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Actual standing waves have nothing to do with causing the output device in a transmitter to fail, in fact it's the reverse or reflected current which helped to create the standing waves which is flowing into the output device that causes the failure, is this true ?

Andrew
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3 Answers3

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This is a bit like saying, "it's not the fall that kills you, but the sudden stop at the end."

Technically, it's not the standing waves per se that cause the damage. However, standing waves imply a mismatched load possibly outside the transmitter's specifications, which implies a potential for excessive heating or voltage in the transmitters, which can lead to damage. Therefore, standing waves imply a potential for damage, even if they aren't the direct cause.

Phil Frost - W8II
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  • can you tell me exactly why a mismatched load can make a transmitter output device fail ? – Andrew Dec 07 '19 at 02:56
  • Didn't you already ask that question? https://ham.stackexchange.com/q/15157/218 – Phil Frost - W8II Dec 08 '19 at 05:41
  • yes i did and the question wasn't answered. – Andrew Dec 08 '19 at 08:53
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    @Andrew it was answered. Operating a device outside its specifications is likely to break it. It's that simple. If you want a more specific answer, then you need to provide specific parameters to the question, like what device, and what operating conditions. You can't reasonably expect a specific answer when your question is effectively as broad as "why do things break when they are abused?" – Phil Frost - W8II Dec 09 '19 at 20:28
  • thanks for the reply, what i meant by the question was, what is the actual reason why the device fails, i think it's obvious that it fails because one or more max ratings have been exceeded. For example, is it because the reflected current flows into the output pin and then damages the device ? or is it because if the load is not 50 ohms then the device becomes less efficient and heats up more ? that sort of thing. I thought my question seemed pretty clear, and the actual physical mechanism involved in causing a device to fail wasn't given in the answers to my question. – Andrew Dec 09 '19 at 21:31
  • saying "The device fails because it's maximum ratings have been exceeded" doesn't answer my question. That's like letting your car engine run out of oil so that the main bearings are destroyed due to the friction and subsequent heat caused by no lubrication, and then giving the explanation for the failure as "oh the oil ran out and then the engine got really hot and it's maximum temperature was exceeded and then it died" ... – Andrew Dec 09 '19 at 21:49
  • @Andrew For the last time, if you want a more specific answer, ask a more specific question. A class C amplifier won't fail in the same way as a class AB amplifier. A tube won't fail the same way as a MOSFET which won't fail the same way as a BJT. The only thing common about how any kind of amplifier might fail is that they are made of parts, and the parts work within certain specifications. Truly, the only way to specifically answer your general question is an entire treatise on electronics. – Phil Frost - W8II Dec 10 '19 at 05:44
  • @Andrew Different types of transistors can have different failure mechanisms – curtis Dec 11 '19 at 18:01
  • If the amplifier involves transistors at all. There might be tubes, a magnetron, or an electromechanical generator. There are all kinds of "transmitting devices". – Phil Frost - W8II Dec 11 '19 at 18:48
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Yes.

If we say there is a standing wave on a feed line, we are saying that the voltage and current along the length of the line have a certain shape in space and over time. But the transmitter doesn't care about any part of the line except the part it's connected to.

You can use the idea of standing waves to help understand and calculate how the end of the line at the transmitter will behave — and in particular, it matters what phase the reflections have, not just that there is reflected power at all — but once you have your answer in terms of current and voltage over time at that point that's all that matters, not what it might be some distance down the line.

Kevin Reid AG6YO
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Standing waves are not what causes the damage.

Heat buildup in the transmitter from accepting some of the reflected power may cause the transmitter to overheat.

But tubes are typically less vulnerable to this and can withstand a higher SWR, and solid state devices typically drop power until the heat buildup is not an issue. Also, it is possible to block the incoming power so the transmitter is not affected. (For instance, a tuner can do this.) Under these conditions, swr around 10 might be acceptable.

Another reason we avoid high swr is to avoid inefficiency. If you have high swr in a feed line with less than perfect efficiency, it will absorb some fraction of the power on each bounce (and heat up slightly) and this will result in less transmitted power. This power drop may or may not be a concern depending on the power level and the feed line efficiency.

But if you have a transmitter that is able to reflect the power back, and transmission line (like windowed ladder line) that is low loss, then the high swr is less of an issue, and eventually the reflected power will reflect again and get into the antenna (perhaps after several reflections). The reflections themselves might cause issues (like ghosting) if the frequency is high enough or the feed line long enough, but at HF this is usually not an issue.

user10489
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    The transmitter doesn't have to accept the power. Just like what happened at the other end of the transmission line, the transmitter can reflect the power. – Phil Frost - W8II Dec 06 '19 at 17:33
  • I said that in my answer already -- there are multiple ways the transmitter can reject the incoming power. – user10489 Dec 07 '19 at 15:03
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    Your statement " the coax typically has a low enough efficency that it will absorb the power " is completely incorrect. – Andrew Dec 09 '19 at 00:29
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    This is also completely incorrect ->> "So the real reason we avoid high swr (other than to avoid the power drop) is to avoid inefficency." You might avoid high SWR for that reason but the rest of the world definitely does not. – Andrew Dec 09 '19 at 00:32
  • I wrote this quickly...adjusted... is that better? If not, please give constructive criticism rather than saying "this is wrong". – user10489 Dec 09 '19 at 12:40
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    Your 2nd sentence implies the transmitter will absorb the power, while it may not. And although you go on to hand-wave some of that away by the 5th paragraph, leading with a common misconception doesn't put the answer on a good start. Tubes are vulnerable to damage due to a mismatch as well, but in practice all tube PAs include a variable matching section at the output so the finals won't see a mismatch if the transmitter is operated properly. Coax isn't really much less efficient: lmr 400 and window line have about the same loss. – Phil Frost - W8II Dec 09 '19 at 20:49