4
\$\begingroup\$

I have a question about how the quality factor is determined for RF antenna matching, especially for PI network matching.

The reason for the question is the Reference Design for antenna matching for the SX1261 LoRa transceiver:

schematic

C9 = C10 = 3.3pF, L5 = 9.1nH.
According to https://www.semtech.com/uploads/documents/AN1200.40_SX1261-2_Reference_Design_Explanation_V1.1.pdf the impedance both left and right is 50 Ohms.

Using a PI matching calculator, I found out they designed for a Q factor of 1:

screenshot from Pi-matching calculator

9.17nH = 9.1nH, 3.67pF = 3.3pF.

My questions are:

  • Why do they even use a matching network, if impedance is 50 Ohms at both input and output?
  • Given Q = f / BW, f = 868 MHz I would expect Q in the range 30 - 50 for LoRa operation. Why did they design for a Q of 1? If I want to calculate my own matching network for a non - 50 Ohm antenna, do I also calculate for Q = 1 and why?

I would love to understand this, but all literature I found online usually calculates Q using the frequency and bandwidth, and in this example BW = 868 MHz clearly makes no sense?

Thanks.

\$\endgroup\$
  • 1
    \$\begingroup\$ +1 very well written question, good examples, clear motivation and prior research. \$\endgroup\$ – Ariser Sep 7 '19 at 19:38
  • \$\begingroup\$ Your high-Q solution could be done. But then are you prepared to add trimming components so that it can be tuned to 868 MHz and do you have the test equipment to properly tune it? And are you prepared to spend $$ on very low-loss PI-network components? \$\endgroup\$ – glen_geek Sep 7 '19 at 23:35
  • \$\begingroup\$ what about temperature variations of inductance and capacitance? \$\endgroup\$ – analogsystemsrf Sep 8 '19 at 2:27
1
\$\begingroup\$

Lets make it more general and say you have the impedance matching network between two devices. Now, as you have said if you could guarantee that both the input and output are flat 50 Ohm across all frequencies, with reactance =0, then sure, you do not need the PI-network. However, this is almost never the case, and to maximise power transfer between two circuits you need to match impedance and hence you need a PI-network. $$Z_1 = R_1+jX_1$$ $$Z_2 = R_2+jX_2$$

What this means then is you need to:

  1. Somehow cancel the reactive part of the two impedances', leaving only the resistive part: $$X_1 = -X_2$$

  2. The resistive part should be the same: $$R_1 = R_2$$

This is refereed to in literature as "Conjugate Matching".

For your question it is important to realise that X_1, X_2 change with frequencies in weird and wonderful ways, so that implies you can only meet point 1 over a narrow range of frequencies. So remember then, maximum power transfer is always unavoidably frequency dependent and is only possible over some range of frequencies.

This is where the concept of bandwidth(hence Q) of a matching network comes into play, as it tells you how effectively such a matching network operates over a wide range of frequencies.

Now using your own formula:$$Q=f/BW$$ Using a small Q implies large bandwidth, hence generally more wide-band operation of the matching network. So you might ask then, why not always design for the smallest Q as possible to maximise frequency range? One answer is, if you play with direct equations a bit more, eventhough designing smaller Q will give you a wider bandwidth of match, at the same time you will notice increasingly impractical values for the inductors and capacitors.

To make sure I am not bluffing, test the above claims using the calculator you posted. For example use: $$F=100 MHz$$ $$RS=50$$ $$XS=-9$$ $$RL=100$$ $$XL=3$$ $$Q= 1.01 \to\ 50$$ Now switch the graph to "Real & Complex" and put the cursor at 100MHz. If the matching network is working well, you should ideally see an impedance that agrees with point 1 and point 2! (In this case you should read around Real = 50, Imag = +9). The graph is reporting what impedance the source sees looking into the PI-network + Load.

EDIT: Other reason determining Q are:

  1. The PI-Network is also clearly a filtering network. So you inherently get some signal filtering characteristic you may (not) be able to tolerate depending on the Q you choose.

  2. In most real world applications, the Q of the network is specified by the Q of the imperfect components, especially the inductor

EDIT 2: You also seem to misunderstand the implication in the datasheet. The manufacturer would certainly love for both sides of the PI-network to be 50 Ohm, but this, as I explained will never quite be the case. For example, the antenna might be designed to only be a good match (and hence absorbs and radiate away RF power!) at its operating frequency and hence will be a bad match at others.

\$\endgroup\$
  • \$\begingroup\$ Thanks! I was looking for some rule of thumb, and "make Q as low as possible, but keep reasonable component values in mind" is something I can live with :) Meanwhile I also found that PI networks without impedance transformation are often used to suppress first and second harmonics. A low Q is also desirable because component tolerances are less relevant and thus cheaper. \$\endgroup\$ – niklas Sep 7 '19 at 22:52
  • \$\begingroup\$ Well careful, there can be other important considerations. See EDIT :) \$\endgroup\$ – Adil Malik Sep 7 '19 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.