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GNSS Antenna Trace Width on 2 LAYER


I've recently been trying to figure out a realistic way to design my traces on my 2 layer PCB (to be made by OSHPark) that can handle GNSS antenna signals (like from GPS satellites). According to the datasheet of BGA725L6 (a GNSS amplifier), it expects an inductor and a capacitor in series of 0402 size between the antenna patch and itself (the GNSS amplifier IC), which implies the trace width for a 50 ohm characteristic impedance would be around 20mil. However I used the calculators online for characteristic impedance for my PCB and instead I got 118mil for a 50 ohm characteristic impedance. Now this would mean I need 1812 components to reduce signal reflecting back caused by the changing trace width from trace to the pads of the passive components (lol V does not equal IR), however this is drastically larger than what the datasheet implies. I wouldn't even be able to avoid the massive trace width jump from the trace to the actual pad of the GNSS amplifier. I have a feeling I'm supposed to use a multilayer board with thinner copper and a thinner dielectric thickness (between layers), but is it even remotely possible to handle GNSS signals well on 2 layer FR4?

Additionally, the datasheet suggests the values of the capacitor and the inductor to be 1nF and 7.5nH, but that places the resonance frequency significantly outside of the 1.1 to 1.6 GHz range that GNSS (like GPS) signals uses. Luckily, the datasheet of the GNSS processing IC I'm using suggests those components to be 120pF and 5.6nH instead which has a resonance frequency right on the target, but both minimum impedances aren't the most optimised (14 ohms, but could be <10 ohms with a better cap inductor choice). Why does the datasheet say the inductor is necessary for impedance matching? Is it worth it going for a 3rd order filter to filter out white (gaussian?) noise better (as white noise is introduced for all frequencies unfiltered)?

Yes, 50 ohm (or whatever) microstrip lines will be thinner track width if you have your signal plane closer to your signal reference / return / ground plane as applicable.

To achieve that you can either
* Use a thinner dielectric between the trace and the return/reference plane
* Use a higher dielectric constant material for the insulator (not typically very adjustable for low cost PCBs)
* If possible reference a physically closer plane for the return/reference plane / path (e.g. L1+L2 instead of L1+L4 for a 4L PCB)
* Use a different transmission line structure rather than microstrip.

In particular if you use CPW (coplanar waveguide) or often better GCPW (ground referenced coplanar waveguide)
the RF track width will be smaller for a given impedance. With GCPW you can have, for instance, the signal on L1
and the return plane on L2 but then you also have grounded pours on L1 next to the RF track on either side, and you can control
the spacing between the L1 signal and L1 ground pours (e.g. possibly as close as 0.1mm if you want) which makes them significantly
closer than maybe your L1 to L2 distance might be.

If you want to use microstrip on a 2L PCB you might want to switch your PCB thickness to 0.8mm or thinner so that your RF trace can be thinner and
the fields will be more contained.

Look at the saturnpcb pcb calculator tool and other such tools for calculating microstrip, stripline, cpw, gcpw, etc.

As for the resonant L/C components, I didn't read the application notes so I can't second guess their optimal or non optimal reference design / application note.
You may be right that the higher Q and more accurately tuned resonance vs. the signal frequency is better.  A technical rationale they might have used would be that the component selection is a trade-off between in-band and out-of-band performance somehow.  Another technical rationale they might have used is empirical optimization -- if they took into account component L/C parasitics due to the tracks, the PCB land patterns, the component construction, et. al. they may have found the resonance "in reality" was more optimum with the given reference components because of the nominal value of the components in conjunction with the package/PCB parasitics.

Non-technically sometimes they use cheaper components for BOM cost reasons even though they could use more precise or higher Q components to make
a higher performance higher precision circuit.

That's the hard one if you can't order a board that is 0.3mm thin (or even less). For a standard cheap 1.6mm board, you need like 2-3mm wide traces and then like 6mm of "nothing" on each side of the trace...

I quasi-solved a similar problem by putting my IC next to the board edge and then connecting it pretty much directly to the SMA connector (for my purpose, there are were no components are in between so it was a bit easier). You might put 0402 elements on as thick trace as you can and make it as short as you can and hope for the best.

Other solution that worked very good on a different project of mine was putting the U.FL connector directly next to the IC (like 0.5mm away) and then using a small U.FL to SMA cable to put the connector where I wanted - I needed the connector on a cable anyways so this was perfect!


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