(Tom Bruhns) wrote in message om...
....
(Example: RG174 at f=30MHz will have a bit more than 3.4dB/100 feet
loss because of R, and probably well under .025dB/100 feet loss
because of G. See Roy's suggested reading for the source of those
numbers.)

It's also worth pointing out that if you use the same dielectric
material in a line with larger diameter, the R loss -- the loss in the
resistance of the wire -- drops, but the G loss stays the same. So
for a 1" diameter solid polyethylene dielectric 50 ohm line with
smooth copper conductors, at 30MHz the attenuation is about
0.214dB/100ft, and the attenuation due to G is perhaps 1/10th that
much. At 3000MHz, the coax probably still propagates pretty well in
TEM mode, and the R and G losses will be nearly the same and each
about 2dB/100ft.

Because of the higher losses in fiberglass-epoxy PC board material,
you can fairly easily end up with stripline which has higher G loss
than R loss, when using that board material in the GHz range. (Howard
Johnson recently had an article about this topic in, um, EDN or
Electronic Design, I can never remember which his column appears in.)

Cheers,
Tom

It occurs to me I might be able to do a whole lot better than actual
measurements. I should be able to simulate a power meter in the SWCadIII
Spice simulator, and do a transient analysis. This will let me go from RF to
DC out. They also have models for lossless and lossy transmission lines,
which should make it possible to see how steady state is reached from
turning the source on at T=0.

Tam/WB2TT

Hi Tam,

Yes, the lower reactance -- lower Q, and the lower frequency, will
both help keep the disturbance from the meter at a reasonable level.
If you simply re-tune the 50pF cap in Cecil's 7.2MHz 'speriment, you
still end up with about 1.2:1 SWR, because it's effectively a "T"
impedance matching network. But the same line at 1.8MHz with C and L
at 50 ohms reactance, re-tuned to resonance after insertion of the
meter, gives about 50dB return loss, and you'd be lucky to resolve
that with a typical SWR meter. Of course, you're stuck with 1.8nF of
capacitance too.

Let us know how it works out when you have time. I like your idea of
peeking inside the bridge; I had the same thought.

Cheers,
Tom

"Tarmo Tammaru" wrote in message ...
"Tom Bruhns" wrote in message
m...

(Tam: my recommendation is to do the test yourself. It will be a lot
easier to play with "what-ifs" and to check out things that don't at
first make sense if you have direct control of the experiment.)

Cheers,
Tom
Tom,

I read you, but first I have to paint the kitchen. I was going to use 50
+/& -j50. I also want to get inside the meter and look at the voltage and
current separately. It's a Kenwood, no sealed slugs. Good point about the
meter changing the reactance; 160 m might be a good place to do this, or I
might use a variable capacitor.


Tam/WB2TT

Tom,

I am making progress with the SWCad model of the power meter. The current to
voltage converter is working, which should be the hardest part. What I like
about doing it this way is that all components have 0 tolerance, and there
is nothing in the circuit that I don't put on the schematic. Unfortunately,
I won't be able to do anything the next couple of days.

Tam/WB2TT

Tom, Cecil, etc

Well, I got the SwCAD model of the SWR/power meter operating. Very
interesting. Learned a lot that I would never have thought of by just
contemplating.

Here is the circuit:

A) An opamp with a gain bandwidth of 10000 MHz that senses the current in
the line. The current to voltage gain conversion constant is 50 I.

B) Another opamp that does the RF sum of K(V + 50 I). I am calling this
output VF.

C) A third opamp that does the RF subtraction of K(V - 50 I). Gee, lets call
this VR.

D) It can be shown that SWR=(VF + VR) / (VF - VR). I love statements like
this, but it is easy enough to prove. Let I=V/RL, and plug the first two
equations into the third.

I did a calibration run at 5W with the source set to 15.811V, ZS=0. With my
K, I get VF=3.13, VR=0, PF=5W

Now for Cecil 1. ZL = 50 - j400. VF=1.62, VR=1.56, SWR=53, PF=1.33W,
PR=1.24W.

Now for Cecil 2. ZL=50-j400, BUT ZS= 0 + J400. VF=11.2, VR=10.9, SWR=73.7,
PF=64W, PR=60.6W. I am at such a high impedance here, that I suspect the 10K
sampling resistors are loading down the circuit somewhat. (I might try 100K
instead).

Note that there is absolutely nothing explicit in the circuit that has
anything to do with transmission lines. All components are perfect; there
are no stray inductances or stray capacitors.

Tam/WB2TT


"Tom Bruhns" wrote in message
m...
Hi Tam,

Yes, the lower reactance -- lower Q, and the lower frequency, will
both help keep the disturbance from the meter at a reasonable level.
If you simply re-tune the 50pF cap in Cecil's 7.2MHz 'speriment, you
still end up with about 1.2:1 SWR, because it's effectively a "T"
impedance matching network. But the same line at 1.8MHz with C and L
at 50 ohms reactance, re-tuned to resonance after insertion of the
meter, gives about 50dB return loss, and you'd be lucky to resolve
that with a typical SWR meter. Of course, you're stuck with 1.8nF of
capacitance too.

Let us know how it works out when you have time. I like your idea of
peeking inside the bridge; I had the same thought.

Cheers,
Tom

"Tarmo Tammaru" wrote in message
...
"Tom Bruhns" wrote in message
m...

(Tam: my recommendation is to do the test yourself. It will be a lot
easier to play with "what-ifs" and to check out things that don't at
first make sense if you have direct control of the experiment.)

Cheers,
Tom
Tom,

I read you, but first I have to paint the kitchen. I was going to use 50
+/& -j50. I also want to get inside the meter and look at the voltage
and
current separately. It's a Kenwood, no sealed slugs. Good point about
the
meter changing the reactance; 160 m might be a good place to do this, or
I
might use a variable capacitor.


Tam/WB2TT

Tarmo Tammaru wrote:
Now for Cecil 1. ZL = 50 - j400. VF=1.62, VR=1.56, SWR=53, PF=1.33W,
PR=1.24W.

Now for Cecil 2. ZL=50-j400, BUT ZS= 0 + J400. VF=11.2, VR=10.9, SWR=73.7,
PF=64W, PR=60.6W. I am at such a high impedance here, that I suspect the 10K
sampling resistors are loading down the circuit somewhat. (I might try 100K
instead).

Note that there is absolutely nothing explicit in the circuit that has
anything to do with transmission lines. All components are perfect; there
are no stray inductances or stray capacitors.

Chipman alludes to such a "phenomenon of resonance" in Chapter 10,
"Resonant Transmission Line Circuits". For instance, at a conjugate match
point where 100+j100 is seen looking in one direction and 100-j100 is
seen looking in the opposite direction, there seems to be a *localized*
exchange of extra energy between +j100 and -j100 that can adversely affect
the value indicated by an SWR meter placed between those two values.
--
73, Cecil http://www.qsl.net/w5dxp



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Goodness, are we talking about energy moving back and forth on a
cycle-to-cycle basis? Instantaneous power? Are you saying boar hog tits
do have a use after all -- in Texas, anyway?

Roy Lewallen, W7EL

Cecil Moore wrote:

Chipman alludes to such a "phenomenon of resonance" in Chapter 10,
"Resonant Transmission Line Circuits". For instance, at a conjugate match
point where 100+j100 is seen looking in one direction and 100-j100 is
seen looking in the opposite direction, there seems to be a *localized*
exchange of extra energy between +j100 and -j100 that can adversely affect
the value indicated by an SWR meter placed between those two values.

Cecil, W5DXP wrote:
"For instance, at a conjugate match point where 100+j100 is seen looking
in one direction and 100-j100 is seen looking in the opposite direction,
there seems to be a "localized" exchange between +j100 and -j100 that
can adversely affect the value indicated by an SWR meter placed between
these values."

If we have a resonant LC circuit, there is only resistance to limit
current. If the resonant circuit is a series combination, we can place a
certain voltage of the resonant frequency across the series combination.
Voltage across either L or C can be much larger than the applied voltage
as the reactive Z`s are equal and opposite. This leaves the applied
voltage equal to (I)(R).

Some day I hope to see Chipman`s analysis. Transmission lines have
distributed inductance and capacitance. A "conjugate match point" seems
an oxymoron to me. A conjugately matched circuit demonstrates this
condition no matter where it is sliced to look in both directions.

A resonant length of transmission line with reflections will have more
loss than a similar matched line simply because the msatched line has no
opportunity to lose some of the reflected energy.

Seems to me, we correct power factor at a load to eliminate reactive
current in the power line. We are resonating the load and eliminating a
reflection from the load.

If loss from reflected power is trivial, we don`t need to worry with
matching at the load and can match at the sending end of the line to get
the power we need.

Best regards, Richard Harrison, KB5WZI

Roy Lewallen wrote:
Goodness, are we talking about energy moving back and forth on a
cycle-to-cycle basis?

We are talking about a localized energy exchange between an
inductive reactance and a capacitive reactance during a cycle -
that third term in your energy equation - and the possible effects
on an averaging RMS wattmeter.
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Harrison wrote:
Some day I hope to see Chipman`s analysis.

I just bought the book on half.com.

Transmission lines have
distributed inductance and capacitance. A "conjugate match point" seems
an oxymoron to me. A conjugately matched circuit demonstrates this
condition no matter where it is sliced to look in both directions.

Nope, it doesn't, Richard. A flat system is conjugately matched, i.e.
you see 50 + j0 in one direction and 50 - j0 in the other direction.
--
73, Cecil http://www.qsl.net/w5dxp



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