Q. 4.2

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) $V_{DSQ}$.

Verified Solution

(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,
$V_{DSQ} = V_{DD} – (R_S + R_D)I_{DQ} = 20 – (5 × 10^3)(1.5 × 10^{-3}) = 12.5 \text{V}$