Question 6 of 8: Line parameters of a parallel-ribbon line

Question 6: Line parameters of a parallel-ribbon line (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A parallel-strip (ribbon) line: two flat conductors of width $w = 2\ \text{cm}$ separated by a dielectric sheet $d = 0.5\ \text{mm}$ thick with $\varepsilon_r = 2.25$. The sheet cross-section quoted in the question, 2 cm by 0.5 mm, confirms that the dielectric exactly fills the space between the ribbons. Since $w/d = 40 \gg 1$, neglecting fringing is well justified.

Find. The distributed capacitance and inductance per metre, the characteristic impedance, and the propagation velocity.

Approach. Treat the cross-section as an ideal parallel-plate capacitor and as an ideal flat current sheet pair, then combine the two distributed parameters in the standard lossless-line relations.

1. Capacitance per metre. A parallel-plate capacitor of plate area $w \times 1\ \text{m}$ and separation $d$: $$C' = \frac{\varepsilon_0\varepsilon_r w}{d} = \frac{(8.85\times10^{-12})(2.25)(0.02)}{5\times10^{-4}} = \boxed{796.5\ \text{pF/m}}$$
2. Inductance per metre. Equal and opposite sheet currents confine a uniform $B$ to the same rectangle, and the flux linked per metre of length gives $$L' = \frac{\mu_0 d}{w} = \frac{(4\pi\times10^{-7})(5\times10^{-4})}{0.02} = \boxed{31.42\ \text{nH/m}}$$
3. Characteristic impedance. For a lossless line $$Z_0 = \sqrt{\frac{L'}{C'}} = \sqrt{\frac{3.1416\times10^{-8}}{7.965\times10^{-10}}} = \boxed{6.280\ \Omega}$$
4. Propagation velocity. $$v_p = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{2.5023\times10^{-17}}} = \boxed{2.00\times10^{8}\ \text{m/s}}$$

Both results carry a useful structural check. Substituting the expressions for $C'$ and $L'$ into the two formulas gives $Z_0 = (\eta_0/\sqrt{\varepsilon_r})(d/w) = (377/1.5)(0.025) = 6.28\ \Omega$ and $v_p = 1/\sqrt{\mu_0\varepsilon_0\varepsilon_r} = c/\sqrt{\varepsilon_r} = c/1.5$. The geometry cancels out of the velocity entirely, as it must for any TEM line: the wave travels at the speed of light in the filling material regardless of the shape of the conductors, and only the impedance depends on the cross-section.

The very low characteristic impedance is not an error but the direct consequence of the aspect ratio. A line 40 times wider than it is thick has a large capacitance and a small inductance per metre, and $Z_0$ falls in proportion to $d/w$. Practical ribbon lines of this kind are used where low impedance and tight field confinement matter — power-distribution bus bars and printed-circuit power planes behave in exactly this way — rather than for signal feeds, which need impedances closer to 50 ohms.