Chapter 4

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}.

4.5
4.6

Step-by-Step

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}