Exercise 6.4 A circuit with an opamp

a)

Derive an equation for (the large signal) $v_{\mathit{OUT}}(v_{\mathit{IN}})$

b)

An answer:
The circuit includes a PMOS transistor. Its SSEC is the same as that of an NMOS transistor.

The combination of the transistor and the resistor forms a CSC that has negative gain. With the feedback to the non-inverting input of the opamp, this gives negative feedback.

c)

An answer:

$$\begin{array}{rcll}& v_X& =V_{\mathit{DD}}+v_{\mathit{GS}}& \\ & & v_{\mathit{GS}}=V_T-\sqrt{\frac{2i_D}{k}}\mathrm{\text{(compared to an NMOS some signs are reversed)}}& \\ & & i_D=\frac{v_{\mathit{OUT}}}{R_1}& \\ & v_X& =V_{\mathit{DD}}+V_T-\sqrt{\frac{2v_{\mathit{IN}}}{k\cdot R_1}}& \end{array}$$

d)

It was assumed that the transistor works in saturation, for which the square law relation is valid. Identify the range of $v_{\mathit{IN}}$ where this assumption is satisfied.

e)

An answer:
$v_X$ will not be in saturation, but in its linear region. The assumed square law then is no longer satisfies resulting in different behavior. For the $i-v$-relation in that region, with $v_{\mathit{DS}}\to 0$, $v_X$ goes to minus infinity.