1
\$\begingroup\$

I'm trying to route a LoRa module to an SMA Antenna on my board (first time tracing RF signal). Shown in image below:

enter image description here

I adjusted my layer stack up for my 2 layer board, 1.6mm thickness and added an impedance profile.

enter image description here

Here, Altium has recommened a 2.64mm trace for a 50 Ohm impedance track. Based on Impedance Matching videos I've seen traces in their circuit have been set to 0.2mm (for 50Ohm matching) (admittedly they were for 4 layer boards) however I just don't know if such a drastic difference is right or I've done something wrong. The image below shows the track (yet to add polygon pour and via stitching):

enter image description here

\$\endgroup\$
6
  • 1
    \$\begingroup\$ The trace is huge because that is a two layer board and the ground plane is the back side of the board. Usually controlled impedance is done on 4 layer boards with a ground plane on the same side of the board as the trace, in which case you will get ~200 um traces. \$\endgroup\$ Oct 6, 2021 at 1:56
  • \$\begingroup\$ So, is it okay to just leave it as a 2 layer board with a trace this wide? Or do I need to use a 4 layer board? \$\endgroup\$
    – Explorex
    Oct 6, 2021 at 1:57
  • 1
    \$\begingroup\$ Most likely it is fine, especially at lower frequencies. Is this a one off design? If so, cost of 4 layers with controlled impedance will be nearly the same as 2 layer. If you're making a lot, might be worth testing the 2 vs 4 layer version. \$\endgroup\$ Oct 6, 2021 at 2:16
  • \$\begingroup\$ Thanks for your response! I appreciate it a lot, comparing the two is a great idea I might just do that. Thanks for taking the time out to give me a reply. \$\endgroup\$
    – Explorex
    Oct 6, 2021 at 2:27
  • \$\begingroup\$ What is the maximum frequency content of the signal and over what length of PCB is the trace? \$\endgroup\$
    – Andy aka
    Oct 6, 2021 at 7:09

1 Answer 1

2
\$\begingroup\$

The LoRa frequency is 915MHz and the PCB trace is roughly 17mm (+-1mm) in the length @Andyaka

915 MHz has a wavelength of 32.8 cm and the fairly widespread rule of thumb as to whether to keep traces/wires at the optimum impedance suggests that if the distance to be covered is less than one-tenth of a wavelength, then it's just not worth doing because the benefits would be so small or insignificant.

Your distance is 1.7 cm and this about one-twentieth of the distance so, just pick tracks widths that are convenient.

\$\endgroup\$
2
  • \$\begingroup\$ Perfect! Thanks for your reply! I would love to know how you calculated the wavelength as a distance, so I can do this for future reference. I assume you possibly used V = f * (lambda) but I'm not sure what you substituted for V? It makes sense why this only matters at high speed circuits then. \$\endgroup\$
    – Explorex
    Oct 7, 2021 at 9:24
  • 1
    \$\begingroup\$ 300 MHz has a wavelength of 1 metre and therefore 915 MHz is shorter proportionately. This assumes V = speed of light. On a circuit board the electric waves travel at about two thirds of c hence, in reality your probable electrical wavelength is about 32.8 cm/1.5 = 22 cm but still more than 10x the rule of thumb limit. \$\endgroup\$
    – Andy aka
    Oct 7, 2021 at 9:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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