Exercise 5.8 An amplifier with 2 NPNs

a)

An answer:

$$\begin{array}{rcll}& R_E& =\frac{V_{E2}}{I_{\mathit{RE}}}& \\ & & I_{\mathit{RE}}=(1+\frac{1}{\mathit{\beta fe}})I_{C2}-I_{B1}& \\ & R_E& =9k\mathrm{\Omega }& \end{array}$$

$$\begin{array}{rcll}& R_{C_1}& =\frac{V_{\mathit{CC}}-V_{B_E,Q2}-V_{E,Q2}}{I_{C,Q_1}-I_{B,Q2}}& \\ & & =15k\mathrm{\Omega }& \end{array}$$

$$\begin{array}{rcll}& R_B& =\frac{V_{R_B}}{I_{R_B}}& \\ & & =360k\mathrm{\Omega }& \end{array}$$

b)

An answer:

$$\begin{array}{lll}\hfill R_{\mathit{in},Q2}=\frac{\mathit{\beta fe}}{g_{m2}}=\frac{120}{0.004}=30k\mathrm{\Omega }& & \hfill \end{array}$$

c)

An answer:

The first stage gives a loaded voltage gain $a_{v1}=g_{m1}\cdot \left(R_{C_1}∕∕\frac{\mathit{\beta fe}}{g_{m2}}\right)$

The second stage gives a voltage gain $a_{v2}=g_{m2}\cdot R_{C_2}$

$$\begin{array}{llll}\hfill a_v& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \hfill & \\ \hfill & =g_{m1}\cdot \left(R_{C_1}∕∕\frac{\mathit{\beta fe}}{g_{m2}}\right)\cdot g_{m2}\cdot R_{C_2}& \hfill & \\ \hfill & =\frac{g_{m1}\cdot R_{C_1}\cdot R_{C_2}\cdot \mathit{\beta fe}}{R_{C_1}+\frac{\mathit{\beta fe}}{g_{m2}}}& \hfill & \end{array}$$

d)

An answer:
Recycle the previously derived equation, rewrite for $R_{C2}$:

$$\begin{array}{llll}\hfill & R_{C2}=2.5k\mathrm{\Omega }& \hfill & \end{array}$$