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

The correct answer is infinite.

To understand why the small-signal current gain of a common-drain (source follower) amplifier is infinite, let us look at the structure and operation of an ideal MOSFET:

1. Input terminal and Gate current:
In a common-drain amplifier configuration, the input signal is applied to the gate terminal of the MOSFET. For an ideal MOSFET, the gate is completely isolated from the conducting channel by a thin, non-conducting layer of silicon dioxide (${\mathrm{SiO}}_2$). Because of this insulating oxide layer, no physical current flows into the gate terminal under static or low-frequency small-signal conditions. Thus, the small-signal gate current (which is the input current, $i_{in}$) is zero:

$i_{in}=i_g=0$

2. Output current:
In a common-drain amplifier, the output is taken from the source terminal. When an input signal is applied, a corresponding change in the drain-to-source current occurs. The small-signal output current ($i_o$) flowing through the load connected at the source terminal is non-zero:

$i_o=i_s\ne 0$

3. Small-signal current gain:
The small-signal current gain ($A_i$) is defined as the ratio of the small-signal output current to the small-signal input current:

$A_i=\frac{i_o}{i_{in}}=\frac{i_s}{0}\to \infty <!--5;</mo-->$

Because the input current is zero, the magnitude of the small-signal current gain is mathematically infinite.