11 Introduction to RF electronics

Q 11.1

(a)

An answer:
From an electrical point of view, a resistor is the only passive component that converts electrical energy (or power) into energy (or power) in another domain (heat, light, etc.).

$$\begin{array}{rcll}& P_{\mathit{diss}}& =i_{\mathit{eff}}^2{R}_a& \\ & & =\frac{v_{\mathit{eff}}^2}{{\left|Z_{\mathit{load}}\right|}^2}{R}_a& \\ & & =\frac{v_{\mathit{eff}}^2}{R_a^2+X_a^2}{R}_a& \end{array}$$

(b)

An answer:
As the antenna is specified to be a monopole, it is basically a conductor with a specific length, connected on one end. An antenna is tuned to the transmit/receive frequency, hence the ”quarter-wave”. At that frequency the antenna has the specified impedance. At $f=\omega =0$ the antenna (monopole or dipole) obeys KCL and KVL and hence is an open: $Z\to \infty \mathrm{\Omega }$ @ 0 Hz.

(c)

An answer:

$$\begin{array}{llll}\hfill P_{\mathit{transmit}}& =\Re \left(i_{\mathit{antenna}}\cdot v_{\mathit{antenna}}\right)& \hfill & \\ \hfill & =i_{\mathit{eff}}^2\cdot \Re \left(Z_{\mathit{antenna}}\right)& \hfill & \\ \hfill & =\frac{v_{\mathit{eff}}^2}{{\left|Z_{\mathit{antenna}}\right|}^2}\cdot \Re \left(Z_{\mathit{antenna}}\right)& \hfill & \end{array}$$

(d)

An answer:

$$\begin{array}{llll}\hfill P_{10+j1000\mathrm{\Omega }}& =\frac{{\widehat{V}}^2}{2\cdot 10^6}\cdot 10=\frac{1}{2}{\widehat{V}}^2\cdot 10^{-5}[\mathrm{\text{W}}]& \hfill & \\ \hfill 1)P_{10\mathrm{\Omega }}& =\frac{{\widehat{V}}^2}{2\cdot 10^2}\cdot 10=\frac{1}{2}{\widehat{V}}^2\cdot 10^{-1}[\mathrm{\text{W}}]\mathrm{\text{ratio }}10^{-4}\mathrm{\text{ or }}10^4& \hfill & \\ \hfill 2)P_{50\mathrm{\Omega }}& =\frac{{\widehat{V}}^2}{2\cdot 50^2}\cdot 50=\frac{1}{2}{\widehat{V}}^2\cdot 2\cdot 10^{-2}[\mathrm{\text{W}}]\mathrm{\text{ratio }}5\cdot 10^{-4}\mathrm{\text{ or }}2\cdot 10^3& \hfill & \\ \hfill & & \hfill \end{array}$$

Q 11.2

(a)

An answer:
$Z_{\mathit{res}}(\mathit{j\omega })=R+\mathit{j\omega l}\cdot 2\mathit{nH}∕\mathit{mm}$ with $l$ the total wire length in mm.

(b)

An answer:
At the limit:

$$\begin{array}{llll}\hfill |Z(\mathit{j\omega })|& =\sqrt{R^2+{\omega }^2{L}^2}& \hfill & \\ \hfill & =\sqrt{2}\cdot R& \hfill & \end{array}$$

leading to:

$$\begin{array}{llll}\hfill & R^2+{\omega }^2{L}^2=2\cdot R^2& \hfill & \\ \hfill {\omega }_{\mathit{max}}& =\frac{R}{L}& \hfill & \\ \hfill f_{\mathit{max}}& =200\mathit{MHz}& \hfill & \end{array}$$

(c)

An answer:
Hmm... that’s pretty close to $R=100\mathrm{\Omega }$!

(d)

An answer:
Hmm... that’s probably larger than anything you’d ever build as audio amplifier.

Q 11.3

(a)

An answer:
$Z_{\mathit{cap}}(\mathit{j\omega })=\frac{1}{\mathit{j\omega C}}+\mathit{j\omega l}\cdot 2\mathit{nH}∕\mathit{mm}$ with $l$ the total wire length in mm.

(b)

An answer:

$$\begin{array}{llll}\hfill Z_{\mathit{cap}}(\mathit{j\omega })& =\frac{1}{\mathit{j\omega C}}+\mathit{j\omega L}& \hfill & \\ \hfill & =-j32\mathrm{\Omega }+j25\mathrm{\Omega }=-j7\mathrm{\Omega }& \hfill & \end{array}$$

(c)

An answer:
About 230 pF.

(d)

(e)

An answer:
About 12 pF.

(f)

An answer:
This yields a negative capacitor... that would be inductive behavior at 100 MHz. Value wise it is just a little below 40 nH.

(g)

An answer:

$$\begin{array}{llll}\hfill \frac{1}{\mathit{j\omega C}}& =\frac{1}{\mathit{j\omega }{C}_{\mathit{target}}}-\mathit{j\omega L}& \hfill & \\ \hfill \frac{1}{\mathit{j\omega C}}& =\frac{1+{\omega }^{2}L{C}_{\mathit{target}}}{\mathit{j\omega }{C}_{\mathit{target}}}& \hfill & \\ \hfill & \to & \hfill & \\ \hfill C& =\frac{C_{\mathit{target}}}{1+{\omega }^{2}L{C}_{\mathit{target}}}\mathrm{\text{with C=100pF and L=40nH}}& \hfill & \\ \hfill C& \approx 39\mathit{pF}& \hfill & \end{array}$$

Q 11.4

(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:

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

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 $a_v=-g_m{R}_L$.