0
\$\begingroup\$

I'm designing a matching network for a 2.4GHz inverted F antenna. I am using an ESP32-D0WDQ6 MCU, and reading from the datasheet, the output impedance for the RF pins is (30+j10) Ohms. (DATASHEET in section 2.5 on page 7)

Should the matching network be set to 50 Ohms? Or should it be set at (30+j10) Ohms?

EDIT: anyone reading this should read the further comments with Andy AKA's answer for completeness.

\$\endgroup\$
1
\$\begingroup\$

If your antenna is 50 Ω and your source is 30 Ω + j10 Ω then, add a series capacitor of -j10 Ω to cancel out the +j10 Ω effect of the inductor.

This now means you are trying to match 30 Ω resistive to an antenna of 50 Ω resistive.

Then use an L-pad calculator like this: -

enter image description here

You can double check the formula derivation on that site.

  • Series inductance needs to be about 1.6 nH
  • Parallel capacitance needs to be about 1.1 pF

Then, if you went back to the start of the problem and analysed what value of inductance is needed to produce the j10 Ω in your driver output impedance, you'd calculate it to be 0.663 nH at 2.4 GHz.

This means that you can actually dispense with the added series capacitor of -j10 Ω (as originally proposed) because you need 1.624 nH from the above calculator. The upshot of this is that 1.624 nH might as well be 1.663 nH so, the external series inductor you need to add is 1 nH.

\$\endgroup\$
17
  • 1
    \$\begingroup\$ If you are not wanting to use matching components you set the antenna to be 30 - j10 to maximise power from a 30 + j10 source but, only if you are not creating a transmission line to connect them i.e. the connection distance is very short. \$\endgroup\$ – Andy aka Jan 10 at 20:28
  • 1
    \$\begingroup\$ Set the antenna to 50ohms, its most common. Primary reason is because if you have any transmission line at all it will typically be 50 ohms, so you want to use that value to make ease of everything matching \$\endgroup\$ – niko20 Jan 10 at 23:30
  • 1
    \$\begingroup\$ I suggest you ask a new question based on your recent comments. Training and teaching is limited (very) in comments and I particularly shy away from that so, compose your new thoughts and ask a new question. Leave me a comment so that I am reminded when I start work tomorrow. \$\endgroup\$ – Andy aka Jan 10 at 23:34
  • 1
    \$\begingroup\$ @wdbwbd1 I didn't see any new question so maybe I missed it? Are we done with this question now or is there something particularly relevant that I may have missed? \$\endgroup\$ – Andy aka Jan 12 at 16:08
  • 1
    \$\begingroup\$ Well, if you’ve finished with this question, you should formally accept the most relevant answer @wdbwbd1 \$\endgroup\$ – Andy aka Jan 14 at 22:33
1
\$\begingroup\$

The antenna may be 50 ohms but you need to check the specs on it. You definitely want to match the antenna so yes you probably want a matching network between the pin and antenna, fire up an Pi matching network app or webpage and type in the parameters to get values back out the you need to match

The datasheet you shared said yes you need a Pi matching network. So im not sure why you are asking, but yes you want 50 ohms out, but like 30-j10 on the input of the Pi network to match the mcu pin (notice the negative j30)

\$\endgroup\$
3
  • \$\begingroup\$ I understand a matching network is required. My question was "Should the matching network be set to 50 Ohms? Or should it be set at (30+j10) Ohms?" Looking into the antenna, should I match to the (30+j10) Ohms point on the smith chart to match the IC, or should I match to the center of the smith chart at 50 Ohms which is normally ideal for transmission lines? What do you mean by "you want 50 ohms out, but like 30-j10 on the input of the Pi network"? \$\endgroup\$ – wdbwdb1 Jan 10 at 8:22
  • \$\begingroup\$ I was wondering if 50 ohms might be more ideal due to normally being the best compromise between low attenuation and power handling. \$\endgroup\$ – wdbwdb1 Jan 10 at 8:33
  • 1
    \$\begingroup\$ You want a matching Pi network that is 30-j10 on the input and 50 ohms on the output. Remember, if you see a J, you want to compenstate and "tune out the j". The chip is +j10 which means you need to "cancel that out" with a -j10 to leave behind only the 30 (30 ohms). If the Pi network does this it will match to the mcu pin exactly. The Pi can have a different input and output impedance, that is its job, so you match on input of the Pi with 30-j10 and simply match the out with 50 (no j) \$\endgroup\$ – niko20 Jan 10 at 23:19

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.