Question 7.SP.11: Move capacitor CS from its parallel connection across RS2 to......

(a) The equivalent circuit (with current-source JFET model) is given in Fig. 7-18.   By KVL,
$v_{g s} = v_{i}  –  v_{L}$                      (1)

Using $v_i$ and $v_L$ as node voltages, we have
$i_{i} = {\frac{v_{i}  –  v_{L}}{R_{G}}}$                (2)

Now let              ${\frac{1}{R_{e q}}} = {\frac{1}{r_{d s}}} + {\frac{1}{R_{S2}}} + {\frac{1}{R_{L}}} = {\frac{1}{30  \times  10^{3}}} + {\frac{1}{1.2  \times  10^{3}}} + {\frac{1}{1  \times  10^{3}}} = {\frac{1}{536}}$

By KCL and Ohm’s law,
$v_{L} = (i_{i} + g_{m}v_{gs})R_{e q}$             (3)

Substitution of (1) and (2) into (3) and rearrangement lead to
$A_{v} = {\frac{v_{L}}{v_{i}}} = {\frac{(g_{m}R_{G}  +  1)R_{e q}}{R_{G}  +  (g_{m}R_{G}  +  1)R_{e q}}} = {\frac{\lbrack(0.002)(1  \times  10^{6})  +  1\rbrack(536)}{1  \times  10^{6}  +  \lbrack(0.002)(1  \times  10^{6})  +  1\rbrack(536)}} = 0.517$

(b)  The current-gain ratio follows from part a as
$A_{i} = {\frac{i_{L}}{i_{i}}} = {\frac{v_{L}/R_{L}}{(v_{i}  –  v_{L})/R_{G}}} = {\frac{A_{v}R_{G}}{(1  –  A_{v})R_{L}}} = {\frac{(0.517)(1  \times  10^{6})}{(1  –  0.517)(1  \times  10^{3})}} = 1070.4$

(c) From (2),                 $i_{i} = {\frac{v_{i}  –  v_{L}}{R_{G}}} = {\frac{v_{i}(1  –  A_{v})}{R_{G}}}$        (4)

$R_{in}$ is found directly from (4) as
$R_{\mathrm{in}} = {\frac{v_{i}}{i_{i}}} = {\frac{R_{G}}{1-A_{v}}} = {\frac{1  \times  10^{6}}{1  –  0.517}} = 2.07\,\mathrm{M\Omega}$

(d) We remove $R_L$ and connect a driving-point source oriented such that $v_{dp} = v_L$.   With $v_i$ deactivated (shorted), $v_{gs} = -v_{dp}$.   Then, by KCL,
$i_{d p} = v_{dp}\left(\frac{1}{R_{S2}} + \frac{1}{r_{ds}} + \frac{1}{R_{G}}\right)  –  g_{m}v_{g s} = v_{dp}\left(\frac{1}{R_{S2}} + \frac{1}{r_{d s}} + \frac{1}{R_{G}} + g_{m}\right)$

and          $R_{o}={\frac{v_{d b}}{i_{db}}} = \frac{1}{\frac{1}{{{R_{S2}}}}  +  \frac{1}{r_{d s}}  +  \frac{1}{R_{G}}  +  g_{m}} = \frac{1}{\frac{1}{1.2  \times  10^{3}} + {\frac{1}{30  \times  10^{3}}} + {\frac{1}{1  \times  10^{6}}} + 0.002} = 348.7  Ω$