# Solving for the Voltage Gain in Cascaded Two-Stage Amplifier

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?

• First, please show us the small signal equivalent circuit. Also, why don't you try another approach. 1 - find the Q1 gain and Q2 gain. Av1 = gm1*Rc1||(Rb2+rpi2) and Av2 = gm2*Rc2. And finally, I can include rpi1 and rpi2 effect on the gain. Av3 = rpi1/(Rb1+rpi1), A4 =rpi2/(Rb2+rpi2). Av = Av1*Av2*Av3*Av4 – G36 Feb 19 '17 at 14:11
• @G36 Edited. My small signal equivalent circuit can now be seen above. – John Smith Feb 19 '17 at 15:11
• Recognize that Rc1 = (Rb2 + Rpi). Current source gmVbe1 splits equally between Rc1 and (Rb2 + Rpi). So your second-stage input voltage Veb2 is gmVbe1/2. Have you run afoul of signs? – glen_geek Feb 19 '17 at 15:38
• @glen_geek Yes, but I don't get why the "second-stage input voltage Veb2 is gmVbe1/2." What do you mean? – John Smith Feb 19 '17 at 15:45
• As glen_geek point out the gm1*vbe1 current will splits equally between Rc1 and (Rb2 + Rpi). This means that Vbe2 = 0.5*(gm1*Vbe1)*rpi2. Or from current divider rule we have (gm1*Vbe1)*(Rc1)/(Rc1+Rb2+rpi2) = I_rpi2 current. I ignore the minus sign, and Vbe2 = I_rpi2*rpi2. – G36 Feb 19 '17 at 16:54