fabricate a somewhat fuller schematic:

^{simulate this circuit – Schematic created using CircuitLab}

(To get their nominal gain of 100, \$R_1=R_3=49.5\cdot R_2\$. And yes, I did take note that they adjusted the gain towards its nomimal value by making changes in \$R_5\$.)

If you take some measurements of the inputs and the output of each of the three opamps while setting \$V_{CM}=0\:\textrm{V}\$ and \$V_D=1\:\textrm{mV}\$, and do this a second time with say \$V_{CM}=100\:\textrm{mV}\$, then you can solve for both the differential gain values as well as the common mode gain values for each of the opamps. And these values will not only be different, they will also differ in sign for \$A_d\$ for each.

The following solution set to use for each opamp would be:

$$\begin{align*}
A_d&=\frac{V_{O_1}\left(V_{\left(+\right)_2}+V_{\left(-\right)_2}\right) - V_{O_2}\left(V_{\left(+\right)_1}+V_{\left(-\right)_1}\right)}{\left(V_{\left(+\right)_1}+V_{\left(-\right)_1}\right)\left(V_{\left(-\right)_2}-V_{\left(+\right)_2}\right)-\left(V_{\left(+\right)_2}+V_{\left(-\right)_2}\right)\left(V_{\left(-\right)_1}-V_{\left(+\right)_1}\right)}\\\\
A_{cm} &= \frac{V_{O_2} \left(V_{\left(-\right)_1} - V_{\left(+\right)_1}\right) + V_{O_1}\left( V_{\left(+\right)_2} - V_{\left(-\right)_2}\right)}{V_{\left(-\right)_1} V_{\left(+\right)_2} - V_{\left(-\right)_2} V_{\left(+\right)_1}}
\end{align*}$$

In the above pair of equations I used the subscript of '1' to indicate the first measurement with, say, \$V_{CM_1}=0\:\textrm{V}\$ and the subscript of '2' to indicate the second measurement with, say, \$V_{CM_2}=100\:\textrm{mV}\$. I suspect they made these kinds of measurements in order to arrive at their values.

Given the above solution equation set and the circuit arrangement, then it's actually true that there will be a negative value for \$A_{d_{A1}}\$ and a positive value for \$A_{d_{A2}}\$ and a negative value for \$A_{d_{A3}}\$, with this arrangement. Nothing magical here.

Yes, this doesn't comport with the usual meaning for the computation of decibels today, where the differential gain is always taken as positive. But in this case, I think they were actually making voltage measurements using a voltmeter and using the above purely mathematical solution approach to computing \$A_d\$. In that case, you can and will get negative values for \$A_d\$, given this topology.

I think that's all it means.

I don't want to try and evaluate a paper I haven't read, though. But at least I can see how they were able to arrive at values such as those you mentioned. The sign is all about the topology here.

To get the equation solutions I gave earlier, let's say we assume there exists a value \$A_d\$ known as the differential gain and that there exists a value \$A_{cm}\$ known as the common mode gain. Let's say that we want to work out what those values are.

Well, we have two unknowns so we will need two equations:

$$\begin{align*}
V_{O_1}&=A_d\left(V_{\left(+\right)_1} - V_{\left(-\right)_1}\right)+\frac{1}{2}A_{cm}\left(V_{\left(+\right)_1} + V_{\left(-\right)_1}\right)\\\\
V_{O_2}&=A_d\left(V_{\left(+\right)_2} - V_{\left(-\right)_2}\right)+\frac{1}{2}A_{cm}\left(V_{\left(+\right)_2} + V_{\left(-\right)_2}\right)
\end{align*}$$

If you simultaneously solve those two equations for \$A_d\$ and \$A_{cm}\$, treating everything else as measurements you made, then you will get the solution equations I provided earlier.

## Best Answer

With ground at two remote points there will inevitably be ground currents (from other equipment also) circulating that cause interfering volt drops between sending end and receiving end. This produces an error and degrades your signal. That’s usually regarded as the primary issue.

A secondary issue is that the impedance seen on hot and common wires from the perspective of electric or magnetic field interference is vastly different. This means interference cannot be coped with to any significant degree. In the original circuit, the impedance to ground on both wires is generally equal.