VSWR doesn't matter?

Your mistake is that you assume the output of the tx is 50 ohms, in the case you stated the transmitter must be matched to the impedance it sees looking into the transmission line.

In my opinion the simplest way to answer your question is that you are assuming that the transmitter is equivalent to a 50 ohm load, which is not true because the transmitter is instead equivalent to the series of a 50 ohm load and a voltage generator.

A simple DC example grossly clarifies thre issue: connect a 12V battery to the series of a 50-ohm load and another 12V battery. How much current flows through the load? Naught (assuming the correct polarity). So no power is dissipated in it.

Tony I0JX

Hi Jimmie,

At the risk of yet another, non-quantitative reply I will repeat:

THAT is true, and it brings us to the point of all this energy sloshing around until the antenna finally dissipates it out into the Æther. It is the reflection off the mismatch of the tuner (the mismatch seen by the antenna as source to the line going back) that prevents energy from presenting any destructive results to the source

- the whole point of using a tuner in the first place.

In message , billcalley writes

It matters when it changes suddenly, like mine did recently on my 70MHz beam, when one of the elements came off in a gale.

Brian GM4DIJ

As pointed out, VSWR does matter. A lot of bouncing means you heat the transmission line with the power instead of radiating the power. 'Doesn't matter', really means it can be tolerated if need be.

Here is what you are missing. In the case of the output, (real/resistive component of the transmitter), seeing the reflected wave, it is _not_ reflecting that power back up the transmission line as you think it is. It would go back to that real impedance and heat the transmitter. Here is what is done with a miss match in the real world.

trans-output -> match -> line -> antenna

The 'match' is where the magic happens. All the energy coming down the line that got reflected from the antenna 'sees' the 'trans-output -> match' as a perfect reflector and gets bounced back[*]. On the other side of the match is the trans->output. There the trans->output sees a perfect impedance, (technically, the conjugate of the trans->output), so that all the power travels through the match toward the antenna.

The magic is that when the match is tuned, both of the above conditions are satisfied.

Best, Dan.

Correct but I just want to remember that the purpose of the tuner is to match the impedance of the transmitter to the impedance of the antenna/ transmission line.The standing waves can be viewed as a reflect voltage, a reflect current or as a reflected impedance. Besides I thought there had been enough quanitative analysis of the question and was hoping a simple answer may be enough to turn on the light bulb for the OP. If he still wanted to know more I figure he would ask.

Hi Tony,

Turn the second battery over. Double the power is dissipated in it.

Phase, you can't live with it, you can't live without it.

Of course. Mine was just a DC example to illustrate things in a simple manner.

When the transmitter is properly tuned, the phase relationship is such that the reflected wave does not get dissipated at all into the 50 ohm output of the transmitter, and is then reflected back to the antenna

Tony I0JX

Kewl, then I'll just run a tap directly to the 5000 ohm plates and start a long chat up ... what the heck is all those pi matching components in the way of the rf? Probably some loss there! ROFLOL!!!

JS

OK ;-))

Saying that SWR doesnt matter is a rather broad statement(like saying never or always) but I have know of antenna systems having an SWR of 30:1 and his was normal. The feedline was balanced line made of 1 inch copper. Of course an SWR lie this on coax could be fatal to coax and equipment. A more common example of this is the 1/4 wl matching section on a J-pole antenna. It matches 50 ohms to a few Kohms so an SWR of 60: 1 or so would not be unusal here.S oas long as the feedline can handle the current and voltage peaks without much los it doesnt matter much as long as the source impedance is matched to the impedance at the input to the transmission line.Im sure there is a practical limit though.

Jimmie

If you want a quick lesson in high vswr find a ham with an old tube transmitter and see if he will hook it up to a mismatched load. The cherry red plates are the reflected energy being absorbed. Transistors will just turn to smoke under the same conditions.

Dave WD9BDZ

Hi david

Wouldnt it be OK to have a high VSWR along the transmission line if the "tank ckt" can be adjusted to match the load to the transmitter output impedance? That is, the VSWR along the transmission could concievely be high, yet, with proper "tank ckt" adjustment that impedance seen by the output circuit (plate) wouldnt result in a "cherry red plate". What I am asking is ? is the transmission line VSWR directly related to "plate reddening"? I'm more asking than *telling*.

Jerry

Unfortunately, you'd be learning the wrong lesson.

The cherry color is due to the transmitter being loaded with an impedance it's not designed for, causing the final to run at low efficiency. You can disconnect the antenna and replace it with a lumped RC or RL impedance of the same value and get exactly the same result. Alternatively, you can attach any combination of load and transmission line which give the same impedance, resulting in a wide variation of "reflected energy", and get exactly the same result. All that counts is the impedance seen by the transmitter, not the VSWR on the line or the "reflected power".

The problem is that the idea of "reflected energy" turning the plates hot is so easy to understand, that people aren't willing to abandon it simply because it isn't true.

See

Roy Lewallen, W7EL

Yes! All that matters to the transmitter is the impedance it sees. It doesn't know or care that you've mathematically separated the delivered power into "forward" and "reverse" components. It doesn't know or care what the SWR is on the transmission line connected to it, or even if a transmission line is connected at all.

Absolutely not.

That's the first step in learning.

Roy Lewallen, W7EL

It appears that you agree with that part of my post but you are drawing an invalid conculsion from it. I never suggested that the passive matching network usually found in a transmitter output is unnecessary.

Bob

snipped-for-privacy@manx.misty.com (Don Klipstein) wrote in news: snipped-for-privacy@manx.misty.com: ...

... I thought we put one of these fires out just a few days ago!

Owen

Jerry;

The point I was trying to make is that the reflected current is disapated as heat in the finals if the transmitter isn't matched to the load. In a tube radio the tank circuit is the equivilent of an antenna match/tuner and converts the 2000 or so ohms at the plate to the 50 ohms of the transmission line and the unknown ohms of the mis matched antenna.

Dave WD9BDZ

As Tim Williams alludes, it depends on the transmitter design. It will often be complex rarther than resistive. Since the active device changes impedance during a single cycle of the RF signal it may not even be adequately described by a single value in ohms for a paticular frequency if you wish to analyse the case of forward and reflected power.

Consider a class C or class E output stage with an output transistor that is low impedance during most of the positive half of a cycle of signal and mostly somewhere near open circuit for the negative half of the cycle. It seems to me that the effect of reflected power is going to be different depending its phase relative to the forward power. I think this also applys to a lesser extent to a class A PA with a nice hi-Q tank circuit.

As usually whan this topic comes up, It don't feel like we have arrived at a usefull and convincing model of what happens, possibly because simple descriptions don't cover everything.

Bob