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I am having some fundamental issues trying to create an impedance matching network for an 868MHz antenna, I have a PI network between the antenna's connector and the antenna feed point

I have measured the characteristic impedance of the antenna using my VNA and this shows me an impedance of 4-j22 ohms. Using the smitch chart matching app at https://www.rfmentor.com/content/smith-chart-matching-app I came up with an L-L network that should get me the match. I first optimised the shunt L that got me to the constant resistance circle in the capacative region, as soon as I try to add a series inductor to bring towards the middle of the chart, it goes off in a completely different direction, it actually began moving further into the capacative region and away from the constant resistance circle.

Could anybody give me some pointers on what could be going wrong here? How could the series inductor be not moving me upwards on the chart? My shunt inductor is on the antenna side and the series is in between the antenna and my VNA.

Here is the proposed matching topology

enter image description here

And here is the smitch chart behind the match

enter image description here

Eoin O Connell
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  • You need an account to access that site. Why not draw out your proposed circuit? – Andy aka Aug 04 '20 at 09:50
  • I have added some updates, basically the series never brings up me towards the 50 ohm point, but the shunt L did get me to the constant resistance circle – Eoin O Connell Aug 04 '20 at 10:09
  • I wouldn't do it this way I guess is the problem. I would convert the series 4 ohm and capacitance of 8.31 pF (4 - j22) to parallel to give 125 ohms in parallel with 8.258 pF then, I'd use an LC pad to make an input impedance of 50 ohms (if 50 ohms is what you are trying to match to). – Andy aka Aug 04 '20 at 10:38
  • MY vna is 50 ohm source impedance, what values and which type of network would you use? Your answer is not very clear – Eoin O Connell Aug 04 '20 at 10:57
  • I haven't given an answer yet because you have another question you raised that needs attending - see my earlier comment on [this question](https://electronics.stackexchange.com/questions/495988/calculate-power-of-an-lte-signal) – Andy aka Aug 04 '20 at 11:07
  • I forgot to upvote your answer! Thanks for that answer – Eoin O Connell Aug 04 '20 at 11:09
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    If the answer was correct and helped, you should formally accept the answer. The upvote is of course very welcome. Answer acceptance helps others see that the answer was useful as seen from the questioner's perspective. [See this](https://i.stack.imgur.com/mKrn6.png) for assistance. – Andy aka Aug 04 '20 at 11:14
  • The components values for your inductors seem to be ok so that makes me think that it may have to do with *how* you're measuring with the VNA. How did you calibrate the VNA? Did the calibration include the connection (cable) between the VNA's port and the antenna's port? If not, you either have to calibrate to include that extra cable or do port extensions--otherwise, the actual impedance of the antenna is being rotated around the smith chart. – Big6 Aug 04 '20 at 15:12

1 Answers1

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MY vna is 50 ohm source impedance, what values and which type of network would you use?

  • First convert 4 - j22 to a parallel capacitor and resistor (8.258 pF and 125 ohms) Try this calculator.
  • Assume an LC pad to will match 50 ohm to 125 ohm (may need a slight mod)
  • Recognize that the LC pad needs too little capacitance compared to 8.258 pF
  • Resonate out most of that capacitance using a parallel inductance of 5.49 nH

If you need the formulas let me know.

Circuit solution

enter image description here

The values above are not far off what were entered into the non-obliging smith chart calculator: -

enter image description here

Result

enter image description here

  • top graph shows that Va is attenuated by 6.019 dB. That's a 2:1 reduction in plain talk.
  • bottom graph shows that Va has a phase displacement of 7.94 thousandths of a degree. That's 0° in plain talk.

Hence the input impedance at Va is 50 ohms and it matches to 125 ohms at the load.

If I substitute the original load of 4 - j22 I get exactly the same result at 848 MHz. If you need a demonstration let me know.

Andy aka
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