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} \$?


1 Answer 1


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}\$.

  • \$\begingroup\$ 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}\$? \$\endgroup\$
    – BowPark
    Apr 19, 2015 at 19:29
  • 2
    \$\begingroup\$ @BowPark: Yes, that's exactly it. The impedance of output 2 is usually more than an order of magnitude greater than the impedance of output 1. \$\endgroup\$
    – Zulu
    Apr 19, 2015 at 19:42

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