0
\$\begingroup\$

I am designing a PCB that will have LoRa communication using the SX1272 , currently I am using the Reference Design for both 868 and 915MHz (image below).

SX1272 modules: SX1272MB2DAS - 868 or 915 MHz - Combined RFI and RFO design

But the BOM attached is missing components for C14, C19 and C20. I couldn't find the values for those components on any app notes or design references, also, I had access to a dev board for the SX1272 and I noticed that both the capacitors C19 and C20 are not mounted on it, so I'm assuming this should be impedance matching components and not needed for 50 Ohm antennas. Am I right to consider it? As for the C14 I still could not find any reference to what value should it have or how can I calculate it based on my PCB design. Any help is appreciated.

\$\endgroup\$
  • \$\begingroup\$ How about a link to the LoRa SX1272 data sheet. \$\endgroup\$ – Andy aka Feb 21 '18 at 18:32
  • \$\begingroup\$ While starting point values can be chosen based on theory and characterization of the chip and antenna, ideal values may require experiment on the actual hardware. At the design stage, you are worried about geometry and footprints; the precise values that go on those footprints will be confirmed after you have a representative prototype. \$\endgroup\$ – Chris Stratton Feb 21 '18 at 18:52
  • \$\begingroup\$ @Andyaka just added it. \$\endgroup\$ – Augusto Feb 23 '18 at 18:19
  • \$\begingroup\$ @ChrisStratton thank you for the answer! I did continue with the design just focusing on geometry and footprint, but I still have no clue as to what value of capacitor should I use for the first prototype. I guess I'll have to go with try and error. \$\endgroup\$ – Augusto Feb 23 '18 at 18:19
1
\$\begingroup\$

The BOM states the following:

"Remark: components not listed in the BOM are not populated on the module."

So you are correct in your observation that they are not mounted. The capacitors are placeholders should you run into compliance issues due to spurious emissions.

\$\endgroup\$

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.