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

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

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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):

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    \$\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\$
    – user1850479
    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
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    \$\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\$
    – user1850479
    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

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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.

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  • \$\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
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    \$\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

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