# 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 $v_{o1} - v_{o2} = (A'_{v,dm} + A''_{v,dm}) v_{icm}$?

## 1 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 $A_{v,dm}=A'_{v,dm}+A''_{v,dm}$.

• 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 $A'_{v,dm} \ll A''_{v,cm}$? – BowPark Apr 19 '15 at 19:29
• @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. – Zulu Apr 19 '15 at 19:42