Exercise 8.10 An opamp circuit schematics

a)

An answer:
A

b)

An answer:
Using a differential input signal: $v_A=V_X+v_{\mathit{in}}∕2$ and $v_B=V_X-v_{\mathit{in}}∕2$, then with the current and voltage conventions in the schematic:

$$\begin{array}{llll}\hfill g_{\mathit{diffpair}}& =\frac{i_x-i_y}{v_{\mathit{in}}}& \hfill & \\ \hfill & i_x=g_m\cdot (v_e+v_{\mathit{in}}∕2)& \hfill & \\ \hfill & i_y=g_m\cdot (v_e-v_{\mathit{in}}∕2)=-i_x& \hfill & \\ \hfill & \to & \hfill & \\ \hfill & v_e=0& \hfill & \\ \hfill & i_x-i_y=g_m\cdot v_{\mathit{in}}& \hfill & \\ \hfill g_{\mathit{diffpair}}& =g_m& \hfill & \end{array}$$

c)

An answer:
$$\begin{array}{llll}\hfill r_{\mathit{in}}& =\frac{v_{\mathit{in}}}{i_{\mathit{in}}}& \hfill & \\ \hfill & i_{\mathit{in}}=\frac{g_m}{\mathit{\beta fe}}\cdot (v_e+v_{\mathit{in}}∕2)& \hfill & \\ \hfill & v_e=0\mathrm{\text{(e.g. from symmetry or from previous question)}}& \hfill & \\ \hfill r_{\mathit{in}}& =2\cdot \frac{\mathit{\beta fe}}{g_m}& \hfill & \end{array}$$

d)

An answer:

$$\begin{array}{llll}\hfill H_x& =\frac{i_z}{i_x}& \hfill & \\ \hfill H_y& =\frac{i_z}{i_y}& \hfill & \\ \hfill & i_{z,x}=i_x\cdot \mathit{\beta fe}\cdot \frac{\mathit{\beta fe}}{\mathit{\beta fe}+2}& \hfill & \\ \hfill & i_{z,y}=i_y\cdot \mathit{\beta fe}& \hfill & \\ \hfill H_x& =\mathit{\beta fe}\cdot \frac{\mathit{\beta fe}}{\mathit{\beta fe}+2}& \hfill & \\ \hfill H_y& =\mathit{\beta fe}& \hfill & \end{array}$$

e)

An answer:
$$\begin{array}{llll}\hfill H& =\frac{i_z}{i_x-i_y}& \hfill & \\ \hfill & i_z=i_x\cdot \mathit{\beta fe}\cdot \frac{\mathit{\beta fe}}{\mathit{\beta fe}+2}-i_y\cdot \mathit{\beta fe}& \hfill & \\ \hfill & i_y=-i_x& \hfill & \\ \hfill & i_z=i_x\cdot \left(\mathit{\beta fe}\cdot \frac{\mathit{\beta fe}}{\mathit{\beta fe}+2}+\mathit{\beta fe}\right)& \hfill & \\ \hfill H& =\frac{i_z}{i_x-i_y}=\mathit{\beta fe}\frac{\mathit{\beta fe}+1}{\mathit{\beta fe}+2}\mathit{\cong \beta fe}& \hfill & \end{array}$$

f)

An answer:
$r_{\mathit{in}}=R_L(1+\mathit{\beta fe})+\frac{\mathit{\beta fe}}{g_m}$

g)

An answer:
$r_{\mathit{in}}\to \infty$

h)

An answer:
$\frac{v_{\mathit{out}}}{v_{\mathit{base}3}}=\frac{R_L(1+\mathit{\beta fe})}{\frac{\mathit{\beta fe}}{g_m}+R_L(1+\mathit{\beta fe})}$

i)

An answer:
$r_{\mathit{out}}\to \infty$.

j)

An answer:

$$\begin{array}{llll}\hfill & g_m=20\mathit{mS}& \hfill & \\ \hfill & \frac{v_{\mathit{out}}}{v_{\mathit{in}}}=g_m\cdot \mathit{\beta fe}\frac{\mathit{\beta fe}+1}{\mathit{\beta fe}+2}\cdot (1+\mathit{\beta fe})R_L\approx g_m\beta {\mathit{fe}}^2{R}_L=\dots & \hfill & \end{array}$$