Audio FET Buffers and Negative Feedback
For the circuit given, if \$R_S = 0\$, then

$$v_{GS} = v_G - v_S = v_{IN} - 0 = v_{IN}$$

However, with a non-zero \$R_S\$, we have

$$v_{GS} = v_G - v_S = v_{IN} - i_S\cdot R_S &lt; v_{IN}$$

Now, since

$$i_S = \beta(v_{GS} - V_P)^2 = \beta_0(1 + \lambda v_{DS})(v_{GS} - V_P)^2$$

we have

$$v_{GS} = v_{IN} - \beta R_S(v_{GS} - V_P)^2$$


  how I am able to ascertain what maths need to be used to work out what
  amplitude signal can be used on the input with a VP as a limit


Solving for \$v_{IN}\$ yields

$$v_{IN} = \beta R_S(v_{GS})^2 + (1 - 2\beta R_SV_P)v_{GS} + \beta R_S(V_P)^2  $$

The bounds on \$v_{GS}\$ (for the \$i_S\$ equation to hold) are

$$V_P \le v_{GS} \le 0$$

so the bounds on \$v_{IN}\$ are

$$V_P \le v_{IN} \le \beta R_S (V_P)^2 $$

Thus, the input voltage range is extended by a factor of \$(1+\beta R_S |V_P|)\$