The following are fairly common knowledge:

  1. Transmitting at full power with too much of an impedance mismatch can damage your finals.
  2. If we do so inadvertently, modern radios drop the power significantly for self-protection.

Yet, it's still very common to hear people suggesting that high SWR risks damaging your transmitter. How much of an issue is this with modern radios? Do we really still need to be concerned about damaging our finals, or is it outdated advice that refuses to die?

In the extreme case, can we freely transmit at (attempted) full-power into an empty jack without shortening the lifetime of the finals?

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    $\begingroup$ previous related question without complete answer. In my experience - I've never damaged an amplifier - handheld, 100 W HF, 50 W VHF, and that's how good the protection should be! $\endgroup$
    – tomnexus
    Commented Nov 30, 2020 at 20:09
  • 1
    $\begingroup$ @tomnexus I did see that. It was interesting reading. Also, lest anyone think this is a duplicate question, that one is about how much the power needs to drop, whereas for this one, I'm wondering how effectively radios do so in practice. For example, is there a delay while the radio detects and reacts to the high SWR that allows a brief burst of high voltage/current to reach the finals? Or, in cases of extreme SWR, is it possible the radio won't reduce power enough? $\endgroup$ Commented Nov 30, 2020 at 20:28
  • $\begingroup$ Sounds like this should be a good separate new question: "What kind of circuitry to modern transmitters use to protect against a high SWR (full open or short) causing damage, and what is the response time of that protection circuitry?" $\endgroup$
    – hotpaw2
    Commented Nov 30, 2020 at 20:55
  • $\begingroup$ I found some related Q&A's by searching for foldback and fold back. But I didn't study them close enough, because I'm working on something else here. $\endgroup$ Commented Dec 1, 2020 at 16:00
  • $\begingroup$ @hotpaw2 Perhaps. Or maybe it could be incorporated into the question? $\endgroup$ Commented Dec 1, 2020 at 16:01

1 Answer 1


I believe the answer currently is: yes it matters less, but it still matters.

Variables include 1) the mode of operation and 2) the speed of the foldback circuit in the radio that provides the trigger for power reduction on a bad match.

While modern transceivers all seem to have a foldback circuit to limit potential damage (for ex, see ic7300 operator manual page 13-4, "The transceiver has a 2 step protection function to protect the final power amplifiers in case the antenna SWR becomes high."), they depend on measuring the reflected power which means it's already inside the radio by the time the circuit can respond.

If operating SSB, that may not be much of a problem since the modulated signal is likely not at full power (probably MUCH less) and doesn't give much opportunity to build-up heat before the foldback kicks in.

If operating CW/digital/any instantaneously full-duty cycle mode, then there is more opportunity for maximized power reflection back into the PA before foldback kicks in... depending on component tolerances, radios from the same manufacturer may even have different foldback activation speed. At least SOME power will make it back to the PA, which are running at full power already... this is not ideal.

So a scenario that I could envision is running a digital mode at high power on a band that the PAs are matched to, heating them up already, then moving to another band with high SWR. Now I've got pre-heated PAs, running at full duty, plus need to dissipate whatever large reflected power is... could be a life-shortener if not a blowout event.

So in short, while not NEARLY as critical as back in the day, it's still important.

I also recognize that this answer is VERY light on referenced material, and ENCOURAGE others to augment this answer (or provide a more robust additional) as they see fit.

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    $\begingroup$ I always thought that the protection circuit measures the voltage and current on the final(s) and their temperature, rather than the reflected power, since it's the voltage, current, and heat that do the damage. Reflected power causes higher voltage, current, and heat on the finals, of course. Or maybe there is no practical difference? $\endgroup$
    – rclocher3
    Commented Apr 15, 2021 at 18:12
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    $\begingroup$ @rclocher3 you make an important point. Since heat transfer capability can also be measured in watts, and the power reflected back into the PAs will be split into 2 components 1) power dissipated as heat in the PAs and 2) power reflected back in the direction of the antenna. Of course the discrete currents and voltages is very important for component analysis, but thinking of it in terms of power makes the block diagram analysis a convenient simplification for me. This is also NOT my area of expertise, and I'd love for others to chime in too. $\endgroup$
    – webmarc
    Commented Apr 15, 2021 at 19:27
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    $\begingroup$ There is no practical difference. An SWR meter, which you might say looks at reflected and forward power, looks at the current and voltage on the line. You can say reflected power causes an impedance other than the characteristic impedance to be seen, or you can say reflected power is a way to decompose (by superposition) the impedance of the line for easier analysis. In the end it's the same thing, the only difference is which variables you think about / calculate first. $\endgroup$ Commented Apr 16, 2021 at 18:47

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