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VSWR doesn't matter?

Discussion in 'Electronic Basics' started by billcalley, Mar 12, 2007.

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  1. billcalley

    billcalley Guest

    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


  2. Tim Wescott

    Tim Wescott Guest

    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
    Wescott Design Services

    Posting from Google? See

    "Applied Control Theory for Embedded Systems" came out in April.
    See details at
  3. Cecil Moore

    Cecil Moore Guest

    Yep, you are missing the "total destructive interference"
    happening toward the source caused by a Z0-match. Here's
    an article that might help:

    A Z0-match eliminates reflected energy from reaching
    the source.
  4. mpm

    mpm Guest

    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.

  5. Bob

    Bob Guest

    No, we are not all told that.
    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.

  6. Hi Bob,

    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?"

    Richard Clark, KB7QHC
  7. Tim Williams

    Tim Williams Guest

    Triode or pentode? ;o)

  8. Two problems:

    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

    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.

    - Don Klipstein ()
  9. For you, cascode, push-pull triodes.

    Richard Clark, KB7QHC
  10. Roy Lewallen

    Roy Lewallen Guest


    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

    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
  11. On a sunny day (Mon, 12 Mar 2007 07:36:25 +0000 (UTC)) it happened
    (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.
  12. Bill,

    an exellent treatment on this question has been published in QEX December 94
    under the title "Where does the power go?"

    73 de Ulrich, DF6JB
  13. Cecil Moore

    Cecil Moore Guest

    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
    100 volts into the same local load.
  14. Cecil Moore

    Cecil Moore Guest

    Yep, 100% efficiency would be quite a benefit.
  15. Cecil Moore

    Cecil Moore Guest

    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:

    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:

    100W------50 ohm---+---Z0>50 ohms-----load
    Pfor1=100W--> Pfor2-->
    <--Pref1=0W <--Pref2

    The power reflection coefficient is 0.5 at point '+'.
    The power reflection coefficient is 0.5 at the load.

    What are the values of Pfor2 and Pref2? What is the physics
    equation governing what happens to Pref2 at point '+'?
  16. Cecil Moore

    Cecil Moore Guest

    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.
  17. Cecil Moore

    Cecil Moore Guest

    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.
  18. 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%.

    - Don Klipstein ()
  19. Cecil Moore

    Cecil Moore Guest

    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?
  20. billcalley

    billcalley Guest

    Thanks guys -- some really great posts here -- it will take me quite a
    while to digest all this!!

    Thanks Again,

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