Exercise 8.2 A current source

The figure below shows a current source configuration that could be used to generate the current $I_{\mathit{TAIL}}$ in the previous exercise. The current $i_C$ through the transistor is given by: $i_C=I_{C0}\cdot e^{\frac{q{V}_{\mathit{be}}}{\mathit{kT}}}(1+\frac{v_{\mathit{CE}}}{v_A})$ with $\frac{\mathit{kT}}{q}=25\mathit{mV}$, $v_A=75V$ while $\mathit{\beta fe}$ is finite.

a)

Derive an expression for the value of $R_E$ needed to generate a current $I_C$. You can assume that $v_{\mathit{BE}}=0.7V$ at this $I_C$.

b)

Calculate the numerical value of $R_E$ for a current of $1\mathit{mA}$.

c)

Draw a small-signal equivalent circuit for the current source and determine the value of the small-signal parameter $g_m$ of the BJT.

d)

Derive an expression for the small-signal output resistance $r_{\mathit{out}}=\frac{v_x}{i_c}$ of the current source. You may assume that $(R_{B1}∕∕R_{B2})\ll \frac{\mathit{\beta fe}}{g_m}$. The output resistance of transistor $T_1$ (between collector and emitter) is $r_{\mathit{ce}}$.

e)

What would be the benefit of placing a capacitor in parallel with $R_{B2}$?