Chapter 13, Problem 13.7P

Pertinent dimensions for the transmission line shown in Figure 13.2 are b = 3 mm and d= 0.2 mm. The conductors and the dielectric are nonmagnetic. (a) If the characteristic impedance of the line is $15\Omega$, find $ϵ\text{r'}$. Assume a low-loss dielectric. (b) Assume copper conductors and operation at $2\times {10}^8$ rad/s. If RC = GL, determine the loss tangent of the dielectric.

To determine

(a)

The dielectric constant ${{\epsilon }^{\prime }}_r$ , so that $Z_0=15\Omega$.

Answer to Problem 13.7P

The dielectric constant ${{\epsilon }^{\prime }}_r$ , so that $Z_0=15\Omega$ is ${{\epsilon }^{\prime }}_r=2.8$.

Explanation of Solution

Given:

The given figure is shown below..

.

   $\begin{array}{l}d=0.2\text{mm}\\ b=3\text{mm}\\ Z_0=15\Omega \end{array}$.

Calculation:

The characteristic impedance of the transmission line is given as:.

   $\begin{array}{l}{Z}_0=\sqrt{\frac{\mu }{{\epsilon }^{\prime }}}\left(\frac{d}{b}\right)\\ Z_0=\sqrt{\frac{\mu }{{{\epsilon }^{\prime }}_r}}\left(\frac{d}{b}\right)\\ {{\epsilon }^{\prime }}_r={\left(\frac{377}{15}\right)}^2\frac{.04}{9}\\ {{\epsilon }^{\prime }}_r=2.8\end{array}$.

Conclusion:

Thus, the dielectric constant ${{\epsilon }^{\prime }}_r$ , so that $Z_0=15\Omega$ is ${{\epsilon }^{\prime }}_r=2.8$.

To determine

(b)

The loss tangent of the dielectric, when $RC=GL$.

Answer to Problem 13.7P

The loss tangent of the dielectric, when $RC=GL$ is $5.85\times {10}^{-2}$ .

Explanation of Solution

Given:

The given figure is shown below:.

.

   $\begin{array}{l}d=0.2\text{mm}\\ b=3\text{mm}\\ Z_0=15\Omega \\ \omega =2\times {10}^8\text{rad/sec}\end{array}$.

Calculation:

For copper conductivity is ${\sigma }_c=5.8\times {10}^7\text{S/m}$ and therefore skin depth can be calculated as:.

   $\begin{array}{l}\delta \text{=}\sqrt{\frac{2}{\omega {\mu }_0{\sigma }_c}}\\ {\sigma }_c=5.8\times {10}^7\text{S/m}\text{.}\\ \omega =2\times {10}^8\text{rad/sec}\text{.}\\ \delta \text{=}\sqrt{\frac{2}{\left(2\times {10}^8\right)\left(4\pi \times {10}^{-7}\right)\left(5.8\times {10}^7\right)}}\\ \delta \text{=}1.2\times {10}^{-5}\text{m}\end{array}$.

Thus, resistance can be calculated as:.

   $\begin{array}{l}R=\frac{2}{{\sigma }_c\delta b}\\ b=3\text{mm}=.003\text{m}\\ {\sigma }_c=5.8\times {10}^7\text{S/m}\\ \delta \text{=}1.2\times {10}^{-5}\text{m}\\ R=\frac{2}{\left(5.8\times {10}^7\right)\left(1.2\times {10}^{-5}\right)\left(.003\right)}\\ R=0.98\text{Ω/m}\end{array}$.

Capacitance can be calculated as:.

   $\begin{array}{l}C=\frac{{\epsilon }^{\prime }b}{d}\\ C=\frac{{{\epsilon }^{\prime }}_r{\epsilon }_{0}b}{d}\\ {\epsilon }_0=8.854\times {10}^{-12}\\ {{\epsilon }^{\prime }}_r=2.8\\ d=0.2\text{mm}\\ b=3\text{mm}\\ C=\frac{2.8\times 8.854\times {10}^{-12}\times 3}{0.2}\\ C=3.7\times {10}^{-10}\text{F/m}\end{array}$.

Inductance can be calculated as:.

   $\begin{array}{l}L=\frac{{\mu }_{0}d}{b}\\ L=\frac{4\pi \times {10}^{-7}\times 0.2}{3}\\ L=8.4\times {10}^{-8}\text{H/m}\end{array}$.

As,

   $\begin{array}{l}RC=GL\\ G=\frac{RC}{L}\\ G=\frac{0.98\times \left(3.7\times {10}^{-10}\right)}{8.4\times {10}^{-8}}\\ G=4.4\times {10}^{-3}\text{S/m}\end{array}$.

Conductivity can also be given as:.

   $\begin{array}{l}G=\frac{{\sigma }_{d}b}{d}\\ {\sigma }_d=\frac{Gd}{b}\\ {\sigma }_d=\frac{Gd}{b}\\ d=0.2\text{mm}\\ b=3\text{mm}\\ G=4.4\times {10}^{-3}\text{S/m}\\ {\sigma }_d=\frac{4.4\times {10}^{-3}\times 0.2}{3}\\ {\sigma }_d=2.9\times {10}^{-4}\text{S/m}\end{array}$.

The loss tangent is given as:.

   $\begin{array}{l}l.t=\frac{{\sigma }_d}{\omega {\epsilon }^{\prime }}\\ l.t=\frac{2.9\times {10}^{-4}}{\left(2\times {10}^8\right)\left(2.8\right)\left(8.854\times {10}^{-12}\right)}\\ l.t=5.85\times {10}^{-2}\end{array}$.

Conclusion:

Thus, the loss tangent of the dielectric, when $RC=GL$ is $5.85\times {10}^{-2}$.