Exercise 4.2 An NMOS amplifier circuit

a)

An answer:
$$\begin{array}{rcll}& I_{D_1}& =\frac{1}{2}K{\left(V_{\mathit{GS}}-V_{\mathit{TH}}\right)}^2& \\ & & V_{\mathit{GS}}=V_G-V_S& \\ & & V_G=\frac{R_{g2}}{R_{g1}+R_{g2}}\cdot V_{\mathit{DD}}& \\ & & V_S=0& \\ & & \Rightarrow & \\ & I_{D_1}& =\frac{1}{2}K{\left(\frac{R_{g2}}{R_{g1}+R_{g2}}\cdot V_{\mathit{DD}}-V_{\mathit{TH}}\right)}^2& \end{array}$$

b)

An answer:

A first step can be replacing impedances of pasives with their value at signal frequencies, and setting DC sources to zero.

After which a second step can be replacing the non-linear components by their SSEC and cleaning up the schematic. This latter usually is unwinding the (virtual) ground node.

c)

An answer:
$$\begin{array}{rcll}& a_v& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & & v_{\mathit{out}}=-g_m\cdot v_{\mathit{gs}}{R}_d& \\ & & v_{\mathit{gs}}=-v_s& \\ & & v_s=v_{\mathit{in}}& \\ & & \Rightarrow & \\ & a_v& =\frac{-g_m\cdot -v_{\mathit{in}}}{v_{\mathit{in}}}{R}_d& \\ & & =g_m{R}_d& \end{array}$$

d)

An answer:
Applying Ohm’s law to the input port and (re)using the input voltage source to drive input port: $$\begin{array}{rcll}& r_{\mathit{in}}& =\frac{v_{\mathit{in}}}{i_{\mathit{in}}}& \\ & & i_{\mathit{in}}=-g_m\cdot v_{\mathit{gs}}& \\ & & v_{\mathit{gs}}=-v_s& \\ & & v_s=v_{\mathit{in}}& \\ & & \Rightarrow & \\ & r_{\mathit{in}}& =\frac{1}{g_m}& \end{array}$$

e)

An answer:
Assuming a current source driving the output port: $$\begin{array}{rcll}& r_{\mathit{out}}& =\frac{v_{\mathit{out}}}{i_{\mathit{out}}}& \\ & & v_{\mathit{out}}=(i_{\mathit{out}}-g_m\cdot v_{\mathit{gs}})\cdot R_d& \\ & & v_{\mathit{gs}}=0& \\ & & \Rightarrow & \\ & r_{\mathit{out}}& =R_d& \end{array}$$