Exercise 8.11 An opamp circuit schematic

a)

Derive the bias currents for all transistors in this circuit schematic, for $v_A=v_B$.
An answer:
Using symmetry, the bias currents in the NMOS transistors is $I_{\mathit{BIAS}}∕2$. The bias currents in the PMOS transistors is then also $I_{\mathit{BIAS}}∕2$; for PMOS transistors this may include a minus sign depending of how strict you follow current direction conventions.

b)

An answer:

$$\begin{array}{llll}\hfill \frac{i_x}{v_A-v_B}& =\frac{1}{2}{g}_m& \hfill & \\ \hfill & g_m\equiv \frac{\partial i_D}{\partial v_{\mathit{GS}}}& \hfill & \\ \hfill & =k\cdot (V_{\mathit{GS}}-VT)=\sqrt{k\cdot I_{\mathit{BIAS}}}& \hfill & \\ \hfill \frac{i_x}{v_A-v_B}& =\frac{1}{2}\sqrt{k\cdot I_{\mathit{BIAS}}}& \hfill & \end{array}$$

c)

An answer:
Using symmetry (remember that the PMOS transistors are identical) it readily follows that

$$\begin{array}{rcll}{i}_{D_{M_4}}=i_{D_{M_3}}=\frac{I_{\mathit{BIAS}}}{2}+i_x& & & \end{array}$$

d)

An answer:

$$\begin{array}{rcll}& v_R& =I_{R}R=\left(\frac{I_{\mathit{BIAS}}}{2}+i_x\right)R& \\ & & =\frac{I_{\mathit{BIAS}}}{2}R+i_{x}R& \end{array}$$

e)

An answer:

$$\begin{array}{rcll}& a_v& =\frac{v_{\mathit{out}}}{v_A-v_B}& \\ & & =\frac{i_x}{v_A-v_B}\frac{v_{\mathit{out}}}{i_x}& \\ & & =\frac{1}{2}\sqrt{k\cdot I_{\mathit{BIAS}}}\cdot R& \end{array}$$