1
\$\begingroup\$

I have a trivial question about EM propagation in a microstrip line.

Why do the electric and magnetic fields have the following lines? Which are the physical laws (Maxwell etc) for determining them?

enter image description here

By looking at the image I think that the electric field lines between the 2 planes are due to the voltage applied to the planes (like those inside the dielectric of a capacitor), while the magnetic closed lines are determined by the current flowing inside the upper strip (but why are the lines of the GND plane not present?).

\$\endgroup\$
3
  • \$\begingroup\$ Can you give a link to where you found the diagram? Which lines are supposed to be E field and which are supposed to be H field? Which mode is this supposed to represent (hint: it's not the quasi-TEM00 mode you normally want to use)? \$\endgroup\$
    – The Photon
    Jun 21, 2019 at 17:32
  • \$\begingroup\$ I'm always amazed when an OP can't be bothered to mark any of the responses as Accepted answer or at least up-vote the ones they found helpful. C'mon! \$\endgroup\$
    – pfabri
    Feb 18, 2021 at 12:32
  • \$\begingroup\$ @pfabri It may happen to forgot doing it. I've asked this question two years ago when I didn't know exactly how this forum worked. Now I don't forget this anymore (except when no answer has helped me, which is quitw rare). Anyway, thank you for the suggestion. I have voted the best answer. \$\endgroup\$
    – Kinka-Byo
    Feb 18, 2021 at 16:09

3 Answers 3

2
\$\begingroup\$

This image is unusual, it presents a half wave resonator which is not connected to anything visible, but has somehow got a wave which reflects forth and back. The line length direction is left-right.

The blue lines present electric field. The transparency presents the E-field cancellation in the middle of the line.

Letter H refers the thickness of the insulation layer.

Maxwell's equations produce with short vector manipulation the vector wave equation. Solving the wave equation in given geometry with given materials gives the possible waves. Which of the possible waves actually exist depend on how they are excited.

There's a place for a big error here. The waves do not occur in the metal, they are around the metal plates, in this case hopefully mostly in the insulation layer. Metal directs the propagation so that hopefully we haven't an antenna, but a transmission line. Of course the fields induce some current in the metal - in simple cases we can follow the wave by thinking only voltages between the conductors and the current in the metal, but the actual energy transfer happens out of the metal by the vector field wave. Forgetting the fields and thinking only the current and voltage is very common for ex. at 50Hz.

\$\endgroup\$
3
  • \$\begingroup\$ Please, can you give me an intuitive idea of how different excitations may change the field lines? \$\endgroup\$
    – Kinka-Byo
    Jun 22, 2019 at 14:01
  • \$\begingroup\$ Rise the frequency => more complex waves become possible. For that reason transmission lines generally are used only in the frequency range where only one waveform is possible. Very short lines can be so regularly made that unwanted waveforms do not exist altough they are possible. Excitation=an incoming wave or locally generated field (=for ex. a microwave transistor starts to conduct , E drops and magnetic field appears). A wave continues along the transmission line if the field directions at least partially fit with a possible waveform. \$\endgroup\$
    – user136077
    Jun 22, 2019 at 14:17
  • \$\begingroup\$ Perfect, thank you very much \$\endgroup\$
    – Kinka-Byo
    Jun 22, 2019 at 16:00
1
\$\begingroup\$

Curved ground lines are not shown because the plane is assumed to be infinite or sufficiently large to be vertical as shown at the ground surface.

\$\endgroup\$
0
\$\begingroup\$

There are only E-field lines drawed in that graphic, because these type of field lines are in right angle to the conductors surfaces.

If there were H-field lines, they would be (mostly) tangential to the conductors surfaces.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.