0
\$\begingroup\$

I'm making a PCB with some high frequency striplines where impedance control is needed. After calculating, the track width according to my stack layer is 0.62 mm. I'm using blind vias to avoid any kind of stub due to excess copper.

I'm aware that vias are relevant in terms of return and insertion loss in high frequency design. After reading some literature I found some inconsistencies.

  • Some places say that the via should be as small (and short) as possible to reduce inductance.
  • Some others say the via diameter should be the same as the width of your track so as to have similar geometry to maintain the impedance. As I understand it, via diameter is via hole + crown. I'm not clear if just using 0.62 mm via diameter, the via hole diameter is not important and I can use 0.1 mm or 0.5 mm or any other as long as is less than the via diameter of course.

Which is the correct approach and what relevance does the diameter of the hole have to the overall diameter of the track?

I'm using Rogers substrate. The design is up to 5GHz but it is a system that sweeps from lower frequencies. Every dB in the final SNR is really important for the final application.

\$\endgroup\$
6
  • \$\begingroup\$ What frequency are you working up to? Via dimensions are pretty moot at (say) 2.4 GHz for non-extreme geometries with that track width. What substrate are you using? If FR4, then its variability will dwarf that of the vias. It's only worth worrying about vias if you need to use RO4350 over FR4. If you are using a via between RF tracks, then you have bigger problems with what the ground plane(s) is/are doing. If it's not RF, but high speed logic, then stop worrying. \$\endgroup\$
    – Neil_UK
    Commented Jun 7, 2022 at 9:18
  • \$\begingroup\$ What's "death excess copper"? Sounds brutal. \$\endgroup\$
    – winny
    Commented Jun 7, 2022 at 9:30
  • \$\begingroup\$ @Neil_UK I'm using Rogers sustrate, the design is up to 5GHz but its a system that sweeps from lower frequencies and each dB in the final SNR is really important for the final application. Anyway more than if I have to be concerned or not for this design I want to know that for get best practice for possible future designs where this will be more critical than now. \$\endgroup\$ Commented Jun 7, 2022 at 10:06
  • \$\begingroup\$ @winny this is excess copper !Valid XHTML. \$\endgroup\$ Commented Jun 7, 2022 at 10:11
  • \$\begingroup\$ For high speed designs, it is not advisable to change layers (except for breakout). The impedance of any two layers is unlikely to be exactly the same so there will be reflections (vendors usually give 10%). If you really need to go through the board them the reference should also go through next to it. \$\endgroup\$ Commented Jun 7, 2022 at 10:13

1 Answer 1

1
\$\begingroup\$

If you're working at 5 GHz with Rogers material, and return loss S11 is very important, then either

  • do not change layers with an RF track or
  • design the whole track layer change area as an RF transition, of which the via diameter is one of the least of your worries

If the RF track changes layer, then either they are referred to the same ground plane, or to different ones.

If they are referred to the same ground plane then they will change height, they will also need to change width. This will introduce a capacitive discontinuity. With the right design, you can introduce a sniff of inductance to turn it into a well matched lowpass filter, by either using a thin via, or relieving the ground plane back from the track judiciously.

If they are referred to different grounds, then you will need to make sure the RF current moves from ground to ground with minimum added inductance. Then you can design the rest of the transition as above.

A 3D RF solver would be useful to visualise just how much violence is occuring to the current flows as a result of the layer change, which will help you with tuning it up.

If you're working at 5 GHz with Rogers material, and through transmission S21 is very important, then you don't have so much of a problem, because you can get quite significant reflections from a single discontinuity with very little impact on the through gain.

If you have two or more bad matches at the right (wrong!) spacing, then resonance between them at specific frequencies can build up to serious through problems. As you've specified a swept measurement, you probably will find the resonant frequency.

\$\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.