Impedance matching in RF Circuit - Determine the required Q

I am working on an RF project and I notice that many sources I have used, recommend using a pi network (CLC), such as the image below, to impedance match the RF trace to the antenna.

I have a really short distance between my RF output and antenna and the impedance is controlled the entire way.

If the antenna impedance is well matched to my characteristic impedance, can I then simply ignore my pi network, open up the caps, and short the inductor ?

If the antenna impedance is does not match, and requires tuning, I've been shown this website (https://home.sandiego.edu/~ekim/e194rfs01/jwmatcher/matcher2.html)

When I enter I know, as seen below, there is one parameter I do not know how to determine - Desired Q. The default is set to 3. In this example, I set the source and load to be 50 ohms.

How do I select an appropriate Q ? What is the general workflow when working with RF that requires impedance matching or tuning ?

• Read this paper from ARRL: Quality Factor, Bandwidth, and Harmonic Attenuation of Pi Networks Commented Nov 15, 2019 at 20:59
• @Captainj2001 will do! Will check it out later tonight. Commented Nov 15, 2019 at 21:00
• how much variation will your RF carrier require? 0.54 to 1.64 MHz ( the entire AM band)? or 313Mhz +- 1MHz ? Commented Nov 16, 2019 at 3:28

If the antenna impedance is well matched to my characteristic impedance, can I then simply ignore my pi network, open up the caps, and short the inductor ?

Quite often not.

Reason: the $$\\pi\$$ network is also a decent "remover" of high order harmonics and can make the difference between a product that is able to pass certification tests and a product that will fail. The chip's output will very likely have unwanted harmonics and it will likely need to be filtered to stop cross-channel interference.

Regards estimating the $$\\pi\$$ filter values, you can also use this calculator. It gives you all the theory behind $$\\pi\$$ networks and shows how you can manipulate values to achieve a tighter bandwidth (aka cutting out those unwanted harmonics).

This is an example of a 10 MHz $$\\pi\$$ network that has 50 Ω input and 300 Ω output impedance and varies the transfer impedance from 10 Ω to 35 Ω: -

The top graph is probably most important in your situation: it shows that the high frequency attenuation can be as much as 7 dB at only 12 MHz. By the time you get to the 2nd harmonic (20 MHz) it can be around 20 dB to 25 dB. It's a 3rd order filter so 18 dB per octave (60 dB per decade). It will be approximately the same for a 50 Ω to 50 Ω converter.

The middle and lower graphs show the phase and magnitude of the input/output impedances.

• Indeed, understanding that this is there for harmonic suppression is key. In practical terms, the asker's approach of trying to determine the Q first is probably mistaken. Rather, what they should do is model the harmonic attenuation. Then verify that the desired frequency range will be passed, though that's fairly likely - it would be hard with ordinary components to build this filter in such a way that would be too narrow for ordinary usage, the question is more will it have sufficient harmonic attenuation. Commented Oct 10, 2020 at 16:54
• The antenna however might only be a good match over too narrow a range, and the network might help with that, but only when considered in the context of the actual antenna. Commented Oct 10, 2020 at 16:55

$$\Bandwidth = \frac{ω0}{Q}\$$ Where $$\ω0\$$ is your center frequency. So choose your $$\Q\$$ to set your bandwidth.

Probably more than you've wanted to know about RF matching.

can I then simply ignore my pi network, open up the caps, and short the inductor ?

Assuming that you are using small components, relative to the structures on your PCB you should be able to do that. Although a 0 ohm resistor probably isn't very good above 1Ghz. I'd look at some way of replicating the structure you have on your board, or do a board spin once you've verified that you don't need the matching network.

If it's an 0402 L, then you should be able to just solder bridge it.