tuner - feedline - antenna question ?
dansawyeror
I am trying to install a trapped 40m dipole in the attic, the antenna is
in place however it is short and resonates at about 7.7 MHz. I decided
to try it by using a tuner close to the transmitter in the shack. The
feedline is 50 ohm coax. On low power the tuner creates a very low SWR.
The transmitter is a solid state 100 watt Heathkit.
However when I transmit according to the SWR Watt meter the system
appears to transmit well over 200 watts. It pined the meter on a 200
watt range. I repeated the test twice and then stopped. When it is
transmitting the SWR reads about 1.1 to 1. The meter works very well and
does not exhibit strange readings on other setups.
My questions are: What is happening? What is causing it?
Thanks - Dan
Cecil Moore
When the transmitter is putting out 100 watts and the
forward power reading is 200 watts, it means that the
SWR is 5.83:1, i.e. the voltage reflection coefficient
is 0.707 and the power reflection coefficient is 0.5.
To get the power delivered to the load, you must subtract
reflected power from forward power. In your case that seems
to be:
Pfwd - Pref = Pload = ~Psource
200w - 100w = 100w = ~100w
--
73, Cecil http://www.w5dxp.com
Richard Clark
<dansa...@comcast.net> wrote:
>resonates at about 7.7 MHz.
>appears to transmit well over 200 watts.
Hi Dan,
The Meter is either not as good as you claim it to be (it doesn't
accurately perform at low power); OR
you have common mode problems (classic).
Do you choke the dipole?
73's
Richard Clark, KB7QHC
Owen Duffy
news:iYGdnQe3abZtOnzY...@comcast.com:
Dan,
I assume in all these scenarios, the VSWR meter is between the
transmitter and ATU, and the ATU is adjusted for low VSWR.
You seem to say raise two issues:
-you have adjusted the tuner for a "very low SWR" on "low power"
(whatever each of those means), and when you transmit at "well over 200
watts" the SWR is 1.1:1; and
- your transmitter rated at 100W indicates "well over 200 watts" into a
1.1:1 load.
Re the first issue: If this is to mean the VSWR is higher on higher
power, the most common reason (but not the only one) that VSWR meters
read better VSWR on very low power is to do with the voltage drop across
the diode(s) in the meter. The scale may be calibrated at higher power
where the diode drop is less significant, perhaps even insignificant, and
when you adjust the meter for maximum sensitivity the diode drop
introduces significant error.
Re the second issue, if the instrument is a typical directional
wattmeter, the power output is calculated by deducting the "reflected
power" from the "forward power", but at VSWR=1.1 the "reflected power is
0.2% of "forward power" and insignificant. Otherwise, it might just have
an RF voltmeter sampling the line and calibrated in watts, and which is
only valid at very low VSWR. Transmitters don't often exceed their rated
power by over 100%, so your reading casts doubt on your meter. It sounds
like you need to make another measurement with another instrument to
locate the problem.
Owen
ml
Cecil Moore <myc...@hotmail.com> wrote:
> dansawyeror wrote:
> > I am trying to install a trapped 40m dipole in the attic, the antenna is
> > in place however it is short and resonates at about 7.7 MHz. I decided
> > to try it by using a tuner close to the transmitter in the shack. The
> > feedline is 50 ohm coax. On low power the tuner creates a very low SWR.
> > The transmitter is a solid state 100 watt Heathkit.
> >
> > However when I transmit according to the SWR Watt meter the system
> > appears to transmit well over 200 watts. It pined the meter on a 200
> > watt range. I repeated the test twice and then stopped. When it is
> > transmitting the SWR reads about 1.1 to 1. The meter works very well and
> > does not exhibit strange readings on other setups.
> >
> > My questions are: What is happening? What is causing it?
>
>
<<two things pop into mind could be your antenna is radiating towards
and into or near your radio and causing some feedback that might effect
the radio and or the meter
there is also the possibility that you have some bad (common mode? etc)
current flowing back down the coax which could also wreck havock
just my guess and double check the above and grounds
put a dummy load into your tuner see if you get proper readings
Jeff
> reflected power from forward power. In your case that seems
> to be:
> Pfwd - Pref = Pload = ~Psource
> 200w - 100w = 100w = ~100w
How can that be? If the meter is basically a directional coupler then the
forward power is just that. Subtracting any reflected power will just give a
stupid answer. The only errors will be due to the directivity of the
coupler, which will give a band of uncertainty which varies with VSWR, and
the error due to the accuracy of the detectors.
My Bird does not subtract any reflected power to give a forward power
reading!! It can't I need to rotate the slug to read reverse power.
73
Jeff
Owen Duffy
news:45e29929$0$26696$834e...@reader.greatnowhere.com:
> How can that be? If the meter is basically a directional coupler then
> the forward power is just that. Subtracting any reflected power will
> just give a stupid answer. The only errors will be due to the
> directivity of the coupler, which will give a band of uncertainty
> which varies with VSWR, and the error due to the accuracy of the
> detectors.
>
> My Bird does not subtract any reflected power to give a forward power
> reading!! It can't I need to rotate the slug to read reverse power.
Jeff, without commenting on whether Cecil's assertions are right or
wrong, you seem to have some misconceptions about what is measured with
your Bird (presumably 43).
The so called "forward power" and "reflected power" are notional values,
but not actual power "components". The only power is the average rate at
which energy passes a point, and it is in one direction or the other.
In fact the power can be calculated taking "forward power" minus
"reflected power", but only in the case where the sampler is calibrated
for Zo being real (as it is in a Bird 43).
My article at http://www.vk1od.net/VSWR/VSWRMeter.htm describes the
operation of a Bruene type of VSWR meter and discusses the power
measurement issue. Though the sampler in the Bird is different to the
Bruene sampler, the Bird samples V and I in a very small region (
regarded a point ) and sums them in the same way as the Bruene circuit.
Owen
Dave Oldridge
news:45e29929$0$26696$834e...@reader.greatnowhere.com:
>> To get the power delivered to the load, you must subtract
>> reflected power from forward power. In your case that seems
>> to be:
>> Pfwd - Pref = Pload = ~Psource
>> 200w - 100w = 100w = ~100w
>
> How can that be? If the meter is basically a directional coupler then
> the forward power is just that. Subtracting any reflected power will
> just give a stupid answer. The only errors will be due to the
> directivity of the coupler, which will give a band of uncertainty
> which varies with VSWR, and the error due to the accuracy of the
> detectors.
Depends if the meter is before or after the tuner. If the meter is after
the tuner, then the tuner is taking reflected power and adding it to the
transmitter's contribution.
> My Bird does not subtract any reflected power to give a forward power
> reading!! It can't I need to rotate the slug to read reverse power.
If you put your bird AFTER a tuner on a line that's near 6-to-1, it will
also show twice as much power as the transmitter is putting into the
tuner. Or a bit less if the tuner is not efficient.
--
Dave Oldridge+
ICQ 1800667
Jeff
> but not actual power "components". The only power is the average rate at
> which energy passes a point, and it is in one direction or the other.
I am sorry, but I disagree, forward power is real and can be measured, or if
you wish separated out with a circulator or isolator. What you are
describing could be called 'transmitted' power or power delivered into a
mismatched load, but that it different from forward power, or the power
delivered by the source.
73
Jeff
Cecil Moore
>> To get the power delivered to the load, you must subtract
>> reflected power from forward power. In your case that seems
>> to be:
>> Pfwd - Pref = Pload = ~Psource
>> 200w - 100w = 100w = ~100w
>
> How can that be? If the meter is basically a directional coupler then the
> forward power is just that. Subtracting any reflected power will just give a
> stupid answer.
Subtracting any reflected power will give the power
being delivered to the load. Power reflected from
the load is power that is NOT delivered to the load.
> The only errors will be due to the directivity of the
> coupler, which will give a band of uncertainty which varies with VSWR, and
> the error due to the accuracy of the detectors.
Consider the following example assuming a lossless
tuner and transmission line.
100W source+tuner--x--50 ohm coax-----291.42 ohm load
Assuming the tuner is properly tuned, what forward power
will a Bird read on the 50 ohm coax at point 'x'? What
reflected power? What is the net power being delivered
to the load?
> My Bird does not subtract any reflected power to give a forward power
> reading!! It can't I need to rotate the slug to read reverse power.
You're right. The operator must measure the forward power
and the reflected power and do the subtraction manually.
In the above example, the Bird will read 200w forward and
100w reflected. The operator must subtract those two values
to determine the net power delivered to the load.
This was my guess as to why the Bird was reading 200w in
the original posting. Note that a Bird between the source
and the tuner does NOT read the SWR on the transmission
line between the tuner and the load.
dansawyeror
I think it may be a common mode problem. There is no choke on the
dipole. I will bring it closer to resonance.
- Dan kb0qil
Cecil Moore
> I think it may be a common mode problem. There is no choke on the
> dipole. I will bring it closer to resonance.
Dan, are you measuring that power on the input or
output of the tuner? If at the output of a transceiver
with autotuner, for instance, the forward power is
the sum of the transmitter power and reflected power.
Cecil Moore
> The so called "forward power" and "reflected power" are notional values,
> but not actual power "components". The only power is the average rate at
> which energy passes a point, and it is in one direction or the other.
That statement depends upon the definition of "power"
being used. The IEEE Dictionary has a different
definition of power than does a physics textbook.
The net power is the average rate at which net energy
passes a point. The net power is the difference between
the forward joules/sec and the reflected joules/sec.
Instead of using the word "power", let's switch over to
the dimensions of power, i.e. "joules/second". Those
joules are indeed *actual energy components*.
The so called "forward power" is a forward traveling
EM wave containing energy moving at the speed of light.
There are indeed actual forward joules/sec moving past
a point on the transmission line.
The so called "reflected power" is a rearward traveling
EM wave containing energy moving at the speed of light.
There are indeed actual reflected joules/sec moving past
a point on the transmission line.
Note that an EM wave cannot stand still. According
to the theory of relativity, EM waves always move at
the speed of light (taking VF into account).
Standing waves consist of a forward traveling wave
containing joules/second and a reflected traveling
wave containing joules/second. The joules/second in
those two waves are supplied during the transient
power-on state. During steady-state, that energy has
not yet reached the load. But the total energy contained
in the transmission line during steady-state is exactly
the amount of energy needed to support the forward
traveling wave and the reflected traveling wave. Standing
waves would not be possible without those two real EM
wave energy components traveling in opposite directions.
At power-down, assuming the source is disconnected
from the transmission line, all of the forward wave
energy and reflected wave energy stored in the lossless
transmission line is eventually dissipated in the load.
That happens during a time when the source is supplying
zero energy.
Richard Clark
<dansa...@comcast.net> wrote:
>I think it may be a common mode problem. There is no choke on the
>dipole. I will bring it closer to resonance.
Hi Dan,
One very simple test is to fire up your rig and note the SWR. Patch
in a short, loose length of transmission line to the existing run
(about an eighth wave) and fire up the rig again.
Did the SWR change?
If so, you are being SWR whipped by Common Mode currents.
You can keep introducing different lengths of line until you find 1:1
(unlikely, but perhaps close) and let it go at that. This will be a
single frequency solution, but take care that it does NOT cure Common
Modality. Instead, it adds a vertical component to your transmission
(maybe).
This is not always good as that same transmission line (plus
extension) may run past RF sensitive devices (like intercoms, VCRs,
light dimmers, etc.) that go whacko. It may also fill in the nulls of
your dipole (sometimes good, sometimes bad).
Richard Clark
>I am sorry, but I disagree
Hi Jeff,
Your appologies aside, it is the convention you are disagreeing with.
The injection of such terminology as
>'transmitted' power
is not part of conventional usage in this discussion. The trap here
of inventing terms is that your term would not account for Ohmic loss
as either forward or reverse power in the balance sheet (and this loss
could well be the source of mismatch); and yet this loss would have a
definite impact on what is "transmitted."
A simple instance proves this. Add a 14 Ohm resistor in series at the
feed to a perfect quarterwave radiator. The reverse reading would be
nada, the forward reading would NOT be "transmitted" power. Further,
the conventional usage of terms seen in this thread would still be
accurate.
Jeff
"> Your appologies aside, it is the convention you are disagreeing with.
I think that most engineers would disagree, what you are describing is the
power delivered into the load, both forward and reverse power exist both by
convention and as a real entity.
> The injection of such terminology as
>>'transmitted' power
> is not part of conventional usage in this discussion.
Perhaps that is sloopy wording on my part, I sould have said 'power
transmitted to the load'; I was not implying radiated power.
73
Jeff
Richard Clark
>what you are describing is the
>power delivered into the load,
Hi Jeff,
Perhaps you should re-read your original complaint.
>Perhaps that is sloopy wording on my part
So it would seem. ;-)
Cecil Moore
> Perhaps that is sloopy wording on my part, I sould have said 'power
> transmitted to the load'; I was not implying radiated power.
Indeed, "transmitted power" could simply mean power
from the transmitter, i.e. "source power".
Owen Duffy
news:45e2b588$0$26698$834e...@reader.greatnowhere.com:
Jeff,
You dropped a number of terms here:
- 'transmitted' power;
- power delivered into a mismatched load;
- forward power;
- power delivered by the source;
The power delivered to a load (of any kind) from a lossless transmission
line section, is the same as the power delivered by the source. In the
case of the lossy line, then the line characteristics and load impedance
also need to be taken into account to calculate the power lost in the
line section, and it is not as simple as using up a dB/100' rating in a
table (unless the line is matched).
You assertion that you have travelling forward and reflected power waves
on the transmission line runs into a problem when you try to analyse the
combination of both at a point (eg the input to the line) as power
doesn't combine vectorially.
When you devise configurations with circulators, isolaters, directional
couplers, hybrids etc to "trap and reroute" reflected power, you have
probably changed the nature of the load on a line section and that
accounts for why the reflected power seems to have been isolated from
forward power.
Owen
Jeff
"
> The power delivered to a load (of any kind) from a lossless transmission
> line section, is the same as the power delivered by the source.
So it is your contention that power is not reflected at a mismatch. The wave
certainly is so the power contained in the reflected portion must be as
well.
> You assertion that you have travelling forward and reflected power waves
> on the transmission line runs into a problem when you try to analyse the
> combination of both at a point (eg the input to the line) as power
> doesn't combine vectorially.
I was not trying to analyse the combination of any wave on the line ("power"
waves, whatever they may be, or anything else), I was merely noting that you
can quantify and measure the power contained the both the forward and
reflected waves and they are real quantities.
Jeff
Cecil Moore
> You assertion that you have travelling forward and reflected power waves
> on the transmission line runs into a problem when you try to analyse the
> combination of both at a point (eg the input to the line) as power
> doesn't combine vectorially.
But it does combine according to the following formula
which is the irradiance equation from the field of
optics.
Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)
where 'A' is the angle between V1 and V2 and V1 is
the voltage associated with P1 and V2 is the voltage
associated with P2.
The first time I saw this equation was in Dr. Best's
Nov/Dec 2001 QEX article on Transmissions Lines. It
really does work for "adding" the two powers in two
coherent waves.
Cecil Moore
> I was not trying to analyse the combination of any wave on the line ("power"
> waves, whatever they may be, or anything else), I was merely noting that you
> can quantify and measure the power contained the both the forward and
> reflected waves and they are real quantities.
The joules/sec are real quantities but whether joules/sec
is power depends upon the definition of "power". Some say
the joules/sec in a reflected wave is not power and they
produce a definition of "power" from a physics book to
prove it, i.e. no work done. To satisfy the purists you
may need to change your statement to: "I was merely noting
that you can quantify and measure the joules/sec contained
in both the forward and reflected waves and they are real
quantities."
Owen Duffy
news:i2HEh.19$iw4...@newssvr23.news.prodigy.net:
> Owen Duffy wrote:
>> You assertion that you have travelling forward and reflected power
>> waves on the transmission line runs into a problem when you try to
>> analyse the combination of both at a point (eg the input to the line)
>> as power doesn't combine vectorially.
>
> But it does combine according to the following formula
> which is the irradiance equation from the field of
> optics.
>
> Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)
>
> where 'A' is the angle between V1 and V2 and V1 is
> the voltage associated with P1 and V2 is the voltage
> associated with P2.
Cecil,
A is not a property of P1 or P2, and cannot be derived from them. I
maintain that you cannot vectorially combine P1 and P2.
Owen
Owen Duffy
news:45e33a9f$0$26696$834e...@reader.greatnowhere.com:
>
> "
>> The power delivered to a load (of any kind) from a lossless
>> transmission line section, is the same as the power delivered by the
>> source.
>
> So it is your contention that power is not reflected at a mismatch.
> The wave certainly is so the power contained in the reflected portion
> must be as well.
The danger in the "power is refelected at a mismatch" explanation, is
that it follows that power reflected at a mismatched antenna flows back
toward the transmitter and is at least partially absorbed in the PA as
heat. Though that is a popular belief, it is not supported by fact.
The power at a point in a transmission line is P=real(V*conjugate(I)).
This expands to four terms, and people arbitrarily allocate the terms
forward power and reflected power to just two of the four terms because
they happen to be VfIf and VrIr.
>
>> You assertion that you have travelling forward and reflected power
>> waves on the transmission line runs into a problem when you try to
>> analyse the combination of both at a point (eg the input to the line)
>> as power doesn't combine vectorially.
>
> I was not trying to analyse the combination of any wave on the line
> ("power" waves, whatever they may be, or anything else), I was merely
> noting that you can quantify and measure the power contained the both
> the forward and reflected waves and they are real quantities.
The Bird 43 does not measure power directly, it responds to Vf or Vr
components at a point as explained in the article I quoted. The article
deals with the conditions under which readings can be converted to power,
and whether forward power or reverse power are of themselves meaninful.
If you have read it and disagree, then thats ok. If you can identify
flaws in the article, constructive feedback is welcome.
Owen
Cecil Moore
> A is not a property of P1 or P2, and cannot be derived from them. I
> maintain that you cannot vectorially combine P1 and P2.
P1 is a property of V1^2/Z0, A is a property of V1.
P2 is a property of V2^2/Z0, A is a property of V2.
There is an unbroken chain of cause and effect.
It is true that one cannot directly vectorially
combine P1 and P2 because P1 and P2 are not vectors.
However, the ability to combine the P1 and P2 of
coherent EM waves dates back to before you were born.
Optical engineers didn't have the luxury of being
able to measure the phase angles. All they could
measure was the total amplitude. Please don't try
to tell us that their total amplitude measurements
were wrong throughout the 20th century and are
still wrong in the 21st century.
The rules for combining P1 and P2 when they are
coherent are known as the irradiance equations in
optics. Dr. Best applied them to RF quantities.
Please reference "Optics", by Hecht, 4th edition,
page 388 and Dr. Best's, "Wave Mechanics of Transmission
Lines, Part 3: ..." in the Nov/Dec 2001 issue of "QEX".
Cecil Moore
> The danger in the "power is refelected at a mismatch" explanation, is
> that it follows that power reflected at a mismatched antenna flows back
> toward the transmitter and is at least partially absorbed in the PA as
> heat. Though that is a popular belief, it is not supported by fact.
That only applies to mismatched systems. For systems
Z0-matched by an antenna tuner, the situation becomes
trivial to understand. The reflected energy is re-
reflected by the Z0-match provided by the properly
tuned antenna tuner. It's all explained in my energy
analysis article at: http://www.w5dxp.com/energy.htm
Owen Duffy
@newssvr14.news.prodigy.net:
> Owen Duffy wrote:
>> A is not a property of P1 or P2, and cannot be derived from them. I
>> maintain that you cannot vectorially combine P1 and P2.
>
...
> It is true that one cannot directly vectorially
> combine P1 and P2 because P1 and P2 are not vectors.
Thanks
...
Cecil Moore
>> combine P1 and P2 because P1 and P2 are not vectors.
>
> Thanks
That doesn't mean that there are not valid rules for
combining P1 and P2. Optical engineers have been doing
it for decades. RF engineers seem to lag behind.
You seemed to be questioning the validity of the power
combination equation. Have you changed your mind?
Owen Duffy
Cecil, it seems that between the two of you, you are constructing a
picture that (in a lossless line for simplicity) if the Bird 43 reads
100W forward and 50 watts reflected, the power radiated (ignoring
antenna ohmic losses) is 100W, but 50W is reflected toward the
transmitter... but that's allright because the 50W will be reflected by
a Zo matched PA, and energy is conserved on the line.
The reality is that the Bird responds to Vf and Vr (depending on the
orientation of the slug), and in the special case where the sampler is
calibrated to respond to |Vf| and |Vr| for a purely real ratio of V/I
(Zn=Rn+j0 which is 50+j0 in the case of the '43), on the line at the
point of the sampler, then the average power passing that point is a
single number, it is |Vf|^2/Rn-|Vr|^2/Rn. The foward and reflected power
readings are not meaningful in themselves, but you can deconstruct rho,
and (knowing Zn) |Vf| and |Vr| from them.
Owen
Cecil Moore
> Cecil, it seems that between the two of you, you are constructing a
> picture that (in a lossless line for simplicity) if the Bird 43 reads
> 100W forward and 50 watts reflected, the power radiated (ignoring
> antenna ohmic losses) is 100W, but 50W is reflected toward the
> transmitter... but that's allright because the 50W will be reflected by
> a Zo matched PA, and energy is conserved on the line.
Please don't insult our intelligence. If the Bird reads
100w forward and 50w reflected, the power radiated by
the antenna is 50w, neglecting losses.
Pload = Pfor - Pref
The situation at the output of a lossless tuner is:
Pfor = Psource + Pref
Please honor the conservation of energy principle in your
postings.
Owen Duffy
@newssvr14.news.prodigy.net:
> Owen Duffy wrote:
>> Cecil, it seems that between the two of you, you are constructing a
>> picture that (in a lossless line for simplicity) if the Bird 43 reads
>> 100W forward and 50 watts reflected, the power radiated (ignoring
>> antenna ohmic losses) is 100W, but 50W is reflected toward the
>> transmitter... but that's allright because the 50W will be reflected
by
>> a Zo matched PA, and energy is conserved on the line.
>
> Please don't insult our intelligence. If the Bird reads
Of course it is nonsense, but it is a logical development based on Jeff's
words "What you are describing could be called 'transmitted' power or
power delivered into a mismatched load, but that it different from
forward power, or the power delivered by the source" and your words "For
systems
Z0-matched by an antenna tuner, the situation becomes trivial to
understand. The reflected energy is re-reflected by the Z0-match provided
by the properly tuned antenna tuner".
The steady state solution of the ratio of V/I at the input to a real
transmission line section (ie lossy) that has a load of some arbitrary
impedance is solved using equations that do not include forward and
reflected power terms. The power at any point is the real part of
V*conjugate(I) at that point, no matter whether it is at the source,
load, line input, output or anywhere along the line.
Owen
Richard Clark
>The danger in the "power is refelected at a mismatch" explanation, is
>that it follows that power reflected at a mismatched antenna flows back
>toward the transmitter and is at least partially absorbed in the PA as
>heat. Though that is a popular belief, it is not supported by fact.
Hi Owen,
This denies experience.
Cecil Moore
> Of course it is nonsense, but it is a logical development based on Jeff's
Please don't blame me for someone else's words.
The irradiance (power) equations from optical engineering
for the combining of electromagnetic power from two EM waves
are accepted as a fact of physics. Do you really think that
the amateur radio community should reject that fact
of physics?
Owen Duffy
news:a3q6u258ajcae732i...@4ax.com:
Do you mean the experience that the PA may (but not necessarily) run
hotter on other than its rated load impedance?
Could that be exlained without introducing transmission lines and the
reflected power issue by examining how the PA works on a load different
to its rated load?
I suggest that if a PA / line / load situation transforms the actual load
to some arbitrary impedance Z at the PA end of the line, the PA will
peform exactly as if the PA were directly loaded by a lumped constant
load of Z. The explanation of any heating effects or changed operating
voltages is in the effect of the load impedance Z (however it is
obtained) on the PA.
Does that make sense?
Owen
Cecil Moore
> I suggest that if a PA / line / load situation transforms the actual load
> to some arbitrary impedance Z at the PA end of the line, the PA will
> peform exactly as if the PA were directly loaded by a lumped constant
> load of Z.
Yes, that is true for the performance of the PA.
Certainly not true for the performance of the
transmission line or antenna.
My favorite quotation by an antenna guru on this
newsgroup is that "a 50 ohm antenna can be replaced
by a 50 ohm resistor without changing anything".
If that were true, we don't need antennas. :-)
I'm going to be away from my computer for 48 hours.
Owen Duffy
> Owen Duffy wrote:
>> Of course it is nonsense, but it is a logical development based on
>> Jeff's words ...
>
> Please don't blame me for someone else's words.
A selective partial quotation to misrepesent what was actually written
Cecil!
Owen
Owen Duffy
> Owen Duffy wrote:
>> I suggest that if a PA / line / load situation transforms the actual
>> load to some arbitrary impedance Z at the PA end of the line, the PA
>> will peform exactly as if the PA were directly loaded by a lumped
>> constant load of Z.
>
> Yes, that is true for the performance of the PA.
> Certainly not true for the performance of the
> transmission line or antenna.
>
Please explain?
Owen
Richard Clark
>Could that be exlained without introducing transmission lines and the
>reflected power issue by examining how the PA works on a load different
>to its rated load?
Hi Owen,
Let's treat this like the Chinese Box problem.
If you didn't know what the load was, could you explain it any
differently? No. Apriori knowledge is not a proof.
Owen Duffy
Richard, I content that:
- the power output of the PA; and
- the efficiency of the PA may be (and usually are) sensitive to the load
impedance.
A steady state analysis is usually adequate for most ham radio
applications, though there may be cases where establishment of steady
state brings its own issues. This discussion is about the steady state.
Though it is often asserted that the PA will get hotter as a result of
"reflected power" being dissipated in the dynamic output impedance of the
PA, and that this may / will damage the PA, the power explanation doesn't
work numerically in the general case.
The constrained ratio of V/I is an aspect of the field setup in a
transmission line, and the launching of a reflected wave to reconcile the
load V/I with the constrained (Vf+Vr)/If-Ir) at the load end of the line
is a solution for what happens on the transmission line. At the source
end of the line, (Vf+Vr)/If-Ir) at that point (a function of (Vf+Vr)/If-
Ir at the load end, and the complex propagation coefficient and Zo of the
line) give us the equivalent (complex) impedance seen by the PA, and we
can predict / explain the behaviour of the PA (maximum power out,
efficiency) on that equivalent load. This method of analysis does work
numerically.
Owen
Richard Clark
>Richard Clark <kb7...@comcast.net> wrote in
>> Let's treat this like the Chinese Box problem.
>>
>> If you didn't know what the load was, could you explain it any
>> differently? No. Apriori knowledge is not a proof.
>
>Richard, I content that:
Contend or offer in contention.
>- the power output of the PA; and
>- the efficiency of the PA may be (and usually are) sensitive to the load
>impedance.
This is not contending nor contention and is content only for a non
sequitur. The line following a tuner exhibits considerable loss (poor
efficiency) that can only occur on the basis of power and mismatch.
You yourself offered in other correspondence that it exceeds cable
attenuation specifications found only in a matching condition. To
suggest that a PA's sensitivity is somehow exhalted in the face of
identical, ordinary behavior of a passive component is hardly
seperable. Consider the simple substitution to your quote:
>- the power output at the terminus of the line; and
>- the efficiency at the terminus of the line may be (and usually are) sensitive to the load
>impedance.
continuing on...
>A steady state analysis is usually adequate for most ham radio
>applications, though there may be cases where establishment of steady
>state brings its own issues. This discussion is about the steady state.
I expressed nothing of transitory behavior.
>Though it is often asserted that the PA will get hotter as a result of
>"reflected power" being dissipated in the dynamic output impedance of the
>PA, and that this may / will damage the PA, the power explanation doesn't
>work numerically in the general case.
Heat is the outward proof of power and is always demonstrable in both
specific and general cases. Occurrences of other, significant
radiation from the source (as long as that source physically occupies
a substantially minor region of wavelength) is exceedingly difficult
to achieve.
You don't offer a numerical proof of a general case, and given that
the general case must allow for the specific cases already allowed in
your discussion above - that may be an untenable assertion for you.
Those specific cases are demonstrably caloric and must follow the same
math you suggest.
I suspect you are trying to argue differences by degree (no pun
intended as to heat); but I seriously doubt you can produce the math
to do that. The arguments that flow from that involve what is called
source resistance, and those arguments are legion in this forum (where
naysayers embrace a refusal to accept or name ANY value - a curious
paradox and an engineering nihilism I enjoy to watch).
Jeff
>
> Of course it is nonsense, but it is a logical development based on Jeff's
> words "What you are describing could be called 'transmitted' power or
> power delivered into a mismatched load, but that it different from
> forward power, or the power delivered by the source" and your words "For
> systems
Only a logical development if you selectively snip Owen.
"What you are describing could be called 'transmitted' power or
power delivered into a mismatched load" was referring to "Pload = Pfor -
Pref".
Ok I admit that 'transmitted' power could have been better phrased.
Power may not actually be dissipated in a lossless line but that does not
detract from the fact that there is current flow and a voltage along the
line produced by two distinct and independent waves travelling in opposite
directions. True that the power can only be realised when it encounters a
load, but it is highly pedantic not to regard the reflected signal as having
'power' until it actually encounters such a load. If you extend this theory
to a radiated signal, you could equally say that there is no power
travelling through the aether until it encounters a receiver.
It is naive to believe that reflected power is not dissipated in a matched
source, or partially re-reflected at the source/line interface is not
matched. Again going back to the optics corollary you would not expect a
reflected light signal not to impinge on, and interact with a source.
If you pad out a source with a sufficiently high attenuator such that the
reflected signal will not have significant effect on it, you will see an
increase in dissipation in the attenuators when the load is mis-matched. I
am confident that an attenuator is not having its "load line" changed such
that its dissipation goes up magically just by the same amount as the power
in the reflected wave!! (Of course the dissipation in the load is measurable
as heat).
Adding a circulator to a system will not change "the load line" (if a
transmission line or circulator can have such a thing), but it will cause
the power in the reflected wave to be separated so that it can be monitored
and measured. Surprisingly power monitored in this way ties up with the
notion that power is reflected at a mis-matched load.
73
Jeff
Ian White GM3SEK
>
>My favorite quotation by an antenna guru on this
>newsgroup is that "a 50 ohm antenna can be replaced
>by a 50 ohm resistor without changing anything".
>If that were true, we don't need antennas. :-)
>
That sounds like a direct misquotation of me.
What I HAVE said - and often - is that if you measure the load impedance
presented by an antenna and feedline at the output socket of the
transmitter, and replace it by the same impedance made from lumped R and
L/C components, then the steady-state operating conditions of the
transmitter will not change.
If the transmitter isn't touched, it will deliver exactly the same power
as before - because that happens to be how much power it can deliver
into that particular load impedance.
That's all the RF power there is. In a lossless system, all of that
power will be radiated from the antenna. With the alternative lumped
load, exactly the same power will be delivered into the resistive part
of the load, and dissipated as heat.
Of course the transmitter is under more stress from voltage, current and
heat when it's operating into an incorrect load impedance (not what it
was designed for) but that's all it is. There is no need to invent
reflected power that is being "dumped" back into the transmitter to
cause this stress.
There is also a strong tendency to invent virtual instruments such as
"directional wattmeters" which do not actually exist. An instrument such
as the Bird 43 is calibrated in watts, but it doesn't actually sample
power. As Owen relates (and so have I) these instrument only sample V
and I on the line - they categorically DO NOT sample power.
The power scale is only a meter calibration - literally, only ink on a
meter scale. It indicates the amount of power delivered into a matched
load, when the "reverse" reading is exactly zero [1]. The instrument was
calibrated under those specific conditions, and the "forward power"
reading is only physically meaningful for that specific case.
For a mismatched load, the meter will read higher in the forward
direction than in the reverse - but that is purely a feature of the
instrument. It all looks so plausible on the meter scale, but those are
not genuine power waves flowing in opposite directions.
Everything that's happening inside the instrument can be completely
explained from the new V and I conditions on the line. Waves of power
simply don't come into it. Most people don't want to go that deep into
the theory... but, regrettably, that may be the only way to understand
that the "power" readings on the meter scale are no longer valid under
these conditions.
What IS physically meaningful is the DIFFERENCE between the forward and
reverse "power" readings. That difference will equal the net power
delivered to the load [1]. But those two readings are only meaningful as
a pair - individually they are only "intermediate results" with no
physical meaning of their own.
[1] Ignoring real-life meter errors such as directivity and scale
accuracy.
>I'm going to be away from my computer for 48 hours.
But you'll be back... :-)
--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek
Cecil Moore
> Cecil Moore <myc...@hotmail.com> wrote innews:jVJEh.2385$m85...@newssvr29.news.prodigy.net:
> > Please don't blame me for someone else's words.
>
> A selective partial quotation to misrepesent what was actually written Cecil!
My true statement was combined with someone else's false statement to
make the combination false. I'm asking politely that not be done
again.
--
73, Cecil, w5dxp.com
Cecil Moore
> Cecil Moore <myc...@hotmail.com> wrote innews:1lKEh.2388$m85....@newssvr29.news.prodigy.net:
If the transmission line and antenna are replaced by a lumped constant
load, the transmission line and antenna cease to function. IMO, that's
not "performing exactly as".
--
73, Cecil, w5dxp.com
Richard Clark
<gm3...@ifwtech.co.uk> wrote:
>As Owen relates (and so have I) these instrument only sample V
>and I on the line - they categorically DO NOT sample power.
Hi Ian,
We've been through this before. No instrument operates in the absence
of power. Simply because you and Owen are graced with instruments
that demand so little, does not negate what power they do rob from
what is available. Even the humble electrometer has to overcome the
force of gravity to open its foil leaves, and climbing that potential
energy hill is work over time - power.
Hammer down the directivity as much as you want, and it will still
resolve to some diminution of power available to the load. To wave a
hand and say NOTHING does not make it so.
Cecil Moore
> Cecil Moore wrote:
> >My favorite quotation by an antenna guru on this
> >newsgroup is that "a 50 ohm antenna can be replaced
> >by a 50 ohm resistor without changing anything".
> >If that were true, we don't need antennas. :-)
>
> That sounds like a direct misquotation of me.
Nope, it wasn't you, Ian. You are usually more careful than that.
> For a mismatched load, the meter will read higher in the forward
> direction than in the reverse - but that is purely a feature of the
> instrument. It all looks so plausible on the meter scale, but those are
> not genuine power waves flowing in opposite directions.
But they are genuine energy waves flowing in opposite directions.
Standing waves require two coherent energy waves flowing in opposite
directions. Can you explain how to create a standing wave without
two energy waves flowing in opposite directions? And remember, the
two EM wave components in the standing wave cannot stand still.
There's truth in what you say, Ian, but it is not the whole truth.
There's
no difference except frequency (and all that frequency implies)
between
an RF electromagnetic wave and a visible light electromagnetic wave.
In fact, RF waves are covered in many physics textbooks whose subject
is light. There is a wealth of information available from the field of
optics
that is applicable to RF waves.
Visible light physicists don't have the luxury of measuring voltage or
directly measuring phase. They have to rely on a power measurement
of irradiance. As a result, visible light measurements are actually
power measurements so we indeed do know how EM waves behave
at the joules/second level.
Visible light physicists found that when they superpose two coherent
light waves, Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A)
where 'A' is the angle between the electric fields of the two waves.
That exact same equation applies to coherent RF waves. Phasor
addition is used for the superposition of two coherent RF voltages.
The power equation is used to find out what happens to the power
during that voltage superposition. P1 = V1^2*Z0 and P2 = V2^2*Z0
The last term in the power equation is known as the interference term
and is either constructive, destructive, or zero.
Since antenna radiation patterns depend upon constructive and
destructive interference of EM waves in space, we hams could
learn a lot from the field of physics known as optics.
> >I'm going to be away from my computer for 48 hours.
> But you'll be back... :-)
Yep, I'm posting from my sister's computer through my Google account.
--
73, Cecil, w5dxp.com
Cecil Moore
> Adding a circulator to a system will not change "the load line" (if a
> transmission line or circulator can have such a thing), but it will cause
> the power in the reflected wave to be separated so that it can be monitored
> and measured. Surprisingly power monitored in this way ties up with the
> notion that power is reflected at a mis-matched load.
Yes, and a little modulation added to the source signal will prove
that the
signal being dissipated by the circulator resistor has made a round
trip
to the load and back. That's hard to explain if reflected energy
doesn't
actually exist.
--
73, Cecil, w5dxp.com
Ian White GM3SEK
Richard, that argument is otiose.
What I said (in full context) was that their principle of operation as
measuring instruments does not involve sampling traveling waves of power
from the line.
Of course they must incidentally consume some power to move the meter
needle, but that is not part of their operating principle.
For that matter, almost all measuring instruments abstract some energy
or power from whatever they are measuring - but that is usually
incidental. It certainly does not make every instrument into a power
meter.
Can you not see this?
>To wave a
>hand and say NOTHING does not make it so.
To wave another hand and say traveling waves of power exist does not
make that so, either.
Jeff
"Cecil Moore" <w5...@hotmail.com> wrote in message
news:1172593889.5...@s48g2000cws.googlegroups.com...
Indeed; TDR's would have areally hard time (;-))
Jeff
>
Owen Duffy
> On Tue, 27 Feb 2007 01:07:12 GMT, Owen Duffy <no...@no.where> wrote:
>
>>Richard Clark <kb7...@comcast.net> wrote in
>>> Let's treat this like the Chinese Box problem.
>>>
>>> If you didn't know what the load was, could you explain it any
>>> differently? No. Apriori knowledge is not a proof.
>>
>>Richard, I content that:
>
> Contend or offer in contention.
Richard
Yes, my spelling mistake.
>
>>- the power output of the PA; and
>>- the efficiency of the PA may be (and usually are) sensitive to the
>>load impedance.
>
> This is not contending nor contention and is content only for a non
> sequitur. The line following a tuner exhibits considerable loss (poor
> efficiency) that can only occur on the basis of power and mismatch.
> You yourself offered in other correspondence that it exceeds cable
> attenuation specifications found only in a matching condition. To
I am being picky, but "it *may* exceed cable attenuation specifications
found only in a matching condition, it may also be lower". If I said it
as you stated, I made an error. The common statement (and I have no doubt
made it) that VSWR exacerbates line loss is actually wrong in the general
case. (Having Googled my own web site I see one statement along those
lines which needs further qualification!)
> suggest that a PA's sensitivity is somehow exhalted in the face of
> identical, ordinary behavior of a passive component is hardly
> seperable. Consider the simple substitution to your quote:
>>- the power output at the terminus of the line; and
>>- the efficiency at the terminus of the line may be (and usually are)
>>sensitive to the load impedance.
I meant the output at the PA terminals where an lumped constant load
would be attached for comparison.
>
...
>>Though it is often asserted that the PA will get hotter as a result of
>>"reflected power" being dissipated in the dynamic output impedance of
>>the PA, and that this may / will damage the PA, the power explanation
>>doesn't work numerically in the general case.
>
> Heat is the outward proof of power and is always demonstrable in both
> specific and general cases. Occurrences of other, significant
> radiation from the source (as long as that source physically occupies
> a substantially minor region of wavelength) is exceedingly difficult
> to achieve.
>
> You don't offer a numerical proof of a general case, and given that
> the general case must allow for the specific cases already allowed in
> your discussion above - that may be an untenable assertion for you.
> Those specific cases are demonstrably caloric and must follow the same
> math you suggest.
>
> I suspect you are trying to argue differences by degree (no pun
> intended as to heat); but I seriously doubt you can produce the math
> to do that. The arguments that flow from that involve what is called
> source resistance, and those arguments are legion in this forum (where
> naysayers embrace a refusal to accept or name ANY value - a curious
> paradox and an engineering nihilism I enjoy to watch).
PAs can be designed to behave as an equivalent fixed voltage or current
source with fixed source impedance of Zo, but HF PAs are not usually
designed in that way.
I know that there is a vein of thought that the process of adjusting a PA
for maximum output always, somewhat magically, creates a match condition
where the source impedance is the conjugate of the load at the PA
terminals, but it is contentious. What of broadband PA designs with no
such adjustment, are they source matched over a broad range of
frequencies? Observations are that experiments to discover the source
impedance by incrementally changing load current can produce a range of
values for the same PA on different frequencies, and at different power
levels. Why do amplifiers with say tetrodes and triodes which exhibit
such different dynamic plate resistance but requiring the same load
impedance deliver the same equivalent source impedance?
I am also aware that supporters of the inherent source match position
assert that you must be selective in choosing tests for source impedance.
It is all rather unconvincing when only some of the implications of a
particular source impedance are effective.
It is my view that modelling the PA as a fixed voltage or current source
with fixed source impedance of Zo, and where reflected waves on a
transmission line are absorbed by the matched source is not a good
general model for HF PAs.
The application of small signal analysis to amplifiers that sweep from
near cutoff to near saturation is suspect.
I believe that it is sound (in the steady state) to resolve the forward
and reflected wave voltages and currents at the source end of the
transmission line, calculate the complex impedance, and predict the
effects of that impedance as a PA load using the same techniques that
were used to design the PA.
Owen
Richard Clark
<gm3...@ifwtech.co.uk> wrote:
>For that matter, almost all measuring instruments abstract some energy
>or power from whatever they are measuring - but that is usually
>incidental. It certainly does not make every instrument into a power
>meter.
>
>Can you not see this?
Hi Ian,
As a trained metrologist, I've seen it far closer up than you. I've
accurately measured the physics of technology out to more decimal
places than most, and there is nothing incidental about it - it is
quite fundamental. In fact I've encountered a spectrum of conflicting
and interfering physical principles to every measurement that had to
be accounted for to obtain that accuracy.
I dare say none here can equal the accuracy of my bench work at RF
Power measurement here, even those with a shelf full of digital meters
used as bookends for their library.
>>To wave a
>>hand and say NOTHING does not make it so.
>
>To wave another hand and say traveling waves of power exist does not
>make that so, either.
Your counter is not an argument for their inexistence.
The examples of "traveling waves of power" abound, even to the trade
craft of antennas. Examples of "traveling wave of power" are classic
in whole to the explanation for the operation of the standard
Directional Couplers (to distinguish them from the Bruene design).
There are more such examples, but the Directional Coupler is quite
suitable to their class, and its operation satisfies the proof of
power flow in both directions. There are other examples outside of
this class that also provide proofs; and they are based on physical
principles as well (polarization and magnetic moment).
Traditionally, these points pass in silence until some later time when
we visit them once again.
Owen Duffy
...
> We've been through this before. No instrument operates in the absence
> of power. Simply because you and Owen are graced with instruments
> that demand so little, does not negate what power they do rob from
> what is available. Even the humble electrometer has to overcome the
> force of gravity to open its foil leaves, and climbing that potential
> energy hill is work over time - power.
>
> Hammer down the directivity as much as you want, and it will still
> resolve to some diminution of power available to the load. To wave a
> hand and say NOTHING does not make it so.
The fact that an instrument may consume power from the circuit does not
imply that it measures power.
For example, a "voltmeter" that samples the voltage and draws a small
current from the cicuit under test does consume some power, but it cannot
"measure" power in the circuit under test without knowledge of the
complex load impedance.
I wrote an analysis of the Breune circuit and show that the meter
deflection is a response to the Vf or Vr component, and I dealt with how
that information can be used (including applicable conditions). If you
(or others) think there are flaws in the article, I welcome constructive
feedback.
Owen
Richard Clark
>> This is not contending nor contention and is content only for a non
>> sequitur. The line following a tuner exhibits considerable loss (poor
>> efficiency) that can only occur on the basis of power and mismatch.
>> You yourself offered in other correspondence that it exceeds cable
>> attenuation specifications found only in a matching condition. To
>
>I am being picky, but "it *may* exceed cable attenuation specifications
>found only in a matching condition, it may also be lower".
Hi Owen,
Lower? That is rather astonishing in light of responding to my
comment.
>If I said it
>as you stated, I made an error. The common statement (and I have no doubt
>made it) that VSWR exacerbates line loss is actually wrong in the general
>case. (Having Googled my own web site I see one statement along those
>lines which needs further qualification!)
This is even more astonishing. Irrespective of you being the source,
why inject this confusing comment? SWR always exacerbates line loss!
Give me any normal line attenuation and SWR at the load, and I will
tell you exactly how much additional loss will occur. There's a
general solution for you.
Something tells me that your comments are based on a confusion between
the power loss of a cable, and its mismatch loss. They are not the
same thing although they are usually tightly twined in discussion. You
later exhibit a confusion between a conjugate match and an impedance
match. They are not the same thing either. The confusion on both
these points have abounded in this group in past "debates."
>I meant the output at the PA terminals where an lumped constant load
>would be attached for comparison.
This then removes the reflection from the argument, doesn't it? It
actually doesn't; but this unwarranted substitution is like Zen
Archery in that the line already demonstrates the validity of
reflected power as distinct from that "power" just being a
mathematical fiction.
Putting the lumped load at the PA terminal merely casts the proof back
into the box, it doesn't negate reflected power. As the proof is
already supported in the line, then removing it is not strictly a
valid counter argument. However, we will explore it further:
>PAs can be designed to behave as an equivalent fixed voltage or current
>source with fixed source impedance of Zo, but HF PAs are not usually
>designed in that way.
OK so we are now in my sidebar of source resistance. Even so, it has
nothing to do with the concept of reflected power except insofar as
that resistance's ability to reveal that power's dissipation.
Other's should ponder how the reflected power has a caloric proof in
the line, and then question why it wouldn't prove out when it arrives
back in the box where the temperature rises on its return.
Same source, same power, same reflection, same loss. The only thing
that varies is the capacity of any point along this signal chain to
support that heat burden. Let's skip these as choices of design.
>I know that there is a vein of thought that the process of adjusting a PA
>for maximum output always, somewhat magically, creates a match condition
>where the source impedance is the conjugate of the load at the PA
>terminals, but it is contentious.
That contention arises out of mistaking Z0 Matches with Conjugate
Matches. This is a common affliction among "debaters" here. Let's
skip their prejudices.
>What of broadband PA designs with no
>such adjustment, are they source matched over a broad range of
>frequencies?
Having had designed broadband amps, this is simply accomplished with
the proper feedback such that, yes, they are matched over a broad
range. The math is quite simple, the cost is another matter. Can you
afford one? Probably not. The lack of commercial examples available
to the Ham is not proof they do not exist. Let's pass on from issues
of economy.
On the other side of the aisle, I've worked with active loads that
will absorb as much power (up to a limit) at any frequency (up to a
limit) that you care to throw at it.
>Observations are that experiments to discover the source
>impedance by incrementally changing load current can produce a range of
>values for the same PA on different frequencies, and at different power
>levels.
This is called "Load Pulling," and is a classic technique to
demonstrate source Z. Thevenin first described it and Norton followed
suit. I cannot, for the life of me, recall any other intellectual
giants that have pulled these apart.
I have done this with my own gear. The variation from a source Z of
50 Ohms wandered the SWR range of 1.5:1 over all bands and most power
levels. Given this conformed to the manufacturer's specification, I
was not particularly surprised. Where it deviated the most, the rig
also operated the worst. What can we say about experience and
performance design converging?
>Why do amplifiers with say tetrodes and triodes which exhibit
>such different dynamic plate resistance but requiring the same load
>impedance deliver the same equivalent source impedance?
A cable connector instead of binding posts? Let's dismiss this as
being obvious.
>I am also aware that supporters of the inherent source match position
>assert that you must be selective in choosing tests for source impedance.
>It is all rather unconvincing when only some of the implications of a
>particular source impedance are effective.
Where is the rig specified to exhibit this condition?
Is your rig a VW that stalls trying to pull a trailer from a stop in
3rd gear? Or is it a Mack truck trying to park in the handicap zone
in an underground mall parking lot? Arguing other's incapabilities is
something I like doing, but with more flair. Let's skip these
Tritonic minnows.
>It is my view that modelling the PA as a fixed voltage or current source
>with fixed source impedance of Zo, and where reflected waves on a
>transmission line are absorbed by the matched source is not a good
>general model for HF PAs.
You have already said as much. I see nothing new so far.
>The application of small signal analysis to amplifiers that sweep from
>near cutoff to near saturation is suspect.
If it is near cutoff or saturation, it is suspect small signal
analysis. Certainly, anyone can conspire to fail gracelessly. Would
you care to elaborate the suspicion beyond the evidence of gross
negligence? Why don't we skip this minor excursion?
>I believe that it is sound (in the steady state) to resolve the forward
>and reflected wave voltages and currents at the source end of the
>transmission line, calculate the complex impedance, and predict the
>effects of that impedance as a PA load using the same techniques that
>were used to design the PA.
Sound though it may be, if I were to line up another transmitter
boresight down the antenna connector of the first, light it up to
provide power with no equivocation of it being fictional; then yes,
all things may appear to be the same. ...and yet I have just
demonstrated reverse power arriving at the antenna terminal (where did
it go?). My having experience in doing just this (aka active load
already described above) fully conforms to your sound idea, and yet,
as for myself, it is not an idea I would rely on to deny the existence
of reverse power nor its capacity to fry the innards of a transmitter
(active loads are heavily heat-sinked and fan driven).
Richard Clark
>The fact that an instrument may consume power from the circuit does not
>imply that it measures power.
Hi Owen,
If that power is from a reflection, then we've come to the logical
conclusion of the argument.
Jim Kelley
Cecil Moore wrote:
> The joules/sec are real quantities but whether joules/sec
> is power depends upon the definition of "power".
In our case here on the internet, it depends on whether or not you
choose to equate 'units of power' with the definition of power.
73 ac6xg
Owen Duffy
Just to deal with this issue, line loss under VSWR, which at first seems
a side issue, but it illustrates one of the problems of a "power
perspective" in analysing a transmission line.
By "line loss" I mean the ratio of power at the load end of the line to
power at the source end of the line, not "forward power" or "reflected
power", but the average rate of flow of energy at those points.
So, to your challenge:
The problem is 1m of Belden 8262 (RG58C/U type) at 3.5MHz with three
loads, 50+j0, 5+j0, and 500+j0.
The loss for 50+j0 load is 0.025dB (equivalent to the Matched Line Loss).
What Line Loss to you get for the other two cases?
(I make it 0.24dB for 5+j0, and 0.014 for 500+j0.)
Owen
Richard Clark
>By "line loss" I mean the ratio of power at the load end of the line to
>power at the source end of the line, not "forward power" or "reflected
>power", but the average rate of flow of energy at those points.
Hi Owen,
What's wrong with conventional terms so that we BOTH know what you
mean? The convention would call this Mismatch Loss. If you dispute
this, then it serves my complaint. Further, convention has no
interest in "forward power" nor "reflected power" except as expressed
as SWR. I thought I was quite terse in this regard.
>So, to your challenge:
>The problem is 1m of Belden 8262 (RG58C/U type) at 3.5MHz with three
>loads, 50+j0, 5+j0, and 500+j0.
Well, as I've pointed out, it is not strictly in the terms of my
challenge, is it?
>The loss for 50+j0 load is 0.025dB (equivalent to the Matched Line Loss).
Sigh... parentheticals?
>What Line Loss to you get for the other two cases?
>
>(I make it 0.24dB for 5+j0, and 0.014 for 500+j0.)
An additional 0.1dB However, this example strains the utility of the
challenge.
Let's try a perverse challenge.
Presume a source of 100+j0 Ohms impedance sees a 50 Ohm line that is
5.35 wavelengths long and is terminated with a load of 200+j0 Ohms.
The normal attenuation of the line is 2.00 dB. What is the loss in
the line?
Can your general solution solve this? It uniquely describes both the
kinetics of reverse power flow AND the impact of source resistance.
No one has every answered this one correctly, by the way (and I can
anticipate you are ready to spring that observation on me with your
source feeding essentialy a voltage oriented high Z load as opposed to
the current oriented low Z load).
Richard Clark
wrote:
>Presume a source of 100+j0 Ohms impedance sees a 50 Ohm line that is
>5.35 wavelengths long and is terminated with a load of 200+j0 Ohms.
>The normal attenuation of the line is 2.00 dB. What is the loss in
>the line?
For the others,
Any who complain about their transmitter having:
1. No source resistance;
2. Not this much resistance:
3. Not this little resistance;
4. None of the above (the usual response).
can take heart that if you simply substitute a tuner which presents
the equivalent SWR at the plane of the input to the line, then you can
progress to the solution with equal fluidity (which is to say like
molasses in December).
This particular example has been around for at least 40 years if not
since WWII. No one has rushed to answer it here in at least a quarter
of that time, I don't expect a cascade of guesses soon either.
Owen Duffy
> On Wed, 28 Feb 2007 01:01:55 GMT, Owen Duffy <no...@no.where> wrote:
>
>>By "line loss" I mean the ratio of power at the load end of the line to
>>power at the source end of the line, not "forward power" or "reflected
>>power", but the average rate of flow of energy at those points.
>
> Hi Owen,
>
> What's wrong with conventional terms so that we BOTH know what you
> mean? The convention would call this Mismatch Loss. If you dispute
> this, then it serves my complaint. Further, convention has no
> interest in "forward power" nor "reflected power" except as expressed
> as SWR. I thought I was quite terse in this regard.
>
>>So, to your challenge:
>>The problem is 1m of Belden 8262 (RG58C/U type) at 3.5MHz with three
>>loads, 50+j0, 5+j0, and 500+j0.
>
> Well, as I've pointed out, it is not strictly in the terms of my
> challenge, is it?
Richard,
Your challenge was "Give me any normal line attenuation and SWR at the
load, and I will tell you exactly how much additional loss will occur."
I didn't state the VSWR, but it is 10:1 in both cases. The "normal line
attenation" you refer to is I expect the Matched Line Loss" which I have
given you.
>
>>The loss for 50+j0 load is 0.025dB (equivalent to the Matched Line
Loss).
>
> Sigh... parentheticals?
>
>>What Line Loss to you get for the other two cases?
>>
>>(I make it 0.24dB for 5+j0, and 0.014 for 500+j0.)
>
> An additional 0.1dB However, this example strains the utility of the
> challenge.
Is that your answer, an additional 0.1db due to the 10:1 VSWR? We do not
agree on either answer.
BTW, my figures were not additional loss, but total Line Loss as I
defined it.
You will note that my calculation for the 5+j0 case is less than the
Matched Line Loss, not higher.
In practical transmission lines, most of the loss is in current flowing
in the R component of an RLGC equivalent of the line, the loss in the
copper conductors forming the line. For VSWR>1, the net current varies
along the line forming the classic standing wave pattern, and the loss in
incremental lengths of the line varies approximately with the square of
current in that increment.
So in the two cases above, even though the load VSWR is the same, the
loss is quite different due to the different current distribution in both
cases, one is near a current maximum, and the other is near a current
minimum. Any adjustment of Matched Line Loss for VSWR>1 using only the
VSWR cannot take the location of the standing wave pattern into account,
and is an inaccurate approximation in some situations.
Many books showing a VSWR based formula for "additional loss due to
VSWR" don't spell out the assumptions underlying the formula. Phillip
Smith does in his book "Electronic Applications of the Smith Chart", he
says "If a waveguide is one or more wavelengths long, the average loss
due to standing waves in a region extending plus or minus a half
wavelength from the point of observation may be expressed as a
coefficient or factor of the one way transmission loss per unit length."
and he gives the ratio as (1+S^2)/(2*S). Though the ARRL shows graphs and
formulas they don't always (if ever) spell out the assmptions.
So, yes I assert that the Line Loss under mismatch conditions may be less
than the Matched Line Loss.
Owen
PS: I calculated my answers using http://www.vk1od.net/tl/tllc.php ,
Dan's TLDETAILS.EXE and ARRL's TLW3.EXE give similar results.
Owen Duffy
Richard,
If you read and understood the article, you would see that the instrument
is based on sampling the V/I ratio at a point, and that being surrounded by
transmission line is not important to the principle of operation, in other
words, it does not directly measure a reflected wave.
Owen
Richard Clark
>You will note that my calculation for the 5+j0 case is less than the
>Matched Line Loss, not higher.
Hi Owen,
I already anticipated that, didn't I? Certainly parsing my last
discussion is hardly necessary.
I note you have no solution that answers for the loss in a fairly
typical instance for a fairly typical line condition. It couldn't
have been any more difficult than your former computations, could it?
(In fact it is, but not conceptually.) And yet the absence of that
effort is notable (OK, so you've been ambushed). I may have stumbled
on a novelty application but I didn't trip over a boulder of a common
usage.
Given this sub-thread flowed from my response that a source does
dissipate a reverse power flow (both of which, the direction and
dissipation, are held in contention); and further given my "perverse"
challenge fully specifies such a condition and has a real solution, it
stands to reason that if your general computation is in fact general,
then it can resolve the contention to one or both of our satisfaction.
All it requires is that your math treatment accepts both directions of
power flow, and loss in the source. This is not unreasonable,
especially when any number of references encompass just such concepts.
Richard Clark
>If you read and understood the article, you would see that the instrument
>is based on sampling the V/I ratio at a point, and that being surrounded by
>transmission line is not important to the principle of operation, in other
>words, it does not directly measure a reflected wave.
Hi Owen,
Can you express it without the presumption and still carry the
argument? I have read and understood many treatments on the topic,
and none claim to be the sole and unadulterated truth as it is
generally understood that many analyses work simultaneously and none
deny the validity of the others.
You have not yet actually offered any treatment that denies the bone
of contention that lies in two subject lines:
1. Reverse power is manifest;
2. The source will absorb and dissipate it.
You may have struggled with others over this in times past, but by
your own descriptions they had little intellectual horsepower, and
less experience in the matter.
I have attended schooling specific to these issues, and have practiced
professionally in their measure to the highest of standards. My peers
have instituted national metrology laboratories in your half of the
planet (OK, so it was Korea).
I have measured SWR with Bruene designs (as vulgar as that is);
Directional Couplers, Slotted Lines; and power with half a dozen
different style of sensors, and as many different methodologies. I
have also calibrated these instruments (all of them including the
vulgar Bruene designs). I can separate out the constituent waves (in
spite of the denial of their existence) by several means - each
appropriate to the problem at hand. I can measure excessively high
SWR precisely where others would shrug and simply call it infinite (it
isn't). I can also reduce residual SWR (anyone know what that is?).
I've done this over a spectrum from nearly D.C. to 12 GHz. And I get
a chuckle out of a claim for 0.014dB loss when I know full well
through experience it is unverifiable, unmeasurable, and hence
unproveable except in a spread sheet as a statistical curiosity.
Owen Duffy
...
> I note you have no solution that answers for the loss in a fairly
> typical instance for a fairly typical line condition. It couldn't
> have been any more difficult than your former computations, could it?
> (In fact it is, but not conceptually.) And yet the absence of that
> effort is notable (OK, so you've been ambushed). I may have stumbled
> on a novelty application but I didn't trip over a boulder of a common
> usage.
...
Richard,
No, it is just that I have not posted a solution as yet.
Your problem was described as "Presume a source of 100+j0 Ohms impedance
sees a 50 Ohm line that is 5.35 wavelengths long and is terminated with a
load of 200+j0 Ohms. The normal attenuation of the line is 2.00 dB. What
is the loss in the line?"
Rather than sit down an write a bunch of stuff to calculate it, I took
the lazy way out and played with a scenario using Belden 8262 (RG58C/U)
in my line loss calculator that was pretty close to your scenario (Zo is
a touch different at 50-j0.24, most lines with loss will have a non-zero
jX component in Zo). My result for 15.8m of 8262 at 67MHz with a 200+j0
load is a transmission line loss (power into the load divided by power
into the line at the source end) is 3.3dB. This value of line loss is
independent of the source equivalent impedance.
If I reproduced the algorithms with the exact propagation constant and Zo
for your problem scenario, the answer would be more accurate, but I think
similar, and still obtained independently of the source impedance, and
not worth the time.
The figures from the calculator are below if someone wants to play with
it.
Now, are you prepared to post your solution?
Owen
Parameters
Transmission Line Belden 8262 (RG-58C/U)
Code B8262
Data source Belden
Frequency 67.000 MHz
Length 15.800 metres
Zload 200.00+j0.00 ?
Yload 0.005000+j0.000000 ?
Results
Zo 50.00-j0.24 ?
Velocity Factor 0.660
Length 1924.75 °, 5.347 ?
Line Loss (matched) 1.997 dB
Line Loss 3.279 dB
Efficiency 47.00%
Zin 30.48+j24.99 ?
Yin 0.019622-j0.016087 ?
VSWR(source end) 2.22
VSWR(load end) 4.00
? 1.46e-2+j2.13e+0
k1, k2 1.30e-5, 2.95e-10
Correlation coefficient (r) 0.999884
Jeff
> Presume a source of 100+j0 Ohms impedance sees a 50 Ohm line that is
> 5.35 wavelengths long and is terminated with a load of 200+j0 Ohms.
> The normal attenuation of the line is 2.00 dB. What is the loss in
> the line?
>
> Can your general solution solve this? It uniquely describes both the
> kinetics of reverse power flow AND the impact of source resistance.
>
> No one has every answered this one correctly, by the way (and I can
> anticipate you are ready to spring that observation on me with your
> source feeding essentialy a voltage oriented high Z load as opposed to
> the current oriented low Z load).
>
VSWR 4.75
Loss 3.55dB
73
Jeff
Owen Duffy
...
>
> You have not yet actually offered any treatment that denies the bone
> of contention that lies in two subject lines:
> 1. Reverse power is manifest;
> 2. The source will absorb and dissipate it.
Richard, if you go back over my postings in this thread, I have not denied
either of these things.
I did comment on 2 as an explanation, one which I think is poor because of
the conclusions that might be drawn from it, eg any mismatch creates
reflected power which must be dissipated in the PA.
I did suggest that in the steady state, in a tx-line-load scenario, the
impedance looking into the line can be found, and that equivalent load
adequately explains the PA's behaviour.
>
> You may have struggled with others over this in times past, but by
> your own descriptions they had little intellectual horsepower, and
> less experience in the matter.
I never said such a thing, if it is your conclusion, I disagree with it.
...
Owen
Owen Duffy
$834e...@reader.greatnowhere.com:
A 200 ohm load on a 50 ohm line is 4:1 at the load end in my view, and it
is lower as you move toward the source.
Owen
> Loss 3.55dB
>
> 73
> Jeff
>
>
>
Jeff
>> VSWR 4.75
>
> A 200 ohm load on a 50 ohm line is 4:1 at the load end in my view, and it
> is lower as you move toward the source.
>
Indeed; 4:1 decreasing to 3.2:1 at the source, BUT only when the source is
50ohms.
Change the source impedance to 100ohms and the picture changes to 5.5:1 at
the load and 4.75:1 at the source end.
By the way the vswr figures will change cyclically with frequency assuming a
fixed length of coax ( 5.35 wavelengths at a fixed frequency), between about
7.5 and 2.5 at the load and 7 and 1.5 at the source.
73
Jeff
Richard Clark
>The danger in the "power is refelected at a mismatch" explanation, is
>that it follows that power reflected at a mismatched antenna flows back
>toward the transmitter and is at least partially absorbed in the PA as
>heat. Though that is a popular belief, it is not supported by fact.
This is the complete quote to which I responded in a recent
side-thread. You have lately expressed:
On Wed, 28 Feb 2007 08:55:24 GMT, Owen Duffy <no...@no.where> wrote:
>> You have not yet actually offered any treatment that denies the bone
>> of contention that lies in two subject lines:
>> 1. Reverse power is manifest;
>> 2. The source will absorb and dissipate it.
>
>Richard, if you go back over my postings in this thread, I have not denied
>either of these things.
What have I missed about "it is not supported by fact?" What do you
mean by "is a popular belief?" I am swayed by facts and I don't
really like general statements that are couched in belief systems.
Richard Clark
>Richard Clark <kb7...@comcast.net> wrote in
>news:1daau2hj7fi70jhgl...@4ax.com:
>
>...
>>
>> You have not yet actually offered any treatment that denies the bone
>> of contention that lies in two subject lines:
>> 1. Reverse power is manifest;
>> 2. The source will absorb and dissipate it.
>
>Richard, if you go back over my postings in this thread, I have not denied
>either of these things.
Hi Owen,
It is surprising the conclusions I've drawn from our correspondence
then. As I've steadfastly expressed nearly every posting in these
terms, you have not exactly responded to my misunderstanding in an
uniform manner.
I shall return to those postings to enquire further rather than
laboring the point here.
>I did comment on 2 as an explanation, one which I think is poor because of
>the conclusions that might be drawn from it, eg any mismatch creates
>reflected power which must be dissipated in the PA.
This is not a denial? I see no positive characteristic you have
derived from 2 as allowing it is acceptable.
>I did suggest that in the steady state, in a tx-line-load scenario, the
>impedance looking into the line can be found, and that equivalent load
>adequately explains the PA's behaviour.
Yes, this allowing reflected power in your terms, allowing you to
express it as a fiction suitable to providing a truth in creating the
lumped equivalent. This may have the heavy hand of my
editorialization, but it is forced by the equivocation I find in your
points I am responding to here.
>> You may have struggled with others over this in times past, but by
>> your own descriptions they had little intellectual horsepower, and
>> less experience in the matter.
>
>I never said such a thing, if it is your conclusion, I disagree with it.
As I have never raised the discussion of "others" or how "they"
developed poor explanations or subscribed to faulty premises; then my
perhaps over-arching characterization is what you are rejecting as
your having said. You may note that at that time I explicitly offered
that their contributions were not germane to the facts.
Richard Clark
>I am also aware that supporters of the inherent source match position
>assert that you must be selective in choosing tests for source impedance.
>It is all rather unconvincing when only some of the implications of a
>particular source impedance are effective.
>
>It is my view that modelling the PA as a fixed voltage or current source
>with fixed source impedance of Zo, and where reflected waves on a
>transmission line are absorbed by the matched source is not a good
>general model for HF PAs.
Hi Owen,
This quote gives me no confidence in what you have offered to me
recently:
On Wed, 28 Feb 2007 08:55:24 GMT, Owen Duffy <no...@no.where> wrote:
>> You have not yet actually offered any treatment that denies the bone
>> of contention that lies in two subject lines:
>> 1. Reverse power is manifest;
>> 2. The source will absorb and dissipate it.
>
>Richard, if you go back over my postings in this thread, I have not denied
>either of these things.
As to point 1 (or 2 it is difficult to determine what you are
responding to specifically), explicitly stated by me, you have
expressed your self in relation to "supporters of the inherent source
match position" without actually identifying if you stand
1. With them;
2. Against them;
3. Indifferent to them.
As to point 2, explicitly stated by me, you have again described
yourself in a negative relation by discussing a model that does not
work.
Perhaps it is this style of ambivalence that clouded my appreciation
of your statement:
>I believe that it is sound (in the steady state) to resolve the forward
>and reflected wave voltages and currents at the source end of the
>transmission line, calculate the complex impedance, and predict the
>effects of that impedance as a PA load using the same techniques that
>were used to design the PA.
where you do allow 1 and 2.
However, I could be mistaken again because you don't actually
acknowledge return power impinges upon the final stage, you transform
it into another solution. Note that I accept such a transformation of
the problem. It is common alternative explanation and perfectly
valid. However, that transformation, in and of itself, does not speak
to the issue of reflected power as a physical fact and a separable
entity. In fact, the development of a lumped equivalent doesn't need
to acknowledge SWR either.
Jim Kelley
Hi Richard,
Not that I dispute anything here necessarily, but I would like to know
how you went about measuring the reflected power dissipated within a
source. Also, how the power being dissipated?
Thanks and regards,
Jim, AC6XG
Richard Clark
>Now, are you prepared to post your solution?
Hi Owen,
Your quick computation of 3.3 dB is suitably close to my reference's
first pass solution (3.27 dB), but it neglects the contribution of the
source's resistance.
The solution is 4.9 dB.
If we were to revisit your 1 meter long cable used in the 80M band and
force the transmitter to be a voltage source through the common
mechanism of adding a substantial resistor, and mismatch the other end
of that 1 meter long cable to the same degree (each end seeing 10K Ohm
for the purpose of this statistical curiosity); then that same cable
will heat up with its contribution of at least 3dB of ADDITIONAL loss.
The 10K Ohm specification is a forced one, but it responds in kind to
the original forced solution too. In fact, it comes close to the
source resistance found in a tube amplifier (a common voltage source)
driving a halfwave element (a common application for such a source)
and demonstrates the common futility of using coax (that I have
already expressed) to accomplish this.
However, we don't have voltage sources to conveniently solve either of
these statistical curiosities. Both the tube transmitter, and the
solid state transmitter employ impedance matching to either draw down,
or pull up the native source resistance to a level suitable for
applying to a transmission line.
I would again point out that reverse power suitably accounts for the
1.6 dB difference between your answer and the solution, it also
accounts for the 3 dB difference between your short cable's example,
and my twist in its application.
All such differences have been described and used in design for quite
a few decades, and they have been couched in exactly the terms I've
used here.
If anyone wants to challenge the 4.9 dB solution, they can impeach my
reference "Reference Data for Radio Engineers," (various editions). I
can supply other references that have been named in this group too,
but I would suggest with tackling one authority at a time.
Richard Clark
wrote:
>Not that I dispute anything here necessarily, but I would like to know
>how you went about measuring the reflected power dissipated within a
>source. Also, how the power being dissipated?
Hi Jim,
Dissipation is caloric, however it can arrive catastrophically by one
of two mechanisms; and they reflect, no pun here, the two types of
phase sense offered by the random opportunity (being phase adding or
subtracting for current or voltage as the occasion demands).
One caloric method is simple in measuring the heat load expressed by
airflow temperature measurements in a confined volume. When I
designed the Flight Recorder, the FAA mandated a heat budget for its
acceptance. This is certainly far afield from the immediate topic,
but it responds to the attention offered in design to this issue. The
point of this sidebar is that efficiency translated immediately into
temperature and this was rigorously anticipated and tested. The same
design philosophy is mandated in RF final design and considerable
attention has been devoted to it in the trade papers.
Returning to our concerns, for certain phase combinations that caloric
solution can arrive suddenly in the form of an arc. Most operators
will immediately act to correct that situation and the heat build up
may not be great, but the damage may still be irreversible. This
harkens back to my discussion of a kitchen table laser cracking a
window pane. Average power may be unspectacular, but instantaneous
power, localized, can be very dramatic and destructive beyond
expectation (it certainly surprised my friend).
For other phase combinations that caloric solution can arrive
gradually (heat soaking); and catastrophe arrives through thermal
runaway. Operators rarely observe this until it is too late.
I hope that the readers can differentiate between these two, and how
certain designs (eg. solid state, and tube design) respond in these
cases and correlate to experience each to their own characteristic
failure mechanism.
Richard Clark
>When I
>designed the Flight Recorder, the FAA mandated a heat budget for its
>acceptance.
Aircraft electronics lives with a common airduct. Your design must
not load the cooling air such that it becomes a flame thrower into the
next instrument in the stack. I won't go into issues of crash
survivability.
>Returning to our concerns, for certain phase combinations that caloric
>solution can arrive suddenly in the form of an arc.
I'm sure most readers who run tube rigs will recognize this situation
immediately. However, there is more than one combination of phases
and currents/voltages. I have also seen heat soaking arrive at a tube
to watch the plates glow cheerily. This, too, is probably an
experience borne by several tube rig operators. In fact, it can be
tolerated far more than a solid state amplifier, and tubes are noted
for their resilience. However, I have also seen the glass envelopes
turned into a taffy consistincy and the vacuum draw them like
heatshrink around the internal structure. Surprisingly, I have also
witnessed that these tubes still worked!
>For other phase combinations that caloric solution can arrive
>gradually (heat soaking); and catastrophe arrives through thermal
>runaway. Operators rarely observe this until it is too late.
The latest generation of solid state components have survivability
design into them such that they are specified to operate into an
infinite mismatch (or some such similar claim). This is suitably
taken care of by being able to withstand more voltage. Other issues
of current crowding, the original thermal disaster for transistors,
has been long solved. That solution revealed how the problem was in
heat confined to a small volume.
>
Finally, my measurements were never pushed to the point of failure.
All may well anticipate that this sudden arrival would preclude any
accuracy in the heat determination to demonstrate a quid-pro-quo of
returned power. Further, once the failure occured, heat is usually
removed by the very failure it brought - it usually removes the source
too. ;-)
Richard Clark
wrote:
>I would again point out that reverse power suitably accounts for the
>1.6 dB difference between your answer and the solution, it also
>accounts for the 3 dB difference between your short cable's example,
>and my twist in its application.
Lest there be any doubt about there being concurrent explanations,
this loss is also expressed in lumped equivalency and circulating
currents. It can be correlated to a very common issue with literal
lumped circuit antenna tuners. It can also be described in terms of Q
and cavities. It can also be correlated to short radiators, radiation
resistance, and Ohmic loss.
Each description is accurate, and as varied as the authors each
offering their interpretations, but no explanation denies the validity
of the other, and reflected power in a line is no exception.
Jim Kelley
Richard Clark wrote:
> On Wed, 28 Feb 2007 07:15:38 -0800, Jim Kelley <jwke...@uci.edu>
> wrote:
>
>
>>Not that I dispute anything here necessarily, but I would like to know
>>how you went about measuring the reflected power dissipated within a
>>source. Also, how the power being dissipated?
>
>
> Hi Jim,
>
> Dissipation is caloric, however it can arrive catastrophically by one
> of two mechanisms; and they reflect, no pun here, the two types of
> phase sense offered by the random opportunity (being phase adding or
> subtracting for current or voltage as the occasion demands).
>
> One caloric method is simple in measuring the heat load expressed by
> airflow temperature measurements in a confined volume. When I
> designed the Flight Recorder, the FAA mandated a heat budget for its
> acceptance. This is certainly far afield from the immediate topic,
> but it responds to the attention offered in design to this issue. The
> point of this sidebar is that efficiency translated immediately into
> temperature and this was rigorously anticipated and tested. The same
> design philosophy is mandated in RF final design and considerable
> attention has been devoted to it in the trade papers.
What I meant was, in what way were you able to attribute and apportion
this heat to its various sources? What evidence were you able to
obtain to show reflected energy re-entering the source output? What
component in the system in fact dissipated the reflected energy? How
were you able to determine the exact source and amount of energy at
any given location within the source? Or did you just presume that
you understood the underlying mechanisms?
Thanks in advance,
Jim AC6XG
Jim Kelley
Cecil Moore wrote:
> On Feb 27, 2:53 am, "Jeff" <j...@local.host> wrote:
>
>>Adding a circulator to a system will not change "the load line" (if a
>>transmission line or circulator can have such a thing), but it will cause
>>the power in the reflected wave to be separated so that it can be monitored
>>and measured. Surprisingly power monitored in this way ties up with the
>>notion that power is reflected at a mis-matched load.
>
>
> Yes, and a little modulation added to the source signal will prove
> that the
> signal being dissipated by the circulator resistor has made a round
> trip
> to the load and back. That's hard to explain if reflected energy
> doesn't
> actually exist.
> --
> 73, Cecil, w5dxp.com
Your example is the same as putting a load resistor on an open
transmission line, measuring the dissipated power, and then claiming
the same thing happens without the load resistor there.
ac6xg
Richard Clark
wrote:
>What I meant was, in what way were you able to attribute and apportion
>this heat to its various sources? What evidence were you able to
>obtain to show reflected energy re-entering the source output? What
>component in the system in fact dissipated the reflected energy? How
>were you able to determine the exact source and amount of energy at
>any given location within the source? Or did you just presume that
>you understood the underlying mechanisms?
Hi Jim,
This knowledge arrived by many avenues.
For one, in a heavily heatsinked design, mapping of temperatures
generally reveal a very diffuse origin. That, of course, is the
purpose of the heatsink. So, in that regard the assignment of where
dissipation occurs is done by induction. You can eliminate a lot
circuitry as being incapable of supporting this dissipation, as it is
both remote from the signal path, and remote physically. The
literature of design reveals much of what is discovered in the field.
That literature reveals the dissipation occurs in the
emitter/collector junction of the finals' transistors. Failures have
been confirmed through post-mortem examination by microscope (no, I
have not done this).
Experience with new designs and frequency of failure (those activities
that I have participated in) lead to the same conclusion. In one
particular case it was a manufacturing/assembly problem of mounting
the transistor to the heatsink. A bur was found in many such mounts
that interfered with a complete mating of surfaces. This raised the
thermal resistance in the path from that same junction to the mating
surface, to the heatsink, to the environment. Knowing each thermal
resistance in that path makes it rather simple to forecast the
junction temperature at the time of failure (or rather, to say failure
which occurred was guaranteed a fatal temperature) when you know the
power consumed by the component. All such "resistance" conform to the
simple math of Ohm's law (once you substitute the necessary units for
heat).
When we return to the design guidelines and this junction, almost
every manufacturer of power transistors specifies a junction
resistance value at rated power. Casting this value through the chain
of transformations and to the antenna connector reveals a value very
nearly 50 Ohms. There are newer power amplification designs today,
and yet the market for Ham gear is dominated by the Class AB design
which is exhibits this property nicely.
Inductive logic leads us to this junction as the principle target of
reflected power (the signal path is symmetric, after all). Experience
has supported this logic. Failures are attributable to design flaw
(or assembly flaw), or poor application (driving a mismatch), or both.
As for tubes, I've already testified to the obvious location for
dissipation. It is far easier to see.
Cecil Moore
Most engineers equate the units of power to power, i.e.
joules/sec = watts and so does the IEEE dictionary. But I
am content to assert that the joules in the joules per
second of a reflected wave is real energy. Do you disagree?
--
73, Cecil http://www.w5dxp.com
Cecil Moore
> Any who complain about their transmitter having:
> 1. No source resistance;
> 2. Not this much resistance:
> 3. Not this little resistance;
> 4. None of the above (the usual response).
Yep, the great majority of amateur radio antenna systems
are matched by a tuner. That act of matching prohibits
reflected load energy from reaching the PA. Except for
overall efficiency, when an antenna system is matched,
the PA impedance doesn't matter. A 5 ohm PA, a 50 ohm PA,
and a 500 ohm PA all output the same power if the output
voltage is the same into the same load.
Cecil Moore
>
>
> Cecil Moore wrote:
> transmission line, measuring the dissipated power, and then claiming the
> same thing happens without the load resistor there.
No, it is more akin to presenting all the evidence.
Your approach is akin to rolling dice in the dark
where you are the only one allowed to report the
results. :-)
Are you willing to assert that the power being
dissipated in the circulator resistor didn't
make a round trip to the load and back even though
the actual delay is easy to measure?
Do the reflected waves that you see when looking
at yourself in the mirror contain any joules/sec?
Cecil Moore
> Change the source impedance to 100ohms and the picture changes to 5.5:1 at
> the load and 4.75:1 at the source end.
Here's the equation for rho at the load.
rho = (Z0-Zload)/(Z0+Zload)
SWR = (1+rho)/(1-rho)
I don't see the source impedance in those equations.
Jeff
>
> rho = (Z0-Zload)/(Z0+Zload)
>
> SWR = (1+rho)/(1-rho)
>
> I don't see the source impedance in those equations.
Your analysis is fine if the source is matched to the coax, but you are
neglecting the mismatch at the source to coax interface.
If you used a TDR, for example, to look at the set-up you would see 2 points
of discontinuity, firstly at the 100 ohm source to 50 ohm cable interface,
and secondly at the cable to 200 ohm load. BOTH of these discontinuities add
to the overall mismatch as seen by the 100 ohm load.
Your application of the above equations neglects the first discontinuity.
73
Jeff
Jeff
">
> The solution is 4.9 dB.
>> If anyone wants to challenge the 4.9 dB solution, they can impeach my
> reference "Reference Data for Radio Engineers," (various editions). I
> can supply other references that have been named in this group too,
> but I would suggest with tackling one authority at a time.
>
> 73's
> Richard Clark, KB7QHC
I have to admit to an error in my analysis of the problem, I made a mistake
with the attenuation of the line.
Re analysing with the correct values gives me the following:
Loss 4.78dB
S11 -5.25dB
Vswr as seen by the source 3.41:1
I think these values are close enough to Richard's answers to make little
difference. I have not got Reference Data for Radio Engineers to hand, but
it may be a graphical solution that could account for the slight
discrepancy. My figures were obtained using Ansoft Designer RF Cad package.
The error crept in because I did the analysis by using 5.35m of coax at 300m
to get the 5.35 wavelengths, and forgot to scale the attenuation value for
the coax.
73
Jeff
Owen Duffy
news:45e68d20$0$26701$834e...@reader.greatnowhere.com:
> If you used a TDR, for example, to look at the set-up you would see 2
> points of discontinuity, firstly at the 100 ohm source to 50 ohm cable
> interface, and secondly at the cable to 200 ohm load. BOTH of these
> discontinuities add to the overall mismatch as seen by the 100 ohm
> load.
Your TDR does not work in the steady state frequency domain space, and is
misleading you.
In the steady state, the (complex) ratio of forward voltage to reflected
voltage is determined solely by the load impedance and characteristic
impedance of the line.
In crude terms, during establishement of steady state, you can view that
a load end reflected wave which is then partially reflected at a
mismatched source end, will reach the load end and be reflected in the
same ratio as the earlier passes. The subsequent round trips as steady
state is approached do not change the (complex) ratio of forward voltage
to reflected voltage in the steady state.
I know you have support here for the assertion that source end mismatch
affects VSWR in the steady state, but you won't find it in reputable text
books.
Owen
Owen Duffy
> On Wed, 28 Feb 2007 08:11:30 GMT, Owen Duffy <no...@no.where> wrote:
>
>>Now, are you prepared to post your solution?
>
> Hi Owen,
>
>
> Your quick computation of 3.3 dB is suitably close to my reference's
> first pass solution (3.27 dB), but it neglects the contribution of the
> source's resistance.
>
> The solution is 4.9 dB.
Reminding you that your question was "What is the loss in the line?",
check your own post.
Well, you posted an answer, not a solution. It wouldn't have been your
solution anyway, because it looks like it is copied straight out of a
book.
Looking at Reference Data for Engineers, Sixth Edition, p24-12, Example 3
(which is the same as the problem you posed), they give the answer as
3.27dB.
I am happy that my answer rounded to 3.3dB is correct.
The source resistance has no influence over the line loss at all.
You posed this problem as difficult and one that no one has ever got
right. No wonder, you have a different answer to the the book!
Owen
Jeff
> Your TDR does not work in the steady state frequency domain space, and is
> misleading you.
>
> In the steady state, the (complex) ratio of forward voltage to reflected
> voltage is determined solely by the load impedance and characteristic
> impedance of the line.
>
What utter rot. The TDR indicates a discontinuity that is IN ADDITION to
the one that gives you your result. Ignoring that discontinuity will
certainly give you the wrong answer, regardless if which domain you are
working in!!!
Jeff
Jeff
>
> The source resistance has no influence over the line loss at all.
>
Well it certainly does not agree with Ansoft Designer, the result it gives
is very close to Richards, and it shows a marked effect when you change the
source resistance.
With a 50 ohm source and 200 ohm load the loss is calculated as is 3.93dB
and VSWR is 3.98:1
Changing to a 100 ohm source the loss is calculated as is 4.78dB and VSWR
is 3.41:1
So a very well respected CAD package agrees with Richard at least!!
73
Jeff
Cecil Moore
>> rho = (Z0-Zload)/(Z0+Zload)
>>
>> SWR = (1+rho)/(1-rho)
> Your application of the above equations neglects the first discontinuity.
The first discontinuity (inside the source) doesn't
have any effect on the SWR on the transmission line.
Tuners at the source present another discontinuity
and have no effect on transmission line SWR.
Jeff
>
> The first discontinuity (inside the source) doesn't
> have any effect on the SWR on the transmission line.
The discontinuity is NOT "inside the source", it is at the source to coax
interface, and as such effects the VSWR that the source sees.
> Tuners at the source present another discontinuity
> and have no effect on transmission line SWR.
There are no tuners involved in the current example.
Jeff
Cecil Moore
> What utter rot. The TDR indicates a discontinuity that is IN ADDITION to
> the one that gives you your result. Ignoring that discontinuity will
> certainly give you the wrong answer, regardless if which domain you are
> working in!!!
When line loss is given in watts, the discontinuity
at the source certainly has an effect.
When line loss is given in dB (10x log of a ratio),
the discontinuity at the source has NO effect.
The discontinuity at the source has NO effect on
the RATIO of two powers.
Cecil Moore
> With a 50 ohm source and 200 ohm load the loss is calculated as is 3.93dB
> and VSWR is 3.98:1
> Changing to a 100 ohm source the loss is calculated as is 4.78dB and VSWR
> is 3.41:1
Looks like your losses and VSWR are taken on the wrong
side of the source resistor.
source R
source V----x--/\/\/\/\/\/\--y----T-line---load
Loss and VSWR should be calculated at 'y', not at 'x'.
We are interested in the VSWR and losses *on the T-line*.
Please change your reference point from 'x' to 'y'.
Cecil Moore
>>> Your application of the above equations neglects the first discontinuity.
>> The first discontinuity (inside the source) doesn't
>> have any effect on the SWR on the transmission line.
>
> The discontinuity is NOT "inside the source", it is at the source to coax
> interface, and as such effects the VSWR that the source sees.
There is usually a piece of coax running from the
source connector back to a filter. I would suggest
that the discontinuity that you are talking about
is indeed some distance "inside the source".
But we users don't care or measure what VSWR the source
sees. Only the PA designer worries about such. We users
only care and measure the VSWR *ON* the transmission line.
Ian White GM3SEK
Owen and Cecil are right: the source (transmitter) has no effect
whatever on the VSWR on the line.
That isn't just an assertion - it is part of the bedrock transmission
line theory. Owen referred to "reputable textbooks", one of which would
surely be 'Theory and Problems of Transmission Lines' by R A Chipman
[1]. This book gains a lot of its reputation from its very complete
mathematical development of the theory, showing all the detailed
working.
Chipman treats standing wave patterns in two different ways: first by
assuming the final steady-state conditions, and then in much more detail
by considering multiple reflections between the load and the source.
Given a sufficient number of reflections, the multiple-reflection model
converges on exactly the same results as the steady-state analysis -
just as it does in the physical world.
VSWR on the line is determined by the ratio |Vmax|/|Vmin|. The complex
impedance that the source sees at the input terminals of the line is the
ratio V/I at that point (where V and I are both vector quantities which
include phase information). An alternative way of calculating either
VSWR or Zin is through the ratio Vforward/Vreflected (again vector
quantities).
All of these approaches are alternative pathways through the same body
of theory. They are all consistent with one another, and there is no
contradiction between any of them.
You will notice that all these standing wave relationships involve
ratios. Chipman's detailed analysis confirms that these ratios are
determined EXCLUSIVELY by the properties of the line and the load -
never the source.
The source properties do determine the magnitudes of all of the
individual voltages and currents - but when you change the source
properties (output voltage and/or impedance) all the individual voltages
and currents on the line and at the load are changed by the same factor.
So when you take the ratio, the source properties cancel right out
again.
All this confirms that, if you sweat out the math in all the different
levels of detail that Chipman did, the source (transmitter) still has no
effect whatever on the VSWR on the line.
[1] Out of print, but well worth searching for: ISBN 0-07-010747-5.
The web bookstores currently have eight copies on offer, at a range of
prices.
--
73 from Ian GM3SEK
Jeff
> on the VSWR on the line.
>
> That isn't just an assertion - it is part of the bedrock transmission line
> theory. Owen referred to "reputable textbooks", one of which would surely
> be 'Theory and Problems of Transmission Lines' by R A Chipman [1]. This
> book gains a lot of its reputation from its very complete mathematical
> development of the theory, showing all the detailed working.
I am sorry but you are not correct, I have not read Chipman so I cannot
comment on his analysis or your interpretation of his results, but my
understanding , practical experiments and CAD analysis would lead me to
disagree.
If we take the situation where the source is matched (50ohms) to the 5.35
wavelength transmission line (lossless to simplify things) with a 100ohm
load, I agree that the vswr is 4:1, unchanging with frequency.
Plotted on a Smith Chart when swept against frequency this gives a circle
centred on 1 (50ohms) with a radius of 4. i.e. on a constant VSWR circle.
Now if we change the source impedance to 100ohms and repeat the same sweep
and re-plot, keeping the chart normalized to 50 ohms, the circle moves on
the resistance axis, still with a radius of 4 and now passing though 2 (100
ohms) resistive. The centre moves to about 0.6 (30ohms). It then becomes
obvious that the locus of the circle is NOT a constant VSWR against
frequency.
You will come to the same conclusion if you normalize the chart to 100 ohms,
the new source impedance and re-plot.
The coax is acting as an impedance transformer, causing a shift along the
resistance axis.
Looking at it another way, the vswr changes sinusoidally with frequency, in
our example, between 2:1 and 8:1. (The same as the Smith chart plot with a
circle of radius 4 centred at about 0.6).
73
Jeff
Cecil Moore
> You will come to the same conclusion if you normalize the chart to 100 ohms,
> the new source impedance and re-plot.
The Z0 of the transmission line has not changed to 100
ohms so normalizing the chart to 100 ohms is not valid.
--
73, Cecil, http://www.qsl.net/w5dxp
Jim Kelley
Cecil Moore wrote:
> Jim Kelley wrote:
> Most engineers equate the units of power to power, i.e.
> joules/sec = watts and so does the IEEE dictionary.
I can't speak for most engineers, but I think the first time I saw it
was in high school physics, and of course later in engineering school.
That was about 35 years ago. I think of it a fundamental concept -
one that I happen to understand very well. Not unlike the
relationship between Joules and electron-volts.
> But I
> am content to assert that the joules in the joules per
> second of a reflected wave is real energy. Do you disagree?
I don't agree that the terms power and energy become interchangeable
by virtue of the fact that their units can both be expressed with the
word Joule in them. One can find himself making unrealistic
predictions if he is not precise in his application of the ideas which
underlie these terms.
73, ac6xg
Jeff
It is just as valid as using 50 ohms, and the result is the same, a changing
vswr.
I see you have not commented on the main point of my post, that being that
the smith chart shows a changing vswr when you change the source impedance.
Hint: transmission line transformers would not work if the vswr did not
change.
73
Jeff
Richard Clark
>Reminding you that your question was "What is the loss in the line?",
>check your own post.
Hi Owen,
Can you offer why I should? Well, I suppose not or you would have.
However, I am one to never turn aside a suggestion and I did review
everything (except my own quote - I've repeated it enough, haven't I?)
and I will respond to that review within the body of this text.
>Well, you posted an answer, not a solution. It wouldn't have been your
>solution anyway, because it looks like it is copied straight out of a
>book.
Does it being someone else's solution make any difference to the
outcome? Owen, your comment reveals a prejudice by implication.
Copying it right out has removed any issue of authority has it not? It
has also removed any issue of accuracy too - if you accept that
authority. Ultimately, having copied it out makes for the best
resolution. Having copied it out, and offering the citation, gives us
both access to the chain of evidence. Did I withhold or otherwise
linger with the citation? You asked for my solution and I immediately
offered both.
Ironically, does your suggestion that
>It wouldn't have been your solution anyway
mean you would suspect I would have come up with a different answer?
That is, ascribing to me the quality of being able to get it right
instead? That would be generous, thank you. However, it appears I
fell short of that mark (and may have been the intent of your
elliptical pat on the back).
>Looking at Reference Data for Engineers, Sixth Edition, p24-12, Example 3
>(which is the same as the problem you posed), they give the answer as
>3.27dB.
A simple review of example four distinctly reveals the details to the
problem I posed; example three contains only some of them. Example 3
is a subset of example 4 (as that example dwells on at great length).
However, example 4 does have one notable difference, it asks:
"What is mismatch loss between the generator and the line?"
for which the answer is:
"1.62 dB"
Ah, the devil is in the details. Continuing from example 4:
"The transducer loss is found by using the results
of 3 and 4 in (4). This is
1.27 + 2.00 + 1.62 = 4.9 decibels"
>I am happy that my answer rounded to 3.3dB is correct.
Congratulations. You may note in my earlier correspondence I allowed
exactly that.
>The source resistance has no influence over the line loss at all.
Upon review of my own reference (not the recommendation you offer
above) I must concur. I was trapped by what I have already described
as being the classic confusion between systems of match and loss. My
solution was not for the loss in the line, but for transducer loss,
and specifically for the inclusion of mismatch loss within the
transducer loss. All caloric, but mis-ascribed to the line loss.
>You posed this problem as difficult and one that no one has ever got
>right. No wonder, you have a different answer to the the book!
Well, in fact my answer conforms exactly to the book. The problem is
not one of inexactitude, it is of poor referencing. As to the matter
of no one else having ever got it right, no one even consulted the
book - even partially. You can count yourself among a population of
one and hashing it through served us well.
Jim Kelley
> On Wed, 28 Feb 2007 13:55:47 -0800, Jim Kelley <jwke...@uci.edu>
> wrote:
>
>
>>What I meant was, in what way were you able to attribute and apportion
>>this heat to its various sources? What evidence were you able to
>>obtain to show reflected energy re-entering the source output? What
>>component in the system in fact dissipated the reflected energy? How
>>were you able to determine the exact source and amount of energy at
>>any given location within the source? Or did you just presume that
>>you understood the underlying mechanisms?
>
>
> Hi Jim,
>
> This knowledge arrived by many avenues.
But primarily, it seems, by speculation. I know how to measure heat,
Richard. What I am asking, and what you have thus far been unable to
answer (which is as I suspected), is how is it that you were able to
ascertain that this heat energy was caused by energy that was
reflected from the load rather than having come directly from the
power supply within the source? How is it that this electromagnetic
energy is so easily reflected from a load, but is utterly immune to
reflection when it encounters the output of a source? I think it's
been fairly well established that the output impedance of these things
is far from 50 ohms. Why should reflected energy not be, at least in
some part, re-reflected back toward the load?
Someone who alleges to be so familiar with load lines should be able
to contend with an increase in dissipation against a mismatched load
without having to explain it as 're-absorbed' reflected energy.
> Inductive logic leads us to this junction as the principle target of
> reflected power (the signal path is symmetric, after all).
Speculation could also lead to that juction.
> Experience
> has supported this logic.
It could be experience coupled with misattributed fact. Possible?
73, Jim AC6XG