Consider this pdf document, pages 58-59-60. It is a differential amplifier with a current mirror as active load.
According to that document, if I take the unbalanced output in the right-hand branch (drain of M2), the transconductance gain is \$ g_m \$, while if I take the unbalanced output in the left-hand branch (drain of M1), the transconductance gain is \$ g_m / 2 \$. It is because the current of M2 and the current of the mirror are both entering the M2 drain, as regards the differential mode signal.
Let \$ v_{o1} \$ and \$ v_{o2} \$ be respectively the M1 drain voltage and the M2 drain voltage.
If \$ R_{out} \$ is the output resistance of this amplifier looking into both \$ v_{o1} \$ and \$ v_{o2} \$, the voltage differential gain is different in the two nodes, being \$ A'_{v,dm} = g_m R_{out} / 2 \$ for \$ v_{o1} \$ and \$ A''_{v,dm} = g_m R_{out} \$ for \$ v_{o2} \$.
First question: Wasn't this circuit perfectly symmetrical?
Moreover: the outputs can be written as
$$v_{o1} = A_{v,cm} v_{icm} + A_{v,dm} \displaystyle \frac{v_{idm}}{2}$$
$$v_{o2} = A_{v,cm} v_{icm} – A_{v,dm} \displaystyle \frac{v_{idm}}{2}$$
where the two input were
$$v_{i1} = v_{icm} + \displaystyle \frac{v_{idm}}{2}$$
$$v_{i2} = v_{icm} – \displaystyle \frac{v_{idm}}{2}$$
(\$ v_{icm} \$ is the common-mode signal component; \$ v_{idm} \$ is the differential-mode signal component)
Second question: What does happen if \$ A_{v,dm} \$ is different between the two output nodes? Should I consider \$ v_{o1} – v_{o2} = (A'_{v,dm} + A''_{v,dm}) v_{icm} \$?
Best Answer
First question: No, the circuit isn't perfectly symmetrical. The current mirror performs a differential- to single-ended conversion. If you wanted a perfectly symmetrical circuit, you would make M3 and M4 current sources, and then use some sort of common-mode feedback to set the appropriate current (so that \$V_{O1}\$ and \$V_{O2}\$ stay in a usable range).
The way it is now, \$A'_{v,dm}=\frac{g_m}{2g_{md}}\$ (where \$g_{md}\$ is the transconductance of the diode-connected device M3), while \$A''_{v,dm}=g_mr_{out}\$. Remember that the impedance seen looking into M3 is simply \$\frac{1}{g_{md}}\$. These two gains are obviously very different: \$A'_{v,dm}\ll A''_{v,dm}\$. The gain \$A'_{v,dm}\$ is so small that it's pretty useless, so the signal \$v_{o1}\$ is usually ignored, and a single, single-ended output on \$v_{o2}\$ is used.
Second question: I already said it, but \$A'_{v,dm}\$ is generally so small that \$v_{o1}\$ is ignored and isn't fed to the next gain stage (or output, or whatever follows). This means that the gain is simply \$A_{v,dm}=A''_{v,dm}\$. If you really want to, though, you can take both \$v_{o1}\$ and \$v_{o2}\$ as outputs to get \$A_{v,dm}=A'_{v,dm}+A''_{v,dm}\$.