Where do you see resonant coupling? Resonant structures can include coupling to an interconnect that is floating and not terminated. The structure does not have to be a uniform transmission line, it can also be a cavity made up of two or more adjacent plates as shown in Figure 1. Commonly, when a signal goes through a cavity and we hit the resonances of the cavity, it absorbs energy and results in narrow dips in the S-parameters.

In Figure 1, the signal goes through a via from top to bottom. The return current has to make the transition from the upper to the lower return plane. If the return path is not well controlled and is not in close proximity to the signal lines, the return current gets coupled into that cavity and we see that the energy of those resonant frequencies sucked out of the signal that gets transmitted.

Let’s look at a real-world example shown in Figure 2.

Figure 2. A printed circuit board with eight symmetrically placed SMA connectors serves as an example to show the effects of several high-Q resonance modes based on the board's dimensions. |

The principal modes are defined by the length and the width of the board. Another resonant mode is defined by the spacings of the SMA connectors, where the connector grounds short the internal planes. Using the given equation, we can calculate the resonant and list the expected frequencies.

__Length (in)__

__Resonant Frequency (GHz)__3.25 0.92

1.187 2.5

0.8 3.75

Now let’s look at the measurement of S21 for this board (in Figure 3).

Figure 3. The measured S21 of the PCB showing 3 principal modes. |

The Q of the resonance is the ratio of the resonant frequency to the width at half the maximum. The resonance at 2.5 GHz has a width of about 0.1 GHz, so the Q is 25. Q’s of more than 10 are considered high.

The depth of the dips due to resonant coupling depends on the degree of coupling. The dip at 0.92 GHz has the smallest dip because the via in the port1 to port 2 path is symmetrically spaced at the center of the board length. Energy propagates equally to the left and to the right and the equal reflections cancel each other. If the via had been at either end of the board, the coupling would be greater. The via is closer to the upper edge of the board, and this asymmetry forces a higher degree of coupling for the resonant dip at 2.5 GHz. There are multiple resonances for the 0.8 inch spacing, and the via is not particularly symmetric with most of them, so the coupling is greater.

Eliminating these dips requires careful design. First, you should try to avoid transitioning the signal between different return planes. If you must, use return vias adjacent to each signal via to suppress resonance.

Watch the on-demand webinar

*Reading S-Parameters Like a Book*by Dr. Eric Bogatin, Signal Integrity Evangelist, for a more in-depth treatment of this topic.

## 2 comments:

Hello,

Thank you for this article!! Can you please explain where the 12GHz in the formula comes from? Or is it explained in the webinar? Thank you again.

Hello DE_T,

Sorry for this long delay in replying!

The formula above is for resonance frequency in GHz, taken from Dr. Eric Bogatin's book "Signal and Power Integrity Simplified" Third Edition, p. 653.

fres = 11.8/sqrt(Dk)*2*Len

where:

--fres - resonance frequency in GHz

--Dk is the dielectric constant of the laminate inside the cavity nominal value of 4 for FR4

--Len is the length of a side of the cavity in inches

11.8 is the speed of light in a vacuum in inches per nanosecond. During his class, Dr. Bogatin rounded the number to 12 in order to simplify the calculation.

Post a Comment