Transmission line without ground

Hi,

There′s various types of transmission line, microstrip, coplanar, slot ... These lines guide the wave between two metallic planes. Now, if you have just a line, without ground (similar to an antenna), can this line guide the wave between two ports ? I did the simulation with CST and found just 1 dB of losses for a lambda/2 line @ 25 GHz and a typical silicon substrate used in 0.13um CMOS technology.
I think this result is very strange. Anyone could explain ?

Thanks

One question is, if you defined the boundary conditions in a way, that they allow a realistic simulation of radiation losses? The other point is, that the silicon substrate can act as a dielectric waveguide and keep most of the energy.

And there's also a contradiction in terms in your description. You are using ports in your simulation. Isn't one terminal of the ports connecting to ground? You didn't mention a differential transmission line.

If you have any "electrical walls" = perfect conductor boundaries, that is where your return current will flow.

It′s true. I forgot that my boundary conditions were electric walls. I replaced by open boundaries. Now I have just -6 dB of transmission.
For the ports, I used waveguide ports. They are not connected to ground.
"the silicon substrate can act as a dielectric waveguide and keep most of the energy". How is it possible ? It occures at dielectric interfaces ?

Thanks a lot

Sure, plenty of examples. Fiberoptic cable has no "ground return", yet propagates electromagnetic signals just fine. Image guide (sending microwaves down a square crossection rod of plastic) works fine too. Quasioptics have similar transmission media.

OK, so the same phenomenon than in fiberoptic can append at microwaves.
I have another doubt about transmission lines. It exists dielectric losses, resistive losses and radiation losses. The dielectric losses are due to transmission of electromagnetic wave in the dielectric and the resistive losses are due to current flowing in the metal. Now imagine that you apply a power to a transmission line with perfect conductor and very poor dielectric. The current will flow without losses and the EM wave will be totally attenuated. What will result in the load ? Same question for a perfect dielectric and a poor conductor.
I know that the two losses must be added to have the total loss. But I can′t understand why. What is the link ?

As far as I know, open boundaries (perfectly matched layers) are implemented by graded resistive sheets, so I would expect that they still provide some lossy return path.

---------- Post added at 19:49 ---------- Previous post was at 19:44 ----------

No, forget the "wave guide" stuff for a moment, that is a different field.

You can understand the transmission line as series of shunt C and series L elements. The capacitance is lossy because of dielectric loss, the inductance is lossy because of conductor loss.

These losses happen in transmission line too. The inductor / capacitor representation is usefull but it doesn′t help me to understand the phenomenum of transmission.

The inductor / capacitor representation with R'C'L'G' is an accurate description used for TEM transmission lines.

Your idea of a "wave" inside the dielectric, and the example of a fiberoptic, are misleading and not applicable to the line types that you described. This would require line diameters that are physically big compared to the wavelength (-> waveguide).

image guide is similar to waveguide, ~λ/2 in the εr of the plastic.

OK, it's different from a waveguide. But, it may exist a lot of similarities.
A type of transmission line is called coplanar waveguide ...
The most part of the electromagnetic field is concentrated between the line and the ground, that guide the wave.
But I know the transmission is different from a waveguide. The current flowing in the line does not exist in the waveguide. This wave is a consequence of this current or the opposite ?