Timeline for Impedance matching with complex impedance source
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 31, 2019 at 11:36 | vote | accept | ultimA | ||
| Jul 23, 2019 at 12:15 | comment | added | user136077 | Not at all correct. The only goal you should have is to make the transmitter see its specified load impedance and that should be achieved with a reactive tuning circuit to minimize power losses. That should be ok also for receiving if your tranceiver is designed properly. Measuring with an analyzer the tranceiver doesn't give info of the internal impedance of the transmitter during it transmits. | |
| Jul 23, 2019 at 11:41 | comment | added | ultimA | To rephrase, I have a transceiver with known impedance ZS, and an antenna with known ZL. There are parasitics and a (electrically long) transmission line between the two. The way to design the matching network, the way I understand now, is to tune the matching components, until I measure ZS* when I replace the transceiver with the VNA, and measure ZL* when I replace the antenna with the VNA. Correct? | |
| Jul 23, 2019 at 11:37 | comment | added | user136077 | @ultimA If your antenna + its cable already happens to be a perfect 50 Ohm resistive load, it should not be especially difficult to design a two reactance matching network which converts that 50 Ohm to (30+10j) Ohms when it's connected between the tranceiver and the antenna cable - at least if you operate at so low frequency that inductors and capacitors are useful. | |
| Jul 23, 2019 at 11:16 | comment | added | ultimA | It is not about measuring the optimal antenna impedance, just matching two known impedances. The antenna itself (without the feed line) is known to be 50Ohm, and the transceiver is specified as above by its manufacturer. My idea was to match the transceiver to the 50Ohm of the antenna. Do you mean it would be easier/better for performance to just tune the "antenna side" of the matching network to be also 35+j10? Given that the antenna is already 50Ohm, isn't that the same as my original intention? Sorry, these things are new to me, you might have confused me here a little. | |
| Jul 23, 2019 at 11:02 | history | answered | user136077 | CC BY-SA 4.0 |