Power’s relationship to Resistance

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

The first thing you should notice is that from a small-signal preceptive point of view \$Q_1\$ \$r_{o1}\$ act just like the emitter resistance for \$Q_2\$

And the small signal mode will look like this

^{simulate this circuit – Schematic created using CircuitLab}

The small-signal parameters are:

\$\large g_{m2} = \frac{0.1\textrm{mA}}{25\textrm{mV}} = 4\textrm{mS}\$

\$\large r_{\pi 2} = \frac{\beta}{g_{m2}} = 25\textrm{k}\Omega\$

\$\large r_{o2} = r_{o1} = \frac{V_A}{\textrm{I}} =\frac{5 \textrm{V}}{0.1\textrm{mA}} = 50 \$

Hence the output resistance will be the same as I show here:

BJT common-base output resistance derivation

finding output resistance of CB amplifier with ro

$$R_{OUT} = \frac{V_X}{I_X} = r_{o2}+\left(1+ g_{m2} r_{o2}\right)\cdot r_{o1}||r_{\pi2} = 3.4\textrm{M}\Omega $$

Also, remember that in the small-signal analysis you can replace PNP with the NPN small-signal model.

^{simulate this circuit}

^{simulate this circuit}