How can the CMRR of an opamp be negative?
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 &ndash; 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:

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

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:

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

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.