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Jan 21, 2021 at 5:29 comment added wdbwdb1 Thanks, I would rather keep reflections and power loss to an absolute minimum. 25% is more than I would like. I posted a new question here: electronics.stackexchange.com/questions/543823/…
Jan 15, 2021 at 9:28 comment added Andy aka If the antenna drive impedance became 50 ohms then, with the current matching network, it won't make a big deal. The reflection coefficient of 50 ohms with a base of 30 ohms is (50-30)/(50+30) = 0.25 so, in effect there is a 25% power loss. That's probably not a big deal provided that reflections and standing waves aren't a big deal. I think that's the bottom line @wdbwbd1 because your t-line length I think you said was quite small.
Jan 14, 2021 at 23:58 comment added wdbwdb1 Ok, best answer given and I'll let you know when I post the new question, might not be today. Thanks again.
Jan 14, 2021 at 23:56 vote accept wdbwdb1
Jan 14, 2021 at 23:41 comment added Andy aka You should probably be aware that this is a q and a site and not a forum where we deal with an evolving scenario of questions or try and train folk in an evolutionary way. You ask a question and, somebody provides an answer. This site is basically a good reference gallery of good questions and good answer. Having said that a certain amount of diversity is tolerated but, in my humble opinion, we are done withe the basic q and a and, you should formulate your new stuff into a new question @wdbwbd1
Jan 14, 2021 at 23:32 comment added wdbwdb1 And what if the antenna doesn't match perfectly to 50 ohms? Normally I believe with a 50 ohm source and 50 ohm antenna, if the antenna isn't exactly 50 ohms then the matching network adjusts for that to match source using custom component values, but if the L pad Network is just changing 30 ohms to 50 ohms instead, that seems to remove the ability to match the antenna to 50 ohms if needed. Is this correct, and how is this addressed?
Jan 14, 2021 at 23:18 comment added wdbwdb1 Actually I'm realizing this may be all I need, but first after removing L components on PCB should I trim the length of my IFA antenna on the PCB until it measures nearest to 50 Ω measured on a VNA at antenna input as in this pic? imgur.com/RbECVR1 I plan on setting the antenna to 50 Ω in the simulator, but drawing it on the PCB with a longer length arm that I will trim to reach nearest to 50 Ω on the actual PCB so it's exact. Is this the right strategy? Then this is the finished circuit? imgur.com/mMQYV32 And 50 Ω transmission line is optimal from RF pin to L network?
Jan 14, 2021 at 22:33 comment added Andy aka Well, if you’ve finished with this question, you should formally accept the most relevant answer @wdbwbd1
Jan 14, 2021 at 21:22 comment added wdbwdb1 @Andyaka Sorry I've just been really busy, I intend to make a new one soon and I will notify you when I do. Thanks a lot for your help.
Jan 12, 2021 at 16:08 comment added Andy aka @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?
Jan 11, 2021 at 0:14 comment added wdbwdb1 Ok thanks I'll do that.
Jan 10, 2021 at 23:34 comment added Andy aka 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.
Jan 10, 2021 at 23:30 comment added niko20 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
Jan 10, 2021 at 21:55 comment added wdbwdb1 Also my excitation port on the simulator is currently set at 50 Ohms.
Jan 10, 2021 at 20:47 comment added wdbwdb1 My transmission line length is less than 1/4 wavelength & more than 1/8, and I'm fine with using matching components. Wouldn't it b more ideal 2 use them, even w/ short line? And should I set the excitation port impedance setting on the simulator to 30Ω + j10Ω or 50Ω, & set the antenna to be 30 - j10, or 50Ω? How would I write that as a single impedance 4 the simulator without the imaginary j? If I'm using matching components, should the antenna always b 50Ω, and matching network at 30Ω - j10Ω? I have my transmission line from IC to ANT set to 50Ω. Just looking 4 most ideal way 2 do everything
Jan 10, 2021 at 20:28 comment added Andy aka 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.
Jan 10, 2021 at 20:06 comment added wdbwdb1 To be complete, I meant when saying "Or set ant imp to 50Ω on smith chart in simulator?", set antenna impedance at 2.4GHz to 50Ω position on the smith chart in simulator, then in real life match to 30Ω + j10Ω (as you mentioned) looking into the antenna on a VNA with the probe positioned before the matching network?
Jan 10, 2021 at 19:56 comment added wdbwdb1 Thank you for that. If I'm designing a microstrip IFA antenna on a simulator, without simulating the matching network, should I set the antenna impedance as close as possible to 30Ω + j10Ω position on the smith chart? And then just match to 30Ω + j10Ω looking into the antenna on a VNA in real life with the probe positioned before the matching network? Or set ant imp to 50Ω on smith chart in simulator? I will b designing the antenna, so the only reason I bring up 50Ω is because I thought it might b more ideal due to 50Ω normally being the best compromise between low attenuation & power handling
Jan 10, 2021 at 11:04 history edited Andy aka CC BY-SA 4.0
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Jan 10, 2021 at 10:58 history edited Andy aka CC BY-SA 4.0
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Jan 10, 2021 at 10:43 history answered Andy aka CC BY-SA 4.0