Question 7.SP.22: In the amplifier of Problem 4.27, let vS = 2 cos ωt V. (a) D......
In the amplifier of Problem 4.27, let v_S = 2 \cos ωt \text{V}. (a) Draw the ac load line on Fig. 4-31. (b) Graphically determine the voltage gain. (c) Calculate the voltage gain using small-signal analysis.
(a) If capacitor C_K appears as a short circuit to ac signals, then application of KVL around the plate circuit of Fig. 4-30 gives, as the equation of the ac load line, V_{PP} + V_{GQ} = i_PR_L + v_P. Thus, the ac load line has vertical and horizontal intercepts
\frac{V_{P P} + V_{G Q}}{R_{L}} = \frac{300 – 4}{11.6 \times 10^{3}} = 25.5\,\mathrm{mA} \qquad \text{and} \qquad V_{P P} + V_{G Q} = 296\,\mathrm{V}
as shown on Fig. 4-31.
(b) We have v_g = v_S; thus, as v_g swings ±2 V along the ac load line from the Q point in Fig. 4-31, v_p swings a total of 2V_{pm} = 213 – 145 = 68 \text{V} as shown. The voltage gain is then
A_{v} = -{\frac{2V_{p m}}{2V_{g m}}} = -{\frac{68}{4}} = -17
where the minus sign is included to account for the phase reversal between v_p and v_g.
(c) Applying (7.9) and (7.10) at the Q point of Fig. 4-31 yields
Plate resistance r_{p} \equiv {\frac{\partial v_{P}}{\partial i_{P}}}\Bigg|_{Q} \simeq {\frac{\Delta v_{P}}{\Delta i_{P}}}\Bigg|_{Q} (7.9)
Transconductance g_{m} \equiv {\frac{\partial i_{P}}{\partial v_{G}}}\bigg|_{Q} \simeq {\frac{\Delta i_{P}}{\Delta v_{G}}}\bigg|_{Q} (7.10)
r_{p} \approx {\frac{\Delta v_{p}}{\Delta i_{p}}}\bigg|_{v_{G} = -4} = {\frac{202 – 168}{(15 – 8) \times 10^{-3}}} = 4.86\,\mathrm{k\Omega}
g_{m} \approx {\frac{\Delta i_{P}}{\Delta v_{G}}}\bigg|_{v_{P} = 180} = {\frac{(15.5 – 6.5) \times 10^{-3}}{-3 – (-5)}} = 4.5\,{\mathrm{mS}}
Then, μ ≡ g_mr_p = 21.87, and Problem 7.21, yields
A_{v} = -{\frac{\mu R_{L}}{R_{L} + r_{p}}} = -{\frac{(21.87)(11.6 \times 10^{3})}{(11.6 + 4.86) \times 10^{3}}} = -15.41
