My question is about another question on this site. On that question Z(in) for small signal analysis with BJT for unbypassed emitter and r0 in place Zin is asked for a common emitter configuration with emitter and ro is in place. My question is about voltage gain Av of the same circuit. In Boylestad and Nashelsky Av is given as below without a proof:

enter image description here

However, I find:

$$A_v=\frac{-R_C(\beta r_o-R_E)}{Z_b(R_E+R_C+r_o)}$$

Can you tell me what am I missing or show me how to find the correct answer?


Here is the circuit:

enter image description here

I used the following equivalent circuit for solution:

enter image description here

Then I defined \$V_o\$ as:

$$V_o=-I_c R_c$$

Then from KVL I obtained:

$$\beta I_b r_o-(I_b+I_c)R_E-R_C I_c-r_o I_c=0$$

$$I_c=\frac{(\beta r_o - R_E)}{R_E+R_C+r_o}I_b$$

Replacing \$I_c\$ from \$V_o\$:

$$V_o=-R_C \frac{(\beta r_o - R_E)}{R_E+R_C+r_o}I_b$$

and finally replacing \$I_b=V_i/Z_b\$:

$$A_v=\frac{-R_C(\beta r_o-R_E)}{Z_b(R_E+R_C+r_o)}$$

  • \$\begingroup\$ If you can post your procedure to get your result, it'll be better to tell where is wrong. \$\endgroup\$
    – diverger
    Nov 9, 2014 at 15:06
  • \$\begingroup\$ @diverger I included my solution. \$\endgroup\$
    – user57823
    Nov 9, 2014 at 15:41

2 Answers 2


Your result has no problem, and your procedure is all right. To verify this, you can substitute the \$Z_{b}\$

$$ Z_{b} = \beta r_{e}+[\frac{(\beta+1)+R_{c}/r_{o}}{1+(R_{C}+R_{E})/r_{o}}]R_{E} $$

into your result and into the textbook's result, and do some simplification, you'll find they give the same result. Cheers!

  • \$\begingroup\$ Are you completely sure? Because calculations get messy after some time and I can't verify this. \$\endgroup\$
    – user57823
    Nov 11, 2014 at 11:50
  • \$\begingroup\$ Yes, it's really a tedious work, but i'm sure you are right, just be confident! \$\endgroup\$
    – diverger
    Nov 11, 2014 at 12:12

The HF gain (\$C_1\$, \$C_2\$ short circuited) of this circuit is much simpler if you use Norton's theorem.

$$V_{out}=I_{cc}(V_{in}) \times R_{out}$$

$$I_{cc}=-\frac{V_{in} g_m r_0}{R_e+r_0(1+g_mR_e)}$$

$$R_{out}= R_c || [R_e+r_0(1+g_mR_e)]$$

$$A_{vHF}= -\frac{g_mr_oR_c}{R_c+R_e+r_0(1+g_mR_e)}$$

NOTE: \$g_m=I_c/V_t\$

\$R_b\$ does not play any role at HF.

the singularities introduced at lower frequencies by the capacitors are:

\$C_1\$ introduces a zero in \$s=0\$ and a pole in \$s=-1/[C_1(Z_b||R_b)]\$

\$C_2\$ introduces a zero in \$s=0\$ and a pole in \$s=-1/[C_2(R_{out}||R_c)]\$

I hope it can help to check your result


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