Although the source follower is not useful as a voltage gain amplifier as it provides a gain < 1, it can still be useful in some applications. It makes a good buffer as it has high input impedance, $r_{in}=\infty$, and low output impedance, $r_{out}\approx \frac{1}{g_m}$.

\[A_v=\frac{g_m}{\frac{1}{R_{oS}}+g_m+g_{mb}}\] \[r_{out}=\frac{1}{g_m+g_{mb}+\frac{1}{R_{oS}}}\] \[r_{in}=\infty\] where \[R_{oS}=r_o || R_S\]

Note that $r_o$ and $R_S$ are in parallel.

Define:

\[R_{oS}=r_o || R_S\]

Write the nodal equation at the output node, $v_{out}$.

\[\frac{v_{out}}{R_{oS}}=g_m\left ( v_{in}-v_{out} \right )-g{mb}\left (v_{out} \right )\]

Do the algebra.

\[v_{out}\left ( \frac{1}{R_{oS}}+g_m+g_{mb} \right )=g_mv_{in}\]

\[A_v=\frac{v_{out}}{v_{in}}=\frac{g_m}{\frac{1}{R_{oS}}+g_m+g_{mb}}\]

Neglecting body effect and output resistance we can view the source follower mosfet as a resistor with resistance of $\frac{1}{g_m}$.

The circuit can be viewed as voltage divider.

Again, Define:

\[R_{oS}=r_o || R_S\]

Write the nodal equation at $v_x$, the output node.

\[i_x+g_m\left ( -v_x \right )+g_{mb}\left ( -v_x \right )=\frac{v_x}{R_{oS}}\]

\[\frac{i_x}{v_x}=\left ( g_m+g_{mb}+\frac{1}{R_{oS}} \right )\]

\[r_{out}=\frac{v_x}{i_x}=\frac{1}{\left ( g_m+g_{mb}+\frac{1}{R_{oS}} \right )}\]

The resistance seen looking into the source of a mosfet is approximately $\frac{1}{g_m}$.

Therefore,

\[r_{out}=\frac{1}{g_m}||R_S=\frac{\frac{1}{g_m}R_S}{\frac{1}{g_m}+R_S}=\frac{1}{\frac{1}{R_S}+g_m}\]