Modeling diode connected transistor in differential amplifier

Question

I have question about the analysis of common-mode gain of BTJ differential amplifier with active load on the Sedra Smith book. In this book, the differential-mode gain \$Ad\$ is derived based on the transconductance \$Gm\$ and output resistance \$Ro\$, \$Av=Gm*Ro\$. These are represented in the figure below.

Notice transistor \$Q_3\$ which is connected as a diode. In the equivalent circuit, \$Q_3\$ is represented as the resistance (\$r_{e3}\$//\$r_{o3}\$). Where \$r_{e3}\$ represents the resistance viewed from the emitter, and \$r_{o3}\$ is the Early resistance of \$Q_3\$.

Now for common-mode analysis, we have the following equivalent circuit for determining the gain. Notice now that the diode transistor \$Q_3\$ is represented by the resistance (\$r_{e3}\$//\$r_\pi\$//\$r_{03}\$), where \$r_\pi\$ is the input resistance of \$Q_3\$.

My questions are: Why is \$r_\pi\$ used to represent \$Q_3\$ in common-mode? Is it ok to use (\$r_{e3}\$//\$r_\pi\$//\$r_{o3}\$) for differential-mode too?

Answer 1

Without more context, I can't answer the question of why different expressions are used but do note that

$$r_e = \frac{1}{g_m}||r_{\pi}$$

so the expressions are, in fact, equivalent.

To see this, recall

$$r_{\pi}= \frac{\beta}{g_m}$$

Thus,

$$\frac{1}{g_m}||r_{\pi} = \frac{r_{\pi}}{\beta}||r_{\pi} = \frac{r_{\pi}}{1 + \beta} = r_e$$