Electronic – finding output resistance of CB amplifier with ro

amplifiercircuit analysiscommon-baseoutputresistance

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.

Best Answer

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

http://www.ittc.ku.edu/~jstiles/412/handouts/5.6%20Small%20Signal%20Operation%20and%20Models/section%205_6%20%20Small%20Signal%20Operation%20and%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