Transmitter Output Impedance
Sal M. Onella
transmission lines, like how to measure, how to match, etc.
Something I haven't seen is a discussion of the source impedance of
the transmitter. My curiosity was piqued today as I took some baby
steps into EZNEC. A particular antenna had such-and-such VSWR if fed
with a 50-ohm cable and a different value if fed with a 75-ohm cable.
While this is hardly news, it got me wondering whether a 75-ohm cable
will load the transmitter the same. Doesn't seem like it.
My point: Using 75-ohm cable to improve the match at the antenna
won't help me ... IF ... I suffer a corresponding loss due to
mismatch at the back of the radio. My HF radios, all solid state,
specify a 50 ohm load. As necessary, I routinely use an internal
autotuner and either of two external manual tuners. (I'm aware of the
published 1/12 wavelength matching method.)
Wisdom in any form would be appreciated. Thanks.
"Sal"
(KD6VKW)
Ralph Mowery
"Sal M. Onella" <sob...@cox.net> wrote in message
news:7634e01d-847f-48ea...@k3g2000prl.googlegroups.com...
A transmitter output impedance is designed for maximum power transfer at a
specific impedance. Most of the the older tube transmitters impedance was
tunable within a range.
In simple terms the impedance of the transmitter tube is the plate voltage
devided by the current. This impedance is then transformed to the nominal
50 ohms of the antenna system. If the transmitter has the usual tune and
load controls, the exect impedance will not mater as you adjust for maximum
transmitter output.
Most of the transistor transmitters are not adjustable so the output
impedance is usually fixed at 50 ohms for maximum power transfer. If the
impedance of the antenna system is not 50 ohms, then the output power will
be less than the designed output. You can use the antenna tuner to adjust
for a match.
Jim Lux
I suspect that most ham transmitters do NOT have a 50 ohm output
impedance. What they do have is a specification that they will
adequately drive a 50 ohm load (and some sort of internal circuitry that
detects an "unacceptable" output condition and turns down the drive).
After all, your transmitter could have an output impedance of zero ohms
(a "stiff" voltage source), and adequately drive your transmission line
and antenna at 50 ohms (yes, this is not the optimum power transfer, but
nobody ever said that ham transmitters are designed for optimum power
transfer... maybe they're perfectly happy with less transfer, but still
operating within their safe area)
ON9CVD made some simple measurements using a couple of resistors and
foudn that a TS440 has a Zout somewhere around 15-40 ohms (depending on
frequency and output power).
http://sharon.esrac.ele.tue.nl/~on9cvd/E-Uitgangsimpedantie.htm
Grant Bingeman also has words on this:
http://www.km5kg.com/loads.htm
Richard Clark
<sob...@cox.net> wrote:
>Something I haven't seen is a discussion of the source impedance of
>the transmitter.
sigh....
>My point: Using 75-ohm cable to improve the match at the antenna
>won't help me ... IF ... I suffer a corresponding loss due to
>mismatch at the back of the radio.
Hi OM,
Look at the prospective SWR and how much is
lost/reflected/absorbed/what-have-you? More heat comes from a less
than optimal system efficiency than what your computation will reveal.
So much that it will swamp it.
But the trick here is that the reflected "power" (arguments turn on
this word) doesn't always mean heat and it could actually cool -
however hot or cold it may alter the situation, that same "power"
never got out into the air.
Now, as for source impedance, that is a subject fraught with denial in
the face of the obvious: Those fins in the back of your rig are to
help bookend your QSL cards into groups (the heat bears no relation to
efficiency nor match loss).
A standard definition (courtesy of Wikipedia) for Return Loss is:
where Zs is the impedance toward the source and
Zl is the impedance toward the load.
and we find from the values you supply that it is
0.20
Of course, such a definition is utterly useless when the concept of Zs
is replaced with (in most cases) "it ain't 50 Ohms, thet's fur
shure").
If, perchance, some brave soul steps into the breach of NOT 50 Ohms to
suggest what Zs is, then we can give it the acid test of engineering
(an act that I am usually reminded is beyond the understanding of
readers and the province of discussion here). Let's be gentle and go
only by an order of two (which is reasonably available and can be
coaxed out of my TS-440). Return loss for a rig exhibiting an Zs of
25 Ohms into the 75 Ohm line (presuming it is infinite in length)
would give us:
0.50
that doesn't look good, so let's try Zs of 100 Ohms:
0.14
that looks better all 'round. Even intuition agrees.
Let's press intuition to the proximal limit and say that Zs is 74 Ohms
(yes, my thumb is on the scale):
0.01
What does intuition affirm? What is preferable?
73's
Richard Clark, KB7QHC
Cecil Moore
> Wisdom in any form would be appreciated. Thanks.
Have you seen these?
http://www.w2du.com/QEXMayJun01.pdf
http://www.w2du.com/Appendix12.pdf
http://www.w2du.com/r3ch19a.pdf
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK
Wimpie
Hello Sal,
Make yourself up for a long discussion (as we had such a thread
earlier).
Most amplifiers are designed just to provide the desired amount of
power into a certain load. The actual output impedance of the
amplifier is not important in many cases. Changing the load impedance
(for example in case of a solid state amplifier) may result in less or
more heat generated in the active device(s). You can see this by
changing the load (for example with an external matcher between power
meter and PA) and watching the output power and current consumption.
If you have an amplifier with tunable output section (vacuum tube PA),
you are within the range of the tuner, and you tune it for maximum
output power (given certain drive), the output impedance equals the
load impedance (or conjugated value in case of non-ohmic load). If
you change the drive (so adjust the output power), the output
impedance may change (due to saturation issues). Same is valid for the
load, if you change the load, voltage across and current through the
active device may saturate, hence changing the output impedance.
Non-tunable amplifiers (for example a 3…30 MHz balanced amplifier)
will mostly not present 50 Ohms to your load (unless specially
designed for that using feedback).
Virtually all high-efficient switching amplifiers do not show 50 Ohms
to the load. If you add an external tuner and match to maximum output
power, you will very likely destroy the amplifier in case of no
supervisory circuits present.
You can do some experiments with your own amplifiers. Just change the
load impedance and see what the forward power indicator on your
reflectometer/VSWR meter does. If it doesn't change, your output
impedance is very close to 50 Ohms.
Measuring the output impedance (for relative small change in load) is
possible, but is not a simple task. Very likely other people will
comment on this.
With kind regards,
Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me very likely.
Cecil Moore
> The actual output impedance of the
> amplifier is not important in many cases.
Yes, consider a source with 70.7 volts at its output terminals
connected to a 50 ohm load through a 50 ohm feedline. Except for
source efficiency, the source impedance simply doesn't matter. Any
70.7 volt source will deliver 100 watts to the feedline no matter what
the source impedance.
There's a simple method called load-pulling. Keeping everything the
same at the source, if the power to the load increases when the load
is changed away from 50 ohms, the source doesn't have a 50 ohm
internal impedance.
Sal M. Onella
Well, I have _now_! 'Twill take just a bit of time to digest them.
Thanks.
"Sal"
Jim Lux
> Measuring the output impedance (for relative small change in load) is
> possible, but is not a simple task. Very likely other people will
> comment on this.
>
ON9CVD's website I linked to has a very simple technique.. 50 ohm dummy
load and a 220 ohm resistor you can switch in.
At 100W (into 50 ohms), the 220 ohms would only dissipate 22W. You
could get some of those non-inductive resistors from Caddock and series
them up to do something like this.
BTW, this is a simplified version of what's called a "load pull" test...
which makes me wonder if one could cobble up a quick test set that could
be controlled by a computer to do automated output Z measurements of an
HF transceiver over a reasonably wide range... One approach would be to
use a RS-232 controlled antenna tuner and, maybe, a antenna relay box
with several different load resistances).
The challenge (having actually looked at doing this with a LDG AT200PC)
is that the Z of the tuner isn't very well defined. It's a pretty big
calibration project in itself.
Maybe, though, one could build a few test dummy loads.. say a 25 ohm and
a 75 or 100 ohm, along with your vanilla 50 ohm, and the antenna switch
(like an RCS-8V). Basically, you're building a "high power resistor
substitution box"
You'd want some sort of nice inline watt meter (like an LP100) to make
the measurements.
Jim Lux
>
> A "match" provides the optimum power transfer from generator to load,
> but that is NOT the maximum load power, nor is it either the maximum or
> minimum power dissipated in the source.
>
> Say I have a zero output impedance on my source and I'm putting out 7
> Volts RMS into a 50 ohm load. That's about 1 watt into the load. Now..
> if I reduce the load impedance to 25 ohms, and since I've got zero
> output impedance, I'm now putting out 2 Watts. The source impedance is
> zero, so I'm not dissipating any extra power in the source, either.
>
> It is true that a "matched load" to my zero ohm source would, in fact,
> be zero ohms, and would have infinite power. Any other load impedance
> would have less power into the load, so the Thevenin theorem is satisfied.
>
>
> Now.. if my generator had a fixed output impedance, it's true that the
> load impedance that will get the most power out is the conjugate of the
> output Z. For resistive sources/loads, here's an example..
>
> You also have to be careful about looking at Thevenin equivalent sources
> (e.g. a ideal voltage in series with a Z, or a ideal current in parallel
> with a Z), because just because *the model* has a resistor in series
> does NOT mean that you're actually dissipating any power in the source.
> (If I had a very efficient op amp, I could simulate any arbitrary output
> impedance, without dissipating any power in the source)
>
>
> Say my generator is 40 ohms, and I'm putting out 7 Volts into a 40 ohm
> load. OK, that means that the imaginary voltage source is putting out
> 14 V. I'm getting about 1.23 Watts into my load. Now, if I decrease my
> load Z to 20 ohms, what do I get? Now, I have 4.67 (=14/3) Volts
> instead of 7, and I get 1.1 Watts. Yep, less.. Thevenin works. Let's
> try increasing the Z to 60 ohms.. Now the voltage on my load is 8.4 V,
> and I'm dissipating 1.18W, again, less than my 1.23.
>
> But here's some weird stuff.. let's look at how much power is dissipated
> in that imaginary resistor (i.e. our source *really is* a ideal voltage
> source in series with a resistor)
>
> At 40 ohm load, we've got 7 volts on the load and 7 volts across the
> resistor, so they both dissipate the same 1.225 Watts. Pload/Pgen = 1
>
> In the 20 ohm load case, we've got 1.1 dissipated in the load and 2.2
> dissipated in the generator. Pload/Pgen = 0.5
>
> In the 60 ohm load case, we've got 1.18 dissipated in the load, and 0.78
> dissipated in the generator. Pload/Pgen = 1.5 (i.e. we dissipate more
> in the load than in the generator... how about that!)
>
> And let's look at "efficiency" of the system (assuming that the total
> power in is the sum of what's dissipated in the generator and the load)
>
> 20 ohm load, 33%
> 40 ohm load, 50% (what you'd expect)
> 60 ohm load, 60% (hey.. it's more efficient, too)
>
> -> take home message... a "good match" is sort of an artificial thing
> from a power transfer standpoint.. it depends on what you're trying to
> optimize for.
>
> -> Where you get bitten is when "match" varies with frequency... now,
> all of a sudden, you have a system that has a response that varies with
> frequency, which is generally undesirable. When you get up into the
> microwave region, where a transmission line is often many wavelengths
> long, that mismatch can result in remarkably wild fluctuations in gain
> with respect to frequency.
>
Vgen Rgen Rload V load I load P load Pgen Pload/Pgen Ptotal Pload/Ptotal
10 40 20 3.33 0.17 0.56 1.11 0.50 1.67 0.33
10 40 40 5.00 0.13 0.63 0.63 1.00 1.25 0.50
10 40 60 6.00 0.10 0.60 0.40 1.50 1.00 0.60
Jim Lux
Wimpie
> "Sal M. Onella" <sob...@cox.net> wrote in messagenews:7634e01d-847f-48ea...@k3g2000prl.googlegroups.com...
Hello Ralph,
The actual output impedance can be anything, but is mostly not 50
Ohms. If you want it close to 50 Ohms, you have to spend money in
components and design time. As 50 Ohms isn't mostly required, one will
not design for that.
Just as an example, take a hard-driven totem pole or push pull stage
with only a series tuned circuit to suppress harmonics (so the LC
circuit shows zero ohms at the carrier frequency). As the active
devices are used a switches, the output impedance of this arrangement
is almost zero (at least far below 50 Ohms).
When you connect a 50 Ohms quarter-wave cable between the output and
the 50 Ohms load, the amplifier-cable combination has very high output
impedance (quarter wave transformer formula).
For power amplifiers, there is no relation between actual output
impedance and efficiency. When an amplifier is designed for 50 Ohms,
it only means that the amplifier will work correctly when terminated
with 50 Ohms. When you deviate from that, output power may decrease
or increase. This may result in more or less stress on the amplifier's
components.
Sal M. Onella
Thanks Richard. Intuition is that the Zs is near 50 ohms for as many
frequencies as the designer can manage. I am on record (including in
this group) of letting intuition lead me down the path to ruin.
I get from you that there's a presupposition that I know the source
impedance or can easily establish it. Hm-m-m ... not so. One big
problem I see is the need to try to measure power delivered in a
non-50-ohm system with my existing instruments that depend on a 50-ohm
system. I don't have a nominal 75-ohm power meter. Won't putting a
50-ohm meter into a 75-ohm circuit not only read wrong but introduce
reflection losses, too?
I think I'd need a collection of non-inductive load resistors and an
accurate rf ammeter. I'd need to connect them and calculate power at
a few points in every band.
Maybe the papers that Cecil cited for me will fill in the gaps or
suggest other approaches.
"Sal"
Richard Clark
wrote:
>(If I had a very efficient op amp, I could simulate any arbitrary output
>impedance, without dissipating any power in the source)
I can see why this is parenthetical, because it covers a lot of sins
of omission.
First we bang up against the wall of Gain Bandwidth Product. If you
are talking about resistive loads at low power DC, then your statement
is trivially valid.
Second, the ability to "simulate" any arbitrary output (or input for
that matter) impedance for an OP AMP is well defined in the closed
loop gain (which robs from the open loop gain for frequency by
proportion to GBP).
Taking the conventional RF Power Deck of any consumer (Ham) product,
the similarity to an OP AMP is wholly foreign, and for good and
commercial reason. If one were motivated to engineer in the necessary
noise amplifier (a term coined by H.W. Bode who defined this topic of
source Z and applied it to the negative gain or feedback path); then
we would find that the exact same loss is exhibited in the exact same
component(s).
However, by virtue of OP AMP characteristics we would benefit to
vastly better distortion figures, far less spurious content, and
virtually no need for either the conventional impedance transformer,
nor the bandwidth filter that follows the same power deck (provided,
of course, that the drive input is sinusoidal - which it never is,
unfortunately, for this scenario). This novel OP-AMP/Power-Deck
redesign would also confer considerable power supply rejection (that
voltage could sag or rise without appreciable effect) and noise
rejection (the internal noise from other circuitry would not migrate
into the signal output). ALL such benefits are strictly derived from
the amount of negative feedback (not to be confused, as are many
readers to this topic, with the rather ordinary compensation cap in
the last stage).
Why isn't this done as a service to the customer?
Cost.
Again, OP AMP design merits are paid for in lost gain and bandwidth.
The price is found in the amount of negative feedback that goes to
lower the overall amplification. Would you pay for this improved cool
performance to run 10W in the 80M band from a formerly crackly and hot
100W 10-80M band capable source?
Wimpie
Hello Jim,
Other method is injecting a slightly off-carrier frequency signal into
the amplifier (this emulates a constant small VSWR shown to the PA
(wtih 50 Ohms load), but with continuous varying phase). Because of
the difference in frequency, one can measure the forward (towards the
PA) and reverse (reflected by PA) signal with a two channel VSA.
This will give you the PA's complex output impedance.
Tom (K7ITM from my head) did this with a HP89410 with couplers.
Richard Clark
<sob...@cox.net> wrote:
>Thanks Richard. Intuition is that the Zs is near 50 ohms for as many
>frequencies as the designer can manage. I am on record (including in
>this group) of letting intuition lead me down the path to ruin.
Well, if you miss the path, you are certain to be reminded where it
is.
>I get from you that there's a presupposition that I know the source
>impedance or can easily establish it. Hm-m-m ... not so.
It is printed in the specifications. There are other ways to derive
it, of course, and they would merely confirm that number.
>One big
>problem I see is the need to try to measure power delivered in a
>non-50-ohm system with my existing instruments that depend on a 50-ohm
>system. I don't have a nominal 75-ohm power meter. Won't putting a
>50-ohm meter into a 75-ohm circuit not only read wrong but introduce
>reflection losses, too?
It is easier to measure voltage and current and use your calculator.
Now having said that, measuring voltage and current is damned hard at
HF. It is achievable with care, but now we are back into your same
question with many hands pointing at that path to ruin.
As both are difficult (power or voltage times current), you could
trust authority (which confirms what is intuitive), or you could
listen to argument (which at the distance of time and recall becomes
murky and opaque).
>I think I'd need a collection of non-inductive load resistors and an
>accurate rf ammeter. I'd need to connect them and calculate power at
>a few points in every band.
Bravo! This reduces complexity because RF Power resistors of high
accuracy and bandwidth are commercially available. You will need to
practice your skill at mounting to a heat sink, however (another
non-trivial achievement).
>Maybe the papers that Cecil cited for me will fill in the gaps or
>suggest other approaches.
Walt and I have been corresponding over these matters just these past
two weeks. Skimming that content will once again confirm intuition.
Jim Lux
> On Tue, 26 Apr 2011 13:12:23 -0700, Jim Lux <james...@jpl.nasa.gov>
> wrote:
>
>> (If I had a very efficient op amp, I could simulate any arbitrary output
>> impedance, without dissipating any power in the source)
>
> I can see why this is parenthetical, because it covers a lot of sins
> of omission.
yep.. not possible to build such a thing, anymore than one can build a
zero ohm output impedance RF source with any signficant power.
Suggested more as an example that the power dissipation in the source
doesn't necessarily correlate with match, load Z, or anything else in
general.
(You can get pretty darn close at powers less than a watt and HF, though..)
>
> However, by virtue of OP AMP characteristics we would benefit to
> vastly better distortion figures, far less spurious content, and
> virtually no need for either the conventional impedance transformer,
> nor the bandwidth filter that follows the same power deck (provided,
> of course, that the drive input is sinusoidal - which it never is,
> unfortunately, for this scenario). This novel OP-AMP/Power-Deck
> redesign would also confer considerable power supply rejection (that
> voltage could sag or rise without appreciable effect) and noise
> rejection (the internal noise from other circuitry would not migrate
> into the signal output). ALL such benefits are strictly derived from
> the amount of negative feedback (not to be confused, as are many
> readers to this topic, with the rather ordinary compensation cap in
> the last stage).
One can also do a lot of this with various clever schemes if the input
to your PA is coming out of some signal processing. Generically,
predistortion, but it can be so much more.
People have literally spent their lives working out ever more
sophisticated approaches
>
> Why isn't this done as a service to the customer?
>
> Cost.
Like race cars... how fast do you want to go..just bring money
Jim Lux
Yes.. that's another way to do it.
I think one would want variable load impedances for the testing in any
case, because I'll bet that most ham rigs have a load dependent Z.
I've occasionally kicked around the idea of what would it take to do it
simply, especially with nifty devices becoming available to help with
the measurements.. $500 VNAs, $400 power meters that directly read
current and voltage, etc.
What I'm not sure about is whether it is "useful" to know. Consider a
ham with a manual or auto tuner. They'll adjust for minimum reflected
power, which is probably as good as anything else. Hams, as a class,
don't much care about "DC power to RF radiated" efficiency (because the
regulatory requirements are imposed at the "RF output" measurement plane..
Power dissipation in the PA is only a second order concern.. can I plug
it into a standard outlet? Will it get too hot in my car?
backpack QRPers are concerned about DC power consumption, but I'm not
sure they're worried about whether their PA is 30% or 35% efficient.
The people designing battery powered tracking transmitters ARE concerned
about this, as are high power broadcasters (since they're both directly
paying for the supply power and have a "radiated RF power" requirement
to meet their need).
Richard Clark
wrote:
>>> (If I had a very efficient op amp, I could simulate any arbitrary output
>>> impedance, without dissipating any power in the source)
>>
>> I can see why this is parenthetical, because it covers a lot of sins
>> of omission.
>
>yep.. not possible to build such a thing, anymore than one can build a
>zero ohm output impedance RF source with any signficant power.
>Suggested more as an example that the power dissipation in the source
>doesn't necessarily correlate with match, load Z, or anything else in
>general.
>
>(You can get pretty darn close at powers less than a watt and HF, though..)
OP AMPs are a constant of my admiration in the possibilities offered.
That and the signal processing you suggest (plus digital oscillators)
"could" change the playing field - if conventional design weren't so
universally fallen back upon.
Cecil Moore
> Other method is injecting a slightly off-carrier frequency signal into
> the amplifier (this emulates a constant small VSWR shown to the PA
> (wtih 50 Ohms load), but with continuous varying phase). Because of
> the difference in frequency, one can measure the forward (towards the
> PA) and reverse (reflected by PA) signal with a two channel VSA.
> This will give you the PA's complex output impedance.
Unfortunately, the impedance encountered by the off-carrier frequency
signal is probably not the same as the impedance encountered by the
carrier frequency so the results don't correlate and are not very
useful. The carrier frequency has interference components that the off-
carrier signal doesn't encounter.
K7ITM
I agree with Jim. While it's true that if a source (transmitter) is
tuned for maximum output, the output impedance must necessarily be the
conjugate of the load impedance, it is NOT generally the case that the
transmitter is tuned for maximum output. Rather, the transmitter is
tuned for an output that won't destroy the output devices and will
result in acceptable distortion (in the case of a linear amplifier).
There are plenty of cases of sources designed to be loaded with an
impedance far different from their output (source) impedance: the AC
power line, audio amplifiers, ... .
A while back, I set up a couple precision high power directional
couplers so I could measure the output impedance of a couple different
ham rigs. In the case of the rig with vacuum tube output stage, if I
operated the output stage with limited grid drive and tuned the plate
tank for maximum output power, indeed the output impedance was 50
ohms, within the tolerance of my ability to adjust the output for
maximum. But if I increased the grid drive for solid class-C
operation and tuned for the rated output power (which is no longer the
maximum possible power), the impedance seen at the output dropped. If
you work through the pi-network transformation back to the vacuum tube
plates, it's apparent that the plates under those operating conditions
represent a considerably higher source impedance than when things are
tuned for maximum available power (as first described).
But coming back to "Sal's" original question, it's always made sense
to me given the availability of inexpensive 75 ohm line with low loss
to go ahead and use it to feed antennas that have a feedpoint
impedance closer to 75 ohms than to 50 ohms. If you need to provide a
bit of matching at the transmitter end so that the transmitter is
operating correctly, it should be straightforward to do that. But
whether the actual source impedance of the transmitter is one value or
another is really of very little importance. The only time I can
think that it would matter is if you're trying to transmit a very
broadband signal and you don't want power that's reflected at the
transmission-line:antenna interface to re-reflect from the
transmitter:transmission-line interface and go back to the antenna,
delayed by enough to cause a "ghost" (in a television picture), for
example. In such a case, you'll be well served by insuring that the
antenna is well matched to the transmission line so there is an
insignificant reflection there anyway.
Cheers,
Tom
K7ITM
Yep, 'twas me. Using the 89410, I can resolve a signal removed from
the transmitter's output by a very small frequency offset, so it's
well within the bandwidth of the transmitter. With synchronous
averaging, the injected signal could be one or two Hz away from the
transmitted carrier, though it helps a lot to have a transmitter
running from a really "clean" (low phase noise) source if you're going
to do that. The injected signal can be many tens of dB lower in
amplitude than the transmitted signal. As I recall, I was using a
signal a kHz or so away from the carrier, still well within the
bandwidth of the RF power amplifier. One thing you have to be careful
about is either disabling ALC, or operating well outside the ALC loop
bandwidth; you don't want the ALC screwing up your results.
As Wim points out, this setup presents a load to the RF power
amplifier that's indistinguishable from an R+jX load that's
continuously varying, tracing a path around a little circle on a Smith
chart (a very tiny circle, when using signals that are very small
compared with the transmitted power). The rate the circles are traced
out is just the frequency offset between the transmitter and the test
signal.
Cheers,
Tom
K7ITM
Wimpie
Helo Cecil,
Depending on the frequency resolution of your VSA, the frequency of
the injected signal can be well within 1 kHz of the carrier, so LC
filters in the PA will not distort the measurement. In case of a 100W
PA and injection of about 100 mW, the difference in wanted signal and
signal to be rejected is 30 dB (not that large).
If you don't have a VSA, you can still do it (with some more
calculations) by using a diode detector, this will give you the
difference frequencies directly and you can observe the phase
differences on an oscilloscope.
There can be a difference between a very slow load variation (for
example manually changing loads and noticing current and voltage
[including phase]) and the VSA method. This is because of decoupling
capacitors in the power supply or bias circuits (for example RC
combination in the grid circuitry to limit the grid current). If you
vary the load about 10..100 times/s, bias and supply voltages don't
have the time to settle to their steady state.
If the VSA method is basically wrong, I would love to hear why.
JIMMIE
Being aware that this was the nature of my solidstate transceiver I
attempted to use a tuner with one to improve my match to my antenna
system. While I didnt damage my transmitter I did notice that the best
settings of the tuner for TX and RX did not coincide. I was wondering
if anyone else has observed this .
Jimmie
Cecil Moore
> Depending on the frequency resolution of your VSA, the frequency of
> the injected signal can be well within 1 kHz of the carrier, so LC
> filters in the PA will not distort the measurement. In case of a 100W
> PA and injection of about 100 mW, the difference in wanted signal and
> signal to be rejected is 30 dB (not that large).
Would any competent optical physicist suggest that it is valid to
study the conditions associated with interfering coherent light waves
inside an interferometer by introducing an incoherent light source
into the system? Why would any competent RF engineer suggest that the
system source conditions associated with interfering coherent RF waves
can be studied by introducing an incoherent test signal?
Richard Clark
<JIMMIE...@YAHOO.COM> wrote:
>Being aware that this was the nature of my solidstate transceiver I
>attempted to use a tuner with one to improve my match to my antenna
>system.
Hi Jimmie,
This is somewhat cryptic, you would be expected to need a tuner.
>While I didnt damage my transmitter
This transcends cryptic. It's like saying that as you came to a stop
sign, you used your brakes and you didn't damage your car.
>I did notice that the best
>settings of the tuner for TX and RX did not coincide. I was wondering
>if anyone else has observed this .
Two different circuits inside, hence two different loads. Two
different loads, hence two different matches. Some receive paths
might be 50 Ohms, but that value is not as essential as for
transmission.
Wimpie
As this slightly off-carrier frequency signal behaves like a load with
very low VSWR with a cable in between that extends with constant
speed. In other words, the amplifier sees a constant VSWR, but with
changing phase. Small frequency difference results in slow phase
change of VSWR.
Maybe you should read the postings from Tom also (K7ITM)
Wim
PA3DJS
www.tetech.nl
Wimpie
Hello Jimmie,
I noticed this also when I was experimenting with CB equipment and
simple antenna experiments.
For several CB transceivers I could get more output by slightly
mismatching the load as seen by the PA (but in many cases with too
much increase in current consumption).
I tried to use a matcher/tuner (and later a high Q resonator) to
reject other stations close by and then I figured out that when
applying mismatch to the receiver, the S-meter moved further.
The above isn't strange. As PA's are mostly not designed to show 50
Ohms, many receivers are also not designed to show 50 Ohms. I am not
discussing wide band receivers (for example digital or analog video).
Most active devices have lowest noise figure when driven from a source
impedance that is far from the input impedance of the active device.
If you want them to be equal, you need to use feedback and that
complicates the design.
Also filters with significant pass band ripple show, even when
designed for 50 Ohms, significant input reflection when referenced to
50 Ohms
Cecil Moore
> In other words, the amplifier sees a constant VSWR, but with
> changing phase. Small frequency difference results in slow phase
> change of VSWR.
From the IEEE Dictionary:
"impedance -
(1)(A) The corresponding impedance function with p
replaced by jw in which w is real. Note: Definitions
(A) and (B) are equivalent.
(1)(B) The ratio of the phasor equivalent of a steady-
state sine wave voltage ... to the phasor equivalent
of a steady-state sine wave current ...
(1)(C) A physical device or combination of devices
whose impedance as defined in definition (A) or (B)
can be determined. Note: This sentence illustrates
the double use of the word impedance ... Definition
(C) is a second use of 'impedance' and is independent
of definitions (A) and (B)."
The pinging experiment seems to be measuring a physical impedance (1)
(C) the nature of which is unclear. When the amplifier is outputting
power, it seems that the source impedance would be a V/I ratio (1)(B)
which doesn't respond to incoherent signals. Seems to me, you guys are
pinging something other than the source impedance.
--
Jim Lux
Interestingly, it's NOT an incoherent test signal. It's a carefully
chosen coherent test signal with a frequency difference. It's the same
idea as having an interferometer with dithering piezo transducer on a
mirror or a modulator in one of the paths.
Jim Lux
> Hello Cecil,
>
> On 27 abr, 20:13, Cecil Moore <w5...@hotmail.com> wrote:
>> On Apr 27, 10:30 am, Wimpie <wimabc...@tetech.nl> wrote:
>>
>>> Depending on the frequency resolution of your VSA, the frequency of
>>> the injected signal can be well within 1 kHz of the carrier, so LC
>>> filters in the PA will not distort the measurement. In case of a 100W
>>> PA and injection of about 100 mW, the difference in wanted signal and
>>> signal to be rejected is 30 dB (not that large).
>> Would any competent optical physicist suggest that it is valid to
>> study the conditions associated with interfering coherent light waves
>> inside an interferometer by introducing an incoherent light source
>> into the system? Why would any competent RF engineer suggest that the
>> system source conditions associated with interfering coherent RF waves
>> can be studied by introducing an incoherent test signal?
>
> As this slightly off-carrier frequency signal behaves like a load with
> very low VSWR with a cable in between that extends with constant
> speed. In other words, the amplifier sees a constant VSWR, but with
> changing phase. Small frequency difference results in slow phase
> change of VSWR.
>
>
controlled line stretcher (a classic automated load pull setup... see
the stuff from Maury Microwave, for instance)
It's a very clever technique. A variant is used in antenna ranges where
you have a probe with a mismatch in it.
Jim Lux
> Being aware that this was the nature of my solidstate transceiver I
> attempted to use a tuner with one to improve my match to my antenna
> system. While I didnt damage my transmitter I did notice that the best
> settings of the tuner for TX and RX did not coincide. I was wondering
> if anyone else has observed this .
>
Not surprising at all.
Consider the whole black art of adjusting the input match for lowest
noise figure, which may or may not coincide with the largest output
signal for a given input.
All those theorems about matching always have an asterisk about the
assumption that they're reciprocal linear devices with constant
impedances, etc. Start putting nonideal active devices in the mix, and
life gets interesting.
Wimpie
Hello Cecil,
You may try to figure out how the signal injection method functions
(it is a form of active load pulling).
Can you agree with:
it doesn't matter whether:
-power reflect towards the amplifier is caused by load mismatch, or
-power is sent towards the amplifier by means of a phase synchronized
source.
This source is phase synchronized with the PA's exciter, so we have a
steady state system.
We assume small load mismatch (or low injected power towards the PA)
so that the operating point of the PA just changes slightly (to allow
linear approximation).
Now we insert a coupler between the amplifier and the load. This
coupler will measure the forward voltage generated by the PA, plus the
reflected part of the voltage that originates from the phase
synchronised source. Depending on the phase relationship, it can be
more or less then the forward voltage of the PA. If the PA shows 50
Ohms, the coupler's output would not change due to the signal
injection (as no signal is reflected by the PA).
We note the forward coupler's output voltage (both phase and
amplitude).
Now we change the phase relationship between the exciter and the
source that transmits some power toward the PA. Lets change 180
degrees and keep the amplitude the same. We again note the coupler's
output voltage (both phase and amplitude).
The voltage that is reflected by the PA equals half the complex
voltage change because of the phase change. Off course you have to
correct the readings because of the coupler loss. If you know the
signal that is send toward the PA, you can now calculate the complex
output impedance of the PA for small load change around 50 Ohms.
Instead of changing the phase of the source manually, you can do that
continuously and note the couplers output continuously. If you change
the phase of a signal continuously (with certain constant rad/s), the
result is a decrease or increase of frequency.
Assuming some reflection by the PA, the complex output from coupler
rotates around a certain point. That certain point is the result of
the PA's output power and the rotating vector is the result from the
injected signal that is reflected by the PA (back towards the load).
With a VSA you can discriminate between the voltage component from the
PA itself and the reflected component (with slightly different
frequency). With a normal spectrum analyser, you can only determine
the magnitude of the PA's reflection coefficient (or VSWR as you
like). Given the dynamic range of today's equipment, you can inject a
very low level signal that may mimic load mismatch well below VSWR =
1.1.
With respect to the impedance concept, we as amateurs do not use
steady state signals, as they contain no information. We modulate them
and are still using the impedance concept, despite the definitiones
you showed.
As long as the signal that is injected is well within the pass band of
the PA and it sufficiently low to allow linear approximation, the
concept of superposition and concept of impedance still holds. But if
you feel more confident with the manual phase change, or using two
known loads with known slight mismatch, I have nothing against it.
But if you have a VSA, some couplers and signal source at hand, it
may save lots of time.
With kind regard,
Wim
PA3DJS
www.tetech.nl
Jim Lux
> With a VSA you can discriminate between the voltage component from the
> PA itself and the reflected component (with slightly different
> frequency). With a normal spectrum analyser, you can only determine
> the magnitude of the PA's reflection coefficient (or VSWR as you
> like). Given the dynamic range of today's equipment, you can inject a
> very low level signal that may mimic load mismatch well below VSWR =
> 1.1.
>
One needs to have an analyzer with a narrow band detector for this, though.
Inexpensive analyzers like the TAPR VNA have untuned detectors, so the
PA's main signal will screw things up.
The N2PK uses a form of direct conversion detector, so your test signal
would have to be far enough away from the main signal so that the LPF in
the detector filters it out. The original N2PK design uses, I think, a
100 Hz filter, and the adc samples at 15.36 kHz with a digital filter.
The overall BW is something like 5 Hz, so putting your test signal 1kHz
away would probably work.
tom
I wonder if this should be applied to these sorts of projects -
I have a 1 watt kit that I haven't put together yet. Several others in
our club have and they love them. Especially nice for CW.
tom
K0TAR
tom
> it doesn't matter whether:
> -power reflect towards the amplifier is caused by load mismatch, or
> -power is sent towards the amplifier by means of a phase synchronized
> source.
>
> This source is phase synchronized with the PA's exciter, so we have a
> steady state system.
>
> We assume small load mismatch (or low injected power towards the PA)
> so that the operating point of the PA just changes slightly (to allow
> linear approximation).
Thanks Wimpie.
It sometimes takes a bit to get me to realize this type of thing isn't
hard, it's actually simple. My old brain tries to obfuscate things from
itself sometimes.
It Calc 100. Make the difference small.
tom
K0TAR
Cecil Moore
> With respect to the impedance concept, we as amateurs do not use
> steady state signals, as they contain no information. We modulate them
> and are still using the impedance concept, despite the definitiones
> you showed.
Trouble is, the impedance in IEEE definition (1)(B) doesn't *cause*
reflections. If the actual source impedance matches IEEE definition (1)
(B), the presumption that source impedance will *cause* a reflection
is invalid.
Walter Maxwell argues that the actual source impedance of an RF
amplifier is in reality a V/I ratio, i.e. it agrees with IEEE
definition (1)(B).
walt
Cecil suggested reading Chapter 19A in Reflections to view the results
of my extensive measurements of the output resistance (impedance) of
RF power amps, but except for Jim and Richard, it appears that the
others have not. Actually, Chapter19A is an addition to Chapter 19,
which when taken completely will provide some information that will
hopefully change some minds concerning the maximum power delivered. It
should be understood that 'maximum' power delivered is that power
delivered with a specified level of drive.
For example, if the drive level is set to deliver a maximum of 100w,
and the pi-network is adjusted to deliver that maximum power into its
load, the source resistance (impedance) will be the (complex)
conjugate of the load impedance. We're not talking here about the
very maximum power that the amp can deliver, with max drive, max plate
current, etc.
If you review the 19A portion of you will see beyond a doubt that the
conjugate match exists between the output of the pi-network and its
complex load impedance, and that the maximum power delivered at the
drive level that allows only 100w to be delivered as the maximum.
Further review of all the data presented there will also show that the
output resistance of the amp is non-dissipative, while the dissipative
resistance is that between the cathode and plate. The reason the
efficiency of the amps can exceed 50 percent is because the cathode to-
plate resistance is less than the non-dissipative output resistance,
where that R = E/I appearing at the output of the pi-network.
Walt, W2DU
I hope the review of my measured data will clear up some of the
confusion concerning the output resistance (impedance) of the RF power
amp.
Walt, W2DU
walt
Cecil suggested reading Chapter 19A in Reflections to view the results
of my extensive measurements of the output resistance (impedance) of
RF power amps, but except for Jim and Richard, it appears that the
others have not. Actually, Chapter19A is an addition to Chapter 19,
which when taken completely will provide some information that will
hopefully change some minds concerning the maximum power delivered. It
should be understood that 'maximum' power delivered is that power
delivered with a specified level of drive.
For example, if the drive level is set to deliver a maximum of 100w,
and the pi-network is adjusted to deliver that maximum power into its
load, the source resistance (impedance) will be the (complex)
conjugate of the load impedance. We're not talking here about the
very maximum power that the amp can deliver, with max drive, max plate
current, etc.
If you review the 19A portion of you will see beyond a doubt that the
conjugate match exists between the output of the pi-network and its
complex load impedance, and that the maximum power delivered at the
drive level that allows only 100w to be delivered as the maximum.
Further review of all the data presented there will also show that the
output resistance of the amp is non-dissipative, while the dissipative
resistance is that between the cathode and plate. The reason the
efficiency of the amps can exceed 50 percent is because the cathode to-
plate resistance is less than the non-dissipative output resistance,
where that R = E/I appearing at the output of the pi-network.
The earlier portion of Chapter 19, that appears in Reflections 2, can
be downloaded from my web page at www.w2du.com, click on 'Read
Chapters from Reflections 2', and select Chapter 19.
Wimpie
> On Apr 27, 6:33 pm, Wimpie <wimabc...@tetech.nl> wrote:
>
> > With respect to the impedance concept, we as amateurs do not use
> > steady state signals, as they contain no information. We modulate them
> > and are still using the impedance concept, despite the definitiones
> > you showed.
>
> Trouble is, the impedance in IEEE definition (1)(B) doesn't *cause*
> reflections. If the actual source impedance matches IEEE definition (1)
> (B), the presumption that source impedance will *cause* a reflection
> is invalid.
Are you familiar with the concept of S-parameters where you determine
impedance by measuring of reflection coefficient?
>
> Walter Maxwell argues that the actual source impedance of an RF
> amplifier is in reality a V/I ratio, i.e. it agrees with IEEE
> definition (1)(B).
> --
> 73, Cecil, w5dxp.com
> "Halitosis is better than no breath at all.", Don, KE6AJH/SK
Wim
Cecil Moore
> Are you familiar with the concept of S-parameters where you determine
> impedance by measuring of reflection coefficient?
Exactly how do you determine the s-parameters for a single-port black
box? It is my understanding that an s-parameter analysis requires an
input port and an output port to be able to measure the parameters.
Where is the input port on an RF source?
What you seem to be measuring is the effect of one or more physical
impedance discontinuities existing in an environment of interference.
Is what you are measuring the actual dynamic source impedance? If I
understand correctly what Walter Maxwell is saying is that whatever
combinations of physical impedance discontinuities from which you guys
are reflecting your test signals, it/they are *not the source
impedance* which is a V/I ratio that originates in the source. A V/I
ratio and a physical impedance discontinuity do not yield the same
reflection coefficients for a 2-port device.
Wimpie
On 2 mayo, 14:37, Cecil Moore <w5...@hotmail.com> wrote:
> On May 2, 5:33 am, Wimpie <wimabc...@tetech.nl> wrote:
>
> > Are you familiar with the concept of S-parameters where you determine
> > impedance by measuring of reflection coefficient?
>
> Exactly how do you determine the s-parameters for a single-port black
> box? It is my understanding that an s-parameter analysis requires an
> input port and an output port to be able to measure the parameters.
> Where is the input port on an RF source?
I used a 2 port VNA frequently for antenna measurements, with the
difference that a single port calibration takes less time than a full
two port calibration.
>
> What you seem to be measuring is the effect of one or more physical
> impedance discontinuities existing in an environment of interference.
> Is what you are measuring the actual dynamic source impedance? If I
> understand correctly what Walter Maxwell is saying is that whatever
> combinations of physical impedance discontinuities from which you guys
> are reflecting your test signals, it/they are *not the source
> impedance* which is a V/I ratio that originates in the source. A V/I
> ratio and a physical impedance discontinuity do not yield the same
> reflection coefficients for a 2-port device.
> --
> 73, Cecil, w5dxp.com
> "Halitosis is better than no breath at all.", Don, KE6AJH/SK
I would recommend you to measure something yourselves, or put it into
a simulation. You will see that it doesn't matter whether you use a
deltaV/deltaI setup (complex values, not magnitudes) or reflection
(time varying phase of VSWR) measurement. Just try to explore other
paths and discover other insights.
It doesn't matter when you are measuring a single port device that
contains a generator in it (as long as your VNA setup is able to
distinguish between the output generated by the single port device and
the reflection towards the VNA).
There is similarity with measuring antenna impedance (single port
measurement) when close to (broadcast) stations. Your antenna is a
generator in that case. You can't use the non-coherent type of VSWR
analyzers as the detector detects the signal from the broadcast
station also. However when using a device with a coherent
(multiplying) detector you can, as the detector doesn't respond to the
output because of the (broadcast) station.
With Tom's HP89410 setup, the injected signal for S11 can be well
within the modulation bandwidth of SSB (that means well within 1 kHz
of the transmitter's carrier frequency).
When you have good understanding of diode detectors, you can even do
it without a VNA by using heterodyning and put your focus on the phase
and amplitude of the beat frequency.
With kind regards,
Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me very likely
Cecil Moore
> I would recommend you to measure something yourselves, ...
Please take a look at the numerous measurements performed by Walter
Maxwell.
Let's take a simple example of the single-port s-parameter results.
There exists a black box with one exposed port. The impedance is
measured to be 50+j0 and s11 is assumed to be zero when driven by a 50
ohm source. There could be a 50 ohm dummy load in the box but there is
not.
Actually, inside the black box is a 1/4WL Z0=100 ohm transmission line
routed to a 200 ohm resistor. s11 is certainly not zero and is
measured, using the 2-port procedure, to be 0.3333.
Which s11 is correct, 0.0000 or 0.3333?
Wimpie
Hello Cecil,
Assuming that your black box is a 100 Ohms quarter-wave line with 200
Ohms termination (all inside the black box), S11 (50 Ohms based) = 0,
as the input impedance of your black box is 50 ohms. If you like, you
may see your room temperature black box as a source with about –174dBm/
Hz output and an output impedance of 50 Ohms.
Wim
Cecil Moore
> Assuming that your black box is a 100 Ohms quarter-wave line with 200
This is what happens when one changes math models in mid-stream. s11
is NOT zero at the mouth of a stub. The impedance looking into a stub
is IEEE definition (1)(B) and cannot have an s11 of zero unless the Z0
of the stub is infinite, which it is not.
Thanks for proving my point. The single-port s11 is completely
different from the dual-port s11 and that is most likely what is
happening with your attempts to measure the source impedance of an RF
amplifier. How can you possibly promote an experimental approach where
s11 changes by an infinite percentage depending on whether one
measures it as a single-port parameter vs a dual-port parameter???
Wimpie
> On May 2, 9:57 am, Wimpie <wimabc...@tetech.nl> wrote:
>
> > Assuming that your black box is a 100 Ohms quarter-wave line with 200
> > Ohms termination (all inside the black box), S11 (50 Ohms based) = 0,...
>
> This is what happens when one changes math models in mid-stream. s11
> is NOT zero at the mouth of a stub. The impedance looking into a stub
> is IEEE definition (1)(B) and cannot have an s11 of zero unless the Z0
> of the stub is infinite, which it is not.
From your text I didn't conlude it is a stub (but just a series line
section that transforms 50 OHms into 200 Ohms).
>
> Thanks for proving my point. The single-port s11 is completely
> different from the dual-port s11 and that is most likely what is
> happening with your attempts to measure the source impedance of an RF
> amplifier. How can you possibly promote an experimental approach where
> s11 changes by an infinite percentage depending on whether one
> measures it as a single-port parameter vs a dual-port parameter???
Please explain (or someone else), as I don't understand anything of
the above with regards to a PA. You described a single-port device and
now starts talking about a two-port device.
If you want to prove that for a two port device the impedance
corresponding to S11 may not be equal to the input or output impedance
of port 1, you are right. When S12*S21 isn't zero, the input
impedance depends on the termination of port 2. This is just S-
parameter math, nothing magic.
For the PA case, the active device is the termination for the two-port
matching network, so in that case S11 measurement equals the input/
output impedance of the PA (just a single port measurement). See it as
the output impedance of the active device is connected to port 2 (of
the matching network), and the VNA (or load) is connected to port 1
(that is the output side of the matching network).
The PA reduces to just a single-port network with a source in it
(compare it with the antenna example where a transmitter is in the
vicinity). As long as the injected signal (or slight mismatch
connected to transmission line with increasing length) is small, you
can apply the small signal approach.
As suggested earlier, you may dive into active load pulling:
http://www.focus-microwaves.com/template.php?unique=232
It doesn't matter whether you show some mismatch to a PA, or a perfect
matched load with a small source in it representing the signal that is
reflected towards the amplifier.
> --
> 73, Cecil, w5dxp.com
> "Halitosis is better than no breath at all.", Don, KE6AJH/SK
Kind regards,
Wim
PA3DJS
www.tetech.nl
K7ITM
> Chapters from Reflections 2', and select Chapter 19.
>
> I hope the review of my measured data will clear up some of the
> confusion concerning the output resistance (impedance) of the RF power
> amp.
>
> Walt, W2DU
When a source is tuned (e.g. through a pi network or any other
matching network between a PA and its load) such that maximum power is
delivered to the load, it's axiomatic that the source impedance is the
complex conjugate of the load. Is there really a need for a whole
chapter for that?
You say, "We're not talking here about the very maximum power that the
amp can deliver, with max drive, max plate current, etc." I beg to
differ. That is EXACTLY what we are talking about. We're talking
about modern amplifiers that would be destroyed if not for protection
circuits, if they were loaded with a load that resulted in maximum
dissipation in the load. We're talking about even old amplifiers with
enough grid drive that they COULD be loaded to higher power output,
but for reasons of wanting the active devices to survive for a
reasonable length of time (or possibly other reasons), are not loaded
so heavily. If you want to exclude such amplifiers from
consideration, then I would hope none would disagree about the
relationship of the load impedance and the source impedance. You
needn't have made any measurements to convince me of that.
Cheers,
Tom
Cecil Moore
> Please explain (or someone else), as I don't understand anything of
> the above with regards to a PA. You described a single-port device and
> now starts talking about a two-port device.
If you recognize the example as a two-port device, you correctly
measure an s11 of 0.3333, (100-50)/(100+50). If you happen to overlook
the second port to which the 200 ohm resistor is attached and treat
the example as a single-port device, you measure an s11 of 0.0000,
(50-50)/(50+50). The example has not changed between those two
measurements so which s11 is correct? Doesn't that fact give you pause
to wonder if you are making essentially the same mistake with the PA
measurements?
Wimpie
On 2 mayo, 23:17, Cecil Moore <w5...@hotmail.com> wrote:
> On May 2, 1:41 pm, Wimpie <wimabc...@tetech.nl> wrote:
>
> > Please explain (or someone else), as I don't understand anything of
> > the above with regards to a PA. You described a single-port device and
> > now starts talking about a two-port device.
>
> If you recognize the example as a two-port device, you correctly
> measure an s11 of 0.3333, (100-50)/(100+50). If you happen to overlook
> the second port to which the 200 ohm resistor is attached and treat
> the example as a single-port device, you measure an s11 of 0.0000,
> (50-50)/(50+50). The example has not changed between those two
> measurements so which s11 is correct?
You want to know the output/input impedance of your black box, this
means you should look into the 100 Ohms line with 200 Ohms
termination. We will see 50 Ohms, no matter the reference impedance of
the VNA. If it was a 75 Ohms VNA, it would read S11 = -0.2. What the
200 Ohms resistor sees is not important if you just want to know the
behavior of the single-port black box.
> Doesn't that fact give you pause to wonder if you are making essentially the same mistake with the PA
> measurements?
I am very sorry Cecil, but I still don't see the point where the
discussed method may go wrong.
The only thing I could think of is that you have in mind a setup where
the input of the PA is port number 1 (and the output is port number
2 ) and you carry out a full-port measurement. For a full class-A,
AB, non-saturated PA, this may give useful results. When S12*S21 <<
1, S22 will equal the output impedance, otherwise you have to do the
match.
In real world many amplifiers do not behave as a linear system (food
for discussion) and then the two-port setup will fail, as during S22
measurement the input port is terminated with 50 Ohms (so there is no
source that provides output).
Therefore carrying out a single-port measurement with a slightly off-
carrier frequency (to create non-coherence) under required output
conditions, will result in a meaningful output impedance. As
mentioned before, doing a (slow) manual load pull measurement may give
different results because of bias and supply voltage variations.
> 73, Cecil, w5dxp.com
> "Halitosis is better than no breath at all.", Don, KE6AJH/SK
Cecil Moore
> I am very sorry Cecil, but I still don't see the point where the
> discussed method may go wrong.
Everyone seems to be charging ahead, willy-nilly, without seeing the
point which is that there are other effects present besides
reflections.
> Therefore carrying out a single-port measurement with a slightly off-
> carrier frequency (to create non-coherence) under required output
> conditions, will result in a meaningful output impedance.
Nope, it won't because virtual impedances don't cause reflections.
Only physical impedance discontinuities cause reflections. The rest of
the redistribution of RF energy is caused by the superposed
interaction between forward and reflected waves, i.e. interference
effects. Most hams do not understand the role of interference in the
redistribution of RF energy. Hope this helps.
http://micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/waveinteractions/index.html
Please pay close attention to the last paragraph. "... when two waves
of equal amplitude and wavelength that are 180-degrees ... out of
phase with each other meet, they are not actually annihilated, ... All
of the photon energy present in these waves must somehow be recovered
or redistributed in a new direction, according to the law of energy
conservation ... Instead, upon meeting, the photons are redistributed
to regions that permit constructive interference, so the effect should
be considered as a redistribution of light waves and photon energy
rather than the spontaneous construction or destruction of light."
You guys are presuming that reflections are the only thing you are
seeing and that is just not true. You are also seeing interference
effects without realizing it so your conclusions are doomed to failure
unless you can differentiate between constructive/destructive
interference and reflected waves. Since there has been no mention of
interference effects, I am forced to conclude that you guys are
ignorant of such effects.
--
Wimpie
> On May 2, 5:23 pm, Wimpie <wimabc...@tetech.nl> wrote:
>
> > I am very sorry Cecil, but I still don't see the point where the
> > discussed method may go wrong.
>
> Everyone seems to be charging ahead, willy-nilly, without seeing the
> point which is that there are other effects present besides
> reflections.
>
> > Therefore carrying out a single-port measurement with a slightly off-
> > carrier frequency (to create non-coherence) under required output
> > conditions, will result in a meaningful output impedance.
>
> Nope, it won't because virtual impedances don't cause reflections.
> Only physical impedance discontinuities cause reflections. The rest of
> the redistribution of RF energy is caused by the superposed
> interaction between forward and reflected waves, i.e. interference
> effects. Most hams do not understand the role of interference in the
> redistribution of RF energy. Hope this helps.
The signal injection is just a way of emulating a non 50 Ohm
termination where you need to change load and or cable length. By
using slightly off-carrier frequency, you emulate a changing phase of
reflection coefficient. That emulated reflected signal goes to the PA
and interferes with the forward signal (produced by the PA). That
interference is required as this modifies current and voltage at the
PA's active device.
>
> http://micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/w...
>
> Please pay close attention to the last paragraph. "... when two waves
> of equal amplitude and wavelength that are 180-degrees ... out of
> phase with each other meet, they are not actually annihilated, ... All
> of the photon energy present in these waves must somehow be recovered
> or redistributed in a new direction, according to the law of energy
> conservation ... Instead, upon meeting, the photons are redistributed
> to regions that permit constructive interference, so the effect should
> be considered as a redistribution of light waves and photon energy
> rather than the spontaneous construction or destruction of light."
To be honest, I, and my ham friends, never had to use photon theory to
solve problems in both amateur and professional RF-Engineering. So I
think this doesn’t contribute to the discussion.
> You guys are presuming that reflections are the only thing you are
> seeing and that is just not true. You are also seeing interference
> effects without realizing it so your conclusions are doomed to failure
> unless you can differentiate between constructive/destructive
> interference and reflected waves. Since there has been no mention of
> interference effects, I am forced to conclude that you guys are
> ignorant of such effects.
Worth to try yourself:
1. Set up a simple linear circuit in simulation (for example AC
voltage source some impedance in series). You can use whatever (free)
program you like. Use a program that allows transient (non-linear)
simulation as you are working with two frequencies at the same time.
I used Beige Bag spice A/D professional version 4 (not free)
2. Determine the output impedance via load pulling (for various loads
if you like)
3. Do the same with the "VNA" setup (with the PA producing RF output).
You need to observe the envelope variations or you have to insert a
narrow band filter to suppress the carrier. I prefer observing the
sinusoidally changing envelope (as this shows distortion
immediately (intermod products) and saves me the settling time for the
filters).
In simulation you have ideal current meters, so if you use voltage and
current directly (and drop the directional couplers) you can skip the
conversion from reflection coefficient to impedance.
Here you will see that the output impedance is independent of VSWR
(as we started with a linear circuit) and both methods give same
results.
If you like, you could now implement a real PA in simulation and do
the simulation again. Note that tuning the amplifier in simulation is
(very) time consuming. You will notice that, depending on the
amplifier you implemented, impedance will change with increasing VSWR
presented to the amplifier (valid for both methods).
I did all the above with various circuits to prove the fitness of the
method. For some circuits there was a difference in result. After
evaluation, that difference was caused by changes in (bias) supply
during "manual" load pulling (that initially did not happen during the
slightly off-carrier signal injection). With stiff bias conditions,
the difference disappeared to within the accuracy limits.
> --
> 73, Cecil, w5dxp.com
> "Halitosis is better than no breath at all.", Don, KE6AJH/SK
With kind regards,
Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me in most cases.
Cecil Moore
> To be honest, I, and my ham friends, never had to use photon theory to
> solve problems in both amateur and professional RF-Engineering. So I
> think this doesn’t contribute to the discussion.
Knowing that EM RF waves must necessarily obey the laws of photon
physics can save one from all sorts of technical blunders. But feel
free to blunder on.
John KD5YI
> 73, Cecil, w5dxp.com
> "Halitosis is better than no breath at all.", Don, KE6AJH/SK
So, a heart attack is better than no heart at all?
Wimpie
Hello Cecil,
Maybe you should change your attitude from "it is wrong" to "it may be
true". This might open your mind to study other insights.
When measuring an RF current in a PA, everybody ignores the
quantisation of charge. Is this lack of knowledge?
The frequency of our waves is sufficiently low to ignore any
quantisation effects due to photons. Is this lack of knowledge? In my
opinion a good Engineer knows what he can ignore and what not to
efficiently solve a certain task.
Starting a discussion about photons, in my opinion, is a way to reduce
the S/N ratio of the discussion.
With kind regards,
Wim
PA3DJS
www.tetech.nl.
Cecil Moore
> Maybe you should change your attitude from "it is wrong" to "it may be
> true". This might open your mind to study other insights.
When an s11 measurement is wrong by an infinite percentage, I cannot
see how it could possibly be "true" - sorry.
Cecil Moore
Don Hubbard was my best friend and that was one of his pet sayings. I
assume he meant it was better to have bad breath than to be an SK,
which he is.
--
dave
> On 3 mayo, 02:09, Cecil Moore <w5...@hotmail.com> wrote:
>
>
>
> > On May 2, 5:23 pm, Wimpie <wimabc...@tetech.nl> wrote:
>
> > > I am very sorry Cecil, but I still don't see the point where the
> > > discussed method may go wrong.
>
> > Everyone seems to be charging ahead, willy-nilly, without seeing the
> > point which is that there are other effects present besides
> > reflections.
I think most of the discussion is ignoring lots of things... first of
all you must answer the question; Does it really matter? If it does,
then i will assume that you are a designer and know how to apply the
tube characteristics to come up with a design that matches your
selected tube to the expected load... but does that process really
ever describe the 'output impedance'?? Then you must also consider
the tuning parameters employed... sure, you can measure the output
impedance at a given operating point, but answer the question again;
Does it really matter? Or do you need to measure it over a wide range
of operating conditions? Every operator i have seen tune up one of my
amps has done it a little bit different... heck, i don't even do it
exactly the same twice in a row i bet. So when i am running an amp
into a switchable set of 7 different antenna combinations on a given
band, can tune from one end of the band to the other without touching
the settings, and can make an infinite number of small adjustments to
the drive, tune, and load settings, and on some bands can tweak a
tuner after the amp to 'make it happier', do I really care what the
'output impedance' really is? As long as the matching network
provides adequate adjustment so i can get out the desired power into
my various loads while keeping the tubes within their operating
limits, do i really care what the 'output impedance' really is at any
one set of conditions that i may never exactly duplicate again? I
think not. So this boils down to an academic discussion, and as in
many cases where no one has, or can, make an exact statement of the
problem the specific answer remains elusive. So consider this, until
you have a complete statement of the problem you will never be able to
derive a value that any two of you will agree on, let alone actually
try to set up a measurement of.
John KD5YI
Hi, Wim -
I believe his confusion is the one-port vs two-port problem. I don't
have a problem with your explanation. But, I think he is throwing in a
port where he should not.
Cheers,
John
Cecil Moore
> So this boils down to an academic discussion, and as in
> many cases where no one has, or can, make an exact statement of the
> problem the specific answer remains elusive.
When one's basic premises violate the laws of physics, it is very
difficult to come up with a valid answer.
Cecil Moore
> I believe his confusion is the one-port vs two-port problem. I don't
> have a problem with your explanation. But, I think he is throwing in a
> port where he should not.
The second port is actually there on the other side of the black box,
but it has been overlooked. Doesn't it give anyone pause to realize
that the s11 parameter changes by an infinite percentage depending
upon whether it is measured as a one-port system or as a two-port
system? If that can happen with a passive black box, consider how much
more complicated it might be with an RF source included in the black
box.
John KD5YI
> This group has presented members with valuable lessons in antennas and
> transmission lines, like how to measure, how to match, etc.
>
> Something I haven't seen is a discussion of the source impedance of
> the transmitter. My curiosity was piqued today as I took some baby
> steps into EZNEC. A particular antenna had such-and-such VSWR if fed
> with a 50-ohm cable and a different value if fed with a 75-ohm cable.
No, the antenna still has the same characteristics. You changed the
impedance at a point remote from the antenna.
> While this is hardly news, it got me wondering whether a 75-ohm cable
> will load the transmitter the same. Doesn't seem like it.
It depends on a number of factors. For example, use a .5 wavelength 75
ohm cable to feed a 50 ohm resistance. Your transmitter happily thinks
it is 50 ohms. Because it is.
> My point: Using 75-ohm cable to improve the match at the antenna
> won't help me ... IF ... I suffer a corresponding loss due to
> mismatch at the back of the radio. My HF radios, all solid state,
> specify a 50 ohm load. As necessary, I routinely use an internal
> autotuner and either of two external manual tuners. (I'm aware of the
> published 1/12 wavelength matching method.)
>
> Wisdom in any form would be appreciated. Thanks.
>
> "Sal"
> (KD6VKW)
Transmitters don't have a clearly defined output impedance and, whatever
output impedance is there, doesn't necessarily mean anything.
Your main concern is to provide a 50 ohm load for the transmitter to see.
Actually, it the feed line and load are both 75 ohms, you will only see
a 1.5:1 SWR. Don't worry about it.
Cheers,
John
John KD5YI
> On May 3, 6:29 pm, John KD5YI<soph...@invalid.org> wrote:
>> I believe his confusion is the one-port vs two-port problem. I don't
>> have a problem with your explanation. But, I think he is throwing in a
>> port where he should not.
>
> The second port is actually there on the other side of the black box,
> but it has been overlooked. Doesn't it give anyone pause to realize
> that the s11 parameter changes by an infinite percentage depending
> upon whether it is measured as a one-port system or as a two-port
> system?
No. The experiment is designed as a one-port experiment. You have broken
the law and must go to jail without passing GO.
"Having cancer of the colon is better than no colon at all."
Wimpie
Hello Dave and John,
When you look to my first posting to this thread, you may conclude
that I agree with you.
We had such a discussion about a year ago where I stated that most RF
amplifiers do not have 50 Ohms output impedance. That statement was
heavily disputed by some persons. I tried to support that statement
with simulations, but without any success.
Regarding the "academic discussion" I also agree. In my professional
career where I designed several RF PA's, only 2 times the output
impedance of the amplifier was of importance. In one of these cases I
couldn't meet the specs and had to insert attenuation (some waste of
power…).
The thing I don't like is that some people criticize methods used by
some of the group members without a solid foundation.
Regarding the two-port single-port issue. One can setup a reasoning
based on a two-port setup, but that significantly complicates the
matter without giving any additional insight. I tried to keep it
simple by referencing to a VSWR measurement (with an antenna analyzer)
of an antenna when a strong transmitter is nearby, but it seems I
wasn't clear enough for all people following this thread.
With kind regards,
Wim
PA3DJS
www.tetech.nl
Remove abc first before setting free the pigeon.
Richard Clark
wrote:
>only 2 times the output
>impedance of the amplifier was of importance.
What a strange ellipsis.
In other words if this statement bears upon the discussion, then it
deserves a fuller context. Why was it of importance?
> In one of these cases I
>couldn't meet the specs and had to insert attenuation (some waste of
>power…).
Ed McMahon: "How much power?"
Just trying to get some perspective on this power sourced in the PA.
73's
Richard Clark, KB7QHC
Cecil Moore
> Regarding the two-port single-port issue. One can setup a reasoning
> based on a two-port setup, but that significantly complicates the
> matter without giving any additional insight. I tried to keep it
"Everything should be made as simple as possible, but not simpler."
Albert Einstein
Here's my earlier example:
Source-----Z0=50----x----1/4WL Z0=100----200 ohm load
s11 is measured at point x equal to 0.3333 and also 0.3333 at the
load. Nowhere is s11 equal to 0.0000. Put everything to the right of
point x into a black box and s11 measures to be 0.0000 under exactly
the same conditions??? And you guys want all of us to trust that
measurement enough to predict the disputed source impedance of an RF
amp when it cannot even predict the load impedance in the above very
simple passive circuit?
There are reflected waves at point x (s11*a1) that are equal in
magnitude and 180 degrees out of phase with the reflected waves
transmitted back from the load (s12*a2). The two waves undergo
destructive interference at point x which creates a V/I ratio of 50 at
point x. But the absence of *net* reflected energy at point x does not
mean that there are no reflections at point x. There are actually two
sets of reflections at point x that mask any attempt to determine the
actual value of the load impedance by measuring s11 when the system is
installed inside a black box. It is foolish to presume that there are
no similar interference patterns inside an RF amp. In fact, the only
condition where there is no interference inside a simple voltage
source is when there are no reflections or the reflections are
orthogonal to the source signal.
There is a good discussion of the role of interference in the creation
of virtual impedances in section 4.3 of "Reflections", by Walter
Maxwell. Even though a lot RF engineers scoff at the laws of EM wave
physics from the field of optics, the best explanation of interference
I have ever read is the chapter by the same name in "Optics", by
Hecht. Another good chapter in "Optics" is "The Superposition of
Waves".
Wimpie
Hello Cecil,
Basically, it doesn't matter what is inside the box. It can be fully
characterized by its impedance versus frequency (small signal
approach).
I am fully aware of that there are (back and forth) reflections in the
quarter wave line inside the black box. However in case of a black
box, you don't know that (black box principle).
Same thing happens in most narrow band antennas. The antenna wire
itself may be subjected to (for example) VSWR = 10, though the
impedance can be 50 Ohm (and therefore doesn’t introduce reflection in
a 50 Ohms feed line). I know that this 50 Ohms is the result of
interfering waves/signals, but an MFJ 259B antenna analyzer, or my
FT7B doesn't (and doesn't need to know it).
One could only guess what is a black box based on its impedance versus
frequency curve, S11 curve or Time domain response (I had to do this
in school).
So if we decide to open the black box and we put the reference plane
inside the box, we get a new thread, something like: "why has a PA
certain output impedance?", or "what is the large signal output
impedance of valves, BJT, FET, etc?".
Jim Lux
>
>
> Transmitters don't have a clearly defined output impedance and, whatever
> output impedance is there, doesn't necessarily mean anything.
>
> Your main concern is to provide a 50 ohm load for the transmitter to see.
actually, to provide an *acceptable* load for the transmitter that
maximizes radiated RF power. That might not be 50 ohms.. and I don't
know that we actually care what it is, unless we're designing amplifiers.
50 ohms is basically a "standard test condition" so that you can compare
amplifiers, and it happens to be convenient that everything is made with
the same standard impedance. That way, your test load (which will
dissipate a fair amount of power in some tests) can be located somewhere
different, and connected by a length of transmission line with the same
impedance, so the test at the amplifier output reference plane is still
valid.
Cecil Moore
> Basically, it doesn't matter what is inside the box. It can be fully
> characterized by its impedance versus frequency (small signal
> approach).
If what is inside the box doesn't matter, why waste time trying to
measure what is inside the RF amp box? What can be "fully
characterized" is the box's relationship to everything outside of the
box. Exactly what is inside the box cannot be ascertained at all while
we handicap ourselves with a blinders. When we actually look inside an
RF amp, we find active non-linear devices upon which we all have been
warned about trying to use linear assumptions.
It doesn't matter what is inside the box when *conditions outside of
the box* are being considered. EE201. What is inside the box matters
considerably for *conditions inside the box*. For instance, replacing
a 50 ohm dipole with a 50 ohm lumped circuit doesn't change things
between the load and the source. But it causes an almost infinite
change from the load out to the rest of reality.
Anything else I could say on this subject would be repetition.
walt
> On May 2, 5:23 pm, Wimpie <wimabc...@tetech.nl> wrote:
>
> > I am very sorry Cecil, but I still don't see the point where the
> > discussed method may go wrong.
>
> Everyone seems to be charging ahead, willy-nilly, without seeing the
> point which is that there are other effects present besides
> reflections.
>
> > Therefore carrying out a single-port measurement with a slightly off-
> > carrier frequency (to create non-coherence) under required output
> > conditions, will result in a meaningful output impedance.
>
> Nope, it won't because virtual impedances don't cause reflections.
> Only physical impedance discontinuities cause reflections. The rest of
> the redistribution of RF energy is caused by the superposed
> interaction between forward and reflected waves, i.e. interference
> effects. Most hams do not understand the role of interference in the
> redistribution of RF energy. Hope this helps.
>
>
> Please pay close attention to the last paragraph. "... when two waves
> of equal amplitude and wavelength that are 180-degrees ... out of
> phase with each other meet, they are not actually annihilated, ... All
> of the photon energy present in these waves must somehow be recovered
> or redistributed in a new direction, according to the law of energy
> conservation ... Instead, upon meeting, the photons are redistributed
> to regions that permit constructive interference, so the effect should
> be considered as a redistribution of light waves and photon energy
> rather than the spontaneous construction or destruction of light."
>
> You guys are presuming that reflections are the only thing you are
> seeing and that is just not true. You are also seeing interference
> effects without realizing it so your conclusions are doomed to failure
> unless you can differentiate between constructive/destructive
> interference and reflected waves. Since there has been no mention of
> interference effects, I am forced to conclude that you guys are
> ignorant of such effects.
> --
> 73, Cecil, w5dxp.com
> "Halitosis is better than no breath at all.", Don, KE6AJH/SK
During 1991 Warren Bruene used the RPG method in which he believes he
measured the source resistance of an RF power amp, which he calls
'Rs'. I have never agreed that his method measures the source
impedance, or that his data has any relevance to anything.
Consequently, I am not impressed with the discussion going on here
concerning applying a signal back into an operating RF power amp to
determine the source impedance. Please define 'source impedance'--
where is it located in the amp? At the plate? At the output of the pi-
network?. And how do you know the data obtained using this method is
correct? Have you verified it by comparing it with data obtained using
another method?
I made a statement in an earlier post that when measuring the output
impedance using the 'load pull' method we're not concerned with the
absolute maximum power that can be delivered, but instead, limiting
the 'maximum' power delivered to that which can be delivered with a
specific level of grid drive, one which allows the power to be
limited to that of a normal operating level. Tom disagrees with this
position, that it is really the ABSOLUTE MAXIMUM power delivery that
should be considered. As you can see, I don't agree with Tom.
I don't know how many on this thread have actually reviewed the
portion of my Chapter 19 that presents the step-by-step procedure I
used in determining the output impedance of the Kenwood TS-830S tx,
which shows precisely the output impedance appearing at the output of
the pi-network.
To summarize the procedure that I maintain will provide an accurate
measurement of the output impedance appearing at the output terminals
of the pi-network is as follows:
1) Adjust the loading and tuning controls of the amp to deliver all
the available power to a complex load in the amount normally used in
operation with the setting of the grid-drive level required to obtain
that output power.
2).Measure the impedance of the complex load.
3) The output impedance, or 'source' impedance of the amp appearing at
the output terminals of the pi-network is the complex conjugate of the
load impedance.
Now, when you measure the source impedance using the externally-
injected signal, does the data from that measurement agree with that
of the load-measuring method? If it does, then I'll agree that the RPG
method is valid. If it doesn't I'll continue to have considerable
doubt as to its validity. But I'd still like to know where the
resistance measured by this method is located in the amp.
Walt
John KD5YI
It seems to me that we maybe should not be talking about source
impedance but maybe about 'regulation' or some other equivalent word.
For an open-loop source made from real components, I think one will
always find that there is a dE/dI number for output loading that we can
call source resistance.
It does not always mean that there is a physical resistance in the
circuit. It is simply a measure of the ability of the device to provide
an unvarying voltage under conditions of varying load.
Does this make any sense?
John
Wimpie
Hello Walt,
Except for bias or supply voltage change due to load change, load
pulling does give similar results as off-carrier signal injection. I
did this in simulation for various circuits (linear and non-linear).
You may remember that I put something in a document (discussion of
last year, it is still on my website).
You can get different results in case of soft power supply or bias
supplies. In case of manual load pulling, bias/supply voltage may
change (think of change in grid current due to change in RF plate
voltage). There is sufficient time for all voltages and currents to
settle. I added a section on the envelope response due to bias
current/voltage change.
When you use the injection method (for example with 130 Hz offset), it
is like load pulling where you switch the load 200 times/s (more
specifically you rotate the phase of the reflection coefficient as
seen by the PA). In such a situation bias and supply voltages will
settle to an average value resulting in (slightly) different results.
While not relevant for here, but nice to mention, stiffness of bias
supplies has influence on IMD also.
Wimpie
Hello John,
You are correct with respect to the physical resistor. If you tune a
real class-C amplifier for maximum output power (conjugated match),
you may not reach the highest efficiency, but at least significantly
above 60%. So this shows that there is no "physical" 50 Ohms in the PA
(assuming 50 Ohms load). If so, the efficiency would never exceed
50%.
When you change the load (after tuning for maximum output) from 50
Ohms to (for example) 30 Ohms and would carry out an impedance
measurement around that 30 Ohms load resistance, you will measure
something completely different. With "around 30 Ohms" I mean using 32
and 28 Ohms (for example).
John KD5YI
> John KD5YI wrote:
>>
>>
>> Transmitters don't have a clearly defined output impedance and,
>> whatever output impedance is there, doesn't necessarily mean anything.
>>
>> Your main concern is to provide a 50 ohm load for the transmitter to see.
>
> actually, to provide an *acceptable* load for the transmitter that
> maximizes radiated RF power. That might not be 50 ohms.. and I don't
> know that we actually care what it is, unless we're designing amplifiers.
Acceptable is what the manufacturer recommends for his gear. What does
this have to do with the device's output impedance?
John
Jim Lux
Absolutely nothing.. which is the point. Manufacturers have to make
their device work into some sort of loads, but can't test every possible
load, nor necessarily optimize the design for every possible load that
someone might use.
So we use, as a convention, that we'll measure the output power or DC to
RF conversion efficiency or Power Added Efficiency or whatever, with a
50 ohm resistive load hooked up to it.
That doesn't mean that we actually want to use a 50 ohm load in real
life. Maybe I have an amplifier designed to drive a loop antenna with a
resistive 10 ohm impedance. I might optimize it for that, but, because
people want to see a spec sheet, I'm going to have to measure the output
into 50 ohms and put that on the data sheet.
Or, perhaps, the active devices I use happen to have a "natural" output
impedance of, say, 8-10 ohms, so I put a 4:1 transformer on the output.
My box, if I were to measure the output Z, would show about 40 ohms.
I'd test it with a 50 ohm load, make sure I could put out 100W into that
load so I can sell it as a "100 W transmitter" and be done with it.
It might happen that you could hook up a 40 ohm load, or a 20 ohm load,
and it will work just fine, and may even put out more than 100W.
The fact of the matter is that ham manufacturers don't actually specify
the output Z: they just give you a range of outputs into which they'll
guarantee the transmitter won't fail. (e.g. say, 25 to 100 ohms..
perhaps specified a bit sloppily as VSWR 2:1 if driven from a 50 ohm source)
Interestingly, my IC7000 manual doesn't even give a output impedance, or
even a range of legal load impedances (page 150 of the manual).
Page 11 does say "Antenna Connector: Accepts a 50 ohm antenna with a
PL-259 connector"
And Page 15 does say "Use a well matched 50ohm antenna" and "An SWR of
1.5:1 or lower is recommended" doesn't say it won't work, doesn't even
say it only develops rated output power (on page 150) into that
particular load. There is a boxed note that says that if the SWR is
higher than approx 2.0:1, the transceiver's output power is reduced.
And, let's look at the service manual.. Spec page is identical to the
one in the user manual. The procedure calls out a whole bunch of
adjustments to set the "% power" to appropriate levels using a 50 ohm
load, but doesn't say what the power into some other load Z would be.
There is a test that when you hook up 100 ohms, the internal SWR meter
reads 2:1, and 50 ohms reads 1:1. But that's not a test of the
transmitter's output Z, more a calibration of the transmitter as RF
ohmmeter for an external component.
The ALC system actually measures the output VOLTAGE (not power) and
matches that up against the setpoint for the power (implying that if you
hook up a 40 ohm load, and you set the radio for 50W, you'll get a bit
more out).
the Automatic Power Limiting is a current sensor: if it exceeds 22A, it
reduces the drive. In fact, it kind of looks like the reflected HF power
(from the sampler at the filter output) isn't used to control the drive
at all.. it's just used to drive the SWR indicator. Kind of tough to
tell, the prose description is a bit unclear, and I haven't followed all
the signals through on the schematic.
but the summary is,
Jim Lux
> It seems to me that we maybe should not be talking about source
> impedance but maybe about 'regulation' or some other equivalent word.
>
> For an open-loop source made from real components, I think one will
> always find that there is a dE/dI number for output loading that we can
> call source resistance.
>
> It does not always mean that there is a physical resistance in the
> circuit. It is simply a measure of the ability of the device to provide
> an unvarying voltage under conditions of varying load.
This is comparable to the concept of "dynamic resistance" and, as well,
to "negative resistance" (as exhibited by gas discharges for instance).
The local slope of the V/I curve is negative, but that doesn't mean that
it has negative power.
John KD5YI
> John KD5YI wrote:
>> Acceptable is what the manufacturer recommends for his gear. What does
>> this have to do with the device's output impedance?
>>
>
> Absolutely nothing.. which is the point.
Are we arguing the same point?
> but the summary is,
That it is a bag of worms? I'm waiting with baited breath...
Richard Clark
wrote:
>That doesn't mean that we actually want to use a 50 ohm load in real
>life. Maybe I have an amplifier designed to drive a loop antenna with a
>resistive 10 ohm impedance.
Interesting. For a community that is so tight-fisted with cash, and
so brag-hearty with power claims, absolutely no one has ever tossed
their hat into the ring that with lower than 50Ohm termination on
their rig (and I'm not talking about 45Ohms) or with a more than 50Ohm
termination on their rig (and this is certainly achievable with a
72Ohm Dipole and 70 Ohm coax) that they have then proclaimed they
substantially exceeded 100W radiated (or lost to heat for that matter)
by the same degree of offset from 50.
>Or, perhaps, the active devices I use happen to have a "natural" output
>impedance of, say, 8-10 ohms,
That must date to the 1950s vintage of solid state. Your own IC7000
certainly doesn't suffer that abysmal "natural" output impedance. The
RD70HHF1 for 97W into 50Ohms exhibits 0.77-j0.22 Ohms @ 30MHz, a far
cry from your supposition. Off by 1000%?
>It might happen that you could hook up a 40 ohm load, or a 20 ohm load,
>and it will work just fine, and may even put out more than 100W.
"May" is lazy, "Does" is more authoritative. Any reports of "Does?"
>And Page 15 does say "Use a well matched 50ohm antenna" and "An SWR of
>1.5:1 or lower is recommended" doesn't say it won't work, doesn't even
>say it only develops rated output power (on page 150) into that
>particular load. There is a boxed note that says that if the SWR is
>higher than approx 2.0:1, the transceiver's output power is reduced.
Which is curious when the manufacturer of the RD70HHF1 power
transistors gives them a "No destroy" rating into a mismatch of 20:1.
Such is the inertia of 1950s design-think with modern components.
J. C. Mc Laughlin
noise of this thread.
I expect to learn something of value from your comments about why twice the
(apparent) output Z of an amplifier was of importance and what you did to
have the amplifiers conform.
73, Mac N8TT
-------------------
"Wimpie" wrote in message
news:d68fc390-da1b-45c7...@u16g2000yqj.googlegroups.com...
Hello Dave and John,
<snip>
Regarding the "academic discussion" I also agree. In my professional
career where I designed several RF PA's, only 2 times the output
impedance of the amplifier was of importance. In one of these cases I
couldn't meet the specs and had to insert attenuation (some waste of
power…).
<snip>
With kind regards,
Wim
PA3DJS
www.tetech.nl
Remove abc first before setting free the pigeon.
--------------
J. C. Mc Laughlin
Michigan U.S.A.
Home: J...@Power-Net.Net
Jim Lux
> On Wed, 04 May 2011 17:54:53 -0700, Jim Lux <james...@jpl.nasa.gov>
> wrote:
>
>> That doesn't mean that we actually want to use a 50 ohm load in real
>> life. Maybe I have an amplifier designed to drive a loop antenna with a
>> resistive 10 ohm impedance.
>
> Interesting. For a community that is so tight-fisted with cash, and
> so brag-hearty with power claims, absolutely no one has ever tossed
> their hat into the ring that with lower than 50Ohm termination on
> their rig (and I'm not talking about 45Ohms) or with a more than 50Ohm
> termination on their rig (and this is certainly achievable with a
> 72Ohm Dipole and 70 Ohm coax) that they have then proclaimed they
> substantially exceeded 100W radiated (or lost to heat for that matter)
> by the same degree of offset from 50.
Maybe that's because of tradition.. it's not only hams who do that. In
the professional world, there's a strong tendency to make the
"interfaces" between subsystems 50 ohms, even if overall system
efficiency might be improved by, say, a 25 ohm impedance. I've worked
on systems where a DAC produced a balanced signal with a 200 ohm Z, and
the vector modulator at the other end had a balanced input with a 200
ohm Z, but everyone insisted we put a 4:1 transformer on each end, so we
could run 50 ohm coax for a <10cm run.
If you're a amp maker, you're probably going to sell a lot more amps if
you specify and design for 50 ohms (the 10 ohm PA for loop antennas
would be a niche market). Likewise, if you're an antenna builder,
designing a product for 50 ohms will sell more than, say, 35 ohms.
In the cellphone world, because the PA and antenna tend to be developed
as an integrated assembly, they're probably a bit more flexible on this.
There's also the issue of wanting to put a power meter in line (but I
think that if you're using a 10 ohm system, you're probably capable of
figuring out how to measure power to make sure that you're not busting
the FCC limits)
>
>> Or, perhaps, the active devices I use happen to have a "natural" output
>> impedance of, say, 8-10 ohms,
>
> That must date to the 1950s vintage of solid state. Your own IC7000
> certainly doesn't suffer that abysmal "natural" output impedance. The
> RD70HHF1 for 97W into 50Ohms exhibits 0.77-j0.22 Ohms @ 30MHz, a far
> cry from your supposition. Off by 1000%?
Just an example. I know that modern parts are MUCH lower. The point is
that if one knew you had a 3 ohm load, you could design an amplifier
that would directly drive it without a transformer or network, and
potentially get better overall efficiency.
>
>> It might happen that you could hook up a 40 ohm load, or a 20 ohm load,
>> and it will work just fine, and may even put out more than 100W.
>
> "May" is lazy, "Does" is more authoritative. Any reports of "Does?"
The link from several days ago for the gentleman who measured the output
Z being around 40 ohms would be a "does"..
>
>> And Page 15 does say "Use a well matched 50ohm antenna" and "An SWR of
>> 1.5:1 or lower is recommended" doesn't say it won't work, doesn't even
>> say it only develops rated output power (on page 150) into that
>> particular load. There is a boxed note that says that if the SWR is
>> higher than approx 2.0:1, the transceiver's output power is reduced.
>
> Which is curious when the manufacturer of the RD70HHF1 power
> transistors gives them a "No destroy" rating into a mismatch of 20:1.
> Such is the inertia of 1950s design-think with modern components.
Yes, but the design limit might not be the transistor mismatch. It
might be a thermal dissipation limit, or some other component that's the
limiting factor.
(Or even, a desire to not have to answer 1000 customer service inquiries
to explain why it really doesn't matter what you hook up, or to explain,
that, yes, you DO need to actually connect an antenna of some sort)
Amateur radio, particularly in the mass market, is pretty slow to move.
Jim Lux
Exactly..
In fact, as interesting as it would be to measure the output impedance
of my radio, I started to think about what it would buy me, and came to
the conclusion, almost nothing (other than satisfying curiosity).
It *might* be interesting to look at (and write an article for QST/QEX
or something) "optimizing radiated power". Answering the question: do
you really want a 50 ohm match on your antenna analyzer, or do you want
maximum net power at the antenna feedpoint, and what that might mean for
typical 100W solid state rigs, antennas, etc.
(as a practical matter, this is what automatic antenna tuners actually
adjust for: either minimum reflected power, or maximum fwd-ref)
but it's possible that deliberately running a mismatch (as shown on your
rig's SWR meter) might actually result in more radiated power. e.g. if
at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W
reflected, so you're actually net 135W vs 100W; assuming your rig
doesn't otherwise have any problems.
Cecil Moore
> Answering the question: do
> you really want a 50 ohm match on your antenna analyzer, or do you want
Strangely enough, for any fixed configuration with reflections, one
way to maximize radiated power is to maximize reflected power at the
antenna feedpoint.:)
Richard Clark
wrote:
>>> That doesn't mean that we actually want to use a 50 ohm load in real
>>> life. Maybe I have an amplifier designed to drive a loop antenna with a
>>> resistive 10 ohm impedance.
>>
>> Interesting. For a community that is so tight-fisted with cash, and
>> so brag-hearty with power claims, absolutely no one has ever tossed
>> their hat into the ring that with lower than 50Ohm termination on
>> their rig (and I'm not talking about 45Ohms) or with a more than 50Ohm
>> termination on their rig (and this is certainly achievable with a
>> 72Ohm Dipole and 70 Ohm coax) that they have then proclaimed they
>> substantially exceeded 100W radiated (or lost to heat for that matter)
>> by the same degree of offset from 50.
>
>
>Maybe that's because of tradition.. it's not only hams who do that.
Well, that and the rest doesn't respond to the observation that NO ONE
makes a claim of more power from a rig that DOES NOT have a source
resistance of 50 Ohm.
>in the professional world, there's a strong tendency to make the
>"interfaces" between subsystems 50 ohms, even if overall system
>efficiency might be improved by, say, a 25 ohm impedance.
Retail engineering is ruled by a market economy. You have offered a
reductio-ad-absurdum of the native components offering ~10 Ohms
without the cost of intervening transformers. Such an "improved
efficiency" at lower cost option has been available for decades and
has yet to appear as a retail product for the tight-fisted Ham market.
Market logic shows by the force of economy that such a solution is not
viable.
>>a far cry from your supposition. Off by 1000%?
>
>Just an example.
An incredibly poor one in the context of 100W transmitters when garden
variety examples abound.
>I know that modern parts are MUCH lower.
That deliver 97W into 50 Ohms at 30 MHZ with a nominal 12VDC supply? I
smell another poor example.
>>> It might happen that you could hook up a 40 ohm load, or a 20 ohm load,
>>> and it will work just fine, and may even put out more than 100W.
>>
>> "May" is lazy, "Does" is more authoritative. Any reports of "Does?"
>
>The link from several days ago for the gentleman who measured the output
>Z being around 40 ohms would be a "does"..
"Does" with how much more power? When I observe the contents of the
page at that link = NOT ONE WATT MORE!
In fact, if we follow the logic of lowered Rs yielding more power, his
own data flips this! He reports the same 100W available at higher Rs.
However, all that aside, when you observe your comment above, the link
"Does not" offer any support to it. In fact on9cvd explicitly offers:
"This is an ambiguous message"
>> Which is curious when the manufacturer of the RD70HHF1 power
>> transistors gives them a "No destroy" rating into a mismatch of 20:1.
>> Such is the inertia of 1950s design-think with modern components.
>
>Yes, but the design limit might not be the transistor mismatch. It
>might be a thermal dissipation limit, or some other component that's the
>limiting factor.
Consult the original spec. It is quite specific. The characteristic
of
Load VSWR tolerance: Load VSWR=20:1(All Phase): No destroy
is not a dissipation reference.
Wimpie
Hello,
Regarding the defined output impedance.
The first time was during my thesis (third harmonic peaking PA). The
stability margin was not very large (expected), and it could be
improved by keeping the impedance seen from the base within certain
limits. My teacher said, be careful, you only have two devices
(BLW76). I tried to make an exciter (based on 2SC1307 BJT) with
defined output impedance, but without success. So in the end I
increased the output power of the exciter and inserted a 3 dB
attenuator, not elegant, but it did the job.
The second time was for H-field generation where wider bandwidth was
achieved by adding a second resonator. When driving from a 50 Ohms
source, it had a nice Chebyshev type pass band. Because of the ripple,
it shows reflection in the pass band (this happens with Chebyshev
response). However when driving from a PA (that was flat within the
pass band when loaded with 50 Ohms), everything went wrong.
I redesigned the filter/coil combination, designed a switching PA
(half bridge in class DE operation) and skipped the 50 Ohms (I
designed around 8 Ohms). The PA drives the filter directly in such
away that most of the time the PA sees a nice (mismatched) load (that
is inductive for harmonics). This resulted in the desired pass band
with significantly increased efficiency. The strength of the H-field
is controlled by varying the supply voltage (PWM circuit).
Leaving out the 50 Ohms in between, saved several capacitors and
inductors.
walt
Hey Mac, are you planning on attending the Chapter 10 QCWA meeting in
Cadillac Saturday? I'm planning on attending, so hope to see you
there!!!
Walt, W2DU
Wimpie
Hello Jim,
I agree with the PA load mismatch issue. It is possible to get (some)
more net power by applying mismatch to a PA stage (from experience).
But frequently it comes with too much increase in input power (so very
hot heatsink).
In case of high efficiency designs, the active devices may communicate
to you by means of smoke or ejection of (hot) particles.
Wim
PA3DJS
www.tetech.nl
In case of PM, tell the pigeon that abc is not in the address.
walt
Hi Wimpie and KD5YI,
Will you please explain how it is possible to get more power delivered
by applying a mismatch to the output of a PA?
And for KD5YI specifically, I believe you have presented some
inaccurate math calculations. You begin with delivering 100w into a
matched load. Then you say you mismatch to 2:1 and get 135w forward
and 15w reflected, leaving 120w delivered. You must be kidding!!
First, with a 2:1 mismatch and 100w delivered by the PA, the reflected
power is 11.111w, which when added to 100w from the source, the
forward power is 111.11 watts. When the reflected power that returns
to the load is subtracted from the forward power, the result is 100w.
You've heard the expression 'there is no free lunch'?
So please explain to me, if you can, how you can deliver 120w when the
source is 100w.
Walt
walt
Ooops, my error, you said you're getting 130w delivered, not 120w as I
said incorrectly.
Walt
John KD5YI
I did not present any numbers, calculations, or assumed conditions,
inaccurate or otherwise. You might want to reply to Jim Lux.
John - KD5YI
Wimpie
No he is not kidding. I observed the same. At that time I was lucky to
have full access to HP, Advantest and Rohde & Schwarz equipment to
double-check everything (I first blamed my own gear). You should
leave the idea that all PA's have 50 Ohms output impedance, then it is
easy to explain yourself.
A certain load that has mismatch referenced to 50 Ohms may have a nice
match to a system with non-50 Ohms output impedance.
>
> First, with a 2:1 mismatch and 100w delivered by the PA, the reflected
> power is 11.111w, which when added to 100w from the source, the
> forward power is 111.11 watts. When the reflected power that returns
> to the load is subtracted from the forward power, the result is 100w.
> You've heard the expression 'there is no free lunch'?
>
> So please explain to me, if you can, how you can deliver 120w when the
> source is 100w.
>
> Walt
The extreme cases where you can get significantly more net output
power by applying mismatch, are PA's with high efficiency (class-E, -
D, -DE, etc). I am currently designing a balanced class-E 500W
stage. It can deliver 1 kW, but within very short time the mosfet's
will explode (if the supervisory circuit doesn't act).
Wim
PA3DJS
www.tetech.nl
walt
Wim, are you saying that by using Class E amps you are able to violate
the Laws of Physics pertaining to the Conjugate Matching Theorem and
the Maximum Power Delivery Theorem? I cannot agree.
A load doesn't care what the source is. If the load impedance is the
complex conjugate of the source, all available power will be delivered
to the load. Then, if the load impedance is either increased or
decreased, the power delivered will decrease. Are you now saying that
the concept I just stated above is no longer true? If you are, please
explain in detail why this is so. How does 'high efficiency' overcome
the requirement for impedance matching in the delivery of power?
And are you agreeing with an earlier poster that with a 100w source
and a mismatch of 2:1 the forward power will be 135w and 15w
reflected, the power delivered to the load will be 130w? If so, will
you please explain in detail how this can occur?
Walt
Wimpie
Walt, Maybe you should familiarize yourself with class E (and other
high efficient topologies). Set up a simulation (or measurement) and
try to apply your "conjugated match" thing. You will find out that you
can't make a class E PA that operates under conjugated match, unless
you are going to play with the power supply's internal resistance (so
the system becomes power supply limited). My current design outputs
500W into 4.5 Ohms, however the output impedance (load pulling) is < 1
Ohm.
The reason for non-conjugated matched operation is that in class-E the
device is in voltage saturation for about 50% of the time. During that
time the device has no gain, so a device used in class-E has less gain
then the same device used in a non-saturated application. In other
words: tuning is designed for highest efficiency, not highest output
power.
>
> A load doesn't care what the source is. If the load impedance is the
> complex conjugate of the source, all available power will be delivered
> to the load.
Quote from text above: "If the load impedance is the complex conjugate
of the source.....". This If-statement has a "false" result for many
PAs, try to broaden your view.
> Then, if the load impedance is either increased or
> decreased, the power delivered will decrease. Are you now saying that
> the concept I just stated above is no longer true? If you are, please
> explain in detail why this is so. How does 'high efficiency' overcome
> the requirement for impedance matching in the delivery of power?
>
> And are you agreeing with an earlier poster that with a 100w source
> and a mismatch of 2:1 the forward power will be 135w and 15w
> reflected, the power delivered to the load will be 130w? If so, will
> you please explain in detail how this can occur?
Yes I agree with him, see my reaction to that statement
My class-E PA design that I am doing now is designed for a 4.5 Ohms
nominal load. If I change that load to 2.5 Ohms (VSWR=1.8), output
increases to > 700W, but it will be destroyed due too non-favourite
combination of Vds and Id.
> Walt
Cecil Moore
> You will find out that you
> can't make a class E PA that operates under conjugated match, unless
As I understand it, Walt's approach is to pick a point inside the
source at which the output becomes linear through filtering and call
the impedance at that point the linear "source impedance" (including
some boundary conditions). It is akin to the motion of a pendulum in a
clock being linear even though the sustaining energy comes in non-
linear pulses.
walt
Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w
forward and 15w reflected with a 100w source. How can a 2:1 mismatch
achieve 130w forward power, when that mismatch can only reflect 0.111
x 100w? Also, how can that mismatch achieve 15w of reflected power
when that mismatch can only reflect 0.111 x 100w? And how can 11.111
watts of forward power add to 100w to achieve 130w? I don't see how
the power reflected at a mismatched load can be affected by the nature
of the source.
But I do understand how with a 4.5-ohm load can deliver 500w and a 2.5-
ohm load can deliver 700w. I now understand what you mean by a
mismatch can increase the power delivery. (Actually, a change in
mismatch)
But Wim, you pulled a fast one one us!!! Until now you didn't tell us
that the source resistance of the source was 1 ohm!!!
You also didn't tell us what the power would be if the load was 1 ohm
(thus matching the load to the source) if the power supply was
sufficiently large so the power delivery would not be limited by the
power supply. So in reality, you really DON'T get an increase in power
delivery by mismatching, but actually a decrease. What you're really
doing is obtaining an increase in power delivery by decreasing the
amount of mismatch from 4.5:1 to 2.5:1.
So now I understand that you haven't violated any laws of physics, but
IMHO, you have been misleading when you say you obtain a increase in
power by mismatching, because that statement isn't really true, is
it?
take care,
Walt
Wimpie
Hello Walt,
Off course I agree with you that you can never have more power then
the conjugated matched power, but many PAs don't operate in this
regime (as my very extreme class-E case). It was the reason for
mentioning:
"A certain load that has mismatch referenced to 50 Ohms may have a
nice
match to a system with non-50 Ohms output impedance."
The problem is in whether you define mismatch based on the ohmic value
printed on the back of the PA, or on the actual output impedance of
the PA (that you mostly don't know in case of many solid state PA's).
As the actual output impedance of the PA may not be 50 Ohms (for
example 100 Ohms), a 50 Ohms load (as mentioned on the back of the PA)
will provide the stated power (for example 100W). By applying 100
Ohms (that is VSWR = 2 for a 50 Ohms reference), the net power will
increase to 112 W. Actually it wasn't me to experience this mismatch
isue first, but my father during the time that CB was very popular
overhere.
The above example assumes that the output impedance of the PA is
independent of load. Of course this is only true when the actual
voltage across the active device and current through it doesn't
saturate the active device too much.
I think the saturation issue is one of the advantages of a linear PA
with accessible R an X tuning. You just tune for maximum output given
a certain load and drive (you may use plate current as a guide also).
As long as your SSB signal's PEP stays below that output power, your
active device will not go into voltage saturation and IMD will likely
be acceptable.
In case of a wide band push-pull PA, VSWR = 2 (with inconvenient
phase) may provide a load where the active devices (mosfet or BJT),
will go into voltage saturation at a net output power below the rated
power of the PA. For constant envelope modulation this isn't problem,
but for SSB/AM it isn't good.
Richard Clark
>Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w
>forward and 15w reflected with a 100w source.
Hi Walt,
Irrespective of who the author is of this statement (I've lost track),
I too would like to see how this is reconciled.
But seeing that you've asked several times....
walt
Thanks for the reply, Wim, but you seem to be evading my questions and
concerns by what appears to be going off on tangents unrelated to my
questions.
Perhaps I need to refresh Electronics 101, because I simply can't
understand the example you provided.
You have a 100-ohm source terminated with a 50-ohm load that provides
100w. But you say with the VSWR = 2 the net power will increase to
112w. How does this happen?
Wim, you are making statements that seem to disagree with known
principles, yet you give no explanation of how these statements can be
justified in relation to the known principles.
And please tell me, Wim, what RF amps do you know of that have the
source impedance indicated on the back? And you refer to solid-state
amps, when you know my discussion (and experience) is with tube amps
with pi-network outputs. I am totally ignorant on the operation of
solid-state amps.
I would like to be learning something from my discussions with you,
but I'm sorry, Wim, you're making me feel more stupid the more we
continue.
Walt
John KD5YI
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From: Jim Lux <james...@jpl.nasa.gov>
Newsgroups: rec.radio.amateur.antenna
Subject: Re: Transmitter Output Impedance
Date: Thu, 05 May 2011 10:23:10 -0700
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Cecil Moore
> You have a 100-ohm source terminated with a 50-ohm load that provides
> 100w. But you say with the VSWR = 2 the net power will increase to
> 112w. How does this happen?
Walt, here is probably what Wim means by that.
Source------1/2WL 50 ohm lossless------RLoad
Vsource = 212v, Rsource=100 ohms
If Rload = 50 ohms, PLoad = 100w
If Rload = 100 ohms, PLoad=112.5w
Wim must be assuming a 50 ohm SWR in both cases.
J. C. Mc Laughlin
electronic problems for a colleague who is particularly skilled at crafting
practice problems for the professional engineering exam (chartered engineer
in Europe).
Your comments about filter response reminds me of the (analog) FM receivers
that had wonderful suppression of the next channel over, but horrible
performance in the presence of noise from ignition - while other receivers
had great performance in both respects. As you know, it had to do with
phase response. Chebyshev is not always desirable.
Warm regards, Mac N8TT
"Wimpie" wrote in message
news:ca14f105-ac19-45b0...@v10g2000yqn.googlegroups.com...
J. C. Mc Laughlin
attend. Sorry not to see you and your bride.
Warm regards, Mac N8TT
"walt" wrote in message
news:27f83cac-2baa-462c...@f2g2000yqf.googlegroups.com...
Hey Mac, are you planning on attending the Chapter 10 QCWA meeting in
Cadillac Saturday? I'm planning on attending, so hope to see you
there!!!
Walt, W2DU