Timeline for How to match RF output of (30+j10) Ohms to 50 Ohm antenna with matching network?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 14, 2021 at 9:54 | comment | added | Andy aka | @wdbwdb1 answered. | |
| Mar 14, 2021 at 9:46 | comment | added | wdbwdb1 | And I don't understand how to factor in an extra +j11 (or whatever the reactive value happens to be) from the antenna into the filter if it’s not right on 50 ohms (which will likely happen), along with +j10 from the RF pin. Since the line has to be 50 ohms, the capacitor at the antenna input won’t be needed, I'm just going to measure on the VNA from before the transmission line and factor the antenna reactive value into the pi filter. | |
| Mar 14, 2021 at 9:46 | comment | added | wdbwdb1 | Hi, I posted another revised version of this if you wanted to have a crack at it, I realized that the datasheet actually says to set the transmission line to 50 ohms, so the antenna will be 50 ohms also, it also says use a pi network close to the chip, and now my transmission line will be less than 15 mm. I need to know if the pi filter will be necessary since the manual specifies it, and they also use it on their modules so that makes me think there is a reason for it, but I don't understand the antenna theory behind the advantages of a pi over an L. Continued next comment... | |
| Jan 21, 2021 at 22:33 | comment | added | wdbwdb1 | Oh whoops yes I did do that LOL, I was thinking wow that was a lot of inductance! Thanks again! | |
| Jan 21, 2021 at 21:52 | comment | added | Andy aka | It should be 0.002268 uH or 2.268 nH AND 1.0958 pF (not 1095 pF). Did you enter frequency as 2400 MHz? I bet you did LOL. It should NOT be 2.4 MHz!! | |
| Jan 21, 2021 at 21:38 | comment | added | wdbwdb1 | Thank you for that great answer, it very much seems like an ideal solution to me. I really appreciate it. One last thing, for the L pad values, I entered 30 ohms for the input and 69 on the output and the calculated values are 2.268 uH series inductor with 1095 pF shunt capacitor. So I would subtract the .663 uH from the series inductor to cancel the +j10 from the MCU to get approx 1.6 uH, and final values for L pad will be 1.6 uH series inductor and 1095 pF for shunt capacitor? And of course the 6.03 pF cap at the antenna input? | |
| Jan 21, 2021 at 21:38 | vote | accept | wdbwdb1 | ||
| Jan 21, 2021 at 9:34 | history | edited | Andy aka | CC BY-SA 4.0 |
added 117 characters in body
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| Jan 21, 2021 at 9:26 | history | answered | Andy aka | CC BY-SA 4.0 |