This is really interesting and represents one of the best explanations of the Smith Chart that I have seen.
One quibble: In the beginning, they make it sound like all refelcted power in a transmission line is lost. That is not really true. Much depends on the frequency, the type of line used and the length of the line. The video presents this "SWR loss" as being very significant, and many hams seem to think that any SWR worse than 1:1 will kill their signal.
Look at the question that I put to Gemini:
I am a radio amateur. My SWR meter shows an SWR of 2:1. I am using 50 feet of RG-58 coax to a dipole antenna. The dipole is cut for 40 meters and I am operating on 40 meters. My rig is putting out 100 watts. How many watts are being reflected? How many watts are being radiated?
Gemini calculated:
Assuming your SWR meter is located at the transmitter and shows exactly 100 watts of forward power, 11.1 watts are being reflected at the meter, and approximately 76.1 watts are actually being radiated by your 40-meter dipole.
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That means that in this scenario, I am putting into the transmission line 100 watts and 76.1 watts are being radiated. That means that the loss resulting from this SWR of 2:1 will be about 1.186 db. With one S unit equaling about 6 db, this is clearly not enough to worry about.
Also, lets remember that we could put a 50 ohm dummy load at the far end of our transmission line and achieve an SWR of 1:1. But that dummy load would not radiate.
Three cheers for Veritasium for doing this Smith chart video.
Thanks to Rogier PA1ZZ and to Mike WN2A for sending this video to me.
Please put below your comments on the video and my observations.
It's not a complete story. being there is a 2:1 SWR, why? Is the dipole not really cut to the proper freq? Is it close to the ground? etc. And what of the coax losses? But more important, what happens to the reflected power? There will be loss in the xmtr final but most should be reflected again.
ReplyDeleteBut the point is correct, don't worry about the the 1.1dB
Good to hear from you Mike. I think of you every time I read about a NASA parachute. I see what you mean. I guess I should not have specified that the antenna was a dipole -- any device that would produce a 2:1 SWR would do. I agree, other then the extent to which the refelctions help heat the coax, there just isn't a lot of loss. The beginning of the video seemed to imply that the reflected RF is just somehow lost. That, as we know, is not true. 73 Bill
ReplyDeleteBill, to get the correct answer to your Gemini question, you need to add
ReplyDelete1) the location of the VSWR measurement , at Transmitter or Antenna. Normally I assume transmitter, but this is not always the case.
2) If you are asking for Radiated Power, you need to include the radiation efficiency of your antenna. Ground loss is a huge factor.
It isn't required for just the Feedpoint Power.
I changed the scenario for Gemini. I put the question this way: I am a radio amateur. My SWR meter located at the transmitter shows an SWR of 2:1. I am using 50 feet of RG-58 coax the antenna. I am operating on 40 meters. My rig is putting out 100 watts. How many watts are being reflected? How many watts are at the antenna feedpoint?
ReplyDeleteThe Gemini answer is: Power at the Antenna Feedpoint. Because 50 feet of RG-58 coax has a matched line loss of 0.525 dB at 7 MHz, some of your power is lost as heat on its way up the tower.To find the power at the feedpoint, we convert this loss to a transmission factor (k = 0.886):Forward Power Reaching the Feedpoint:Of the 100 watts sent from the rig, 88.6 watts physically reach the antenna feedpoint.Power Accepted at the Feedpoint (to be radiated):Because of the impedance mismatch at the antenna (which is a true SWR of ~2.21:1 at the feedpoint once feedline loss is accounted for), 12.5 watts of that 88.6 watts is reflected back down the coax toward the transmitter.This leaves a net power of 76.1 watts actually accepted by the antenna feedpoint and radiated into the air.
So still, about 1.186 db down. I don't know why they are saying that all this power is "radiated into the air. "