First, I drew the cascaded small signal equivalent circuit (unilateral hybrid pi transformation) for Figure 3 as:
After drawing the cascaded small signal equivalent circuit (unilateral hybrid pi transformation) of Figure 3, I determined that the output resistance \$R_o=R_{C2}\$ and the input resistance \$R_i=R_{B1}+r_{\pi}\$. Meanwhile, to find the voltage gain \$A_v=\frac{V_o}{V_i}=-G_mR_o\$, I first had to find an expression for the transconductance \$G_m\$. I began by writing an expression for the output current \$i_o=-g_mV_{EB2}\$. I needed to express \$V_{EB2}\$ in terms of \$V_i\$ and I did so using the following circuit analysis equations:
1) \$g_mV_{BE1}-\frac{V_{EB2}}{r_{\pi}}+\frac{V_{C1}}{R_{C1}}=0\$
2) \$\frac{r_{\pi}}{R_{B2}+r_{\pi}}V_{C1}=V_{EB2}\$
3) \$V_{BE1}=\frac{r_{\pi}}{r_{\pi}+R_{B1}}V_i\$
Eventually, \$V_{EB2}=-g_m(\frac{r_{\pi}}{r_{\pi}+R_{B1}}V_i)(\frac{r_{\pi}R_{C1}}{R_{B2}+r_{\pi}-R_{C1}})\$.
Plugging this into \$i_o=-g_mV_{EB2}\$ and solving for \$\frac{i_o}{v_i}=G_m\$, \$G_m={g_m}^2\frac{{r_{\pi}}^2R_{C1}}{(r_{pi}+R_{B1})(R_{B2}+r_{\pi}-R_{C1})}\$ so that \$A_V=-G_mR_o=-\frac{{g_m}^2{r_{\pi}}^2R_{C1}R_{C2}}{(r_{pi}+R_{B1})(R_{B2}+r_{\pi}-R_{C1})}\$ where \$r_{\pi}=\frac{\beta}{g_m}=\frac{100}{0.01{\Omega}^{-1}}=10000\Omega\$. My problem is that the denominator factor \$(R_{B2}+r_{\pi}-R_{C1})=30000\Omega+10000\Omega-40000\Omega=0\Omega\$ so that the gain \$A_v\$ becomes negative infinity. Is there something wrong in my process? Any help please?
