# Differential gain of amplifier with current mirror

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} \$$?

• Downvoting because the the link to the schematic is broken Commented Apr 11, 2022 at 14:10
• courses.ece.wpi.edu/ece4902/lectures/DiffAmp_rev_slides.pdf Commented Jun 22, 2022 at 20:04
• @BrianCannard Thank you so much for the link! Commented Jun 22, 2022 at 21:05

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 $A_{v,dm}=A'_{v,dm}+A''_{v,dm}$.
• Thank you for your very useful answer. I didn't provide an expression for $R_{out}$, but we can state that looking into output 2 it is about $r_{o4} || r_{o2}$, the output resistances (due to channel length modulation) of M4 and M2. It could be in kOhms. The $R_{out}$ relative to the output 1 instead could be $r_{o1} || 1/g_{m2} || r_{o2}$ and the total resistance could be $\simeq 1/g_{m2}$, so much smaller than kOhms. Is this the reason why you state that $A'_{v,dm} \ll A''_{v,cm}$? Commented Apr 19, 2015 at 19:29