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First, I drew the cascaded small signal equivalent circuit (unilateral hybrid pi transformation) for Figure 3 as: enter image description here

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?

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  • \$\begingroup\$ 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 \$\endgroup\$ – G36 Feb 19 '17 at 14:11
  • \$\begingroup\$ @G36 Edited. My small signal equivalent circuit can now be seen above. \$\endgroup\$ – John Smith Feb 19 '17 at 15:11
  • \$\begingroup\$ 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? \$\endgroup\$ – glen_geek Feb 19 '17 at 15:38
  • \$\begingroup\$ @glen_geek Yes, but I don't get why the "second-stage input voltage Veb2 is gmVbe1/2." What do you mean? \$\endgroup\$ – John Smith Feb 19 '17 at 15:45
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    \$\begingroup\$ 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. \$\endgroup\$ – G36 Feb 19 '17 at 16:54
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Givem gm is 0.01 mho [or 100 ohms emitter 'reac'], knowing gm = 1A/V at 26mA, we know your Ic is 0.26mA in each transistor. With beta=100, the rin of each transistor is (100=B * 100=reac) = 10,000 ohms.

The load on collector of Q1 is 40Kohm || (30Kohm + 10Kohm) = 20Kohm. Gain of stage1 is gm * effectiveRC (ignoring Vearly effects), thus 0.01 * 20K == 200x. Scaled by voltage division of Rb1 and beta*reac = 4:1, thus only 50X.

The load on collector of Q2 is 4Kohm. Av is gm*Rc = 0.01 * 4K = 40X. But again we have that input 4:1 divider, so only 10X.

Total gain is 50 * 10 = 500x.

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