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enter image description here

Attached is an image of a microstrip PCB I am working on. In the image the orange is the copper trace and the blue is the Rogers4350 substrate. Underneath the blue substrate is the ground plane. The imaged circuit is a power divider which takes the RF signal and divides it equally between two directions. Initially the microstrip line has 100 ohm impedance. At the T junction the microstrip line increases to 200 ohms (200ohms in parallel with 200 ohms is 100 ohms). From a circuit standpoint I can understand how and why the power flows from the 100 ohm line to the 200 ohm lines.

Where I am confused though is when I view the signal as an electromagnetic wave. I struggle to understand why the wave would change direction and flow into the 200 ohm lines as opposed to simply reflecting from the open in the microstrip? Does the wave diffract from the corners of the T junction? Is the fact that the wavelength of the EM wave is much larger than the T junction a relevant fact? My rational here being that due to a larger wavelength there is charge which is being forced to pile up at the T-junction. Eventually this charge must redistribute and flows towards the 200 ohm lines?

Simply put I am trying to better understand the mechanism by which the majority of the wavefront is able to squeeze from the wide 100 ohm trace into the two much thinner 200 ohms traces with less than -20 dB of reflected power.

Thank You

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  • $\begingroup$ The copper forms a waveguide. It guides the wave. A right angle-bend in a microstrip with 100 ohm geometry on both sides will also guide the wave into a right angle turn without it reflecting (very much) off the bend. The same principle applies here. $\endgroup$
    – The Photon
    Commented Jan 19, 2022 at 1:04

2 Answers 2

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Huygens' principle applies as much to RF waves propagating in waveguides as it does to optical waves passing through apertures. Each point on the wavefront can be treated as a source of spherical waves and the superposition of the waves radiating from all these sources will give the further propagation of the wave.

In this case, the wave encountering the "end" of the 100-ohm line segment radiates in all directions. In the "forward" direction it can't propagate further because the copper structure that guided it ends. But to the left and right it can propagate into the 200-ohm line segments. So that is what it does.

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  • $\begingroup$ Huygen explains how the wave could enter the 200-ohm lines. But does not explain why the wave doesn't reflect from the end of T-junction. Quantitatively I could view the 100-ohm section just before the 200-ohm section as an aperture. Calculating the field from this aperture at all angles and integrating I could calculate the percent of power that goes to the back wall vs into the 200 ohm line. Treating the fields at the T-end as an aperture I could repeat the process. Comparing the 100 ohm width to the 200 ohm width I expect the calculation would result in more power being reflected $\endgroup$
    – CMH12
    Commented Jan 19, 2022 at 1:51
  • $\begingroup$ @CMH12, in another comment you said you're modeling this in HFSS. So does HFSS show the energy reflecting back up the feed line, or being transmitted into the 200-ohm lines? I would expect the reflection is non-zero, but the majority of the input power should end up in the output lines. $\endgroup$
    – The Photon
    Commented Jan 19, 2022 at 2:00
  • $\begingroup$ yes I could calculate the S11 and S21/S31 for this circuit as well as plot any fields. I have no doubt that as you say most of the energy will be delivered to ports 2 and 3 (S11 is -20dB) but I am still working to understand the physics of why this happens. $\endgroup$
    – CMH12
    Commented Jan 19, 2022 at 2:09
  • $\begingroup$ My point in invoking Huygens' principle is that waves don't propagate in straight lines and you shouldn't expect them to. When the wave reaches the T junction, there is just as much reason for it to start propagating "sideways" as to continue propagating downwards, because the "sources" at the wavefront in the Huygen's view are spherical---there is no preferred direction for the onward propagation. $\endgroup$
    – The Photon
    Commented Jan 19, 2022 at 2:12
  • $\begingroup$ I agree, the wave has no preferred direction, it cannot be assumed that the wave prefers to propagate downwards. From simulations it is clear that the wave prefers to propagate sideways. I am trying to come up for a justification of why this would be the case as Huygens principle alone does not suggest either sideways or downwards propagation. $\endgroup$
    – CMH12
    Commented Jan 19, 2022 at 3:34
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If you have not yet learned it then you will soon that discontinuities on a transmission line, such as sudden change in dimension or dielectric/magnetic properties, etc., correspond to a a network of lumped element circuits, specifically capacitors and inductors, and if the metallic/dielectric losses are also taken account then also including resistors. These lumped element circuit elements cause the (q)TEM wave (q-quasi: because it is not a pure TEM for microstrip) propagating along the $R_1=100\Omega$ transmission line would be reflected as you have expected it by the junction where it meets the two $R_2=200\Omega$ lines.

Notice that in the drawing those two high impedance lines are shaped so that they flare out going away from the junction. Because of the flare the wave impedance of the nominally $R_2=200\Omega$ line is actually higher than $R_2$ near the junction and then gradually after a couple or so wavelengths becomes $R_2$ towards the load. The purpose of that shaping is to tune out the mismatch caused by the junction (lumped element L/C circuit, losses are ignored), as much as possible. The longer is the flare the broader the tuning bandwidth is. If there is not enough room for the flare then tuning can also be accomplished by smaller reactive load(s) directly on the junction. In metallic waveguides screws and other metallic plates (iris) can be inserted to act as reactance(s). On a microstrip line you can put notches, slits, etc., to match the junction.

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  • $\begingroup$ I am aware that the discontinuity could be modeled as a lumped element circuit, but I am not interested in this method as does not consider the fundamental wave phenomena that allows the EM energy to propagate down the transmission line. Also this is a circuit I am designing in HFSS. At the junction the trace width is the width for 200 ohms. The taper is actually there because the 200 ohms lines quickly return back to 100-ohm lines after the junction. $\endgroup$
    – CMH12
    Commented Jan 19, 2022 at 1:58

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