Question Details

For an ideal MOSFET biased in saturation, the magnitude of the small signal current gain for a common drain amplifier is

Options

A

infinite

B

1

C

100

D

0

Correct Answer :

infinite

Solution :

The correct option is infinite.

Let us analyze the small-signal behavior of a common-drain (source follower) amplifier using an ideal MOSFET biased in saturation to understand why the magnitude of its small-signal current gain is infinite.

First, recall the basic structure of a common-drain amplifier. In this configuration, the input signal is applied to the gate terminal, the output is taken from the source terminal, and the drain terminal is connected to a constant DC voltage (which acts as an AC ground for small-signal analysis).

Next, we examine the input of the ideal MOSFET. The gate of an ideal MOSFET is insulated from the channel by a layer of silicon dioxide (SiO2). Consequently, no physical current flows into the gate terminal under DC or low-frequency AC small-signal conditions. Thus, the small-signal input current (iin) entering the gate is zero:
iin=0
This corresponds to an infinite input impedance at the gate of the MOSFET.

On the other hand, the small-signal output current (iout) is the current flowing through the load connected at the source terminal. Due to the field-effect operation, a change in the gate-to-source voltage (vgs) modulates the channel charge, producing a non-zero small-signal drain-to-source current controlled by the transconductance (gm):
iout=gmvgs
Since the transistor is biased in saturation and operating normally, this small-signal output current is non-zero (iout0).

The small-signal current gain (Ai) of an amplifier is defined as the ratio of the small-signal output current to the small-signal input current:
Ai=ioutiin
Substituting the values we obtained:
Ai=iout0
Therefore, the magnitude of the small-signal current gain for a common-drain amplifier using an ideal MOSFET is infinite.

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