Tom Rauch wrote:
>>>But can I ask, "What is the difference in perception at
>>
> the receive end of
>
>>>a HF wireless signal transmitted with an antenna measured
>>
> that has a 2:1
>
>>>(impedance either high or low) SWR versus an antenna that
>>
> has a measured
>
>>>SWR of 1:1.2 or so - given you are working with the same
>>
> antenna AND given
>
>>>your modern rig does not power reduce prior to 2:1 SWR?"
>>
>>With a 2:1 mismatch at the far end the return loss is 10
>
> dB. This is
>
>>another way of saying that for every 1000 watts you send
>
> to the antenna,
>
>>100 watts is reflected back towards the transmitter. The
>
> radiated power is
>
>>reduced from 1000W to 900W (about 0.5 dB decrease)
>>
>>With 1.2:1, the return loss is 26.4 dB: only 2.2 Watts
>
> gets reflected back.
>
>>(0.01 dB decrease).
>>
>>In reality, it's not quite that bad, because most
>
> transmitters don't
>
>>present a perfect match to the reflected wave anyway. So
>
> in the 2:1 case,
>
> <snip>
>
> Actually not.
>
> Reflected power and return loss is a model that represents
> mismatch. The reflected power can only be taken as a "loss"
> when the source appears to be a "fixed" dissipative
> resistance that looks like 50 ohms, like a signal generator
> through a large attenuator pad would appear.
>
> Real transmitters almost never behave that way, and if the
> transmitter has a tuner it absolutely doesn't behave that
> way. Most SSB transmitters actually favor a load impedance
> higher than 50 ohms, that's to increase linearity.
>
> In any event, it is always considered that the reflected
> power makes only one trip, and that 100% of the reflected
> power returns to the antenna from the transmitter.
>
> The end effect of all this is the VAR (volt-amperes
> reactive) power in the transmission line increases. As far
> as we are concerned the only thing that happens is
> transmission line loss increases slightly. Since HF losses
> are predominantly resistive losses in the conductors, unless
> the line is a significant fraction of a wavelength long
> losses can actually DECREASE when the mismatch is in a
> direction that reduces current!!
>
> This is why open wire line, even with very high standing
> wave ratios and extremely high return loses, "return losses"
> that would imply only a few percent of power is transferred
> often operate with very high efficiency.
>
> Now if the source is a generator or transmitter through a
> very large attenuator pad of an impedance equal to the
> impedance you are considering "return loss" as being
> referenced, you could consider the "return loss" as a direct
> reduction of power. Of course that loss would have to be
> added to the increase in loss in the transmission line
> caused by the termination mismatch.
>
> For amateurs, in order to not be confused by models of
> "reflected power" and "return loss", it's best to just
> consider the additional loss from the increase in SWR. We
> can only do that if the line is near 1/4 wl long or longer.
> When the line is short, losses can actually DECREASE with
> increased return loss if the current integrated over the
> length of the line decreases!!
>
> So if your line is fairly long compared to 1/4 wl and if you
> have SWR, the best advice would be to ignore everything else
> and look at a table showing the increase in line loss caused
> by an increase in SWR. If the transmitter does not fold
> back, that is pretty much the only loss you will have and it
> can be pretty small.
>
> When MFJ wanted to add "return loss" to the MFJ analyzers I
> objected, because return loss is very misleading. It really
> is just another way of expressing SWR (another SWR model),
> and doesn't mean that amount of power is "lost" or
> unavailable.
>
> The suggestion that HF PA's become "unstable" because of
> mismatch is also generally untrue in the case of any PA with
> an adjustable tank. There are some solid state PA's that
> suffer instability with certain loads, but that generally
> has little to do with load impedance at the operating
> frequency. That problem will just as well manifest itself
> with a PERFECT SWR on the operating frequency when the load
> presents a critical impedance on any frequency passed by the
> output filters of the PA. An example of this is a solid
> state mobile amplifier driving a mobile antenna on 160 or 80
> meters. The antenna changes impedance so rapidly with
> frequency the amplifier can be considered to be operating
> into a nearly pure
> reactance, and that reactance can change phase at the output
> of the PA so much that the negative feedback circuits that
> normally reduce gain now increase gain. The result is an
> oscillation that is out-of-band. It has nothing to do with
> SWR or return loss on the operating frequency, but rather
> mistermination of the system out of band whether
> transmitting or not!
>
> That is actually a MAJOR design headache solid state
> amplifier designers (or any amp with heavy feedback from
> output to input) must face.
>
> Don't get caught up in return loss or reflected power unless
> you understand it and how systems behave totally, otherwise
> it will really mislead you!!!!
>
> 73 Tom
>
I agree with Tom about reflected power not being dissipated in the
transmitters final. For all practical purposes, it all gets returned to
the antenna and radiated except for the small amount of loss the
transmission line causes from the mismatch.
The exception would be as Tom described an attenuator at the transmitter
absorbing the reflected power. Another way to look at it would be if you
had an isolator at the transmitter output. Then any reflected power
would not get re- reflected back to the antenna but all reflected power
would get dumped to the load on the isolator.
If a tuner was in line and matched the transmitter would never see any
reflected power. The tuner does not absorb the reflected power but
reflects it back to the antenna.
I don't understand what is meant by "losses can actually decrease when
the mismatch is in a direction that decreases current"? I don't see how
that can happen. How can current ever be less than 1 unless you change
the characteristic impedance of the transmission line? Unless it is so
short that it is no longer acting like a transmission line.
Open wire line operates efficiently with high standing waves (high
reflected power) because its characteristic impedance is high and
therefore the current is low compared to a 50 ohm coax line. Much less
current in the open wire line even with high swr and therefore lower I
squared R loss. I know that Tom knows that but the way it is written it
may be misleading.
73
Gary K4FMX
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