Exactly why does an RF output transistor fail if the SWR is bad ?

Is it A because the impedance mismatch between the transmission line and the antenna results in maximum power not being transferred to the antenna but rather being converted to heat inside the transistor ?

Or is it B because the reflected AC current caused by the impedance mismatch from the antenna goes back into the transistor output somehow ?

If the answer is B, then what stops the output signal produced by the RF output transistor from going back into the transistor in the same way as the reflected signal caused by a bad SWR ?


A high SWR implies the load impedance is significantly different from 50 ohms, violating the design specifications of the amplifier.

Yes, there is power reflected in a feedline. But it's not this power per se that causes problems, it's simply that the impedance seen by the transmitter isn't the 50 ohms it was designed for. In other words, the transmitter can't tell the difference between reflected power and simply the wrong impedance connected directly at the antenna connector with no feedline.

With the design assumptions violated, all the design specifications (like not catching fire) are off the table. Excessive current may lead to overheating. Or there may be too much voltage across them leading to avalanche breakdown. These are just examples: other failure modes are possible depending on the design of the amplifier.

  • $\begingroup$ So why aren't all transmitters simply designed for loads from 0 to infinite resistance (or reactance) to prevent this problem? $\endgroup$ – hotpaw2 Aug 21 '19 at 18:33
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    $\begingroup$ @hotpaw2 cost.. $\endgroup$ – Phil Frost - W8II Aug 21 '19 at 19:25
  • $\begingroup$ @phil - the answer to my question is not that the 50 ohm design of the amp has been violated. I want to know why EXACTLY the output transistor fails. Why exactly. I already know it fails because it gets too hot. Why does it get hot when the SWR is bad ? is it because reflected AC current flows into the collector ? whereas AC current is meant to flow out of the collector. $\endgroup$ – Andrew Aug 27 '19 at 22:05
  • $\begingroup$ @Phil Or is it because the incorrect load impedance results in power transfer from source to load being less than if the SWR was perfect, and so more ;power is dissipated inside the output transistor rather than being transferred to the load ? or something else ? $\endgroup$ – Andrew Aug 27 '19 at 22:05
  • $\begingroup$ @Andrew I'm afraid to know EXACTLY why the transistor fails would require an exact specification of the transistor, and the amplifier, and the load impedance. Thermal overload is just one possible failure mode. It might not fail at all. $\endgroup$ – Phil Frost - W8II Aug 28 '19 at 2:53

I like to turn the question around.

Why does a transmitter amplifier not burn up or fail? There's power coming into the transmitter from the power supply. Why doesn't it melt the transmitter?

Well, some of that energy (power over some period of time) goes into warming up heat sinks and eventually the air around your locale. But a transmitter isn't (designed as) a toaster or room heater. It's designed as part of a system to radiate RF energy. So a transmitter is designed for (a large fraction of) the energy to exit your locale as electromagnetic radiation.

But a high SWR means a large portion of the power being sent up the feedline isn't being electromagnetically radiated. Thus, energy has to be dissipated somewhere. So, if the transmitter doesn't fold back the power level to below what the feedline and the heatsinks can radiate away, something will melt or otherwise stop functioning due to overheating.

Added: Alternatively the power reflected back by a high SWR can be reflected back up the transmission line by yet another impedance mismatch at the transmitter. But that reflection will either increase the voltage or current at the reflection point, depending on the impedance mismatch. So again, you either dissipate, arc, or melt something.

Also, a non-resonant antenna system is a reactive load. One thing about reactive systems is that they store energy. If you pump more power into a reactive system than it dissipates, the system will store energy in the form of increasing voltages and/or currents across or within components. If you keep on increasing voltages or currents without bound, eventually something will melt or arc over. An antenna system with a nice low SWR and moderate resistance will radiate most of that energy away (as RF or heat), rather than turning the transmitter into a blown fuse.

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    $\begingroup$ High SWR does not mean the energy isn't being radiated. It just means it makes more than one trip down the feedline before it finds a way out. What fundamental law says it has to be absorbed by the transmitter, versus making another pass at the antenna? $\endgroup$ – Phil Frost - W8II Aug 21 '19 at 23:02
  • $\begingroup$ It only makes a second trip up the feed line if the transmitter reflects it due to yet another impedance mismatch. $\endgroup$ – hotpaw2 Aug 21 '19 at 23:20
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    $\begingroup$ And what's to say there isn't such an impedance mismatch? $\endgroup$ – Phil Frost - W8II Aug 21 '19 at 23:54
  • $\begingroup$ Most of this to me seems not relevant to my question. $\endgroup$ – Andrew Aug 30 '19 at 0:38
  • $\begingroup$ This comment is getting closer -> "Thus, energy has to be dissipated somewhere. So, if the transmitter doesn't fold back the power level to below what the feedline and the heatsinks can radiate away, something will melt or otherwise stop functioning due to overheating." $\endgroup$ – Andrew Aug 30 '19 at 0:39

When there is a mismatch between the output stage and the antenna, a certain amount of power is reflected back to the transistor. As the mismatch becomes larger, so does the reflected power, and this generates heat in the transistor.

If you put a large heatsink on the transistor, and keep liquid nitrogen circulating around the heatsink, you might be able to operate for an indefinite time with an infinite SWR, as the heat generated by the reflected power is being taken away by the heatsink surrounded by liquid nitrogen.

The main problem with semiconductors in this scenario is that if you heat them up, their internal resistance goes down, which causes more current to flow for a given voltage (which in turn generates more heat, etc.). This can lead to thermal runaway, which itself usually results in the destruction of the transistor.

Related - see also this answer: Why are vacuum tubes still used in amateur radios?

To answer the question directly, though: an RF transistor fails when the SWR is too high because it can not dissipate the heat generated by the reflected power, and this can cause the thermal breakdown of the transistor. Usually with a bang, and letting out the magic smoke.

Some radios have a heat sensor and shut down the PA stage when the PA transistors reach a threshold temperature.

  • $\begingroup$ why does reflected power generate heat in the transistor ? $\endgroup$ – Andrew Aug 21 '19 at 5:41
  • $\begingroup$ The power has to be dissipated somewhere. It gets sent along the transmission line to the antenna, and some of it is reflected back. The transmission line will convert some of the power to heat, but the lion's share of it goes right back to the output stage and PA transistor $\endgroup$ – Scott Earle Aug 21 '19 at 7:09
  • $\begingroup$ then what stops the output signal produced by the RF output transistor from going back into the transistor in the same was as the reflected signal caused by a bad SWR ? $\endgroup$ – Andrew Aug 21 '19 at 10:39
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    $\begingroup$ Why does the antenna get to reflect the power but the transistor has to absorb it? $\endgroup$ – Phil Frost - W8II Aug 21 '19 at 15:50
  • $\begingroup$ @PhilFrost-W8II the antenna absorbs some, but antennas are made of conductors and transistors are made of semiconductors. $\endgroup$ – hobbs - KC2G Aug 24 '19 at 15:56

When an amplifier of a fixed output impedance (50 ohms) is tuned for minimum VSWR into a load (your antenna), then maximum power is transferred from that device to be radiated into the ionosphere.

Let's assume that our RF amplifier feeding  our antenna (and in the interest of simplification is 100% efficient) has 5 volts applied to it, and the current is 20 amps. 5 * 20, that's 100 watts.

The SWR is 1:1, so in this ideal amplifier there are 100 watts delivered to the antenna, and the RF power amplifier transistor doesn't get hot.

(But, of course, in the real world, there is no such thing as an RF amplifier device (either transistor or vacuum tube) that is 100% efficient. Therefore, those devices will get warm to some degree, even when perfectly matched at a 1:1 VSWR.)

Now, suppose that we deliberately change things and mistune our antenna so that its feedpoint impedance is lower or higher than 50 ohms.

 - Less power will be radiated by the antenna.

 - The current through the output transistor may change up or down slightly, but now more power will be dissipated in the transistor, making it hotter.

 - As we continue to detune our antenna, even less power will make it to the antenna and more power will be dissipated in the transistor. Eventually, so much power will heat up the device to the point where it is destroyed.

Higher SWR = less power to the antenna to be radiated into space = more power to heat up and destroy our device .

  • $\begingroup$ can you tell me exactly why there is "more power to heat up and destroy our device" ? and where does that power come from ... $\endgroup$ – Andrew Sep 9 '19 at 5:18
  • $\begingroup$ @Andrew That power comes from the DC power supply. $\endgroup$ – Mike Waters Sep 9 '19 at 12:59

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