Exercise 8.2 A current source

a)

An answer:

$$\begin{array}{rcll}{R}_E=\frac{V_{\mathit{BB}}\cdot \frac{R_{B2}}{R_{B1}+R_{B2}}-\frac{I_C}{\mathit{\beta fe}}\frac{R_{B1}{R}_{B2}}{R_{B1}+R_{B2}}-0.7}{\frac{\mathit{\beta fe}+1}{\mathit{\beta fe}}{I}_C}& & & \\ \cong \frac{V_{\mathit{BB}}\cdot \frac{R_{B2}}{R_{B1}+R_{B2}}-\frac{I_C}{\mathit{\beta fe}}\frac{R_{B1}{R}_{B2}}{R_{B1}+R_{B2}}-0.7}{I_C}& & & \end{array}$$

It is given that $\mathit{\beta fe}=100$, $R_{B1}=R_{B2}=10k\mathrm{\Omega }$, $V_{\mathit{BB}}=V_{\mathit{CC}}=6V$.

b)

c)

An answer:

$$\begin{array}{rcll}{g}_m=40I_C& & & \end{array}$$

d)

An answer:

$$\begin{array}{rcll}{r}_{\mathit{out}}=R_E∕∕\frac{\mathit{\beta fe}}{g_m}+(1+g_m(R_E∕∕\frac{\mathit{\beta fe}}{g_m})r_{\mathit{ce}}& & & \end{array}$$

e)

An answer:
In practice this could help to keep $V_B$ more constant and thereby to keep $I_C$ more constant.