Exercise 11.4 An RF amplifier

a)

An answer:
$a_v=-g_m{R}_L$

b)

An answer:
$a_v=-\frac{g_m{R}_L}{1+g_m{L}_{\mathit{wire}}}$

c)

An answer:

$$\begin{array}{llll}\hfill I_C& \approx 5\mathrm{\text{mA}}& \hfill & \\ \hfill g_m& \approx 190\mathrm{\text{mS}}& \hfill & \end{array}$$

d)

An answer:

$$\begin{array}{llll}\hfill L_{\mathit{wire}}& =10\mathrm{\text{nH}}& \hfill & \\ \hfill Z_{\mathit{Lwire}}& =2\pi \mathrm{\Omega }& \hfill & \\ \hfill g_m& \approx 190\mathrm{\text{mS}}& \hfill & \\ \hfill a_{v-a}& =-g_m{R}_L& \hfill & \\ \hfill a_{v-b}& =-g_m{R}_L\cdot \frac{1}{1+g_m{L}_{\mathit{wire}}\cdot (1+1∕\mathit{\beta fe})}& \hfill & \end{array}$$

From this it follows that the decrease in voltage gain when including 1 cm wire at 100 Mhz is a factor 2.2.

e)

An answer:
Using an ideal driving source, it is still $a_v=-g_m{R}_L$.