Assume a self- biased JFET amplifier has IGSS = 10 nA, VGS = 1 V, and RG = 10 M▲. The input resistance, RIN, of the amplifier is
A)10 M▲
B)9.1 M▲
C)9.5 M▲
D)8.6 M▲

Question

Quizplus · Accepted Answer

The input resistance of a self-biased JFET amplifier is given by:RIN = RG || (1/gm)where:RG is the gate resistorgm is the transconductance of the JFETThe transconductance of a JFET is given by:gm = 2 * IDSS * (1 - VGS/VP)where:IDSS is the drain-source current at VGS = 0VGS is the gate-source voltageVP is the pinch-off voltageIn this problem, we are given that IDSS = 10 nA, VGS = 1 V, and RG = 10 MΩ. The pinch-off voltage of a JFET is typically around -4 V.Substituting these values into the equations above, we get:gm = 2 * 10 nA * (1 - 1 V/-4 V) = 30 μSRIN = 10 MΩ || (1/30 μS) = 9.1 MΩTherefore, the input resistance of the amplifier is 9.1 MΩ.

Assume a Self- Biased JFET Amplifier Has IGSS = 10