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I'm using a directional coupler for a radar application using frequencies of 5.3-5.9GHz. The directional coupler linked above is matched to a 50ohm characteristic impedance.

How do I terminate the isolation port to prevent reflections? Various sources online just say that I should terminate with the characteristic impedance, but is it sufficient to place a short microstrip at the isolated port and then via to a common ground plane? Is it bad to use the common ground plane? Do I need to worry about RF vias here, or is a regular via sufficient? I've seen a design that uses a 0402 49.9ohm resistor between the isolated port and a via to ground (connected by short microstrips). Does that provide any additional protection?

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How do I terminate the isolation port to prevent reflections? Various sources online just say that I should terminate with the characteristic impedance

And that would be correct, any output with a characteristic impedance can only be terminated properly by terminating it with a resistor to ground with the same value as the characteristic impedance, so 50 ohms in this case.

The ground should be the same ground as the shield of the signal output. Use any other ground and the signal path will be longer which adds inductance to the 50 ohms which will prevent a proper termination.

What if you do not properly terminate the signal? Then the output signal at that port will reflect back into that same port. Terminate it properly and all signal will be converted to heat by the resistor. Then (almost) no signal will reflect back.

but is it sufficient to place a short microstrip at the isolated port and then via to a common ground plane?

A microstrip is just a path for the signal to a different location. It is not a proper termination. At the end of the (50 ohm) microstrip you would still need a 50 ohm resistor.

Is it bad to use the common ground plane?

It depends on your design, if you are making a sensitive (easily disturbed) RF circuit then you don't want to share the same ground (plane) with some digital circuits. That would increase noise levels. If your RF signals are quite large it might not matter.

Do I need to worry about RF vias here, or is a regular via sufficient?

My guess is that an RF via has less inductance. Less inductance is almost always a good thing in RF design, so if you have the option: yes, use RF vias.

I've seen a design that uses a 0402 49.9ohm resistor between the isolated port and a via to ground (connected by short microstrips). Does that provide any additional protection?

Protection for / against what? This is the proper way to terminate a 50 ohm (49.9 ohms is close enough to 50 ohms) port. Just do the same!

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The matched termination that you provide on the board is vital for the specification of the coupler.

If you leave it reflective, it will reflect the energy coupled from the backwards direction through the coupler and send it out through the coupled port. You will then have lost all your directivity, the reason you are using a coupler instead of the simpler and cheaper voltage pickoff (which has no directivity).

Take a microstrip far enough away from the package so you can handle the footprint easily, then put a resistor to ground. Usually at this frequency, I'd use two 100ohm resistors in parallel, which tends to provide a better load than a single 50ohm at these frequencies and component sizes. However, the specifications of this coupler are so poor that there's little to be gained by that refinement.

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  • \$\begingroup\$ Any recommendations on where to find an smd coupler with better specs? A quick look on digikey/octopart didn't yield much for my frequency requirements and coupling factor. I considered creating a footprint myself, but I imagine that often generates worse performance and I don't have any experience doing that. I also don't have access to any proprietary software which I believe people use for this sort of thing. \$\endgroup\$ – MattHusz Jan 8 at 14:43
  • \$\begingroup\$ Also, what's bad about the specs? Given my inexperience with RF I didn't notice any significant issues. \$\endgroup\$ – MattHusz Jan 8 at 14:49
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    \$\begingroup\$ @MattHusz I worked in instrumentation, where we always strove for typically 10dB better on everything than a commercial product, so maybe I'm being too dismissive and it's not bad for an available SMD part. Band-centre the directivity is pretty good, but the isolated and coupled ports have marginal S11. I'm sure you could match this performance with a printed device, but it's a bit scary if you've no experience in doing that, and it does put you at the mercy of the board fab to be repeatable. \$\endgroup\$ – Neil_UK Jan 8 at 14:57
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You need to connect it to something that both:

1) matches the impedance (ie doesn't provide a reflection at the connection) and

2) absorbs any signals that come out.

A microstrip terminated in a via to ground does 1), but reflects the signals straight back in, so doesn't do 2).

A microstrip terminated in a resistor will (if the resistor functions like a resistor at the frequencies you're operating at) will do both.

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  • \$\begingroup\$ a microstrip terminated in a via to ground does not do 1), because it doesn't do 2). They are equivalent. \$\endgroup\$ – Neil_UK Jan 8 at 14:10
  • \$\begingroup\$ Consider the thought experiment of 7,500km of impedance matched transmission line with a short at the end of it, would you say the same thing about that? why is one inch of transmission line different? \$\endgroup\$ – james Jan 9 at 11:31
  • \$\begingroup\$ assuming electrical lengths (assuming lightspeed propagation) and lossless cable, 15000km round trip takes 50mS for light. Assuming you do all your measurements within 50mS before the reflection gets back, then yes, it's matched, and no reflection comes back. After 50mS, you'd get interference,a 20Hz raster on your frequency measurements as things go in and out of phase. A real line with attenuation even a few m long will have enough loss to be a reasonable load, however it's terminated, all that's needed is round trip loss >> 10dB, ideally > 20dB. \$\endgroup\$ – Neil_UK Jan 9 at 12:24
  • \$\begingroup\$ For a 1" cable, there's unlikely to be enough loss to get enough round trip attenuation unless it's made from wet string rather than copper. The reflection comes back in less than 1nS, so you'd need to be pretty nifty with the measurement. There are things that make a measurement that fast, a TDR, Time Domain Reflectometer. The old ones sent a fast pulse down a cable and sampled the return. Newer ones do a frequency domain measurement of the cable's reflection, and FFT it. Google for TDR and study the theory, it will inform your 1" and 7500km cable conundrum. It's good to ask these questions \$\endgroup\$ – Neil_UK Jan 9 at 12:27

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