Question 7.SP.12: Use the small-signal equivalent circuit to predict the peak ......
Use the small-signal equivalent circuit to predict the peak values of i_d and v_{ds} in Example 4.3. Compare your results with that of Example 4.3, and comment on any differences.
The values of g_m and r_{ds} for operation near the Q point of Fig. 4-6 were determined in Problem 7.1. We may use the current-source model of Fig. 7-1(a) to form the equivalent circuit of Fig. 4-5. In that circuit, with v_{gs} = \sin t \text{V}, Ohm’s law requires that
v_{d s} = -g_{m}v_{g s}(r_{d s}||R_{D}) = \frac{-g_{m}r_{d s}R_{D}v_{g s}}{r_{d s} + R_{D}} = \frac{-(1.5 \times 10^{-3})(75 \times 10^{3})(3 \times 10^{3})v_{g s}}{75 \times 10^{3} + 3 \times 10^{3}} = -4.33v_{g s}
Thus, V_{dsm} = 4.33V_{gsm} = 4.33(1) = 4.33\,\mathrm{V}
Also, from Fig. 7-1(a),
i_{d} = g_{m}v_{g s} + {\frac{v_{d s}}{r_{d s}}}
so I_{dm} = g_{m}V_{g s m} + \frac{V_{d s m}}{r_{ds}} = (1.5 \times 10^{-3})(1) + \frac{1}{75 \times 10^{3}} = 1.513\,\mathrm{mA}
The ±1-V excursion of v_{gs} leads to operation over a large portion of the nonlinear drain characteristics. Consequently, the small-signal equivalent circuit predicts greater positive peaks and smaller negative peaks of i_d and v_{ds} than the graphical solution of Example 4.3, which inherently accounts for the nonlinearities.
