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I'm using an "IOT" radio module designed with a 50 ohm single ended antenna output. The working frequency is about 868 MHz, the maximum transmit power is 14dBm. The module antenna output is connected to a 50 ohm PCB transmission line and some RF switches and ends in a SMA female connector. The connector is meant for attaching an external 50 ohm antenna. In the path is also place for PI matching network.

I know how to correctly match a path and SMA including attached antenna (PCB or external) with a VNA - many explanations can be found online for this task.

In my current case the antenna is not yet known, but I want to make sure that any antenna would see 50 ohm on the SMA connector.

How can I match just the RF path + SMA connector to 50ohm instead of matching the RF path, SMA connector, and attached antenna?


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Hello Lorenzo,

thanks for your answer. I see your point that everything should be 50ohm. But my "own" antenna is not pure 50 ohm and needs some tuning.

Let me show you my situation in more detail: I use a custom L pcb antenna attached to a std. 50 ohm rf connector (mmcx in my case). The unmatched or single network matched bandwidth of the L-monopole is not enough to cover 791 to 960 MHz, needed for LTE Bands 8, 20 and 868 MHz SRD. This is why I added four individual matching networks using rf switches (see image) that switch according to the used frequency. My custom antenna is obviously not 50ohm but most likely no to far of since it's a monopole.

So I think my actual question is: How to determinate how much the match alters the 50 ohm impedance on the rf connector if one attaches an external antenna that has been tuned to 50 ohm? From my understanding i should measure the transmission line including the match to the rf connector and "hope" that the match for my pcb antenna does allow also the usage of an external antenna.

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  • \$\begingroup\$ What is a RF path from your point of view? \$\endgroup\$ Oct 14 at 6:58
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    \$\begingroup\$ "Any antenna" does not mean 50 ohm. In fact most antennas will not be 50 ohm. A quarter wave monopole is 37.5 ohm resistive and twenty odd ohms capacitive. A dipole is twice this amount. You need to be clear and think more about this. \$\endgroup\$
    – Andy aka
    Oct 14 at 7:15
  • \$\begingroup\$ Your best bet is to tune everything up to the SMA to 50 ohm. Then you put a compensating element at the base of your antenna (maybe a simple element will suffice from 791 to 960Mhz) to obtain an 'antenna assembly' tuned to 50 ohm and add coax from there. Otherwise the resulting impedance will depend on the length of the feed line which is ugly (it actually has uses in microwave:D). Search for multi-junction transmission lines for the ugly math \$\endgroup\$ Oct 14 at 7:33
  • \$\begingroup\$ Theo - Hi, On Stack Exchange, important new information & clarifications to the question should always be added into the question as an edit. Since your reply (originally an "answer") to Lorenzo included a new photo and clarification of your actual question which will be important to anyone else who is trying to help, and since it was not an actual answer (by Stack Exchange rules), I have moved it to become an edit to the question. There is more explanation of site rules & etiquette in the tour and help center. Also see my comment on Lorenzo's answer here. Thanks. \$\endgroup\$
    – SamGibson
    Oct 14 at 13:51
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It's really a non issue… the SMA connector is 50 ohm, the antenna would be 99% 50 ohm and the coax in between will be 50 ohm. So place your reference plane at the SMA connector and match up to that; the rest is 'assumed' to be already matched.

If for some strange unusual and incomprehensible reason your antenna will not be 50 ohm then you'll have to match from the coax feed to the antenna input at the attach point of the antenna. Ham antennas are routinely matched in this way.

It would be wrong to match on the board for an unmatched antenna since the connector and the coax would not be matched: you have to try to keep 50 ohm at every point.

Of course if the feed line is extremely short you could determine that the mismatch has no influence on the VSWR but that need to be verified case by case.

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  • \$\begingroup\$ Thank you. Since I don't know how to post images in a comment I posted my reply below as another answer. \$\endgroup\$
    – Theo
    Oct 14 at 7:25
  • \$\begingroup\$ @Theo - Hi, "Since I don't know how to post images in a comment I posted my reply below as another answer." People must not do that on Stack Exchange. Unlike forums with their sequential flow of new posts, on SE the box labeled "Your Answer" must only be used for entering answers. As you asked the question, the only time you would enter an answer is if you solved the question yourself, not following an existing answer, and you wanted to close the topic (a "self-answer"). || Since you are adding more info, I moved your "answer" to be an edit to the question (standard etiquette on SE). \$\endgroup\$
    – SamGibson
    Oct 14 at 13:28

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