I'm trying to do impedance matching between my PCB antenna and my circuit. My PCB antenna has a measured impedance of 49.4 - j8.1 ohms. The operating frequency is 915 MHz. The goal of matching is to move the circuit impedance as close to 50 ohm as possible.

The designed matching network is Pi-type as attached here.

My process is as follows and it follows the instruction here: (http://wisp.wikispaces.com/WISP+Measurement+Procedures?responseToken=016c891618c82df3ef37a9c2b69cf236)

Step 1: Shorten Lm1 by a small wire (perhaps it's inductance is 2 nH?). Leave other components blank.

Step 2: Measure the input impedance of the circuit afterwards by a Network Analyzer. It gave 0.355 + j42 Ohm.

Step 3: Use that value as the load impedance to find an appropriate network topology by using Smith V4.0, a free software. I've tried this topology enter image description here

So Lm1 was replaced by a resistor of 50 Ohm, while Cm2 was replaced by an inductor of 23.5 pH. No Cm1.

Step 4: Start to replace the components as designed in Step 3. Firstly add Cm2, then add Lm1. Do Step 2 after replacing any component (for example, measure the input impedance again after replacing Lm1).

All measurements were performed as 1-port measurement on VNA (Vector Network Analyzer).

There are two problems:

1/ Step 4 problem: Anytime I replaced any component, the measured input impedance was totally different than it should be by being simulated in Smith V4.0. For example, if I add a shunt capacitor of 2 pF, the measured Zin was not be 1.3 + j81.2 Ohm as found by using Smith V4.0 software. I did not record the result so I do not remember the exact result, but I can say that it was totally different both in values and the direction that the impedance should move. I'll get back this situation with numerical values.

2/ Matching performance: Even I finally got the measured Zin ~ 50 Ohm, not by designing in software but by replacing and guessing and replacing again the components, the delivered power after matching was lower than before matching! In theory, the power must be higher because less power mismatch!

My friend got the same 2nd problem when he tried to match his transducer. The signal he got after matching was more noisy than before.

Could anyone help me explain these things? Any suggestion/comment is highly appreciated. Thank you for your time.

------- Updated for response for Mr. Glenn W9IQ-------------------------------

Thank you so much for your comments and your kind explanation @Glenn W91Q. I'll keep in my mind the correct terms. Let me explain more about my measurement set up.

1/ Calibration of the test cable: I used a SHORT-OPEN-LOAD adapter and a coaxial cable (Pic 1). The cable is terminated by a kind of SMA adapter (I don't know how to name it). I need that adapter to connect my antenna/circuit to the cable. So, I attached the SHORT-OPEN-LOAD adapter to the cable through that adapter for calibration. So, it means I calibrate both the cable and that adapter. Am I correct? I followed the calibration procedure of this video https://www.youtube.com/watch?v=GKHHy61qRak&t=18s. But I chose 1 port calibration instead. I used Agilent E5062A VNA model.


2/Set up for antenna input impedance measurement: I connected my dipole antenna to the cable as follows. The SMA connector attached to the antenna is used to connect the antenna to the circuit. The antenna will be a receiving one, not a transmitting one. It was designed using HFSS tool kit. The simulated reflection coefficient is -22 dBm @912 MHz.

enter image description here

By using S11 measurement mode, Smith Chart display tool of the VNA, I found the impedance of (antenna + SMA connector). It must be INPUT impedance, right?

To get the impedance of as close to 50 ohms as possible, the antenna was tuned by cutting its length. I also added a matching circuit (plz see below). The final impedance I got, using this set up, was 49.4-j8.1 ohms.

enter image description here

3/Set up for receiver input impedance measurement: The same set up as above was applied for this.

Can I ask you for more clarifications?

1/ I thought my measurement set up as illustrated above was wrong, because I did not connect the antenna/receiver ground to the shield of the VNA cable as you mentioned earlier. And I still cannot imagine how to do this. If you've got some images for this, I'd highly appreciated it if you can share some.

The VNA must have two connections to the antenna under test. This perhaps is basic but I have seen students completely ignore the ground (shield) connection to the VNA cable or they simply connect it to any convenient ground point on the circuit, treating it like a basic DC ground.

2/ The impedance I got from the set up of measuring input impedance of the receiver was far from 50 ohms. So, I need another matching circuit to move it close to 50 ohms, right?

Thank you again for your time.

  • $\begingroup$ You say the antenna had an impedance of 49.4 - j8.1 ohms and then your first VNA measurement says it was 0.355 + j42 ohms. Can you please explain further? $\endgroup$
    – Glenn W9IQ
    Commented Jun 9, 2018 at 7:58
  • $\begingroup$ Thanks for your comment. I mean the input impedance (if I understand correctly) of the circuit was 0.355 + j42 ohms, while that of antenna was 49.4 - j8.1 ohm. All of them were measured using 1 port of VNA. $\endgroup$
    – Minh Lam
    Commented Jun 9, 2018 at 10:34
  • $\begingroup$ The results of your measurement of the transmitter ouput impedance indicates an error with your measurement technique. You should simply use the specified impedance from the transmitter spec sheet which is likely to be 50 ohms. $\endgroup$
    – Glenn W9IQ
    Commented Jun 9, 2018 at 11:09
  • $\begingroup$ My circuit is a receiver. It is quite complex. It's comprised of a rectifier, a supercapacitor, a buck-booster circuit, a MCU and some sensors. Lots of inductors and capacitors. So I cannot guess its input impedance from datasheet. Actually, I followed this website to get the input impedance of the circuit and do the matching network. The difference was that the calibration tools were the commercial one (as I added in my question), not the modified ones. The link: wisp.wikispaces.com/WISP+Measurement+Procedures $\endgroup$
    – Minh Lam
    Commented Jun 9, 2018 at 15:44

2 Answers 2


Please take the following commentary not as a hollow or haughty criticism but rather to help your technical learning and to help you to communicate more effectively.

When communicating in a technical field, it is important to be as concise and as accurate as possible. While corrections can be inferred in some cases, this is not optimal. In a half duplex forum such as this, it also helps you to get the answer you desire in the most expedient fashion. There are a few things in your question, in addition to my comment to your question, that don't seem to be correct:

perhaps it's impedance is 2 nH

An impedance is given in ohms. 2 nH is short for 2 nanohenries, a unit of inductance. You probably mean to say "perhaps it is an inductance of 2 nH".

Lm1 = 50 Ohm

The schematic designation of Lm1 indicates that this component is an inductor. The 50 ohms indicates an impedance. Normally this equality should be set to a component value (e.g. Lm1 = 2 nH) unless it is a statement or equation giving the impedance of Lm1 (e.g. the impedance of Lm1 at 915 MHz is 11.5 ohms). This is a case where your statement is ambiguous.

if I add a shunt capacitor of 2pH

The Farad (F) is the base unit of measure for a capacitor. The base unit of measure for an inductor is Henry (H). This is again an ambiguous statement.

For example, if I add a shunt capacitor of 2pH, the measured Zin will not be (1.3 + j*81.2) Ohm. The result I got is totally different. I don't know why.

There is no analysis of this possible without you giving the measurement you observed. "Totally different" is not a quantification.

the measured Zin will not be (1.3 + j*81.2) Ohm

This one is minor but the complex impedance you referenced (and all other such impedance references) should be written in the form of 1.3+j81.2. Notice there are no parenthesis or asterisk (*) in the nomenclature.

Measurement Technique

When working at these high frequencies, correct measurement technique is essential.

The VNA must be correctly calibrated at the frequency of measurement and using the test cables and connectors that will be used for the measurements. This typically consists of an iterative technique of sequentially placing shorted, open, and resistive loads at the end of the measurement cable following the prompts of the VNA calibration procedure. This has the effect of removing the transmission line effects of the test cable and its connectors.

The antenna under test must be oriented as it would be used in the application. All test equipment, test benches, fixtures, cables, connectors, etc. must not be placed in close proximity to the antenna as these will couple to the antenna and affect the results.

The length and orientation of traces on the PCB that connect the VNA cable to the antenna can alter the measured antenna impedance. For example, at 915 MHz, a wire or PCB trace length length of more than 20 mm can have significant implications.

The VNA must have two connections to the antenna under test. This perhaps is basic but I have seen students completely ignore the ground (shield) connection to the VNA cable or they simply connect it to any convenient ground point on the circuit, treating it like a basic DC ground.

Interpreting the Measurements

My PCB antenna has a measured impedance of (49.4 - j*8.1) ohm

If you have an antenna with an impedance of 49.4-j8.1 ohms and your transmitter has an output impedance of 50 ohms, there is no point in attempting to get a better match. Any losses that may occur as a result of this slight mismatch are decimal dust. A matching circuit would likely introduce more losses than it would remove.

Measure the input impedance of the circuit afterwards by a Network Analyzer. It gave (0.355 + j*42) Ohm

If you have an antenna system with an impedance of 0.355+j42 ohms and the goal is a 50 ohm antenna system, you have a serious problem that needs correcting. Either the measurement is wrong or your antenna system is improperly designed/constructed. If the antenna (without the matching network) has this impedance, then the realized antenna design is incorrect.

Except in the most dire of circumstances, the antenna impedance should have a resistive component (the part before the +/- j) of at least a few ohms. Anything less than this tends to make the matching circuit have more loss and it makes the antenna itself less efficient. In general, transmit power is turned to heat instead of radiated as intended in this case.

Even I finally got the measured Zin ~ 50 Ohm, not by designing in software but by replacing and guessing and replacing again the components, the delivered power after matching was lower than before matching!

This conclusion is difficult to resolve given all of the above mentioned issues. It would be helpful if you stated what components you finally selected and in what configuration you placed them.

If your measurement technique is not correct, you may not have a 50 ohm input impedance to your antenna system. Alternatively, you may not be correctly measuring the transmit power in both cases. Some explanation of the equipment and technique used for this measurement would be helpful.

Better Answer

[Edit: This section has been revised based on the additional information provided by the OP]

Thank you for posting the pictures and additional commentary. Your situation is now much clearer.

The calibration fixture that you used for the VNA is fine. Well done.

The connection to the antenna is fine. The connector will take care of completing the shield (ground) connection to one side of your dipole and the center connection to the other side of the dipole.

Make certain that when you are testing the dipole, that it is away from other objects and that the test cable is routed at a right angle to the antenna for at least two antenna lengths.

Trimming the dipole as you described is a good way to tune it. Well done.

Regarding the measurement of the input impedance of your receiver. The design is unique in that the diode circuits will make the input impedance non-linear when compared to input power. Most calculations for matching networks assume a linear response for a given frequency. Thus your approach to try different values until the input impedance matches ~50 ohms may be valid but as the input power level varies, the results may vary as well. You may find better results if you reduce the power level of the VNA to more closely match that which is expected from the actual WISP signal. From this, you can either better calculate the matching values or better validate your 'cut and try' values.

enter image description here

You should also check that you have correctly installed the Q1 and Q4 circuitry. Any errors in this section will effectively short out the antenna input rendering your VNA measurements and impedance matching efforts invalid. By the same token, make sure that 'Transmit_RFID' is near ground when making your measurements. If this is not connected, I recommend grounding it during your testing to ensure that the transistors are biased to an off state.

  • $\begingroup$ Thank you very much for your explanation. I did not know how to clarify my situation with more pictures, this is my first time to join this community. So I edited my questions with my clarifications and more questions in my first question. $\endgroup$
    – Minh Lam
    Commented Jun 9, 2018 at 15:40
  • $\begingroup$ @MinhLam Your edits are well done. Welcome to the site. $\endgroup$
    – Glenn W9IQ
    Commented Jun 10, 2018 at 16:37
  • $\begingroup$ Thank you very much for your clear edition. I have already reduced the power level of VNA from 0 dBm as default to -9 dBm. I was surprised that it did not make any difference when I measured the input impedance of the circuitry! I've also programmed the MCU to sleep before the measurement. Does it affect the measurement? And actually I made another antenna for RFID transmission. The 1st one is used only for energy harvesting. So the problem of matching doubles! Could you please recommend me any method for matching 2 antennas to one transceiver? Thank you so much again. $\endgroup$
    – Minh Lam
    Commented Jun 20, 2018 at 6:56
  • $\begingroup$ @MinhLam If the MCU is sleeping, the transistors that I mentioned earlier may not be properly biased, particularly in the presence of RF. Regarding matching the antennas, simply have a matching network for each one and match them individually. Naturally, you should not connect both antennas at the same time once you have them matched. $\endgroup$
    – Glenn W9IQ
    Commented Jun 27, 2018 at 15:32
  • $\begingroup$ W91Q Thank you very much for your recommendation. $\endgroup$
    – Minh Lam
    Commented Jul 3, 2018 at 11:12

Remember the time-honored engineering adage: Better is the enemy of Good Enough.

It is difficult to imagine any practical value to inserting a matching network between your generator or feedline and your antenna: 49.4 - j8.1 ohms equates to an SWR of less than 1.2:1 on a 50$\Omega$ line, already a very good match. Given the relatively large tolerances and potential losses of the components you would use in a matching network, whether lumped or distributed, it is likely that adding a matching network would actually reduce the amount of power that is radiated by the antenna.

If, however, you need to increase the match bandwidth - the range of frequencies over which the match does not exceed an upper limit on SWR - a properly engineered matching network could deliver improvement.


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