I know it's important to tune an antenna to the same impedance as the transmitter for maximum power transfer. But I've read that if you don't do so, you not only get less radiated power, but you can damage your transmitter, due to standing waves and reflected energy. I'm having trouble understanding what that means, and would like some help.

Below is what I know:

Antenna impedance is largely due to radiation resistance: a center fed half wavelength dipole has a radiation resistance of about 73 ohms. Resistance due to heat loss is negligible, and I believe reactance negligible as well.

Let's say we have an oscillator consisting of a 1 MHz voltage source (30V RMS) in series with a 75 ohm resistor. This is center fed to the above half wavelength dipole antenna. Total impedance in the system is about 150 ohms (75 + 73), current 200 mA, total power 6 W, with 3 W loss internal to the oscillator and 3 W radiated into space. So far, so good.

Now, let's say we modify the antenna so its radiation resistance is now 300 ohms. Current drops to 80 mA, total power to 2.4 W, and radiation to 1.9 W. I understand that we're now transmitting less power - but how does that damage the transmitter? In what way does the voltage or current at the oscillator change to be harmful? And what does that have to do with standing waves?

Or, let's say we modify the antenna so its radiation resistance is only 15 ohms. Current increases to 330 mA, total power to 10 W, but only 1.7 W is radiated, with remaining 8.3 W lost as heat. I understand that this heat could damage the transmitter, but that doesn't seem to have anything to do with standing waves or reflected energy.

  • $\begingroup$ Good question, and one that's been answered here before. $\endgroup$ Commented Nov 1, 2023 at 19:53
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    $\begingroup$ @MikeWaters Before posting, I spent several hours reading this site. I learned enough to write the question above - but not enough to answer it. (If helpful, I could document all the posts I read, what I learned from them, and what remains unclear - but then again, my question is already long). If there's a particular resource I missed, please direct me to it. $\endgroup$ Commented Nov 1, 2023 at 21:34
  • $\begingroup$ @SRobertJames Sorry, I was too lazy to search for one. ;-) Maybe when I get alert. $\endgroup$ Commented Nov 2, 2023 at 11:22
  • $\begingroup$ See also this excellent exploration of VSWR by Phil $\endgroup$
    – webmarc
    Commented Nov 2, 2023 at 14:53

2 Answers 2


There are a few parts to this, it's good to keep them separate.

1. Antenna impedance

Antennas are complicated beasts, they connect the circuit world of voltage, current and power, into the electromagnetic world of Electric and Magnetic fields, and EM waves.

The antenna impedance is made up of the radiation resistance, loss resistance, standing and travelling waves on the antenna, reflections from nearby objects, the matching circuit, interaction with the ham holding the mast up, etc. All of these things change quickly with frequency.

But for this discussion, all you need is that at some frequency, the antenna has a connector or terminals, and looks like an RL or RC circuit. Its impedance is fixed and is expressed as R+jX. (Real and Imaginary parts).

2. Standing waves

Are an effect on transmission lines. When the termination impedance is not the same as the characteristic impedance, the travelling waves are reflected by the termination, This causes standing waves, places where the magnitude of the RF voltage is larger and smaller. This effect is stationary on the line (because the phase relationship between the forward and reverse waves is fixed).

The standing waves themselves have only one interesting effect: they lead to localised heating in regions where the current is larger (and sparking will happen first at the voltage maxima).

The presence of the reflected wave also means that the impedance looking into the start of the line is not Z0 but something else, and there are formulas for that.

We often describe the degree of mismatch by the Voltage Standing Wave Ratio. For example:

  • on a 50 ohm line
  • with a terminating impedance of 25 ohms
  • VSWR will be 2:1
  • after some length of transmission line, impedance will be 25 ohms, 100 ohms, or any complex impedance between these.

3. Transmitters

A transmitter is (for now) a circuit of lumped components (transistors, resistors, capacitors etc, but all obeying normal circuit rules), designed to with a 50 ohm load at the antenna connector. It will also work with 40 ohms or 60 ohms, but may not work with 1 ohm or 10k ohms.

The question is about what impedance variation is OK, and how the amplifier might be damaged outside of that range.

One can certainly buy amplifiers that are safe and work without "cutting back" or self-protection, into any impedance. But these are not optimal for most users. This one is 10 watts out, consumes 400 watts, weighs 28 kg, and probably costs $50k.

So the designer optimises the amplifier for power and efficiency, in limited frequency bands, and one of the results is that the amplifier can't be operated at full power, with any impedance load. A short or open circuit will damage it.

This happens because the final transistors in the amplifier have a maximum voltage, current, and power dissipation, and operating them out of this envelope will quickly damage them. Of course you could get bigger transistors, but they're more expensive, larger, have more capacitance, so the designer makes a reasonable choice based partly on the range of load impedance that is expected. So an amplifier that delivers full power at better than 2:1, will be cheaper and smaller than one that tolerates 4:1.

Because transistors will fail almost instantly, there are usually protection circuits to detect these conditions and quickly reduce or trip the power, so it's safe but not really transmitting any more.

An aside - there is no 50 ohm series resistor in the amplifier, in fact its output impedance is probably much less than 50 ohms. This means that it can probably tolerate high impedances better than low impedances... VSWR is a very blunt measure of impedance, but it's easy to measure. Some discussion in this answer.

How these all come together

In practice, the user starts with (1) a somewhat mismatched antenna. Connects a long transmission line to it, which changes the impedance (2). Connects this to the amplifier (3), which sees the transformed impedance at its terminals.

The question is whether it's OK with this impedance, or it's too far from the design value and it is damaged (or cuts back).

I think it's important to frame it like this, not as "reflected power coming back and damaging the amplifier". The amplifier is self-destructing when connected to the wrong load impedance. The load happens to be an interesting electrically-large circuit.

  • $\begingroup$ Fascinating. Would it be correct to say: Transmitters, like all circuits, can handle some loads but not others. A load of too low impedance (e.g. a short circuit) will damage it, and even one that's too high can as well. This has nothing to do with standing waves or reflection, but rather is the same mechanism in any circuit. The best circumstance (for any circuit) is when the load impedance is the complex conjugate of source impedance. Since we can't easily measure load impedance, we measure SWR, and if it's low, we know the load is within the transmitter's design range. Correct? $\endgroup$ Commented Nov 3, 2023 at 0:25
  • $\begingroup$ The first part is correct. The second, about conjugate matching, is not, and I mentioned the source impedance above. Maximum power transfer theorem is its own topic and surprisingly not really relevant to the transmitter (but it is at the antenna) . The third is true, low SWR = close to design load impedance, but the converse, high SWR = damage, is not necessary true. Eg it's possible the transmitter will tolerate 10:1 SWR if the impedance is high, but not 3:1 if it's low. i.e. the safe impedance range may not be symmetrical around 50 ohms. A VSWR limit is a kind of lower bound of safety. $\endgroup$
    – tomnexus
    Commented Nov 3, 2023 at 5:39

– let's say we modify the antenna so its radiation resistance is only 15 ohms. Current increases to 330 mA, total power to 10 W, but only 1.7 W is radiated, with remaining 8.3 W lost as heat. I understand that this heat could damage the transmitter, …


… but that doesn't seem to have anything to do with standing waves or reflected energy.

If you attach the antenna directly to the transmitter, then perhaps it does not. But if you have a feed line, then depending on the length of the feed line, then you might just as likely find a condition where your incorrectly low impedance antenna is transformed into an incorrectly high impedance antenna. In general, the reflected wave appears to the transmitter's antenna port, for circuit analysis purposes, as an AC voltage source connected to it, whose magnitude and phase depend on the impedance at the far end of the transmission line and on the length of the transmission line, and this will alter the impedance observed by the transmitter.

So, the reflected wave is the direct cause of the over-current (or over-voltage) condition, regardless of which part of the system causes the reflection. The reflected wave is equal to the unintended additional current or voltage appearing at the transmitter's antenna port; if there is no reflected wave (in a nonzero length of feed line), then there is no possible damage (and the antenna system is either properly matched or lossy enough you can't tell the difference, and appears as a resistive load).

And if you do have a zero-length feed line then, yes, it is somewhat silly to say that there are any standing waves because it's too short to observe standing-wave behavior, and even saying there's a reflected wave is roundabout compared to a lumped circuit analysis — but the physics doesn't abruptly change.

In another extreme case, if you imagine having an extremely long feed line — one in which the round-trip time of the reflection is significant — then you will no longer see any consistent impedance at the transmitter end of the feed line, because the transmitter is of course modulating the signal, and what's coming back is a time-delayed version of that signal, and the instantaneous voltage will depend on what signal you've been transmitting, not just on the system's physical characteristics. When you start transmitting, you'll always measure 50 Ω (or whatever the characteristic impedance is) and a 1:1 SWR, right up until the first reflection comes back, and then it's messy.

(Disclaimer: I am not actually an electrical engineer, and this is largely a “qualitative” not “quantitative” explanation. There may be some inaccuracies. But I think the overall point is solid.)

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    $\begingroup$ Thank you. "The reflected wave appears to the transmitter's antenna port... as an AC voltage source... whose magnitude and phase depend on the impedance at the far end of the transmission line and on the length of the transmission line" - this is new information. Can you elaborate on this (or point me to a reference). If I know the impedance and length, what will the AC voltage source look like? $\endgroup$ Commented Nov 2, 2023 at 2:50
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    $\begingroup$ @SRobertJames how familiar are you with the term Transmission Line Theory or with Smith Charts? $\endgroup$ Commented Nov 2, 2023 at 3:45
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    $\begingroup$ Either way, whether you've heard of that before or not, it might be an excellent idea to ask the question you've just asked in the comment as an actual new question post $\endgroup$ Commented Nov 2, 2023 at 10:09
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    $\begingroup$ Regarding "I think this is muddying the discussion" — My intent is to clarify the origin and rationale of the existing muddy terminology. $\endgroup$
    – Kevin Reid AG6YO
    Commented Nov 2, 2023 at 17:19
  • $\begingroup$ @tomnexus Your comments are very intriguing, but I can't say I follow them. Would you consider writing them up as a full answer? I'm eager to get to the bottom of this, and your comments suggest a path. $\endgroup$ Commented Nov 2, 2023 at 19:28

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