350 Solved Electrical Engineering Problems: For the FE/PE Exams in Electrical Engineering

2.11: Common gate amplifier

2.11 Common gate amplifier

The common gate FET circuit shown in Figure 2.11 has the following parameters. Forward transadmittance, y fs, equals 1.5 10 ?3 siemens. Output conductance, y os, equals 10 10 ?6 siemens. Gate-to-source resistance, r gs, equals 1 10 8 ohms. Gate-to-drain resistance, r gd, equals 1 10 8 ohms. Calculate the (a) Z in, (b) Z out, and (c) A v respectively.


Figure 2.11: Common gate amplifier

Solution:

r d = 1/y os = 1 10 5 ohms

g m = y fs = 1.5 10 ?3

? = y fsr d = g mr d = 1.5 ?3 1 10 5 = 150

  1. Z in = (r d + R D)/( ? + 1) R s

    Z in = (1.1 10 5/151) (1.5 10 3) = 728 (1.5 10 3) = 490 ohms

  2. Z out = [r d ( ? + 1)R s] R D = [1 10 5 + (151)1.5 10 3] (1 10 4)

    Z out = (3.265 10 5) (1 10 4) = 9,969 ohms

  3. A v = ( ?

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