I've been struggling on how to design (and develop in practice) an impedance matching network with some traces/tracks/paths (whatever you wanna call it) and some connectors in between. So here is the architecture:

Antenna (50 Ohm) + path (50 Ohm) + Matching network + path & SMA connector (?) + SMA connector & path (?) + Rectifier & load (Z_L known)


simulate this circuit – Schematic created using CircuitLab

The perfect matching would be if right after the last component of the matching network comes the first component of the rectifier, but it is not the case by far, since there are paths and connectors in between.


  • Should I count the impedance of the load starting from the connector? --> then... probably there are reflections between the connector&path and the rectifier.
  • Should I characterize the rectifier with a TRL calibration (having out path and connector) and count it as part of the matching?? --> I don't have a clue how to do the matching like that... with ADS or directly with the Smith Chart I haven't arrived that far.
  • Looks like it is not possible and what I have to do is include the rectifier in the PCB1 and the rest of the load in the PCB2... But again... even the small path connecting the last component from the matching and the first from the rectifier could have an impact. How could I include this into it? (it would be the same case as the current one though).
  • Any other solution? :C

If somebody could throw some light there, I'd be eternally grateful.


  • \$\begingroup\$ Note sure what you are trying to match here, at every connector you should have a defined impedance until to the parts that "translate" it \$\endgroup\$
    – PlasmaHH
    Aug 17, 2018 at 12:51
  • \$\begingroup\$ Do you know your \$Z_L\$? If so, you're just trying to create a matching network that transforms \$50\Omega\$ into your \$Z_L\$. Am I understanding your issue correctly? Also, I assume you're looking for a narrowband match? \$\endgroup\$
    – Shamtam
    Aug 17, 2018 at 12:57
  • \$\begingroup\$ a schematic and a drawing of what you're trying to do seems a little necessary... \$\endgroup\$ Aug 17, 2018 at 14:12
  • \$\begingroup\$ Thank you guys for your comments, I just edited the question with a bit more of explanation. \$\endgroup\$
    – emece
    Aug 17, 2018 at 14:56

1 Answer 1


It's tough understand exactly what you're trying to do or where everything is located, but in general your matching network will sit between two different blocks (let's call them \$Z_{source}\$ and \$Z_{load}\$). If you know the impedance looking in at both of your blocks, then your matching network should simply transform \$Z_{source}\$ into \$Z_{load}\$.

From your question and your wording, I'm assuming you're doing matching on a PCB or similar (because you use the word "path"). It seems to me that the first impedance is the impedance looking into the trace leading to your antenna, which you mention is \$50\Omega\$. So let's say \$Z_{source}\ = 50\Omega\$.

The second impedance to consider is what you see looking towards your rectifier/load from the matching network. It looks to me like that would be your "path and connector" as well as the rectifier.

Keep in mind, if your "path and connector" impedance isn't well matched to your rectifier at that reference plane, you'll be getting reflections at that interface that your matching network won't be able to help with. As such, I think your question looks strange because the "matching network" should be next to your rectifier and load, converting its impedance into \$50\Omega\$, then feeding that into the first "path and connector" and using standard \$50\Omega\$ coax, lines, etc. all the way to the antenna.

Edit (post-update of question): Your matching network should be on PCB2, just between "SMA & Path" and "Load." Transform the load to \$50\Omega\$, then just calculate and use controlled \$50\Omega\$ traces all the way to your SMA connectors on both PCBs.

  • \$\begingroup\$ Yes, exactly. The point is if: - Counting the "load" from the last component of the matching, the path and connectors will be also matched, but... will be reflections between them and the rectifier?? - Counting the "load" from the rectifier, obviously the matching will be incorrect, since the impedance of the paths and connectors are not taken into account. Thank you Shamtam for your answer! \$\endgroup\$
    – emece
    Aug 17, 2018 at 14:56
  • \$\begingroup\$ And also... if the rectifier was right after the matching but the path was not "small enough". How to take it into account for the matching? Should be added in series with the rectifier impedance? Wouldn't be reflections between this path and the rectifier? \$\endgroup\$
    – emece
    Aug 17, 2018 at 15:03
  • \$\begingroup\$ @emece Correct, if the "path" isn't small enough, then you'll get reflections at that interface. Part of the matching network can be a tapered feed line or similar, though, if for some reason you need some extra space between your load and your matching network. \$\endgroup\$
    – Shamtam
    Aug 17, 2018 at 15:12
  • \$\begingroup\$ Got it, thank you! I guess I have to move the rectifier to PCB1. But still, the small path connecting matching and rectifier, which impedance should it have?? (because I can change the width and length for that tuning) (assuming it was big enough for having an influence). \$\endgroup\$
    – emece
    Aug 17, 2018 at 15:21
  • \$\begingroup\$ You want the impedance to be as close to the impedance of the load as reasonable. \$\endgroup\$
    – Shamtam
    Aug 17, 2018 at 15:23

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