I have this CB amplifier circuit below

enter image description here

This is the small signal equivalent circuit arranged to find the output resistance R_out with the r_o included

enter image description here

Here is my analysis below to find the output resistance R_out

\begin{equation} v_x\:=\:r_o\left(i_x-g_mv_{\pi }\right)+R_e\left(i_x+\frac{v_{\pi }}{r_{\pi }}\right) \end{equation} \begin{equation} v_x\:=\:i_x\left(r_o+R_e\right)+\left(\frac{R_e}{r_{\pi }}-g_mr_o\right)v_{\pi } \end{equation} \begin{equation} v_{\pi }+R_e\left(i_x+\frac{v_{\pi }}{r_{\pi }}\right)\:=\:0 \end{equation} \begin{equation} therefore\:\:v_{\pi }\:=\:-\frac{r_{\pi }}{r_{\pi }+1}R_ei_x \end{equation} after some rearragements \begin{equation} \frac{v_x}{i_x}\:=\:R_{out}\:=\:r_o+R_e+\left(g_mr_o-\frac{R_e}{r_{\pi }}\right)\left(\frac{r_{\pi }R_e}{r_{\pi }+1}\right) \end{equation} I tried to manipulate my results but it does not come out to the same formula in the book which is below: \begin{equation} \frac{v_x}{i_x}\:=\:R_{out}\:=\:r_o+R_e\backslash \backslash r_{\pi }+\left(R_e\backslash \:\backslash \:r_{\pi \:}\right)g_mr_o \end{equation}

can somebody please tell what am doing wrong?? thanks in advance for all the help.


2 Answers 2


For those who are not familiar with a small-signal linearized T-model take look here: http://www.ittc.ku.edu/~jstiles/412/handouts/5.6%20Small%20Signal%20Operation%20and%20Models/The%20Hybrid%20Pi%20and%20T%20Models%20lecture.pdf


You made a mistake here

$$v_\pi+R_e \left( I_x+ \frac{v_{\pi}}{r_\pi} \right) = 0$$

Solve for \$v_\pi\$

$$v_\pi+R_eI_x+ \frac{R_ev_{\pi}}{r_\pi} = 0$$

$$v_\pi r_\pi + R_e I_x r_\pi + R_e v_{\pi} = 0$$

$$v_\pi r_\pi + R_e v_{\pi} = - I_x R_e r_\pi$$

$$v_\pi (r_\pi + R_e) = - I_x R_e r_\pi$$

$$v_\pi = - I_x \cdot \frac{ R_e \cdot r_\pi}{r_\pi + R_e} = -I_x\cdot R_x$$

Also, you can look here:

BJT common-base output resistance derivation

Calculation of output impedance of CE emitter bias configuration( unbypassed) with r_0

  • \$\begingroup\$ Woaw, comic sans. Nice. \$\endgroup\$ Jun 2, 2018 at 9:41
  • \$\begingroup\$ @G36, thank you so much for your help. I knew I was heading in the right direction. Turned out it's a simple algebra mistake. Thank you very much for spotting the error and appreciate your help. \$\endgroup\$
    – JordenSH
    Jun 2, 2018 at 13:46

Your dependent current source is in the wrong place in your AC equivalent diagram, it should be in parallel with r_o, going from C to E junction.

  • \$\begingroup\$ This is a screenshot from one of the most famous electronics textbooks, Sedra Smith. What are you talking about?? plus your reasoning is exactly what is on the diagram. \$\endgroup\$
    – JordenSH
    Jun 2, 2018 at 1:36
  • \$\begingroup\$ Accidents happen. If you look at the hybrid pi model even here on Wikipedia, you can see that the dependent current source in the equivalent circuit of a BJT goes between the C and E junctions, not C and B. \$\endgroup\$ Jun 2, 2018 at 1:58
  • 1
    \$\begingroup\$ But this is the T-model, and this is the convention used throughout the book so far. \$\endgroup\$
    – JordenSH
    Jun 2, 2018 at 2:02
  • \$\begingroup\$ @Raykh I don't have sedra/smith, nor have I read it. By "T-model" do you mean a linearized version of the "transport model" as shown here at three equivalent Ebers-Moll models of the BJT? Just curious. \$\endgroup\$
    – jonk
    Jun 2, 2018 at 2:43
  • \$\begingroup\$ @jonk these look like more of a device physics models. The T-model is another version for the hybrid pi model \$\endgroup\$
    – JordenSH
    Jun 2, 2018 at 3:10

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