Question 4.2: In the amplifier of Fig. 4-5(a), VDD = 20 V, R1 = 1 MΩ, R2 =......
In the amplifier of Fig. 4-5(a), V_{DD} = 20 \text{V}, R_1 = 1 MΩ, R_2 = 15.7 MΩ, R_D = 3 kΩ, \text{and} R_S = 2 kΩ. If the JFET characteristics are given by Fig. 4-6, find (a) I_{DQ}, (b) V_{GSQ}, and (c) V_{DSQ}.
(a) By (4.3),
R_G = \frac{R_1R_2}{R_1 + R_2} \quad \text{and} \quad V_{GG} = \frac{R_1}{R_1 + R_2} V_{DD} (4.3)
V_{GG} = \frac{R_1}{R_1 + R_2} V_{DD} = \frac{1 × 10^6}{16.7 × 10^6} 20 = 1.2 \text{V}
On Fig. 4-6(a), we construct the transfer bias line (4.4); it intersects the transfer characteristic at the Q point, giving I_{DQ} = 1.5 \text{mA}.
i_D = \frac{V_{GG}}{R_S} – \frac{v_{GS}}{R_S} (4.4)
(b) The Q point of Fig. 4-6(a) also gives V_{GSQ} = -2 \text{V}.
(c) We construct the dc load line on the drain characteristics, making use of the v_{DS} intercept of V_{DD} = 20 \text{V} and the i_D intercept of V_{DD}/(R_S + R_D) = 4 \text{mA}. The Q point was established at I_{DQ} = 1.5 \text{mA} in part a and at V_{GSQ} = -2 \text{V} in part b; its abscissa is V_{DSQ} = 12.5 \text{V}. Analytically,
V_{DSQ} = V_{DD} – (R_S + R_D)I_{DQ} = 20 – (5 × 10^3)(1.5 × 10^{-3}) = 12.5 \text{V}