We are all told that VSWR doesn't matter when using low loss transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? For the RF to bounce back and forth, wouldn't the transmitter's impedance have to be very, very high (or low) when the reflected RF energy hit its output stages? I know I'm missing something vital here...
That's assuming you use an antenna tuner. The tuner will transform the transmitter's output impedance* just as it transforms the line. Were the transmitter output impedance actually at 50 ohms, on the other side of the tuner it would have the same VSWR as the line when everything was tuned up.
Having said that, the VSWR _does_ matter somewhat when using low loss lines, both because the line loss is low but not zero, and the tuner loss will tend to go up as you correct for higher and higher VSWR.
I am _not_ going to start the Big Transmitter Output Impedance Debate. sed denizens -- just don't comment on what a transmitter's "actual" output impedance may be, lest you start a flame war.
--
Tim Wescott
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Well, I was never taught that -- which is a good thing because it is not true.
The most intuitive way to prove this (antenna tuner or not), is to simply supply more and more power to the load. You will eventually burn it up.
Now, had the VSWR been better (ie. a better network Return Loss), it would have taken even more power to achieve the same "charcoal" results. So VSWR matters.
The active part of the transmitter output isn't 50 ohm. That would cause half the power to be lost as heat in the output stage. It's only 50ohm once it becomes a moving wave in the transmission line.
Well, aside from the initial misunderstanding of how power gets to the load (much less back, and then to the load again); I will put to you a question that has NEVER been answered by those who know what the transmitter output Z ISN'T: "What Z is it?"
1) The transmitter may well have output impedance matching the characteristic impedance of the transmission line. RF power reflected back in this case gets converted to heat in the output stage of the transmitter, in addition to whatever heat the output stage already has to dissipate.
1a) The reflection may increase requirement of the output tubes/transistors to both drop voltage and dissipate power. This can be a problem for many transistors, especially a lot of bipolar ones. It is not necessarily sufficient to stay within power, current, voltage and thermal ratings. Many bipolar transistors have reduced capability to safely dissipate power at voltages that are higher but within their ratings - sometimes even at voltages as low as 35-50 volts. This problem tends to be worse with bipolar transistors that are faster and/or better for use with higher frequencies. The keyphrase here is "forward bias second breakdown", a problem of uneven current distribution within the die at higher voltage drop.
2) It appears to me that transmitters can have output stage output impedance differing from the intended load impedance. An analog is common practice with audio amplifiers - output impedance is often ideally as close to zero as possible, as opposed to matching the load impedance.
If zero output impedance is achieved in an RF output stage, I see a possible benefit - reflections do not increase output stage heating but get reflected back towards the antenna. Then again, the impedance of the input end of the transmission line could be low or significantly reactive depending on how the load is mismatched and how many wavelengths long the transmission line is, and that can increase heating of the output stage. In a few cases transmitted power can also increase.
Not only is increased output stage heating possible and maybe fairly likely, high VSWR also causes a high chance of the output stage seeing a partially reactive load. RF bipolar transistors often do not like those due to increased need to dissipate power with higher voltage drop. As I said above, RF bipolar transistors are likely to really dislike simultaneous higher voltage drop and higher power dissipation.
The same misconceptions keep coming up, as they have countless times on this newsgroup and I'm sure they will for decades or perhaps centuries to come. After one of the many previous discussions, I wrote a little tutorial on the topic. Originally in the form of plain text files, I've combined it into a pdf file for easier viewing. You can find it at
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On page 8 you'll find the statement "THE REVERSE POWER IS NOT DISSIPATED IN OR ABSORBED BY THE SOURCE RESISTANCE". Above it is a chart of several examples which clearly show that there's no relationship between the "reverse power" and the source dissipation. The remainder of the tutorial explains why.
Any theory about "forward" and "reverse" power, what they do, and their interaction with the source, will have to explain the values in the example chart on page 8. Does yours?
Roy Lewallen, W7EL
D>> We are all told that VSWR doesn't matter when using low loss
On a sunny day (Mon, 12 Mar 2007 07:36:25 +0000 (UTC)) it happened snipped-for-privacy@manx.misty.com (Don Klipstein) wrote in :
All true. Also normally, there is a pi type filter (to prevent harmonics), between amplifier and antenna. This filter _WILL_ match the antenna to the output impedance of the transmitter, so _even_ if the transmitter output impedance is very very low (low voltage high current output stage for example), the reflected power will be nicely converted to match the transmitter, and heat up the output amp, with its possible destruction as result.
It really doesn't matter except for overall efficiency. A 10 ohm source outputting 100 volts into a local load of R +/- jX sources the same amount of power as a 100 ohm source outputting
Mine does. All of your values can be understood by looking at the destructive and constructive interference and applying the irradiance (power density) equations from the field of optics. You see, optical engineers and physicists don't have the luxury of measuring voltage and current in their EM waves. All they can measure is power density and interference and thus their entire body of knowledge of EM waves rests upon measurements of those quantities. Those power density and interference theories and equations are directly applicable and 100% compatible with RF theories and equations. Any analysis based on power density and interference will yield identical results to the ones you reported in your "food for thought" article which includes the following false statement:
"While the nature of the voltage and current waves when encountering an impedance discontinuity is well understood, we're lacking a model of what happens to this "reverse power" we've calculated."
We are not lacking a model of what happens to this "reverse power" we've calculated. The model is explained fully in "Optics", by Hecht. When one has standing waves of light in free space, it is hard to hide the details under the transmission line rug.
In general, it is just as easy, and sometimes easier, to deal with the energy values and then calculate voltage and current as it is to start with voltage and current and then calculate the power.
All this is explained in my WorldRadio article at:
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The great majority of amateur antenna systems are Z0-matched. For such systems, an energy analysis is definitely easier to perform than a voltage analysis. Here's an example:
Some gurus will say that it's the voltage and/or current that is destroying the final, not the reflected energy. They have yet to explain how those dangerous voltages and/or currents can exist without assistance from the ExH joules/second in the reflected energy wave. Depending upon phase, the E in the ExH reflected wave is what causes the overvoltage due to SWR. The H in the ExH reflected wave is what causes the overcurrent due to SWR.
The impedance seen by the source is
Z = (Vfor+Vref)/(Ifor+Iref)
Where '+' indicates phasor (vector) addition.
The above equation also gives the impedance anywhere along the transmission line and anywhere along a standing-wave antenna.
Unfortunately, that article doesn't explain where the power does go. A much better treatment of the subject is in "Optics", by Hecht. To understand where the power does go, one must understand destructive and constructive interference. Please see my transmission line example in another posting.
The energy content of a transmission line during steady-state is always exactly enough to support the forward traveling wave and the reverse traveling wave without which there would be no standing wave.
There are audio amplifiers with output impedance around .1 ohm, driving
8 ohm speakers, and having efficiency nowhere near 80/81. The theoretical limit for efficiency of a class B amp driving a resistive load with a sinewave is 78.54%.
Of course, that was a tongue-in-cheek posting. But if you could design a Thevenin equivalent source with a 0.1 ohm source impedance, wouldn't the efficiency calculate out to be pretty high?
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