Question 7.6: The triode amplifier of Fig. 4-14 has VGG, VPP, RG, and RL, ......
The triode amplifier of Fig. 4-14 has V_{GG}, V_{PP}, R_G, and R_L, as given in Example 4.7. If the plate characteristics of the triode are given by Fig. 7-8 and v_S = 2 \sin ωt \text{V}, graphically find v_P and i_P.
The dc load line, with the same intercepts as in Example 4.7, is superimposed on the characteristics of Fig. 7-8; however, because the plate characteristics are different from those of Example 4.7, the quiescent values are now I_{PQ} = 11.3 mA and V_{PQ} = 186 \text{V}. Then a time axis on which to plot v_G = -4 + 2 \sin ωt \text{V} is constructed perpendicular to the dc load line at the Q point. Time axes for i_P and v_P are also constructed as shown, and values of i_P and v_P corresponding to particular values of v_G(t) are found by projecting through the dc load line, for one cycle of v_G. The result, in Fig. 7-8, shows that v_P varies from 152 to 218 V and i_P ranges from 8.1 to 14.7 mA.
The following treatment echoes that of Section 6.2 For the usual case of negligible grid current, (4.7) degenerates to i_G = 0 and the grid acts as an open circuit. For small excursions (ac signals) about the Q point, Δ_{iP} = i_p and an application of the chain rule to (4.8) leads to
i_{p} = \Delta i_{P} \approx d i_{P} = \frac{1}{r_{p}}\,v_{p} + g_{m}v_{g} (7.8)
i_{G} = f_{1}(v_{P},v_{G}) (4.7)
i_{P} = f_{2}(v_{P},v_{G}) (4.8)
where we have defined
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)
Under the condition i_G = 0, (7.8) is simulated by the current-source equivalent circuit of Fig 7-9(a). The frequently used voltage-source model of Fig. 7-9(b) is developed in Problem 7.19.
