Question 6.SP.5: The CE tee-equivalent circuit of Fig. 6-10(b) is suitable fo......

(a) The appropriate small-signal equivalent circuit is given in Fig. 6-12.    By KVL around the B,C loop, with r_m = αr_c (from Problem 6.3),
v_{B} = r_{b}i_{b} + r_{m}i_{b} + (1  –  \alpha)r_{c}(i_{b}  –  i_{e}) = (r_{b} + r_{c})i_{b}  –  (1  –  \alpha)r_{c}i_{e}             (1)

Application of KVL around the C, E loop, again with r_m = αr_c, gives
0 = r_{e}i_{e}  –  r_{m}i_{b}  –  (1  –  \alpha)r_{c}(i_{b}  –  i_{e}) + \frac{R_{E}\,R_{L}}{R_{E} + R_{L}}\,i_{e} = -r_{c}i_{b} + \left[r_{e} + (1  –  \alpha) r_{c} + \frac{R_{E}R_{L}}{R_{E}  +  R_{L}}\right]i_e       (2)

By Cramer’s rule applied to the system consisting of (1) and (2), i_e = \Delta_{2} /\Delta, where
\Delta = r_b \left[r_{e} + (1  –  \alpha) r_{c} + \frac{R_{E}R_{L}}{R_{E}  +  R_{L}}\right] + r_c \left(r_{e} + {\frac{R_{E}R_{L}}{R_{E}  +  R_{L}}}\right)
\Delta_2 = r_c v_B

Now, by Ohm’s law,
v_{L} = (R_{E}||R_{L}) i_{e} = \frac{R_{E}R_{L}}{R_{E} + R_{L}} \,\frac{\Delta_{2}}{\Delta}

Then                 

(b) The input impedance can be found as Z_{\mathrm{in}} = R_{B}\parallel (v_{B}/i_{b}).    Now, in the system consisting of (1) and (2), by Cramer’s rule, i_b = \Delta_{1} /\Delta, where
\Delta_{1} = \Big[r_{e} + (1  –  \alpha)r_{c} + \frac{R_{E}R_{L}}{R_{E}  +  R_{L}}\Big] v_{B}

Hence,                Z_{\mathrm{in}} = R_{B}\parallel \left({\frac{\Delta}{\Delta_{1}}}\,v_{B}\right) = \frac{R_Br_b \Big[r_{e}  +  (1  –  \alpha)r_{c}  +  \frac{R_{E}R_{L}}{R_{E}  +  R_{L}}\Big]  +  R_B r_c\Bigl(r_{e}+{\frac{R_{E}R_{L}}{R_{E}+R_{L}}}\Bigr)}{(R_B  +  r_b) \Big[r_{e}  +  (1  –  \alpha)r_{c}  +  \frac{R_{E}R_{L}}{R_{E}  +  R_{L}}\Big]  +  r_c\Bigl(r_{e}+{\frac{R_{E}R_{L}}{R_{E}+R_{L}}}\Bigr)}