Question 7.SP.7: Suppose capacitor CS is removed from the circuit of Problem ......

(a)  The voltage-source small-signal equivalent circuit is given in Fig. 7-14 (the current-source model was utilized in Problem 7.4).    Voltage division and KVL give
v_{g s} = {\frac{R_{G}}{R_{G}  +  r_{i}}}\,v_{i}  –  i_{d}R_{S}          (1)

But by Ohm’s law,
i_{d} = \frac{\mu v_{g s}}{r_{d s}  +  R_{S}  +  R_{D}\|R_{L}}          (2)

Substituting (2) into (1) and solving for v_{g s} yield
v_{g s} = {\frac{R_{G}(r_{d s}  +  R_{S}  +  R_{D}\| R_{L})v_{i}}{(R_{G}  +  r_{i})[r_{d s}  +  (\mu  +  1)R_{S}  +  R_{D}\|R_{L}]}}                  (3)

Now voltage division gives

v_{L}=-\frac{{ R}_{D}\|R_{L}}{r_{d s}+{ R}_{S}+{ R}_{D}\|R_{L}}\;\mu\upsilon_{g s}              (4)

and substitution of (3) into (4) and rearrangement give
A_{v} = {\frac{v_{L}}{v_{i}}} = {\frac{-\mu R_{G}R_{D}R_{L}}{(R_{G}  +  r_{i})\{(R_{D}  +  R_{L})[r_{d s}  +  (\mu  +  1)R_{S}]  +  R_{D}R_{L}\}}}              (5)

With \mu = g_mr_{ds} and the given values, (5) becomes
A_{v} = {\frac{-(2  \times  10^{-3})(30  \times  10^{3})(160)(2)(2)}{(160  +  5)\{(2  +  2)[30  +  (60  +  1)3]  +  (2)(2)\}}} = -0.272

(b)  The current gain is found as
A_{i} = {\frac{i_{L}}{i_{i}}} = {\frac{v_{L}/R_{L}}{v_{i}/(R_{G}  +  r_{i})}} = {\frac{A_{v}(R_{G}  +  r_{i})}{R_{L}}} = {\frac{(-0.272)(160  +  5)}{2}} = -22.4

(c)  R_L is disconnected, and a driving-point source is added such that v_{dp} = v_L.   With v_i deactivated (short-circuited), v_{gs} = 0 and
R_{o} = R_{D}\|(r_{d s} + R_{S}) = {\frac{R_{D}(r_{d s}  +  R_{S})}{R_{D}  +  r_{d s}  +  R_{S}}} = {\frac{(2  \times  10^{3})(30  \times  10^{3}  +  3  \times  10^{3})}{2  \times  10^{3}  +  30  \times  10^{3}  +  3  \times  10^{3}}} = 1.89  \mathrm{kΩ}

Note that when R_S is not bypassed, the voltage- and current-gain ratios are significantly reduced.