# Modeling diode connected transistor in differential amplifier

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?

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,

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