The basics of VSWR and Return Loss
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- เผยแพร่เมื่อ 11 มี.ค. 2023
- Reflected RF power and VSWR (Voltage Standing Wave Ratio) are two related concepts in amateur radio. This reflected power can cause damage to the source and reduce the efficiency of the system. VSWR is a measure of the degree of impedance mismatch and is defined as the ratio of the maximum voltage to the minimum voltage in a transmission line. A high VSWR indicates a greater degree of mismatch and therefore a higher level of reflected power. A low VSWR is desirable for optimal power transfer and system performance.
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• Understanding VSWR and...
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I liked the way you presented this info.
Nice talk, thanks
Also reflected power is not the same as common mode current. Reflected power is caused by an impedance mismatch where common mode current is caused by unbalance in the load - for example a dipole where one side is not the same length as the other.
Absolutely agree ,I've seen videos and comments recently that misunderstand this science
Nice job. Never thought of a dummy load on 2nd antenna input. Good idea
Rhode and Schwartz have monthly videos on a lot of good topics. You can sign up for notification. Sometime they go over your head but many make you think and research and that is a good thing. This hobby is not about making QSO's its all about leaning and communicating with people. 73
Thanks Jim.
Fine business OM
This is mostly incorrect. "with moderate lengths of low-loss coax, such as we
commonly use for feed lines, loss of power because
of reflected power in the hf bands can be
insignificant, no matter how high the SWR. For
example, if the line SWR is 3, 4, or even 5 to 1 and
the attenuation is low enough to ignore the
reflected power, reducing the SWR will yield no
significant improvement in radiated power because
all the power being fed into the line is already
being absorbed in the load." That from antenna engineer Walter Maxwell who has written extensively on the subject of reflected power. If you transmit 50 watts to the antenna and 25 watts is reflected back, that in turn is reflected back to the antenna. So now you have 75 watts (50 watts + 25 watts) headed to the antenna, and the antenna radiates full power of 50 watts. The only important thing is to use an antenna tuner so the transmitter sees about 50 ohms or an SWR of no more than about 1.5 to 1.
I agree. Antenna tuners are often misunderstood. The impedance they present on the antenna side forms a resonant tank circuit with the feed line/antenna mismatch. So like any resonant circuit, power flows back and forth with no loss at the ends. However, because of I^2*R loss in the feed line, the reflected + transmission power results in more feed line power loss than would occur without reflected power.
Some of that signal is lost as heat though, correct? It's not a perfect system.
@@FEPLabsRadio Yes, correct. The standing waves are composed of the constructive interference of the forward and reverse signals, so they are larger than the normal waves. Per the standard power formula, power loss = I^2*R. So if the standing waves are twice as large as the normal waves, the power loss from them is (2)^2 = 4 times as much. This loss heats the cable at the locations of the standing waves. In extreme cases where you're running high power, and the SWR is high, and you remain on the same frequency with the same standing wave locations, there's the possibility of damaging the cable at the standing wave locations due to the heat. So the transmitter output power is combined with the reflected power reaching the tuner, but as this power increases, so does the heat loss in the cable which eventually limits how much power gets to the antenna junction. The point that many people don't understand is that most of the reflected power from the feed-line/antenna mismatch is reflected back to the antenna by the tuner.
Agreed. I think that the amateur radio community should read Reflections III by M. Walter Maxwell, W2DU. The book really exposes the myths surrounding SWR. Simply having a low SWR does not mean you have an efficient antenna system.
Thanks Jim, you must have a bad TH-cam vswr cause your audio was too low. :)
I can't find the DX Commander 1Million!... Is Callum only selling it to the TH-cam elite?😆🍻
Yes and yes! You have to be in the secret YT club to get the DXCC Platinum 1M Ultra DX! ;)
"If you have infinite impedance, you have 100 percent return loss (10:24)," or is it 100 percent return? ....or 100 percent insertion loss. I was always taught the higher the return loss, the lower the SWR and the lower the insertion loss.
Hmm, I probably said something backwards and got myself tripped up with return loss vs insertion loss. I need to go back and watch that video.
sorry , voice volume is so low difficult to understand, other wise it looks like very interesting. AG6JU
Nope coz the 7300 has gone into standby mode lol
Thanks. FYI, your audio level is low.
Sorry about that - older video. Much progress since then. THanks for watching!
I stopped watching at the point you declared reflected power as common mode current, please learn the simple science that reflected power from an antenna feedpoint mismatch does not travel on the outside of the braid of coax ,that is totally reserved for common mode current from an antenna imbalance