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Re: O S L on antenna side of a balun / choke with stud terminals?

 

I've got a 5hp motor sitting in the basement.
Has both a starting cap and a run cap, a centrifugal switch
removes the starting cap from the circuit once it's running.
My supposedly "9000 Starting Watts" generator can't get it moving,
there's a quick twitch of the rotor and then the generator cuts out.
Starting currents can be up to 20 times what you might expect
from the horsepower rating, so maybe up to 5hp*746*20=74.6kw.

Correct, reactance in itself can not produce harmonics. But the
interaction of the offset between voltage and current and reflected power
can.
Again, I'm thinking linear vs non-linear.
And I don't see anything non-linear in reflected power that might
cause harmonics.

Jerry, KE7ER



On Thu, Aug 20, 2020 at 03:46 PM, David Eckhardt wrote:


Yes, many are starting capacitors that allow max current into the windings
to get the thing rotating. Pretty much a short on the grid for a short
duration.

Correct, reactance in itself can not produce harmonics. But the
interaction of the offset between voltage and current and reflected power
can.

Dave


Re: What is the power rating of the 50 Ohm dummy load provided with the nanoVNA?

 

look on eBay for cheap high power low inductance load resistors for RF.

On Thu, 20 Aug 2020, 21:05 Jim Lux, <jimlux@...> wrote:

On 8/20/20 11:39 AM, rahul@... wrote:
Can I use it for a 5 W radio to calibrate it?
< 1/2 Watt is typical.

the ones with a "hat" might be rated higher (1 or 2W) but probably not.

You really don't want to overheat a cal standard.








Re: O S L on antenna side of a balun / choke with stud terminals?

 

Yes, many are starting capacitors that allow max current into the windings
to get the thing rotating. Pretty much a short on the grid for a short
duration.

Correct, reactance in itself can not produce harmonics. But the
interaction of the offset between voltage and current and reflected power
can.

Dave

On Thu, Aug 20, 2020 at 10:12 PM Jerry Gaffke via groups.io <jgaffke=
[email protected]> wrote:

Yes, PFC is an issue at 60hz.
Just one nit to pick:

Harmonic content is a signature of a reactive load - a
purely resistive load can not generate harmonics. here.
An inductive or capacitive load is still linear, it does not generate
harmonics.
All it does is cause the current through the device to be out of phase
with the voltage.

Small switching power supplies used to be just a bridge rectifier on the
AC
feeding a big cap, the high voltage from that was then brought down
to 12vdc or whatever using some switch mode topology, perhaps a flyback
converter.
But the bridge rectifier only has current flowing in spikes at the
positive and negative
voltage peaks of the incoming 60hz voltage sine wave, and that definitely
does
create harmonics. A number of power factor correction schemes have been
developed to smooth out the current draw from the mains for this
application.

In some other thread, we spoke of using a common mode choke in the antenna
feed line to keep household noise from sneaking into the receiver,
assuming the antenna itself is far from those noise sources. Those nasty
little wall warts without PFC would be a prime example of why this might
be needed.

I don't know a lot about the various kinds of electric motors, but many
have a "run capacitor" in there to balance out the inductive load for
some crude PFC. Not just to be nice to the utility company,
but to be able to get some torque out of the motor.

Jerry, KE7ER


On Thu, Aug 20, 2020 at 02:24 PM, David Eckhardt wrote:

Some excellent points, Jim!

Let me take this concept of reflected power to something we are all more
familiar with: The AC power sockets in our walls. Whether its 120 vrms,
220 vrms, 308 vrms, 408 vrms, or 480 vrms, at 60 Hz, the power source
assumes a purely resistive load. Where else have we run into this
concept? Definition of resonance: no reactive component to the load (or
source). Why does the power provider prefer resistive-only loads? Their
generators run cooler and the whole grid distribution system operates
more
efficiently. They are a resonant system. Now what happens if your
appliance, not a filamented light bulb, your toaster, your electric oven
which are purely resistive, happens to not resonate at 60 Hz. Heavy
inductive motor start-up is a good example. Note how your lights dim when
these items are first started? Reactive loads reflect power back onto the
power grid and ultimately the generators. Those non-resistive appliances
reflect some complex portion of their load onto the power grid and
ultimately the generators. These reactive loads many times require what
is
called power factor correction (PFC). This is nothing more than a
matching
network to cancel the complex portion of the load which those reactive
loads present to the power grid and generators to keep the whole system
cooler.

By law, appliances less than 70-watts power consumption are exempted from
PFC (Power Factor Correction). Above that, laws driven into the systems
by
the power utilities, the FCC, and the EU require PFC. The EU requires (by
law) limits on harmonic output from appliances and power consuming
installations. Harmonic content is a signature of a reactive load - a
purely resistive load can not generate harmonics. Germany is a stickler
for this. However, large consumers of power like large factories (think
Detroit auto manufacturers) are exempted from PFC. If California required
PFC on all power consuming installations, I'm willing to bet there might
be
no need for rolling power outages. I can't prove that statement, but I'm
sure requiring PFC would alleviate much of the problem Cal. has every
summer. Of course, the power providers make more money in supplying more
power to deal with the reflected waves from complex loads, so, they may
prefer no PFC (California power lobby?). Money talks.

These PFC circuits are nothing more than matching circuits. They are
designed to provide to the power grid as much as possible a purely
resistive load from a complex power consuming appliance. Since the power
grid is pretty much zero-ohms, non-reactive, any resistive load can be
handled with grace and no need for generating extra power to compensate
for
reflections caused by reactive loads to the grid.

Matching loads to sources doesn't apply just to RF energy. It's
everywhere. Bet you weren't aware of the PFC and power grid matching?


--
*Dave - W?LEV*
*Just Let Darwin Work*


Re: O S L on antenna side of a balun / choke with stud terminals?

 

On 8/20/20 2:24 PM, David Eckhardt wrote:
Some excellent points, Jim!
Let me take this concept of reflected power to something we are all more
familiar with: The AC power sockets in our walls. Whether its 120 vrms,
220 vrms, 308 vrms, 408 vrms, or 480 vrms, at 60 Hz, the power source
assumes a purely resistive load. Where else have we run into this
<snip>

PFC correction on loads >70W but smaller than, say, a few kVA, was driven in a lot of cases by the problem with neutral currents on 3phase Y office loads. All those switching power supplies which have a diode feeding a capacitor input filters have high peak currents at light loads. If you have them on three phases, those peaks occur 60 degrees apart - so the load isn't nicely balanced, and the neutral carries 3x the current of the phase legs.

It's the industrial consumers with motors (inductive load when lightly loaded), induction furnaces, and things like "magnetic ballasts" on lighting loads that have the classic lagging power factor which requires correction with capacitor banks or synchronous condensers.

Modern power supplies with "PFC correction" are really doing more about harmonics than actual lagging power factor. In any case, what they do is have a switching regulator that draws "the right amount" of current from the line in phase with the voltage.


On a grid basis, the "impedance matching" issue is mostly about power flow and transients. Utilities set the current phase relative to the voltage to control whether power flows one way or another - in the very short term they do that by switching in capacitors and inductors as needed. In the medium term, they do it by adjusting the generator rotational speed relative to the grid. If you load a generator heavily, it slows down, and the faster generators that are tied to it tend to pick up the load. These relationships are non-linear.

1000 km is 10 milliseconds light time (a big fraction of a cycle at line frequency), and a lot slower in a power line with distributed L and C, so the whole "relative phase management" is very, very complex, and difficult to stabilize. (My father's PhD committee chairman developed a lot of the computer codes to do this in the 60s) This is why they love DC interties - at one end you set constant voltage, and the other end you set constant current, and the system is stable.


So, PFC (and harmonic management) in office and light industrial applications is addressing a different problem than the PFC compensation in heavy industry and power system management.

But yes, an understanding of phasors is really useful for power systems design and operation.

An excellent textbook for this is by Theodore Wildi,"Electrical machines, drives, and power systems" that doesn't get into too much "rotating magnetic field design" for motors and such. Lots of useful stuff on synchronous motors/generators vs induction motors vs DC machines, and also on power systems design.








These PFC circuits are nothing more than matching circuits. They are
designed to provide to the power grid as much as possible a purely
resistive load from a complex power consuming appliance. Since the power
grid is pretty much zero-ohms, non-reactive, any resistive load can be
handled with grace and no need for generating extra power to compensate for
reflections caused by reactive loads to the grid.
Matching loads to sources doesn't apply just to RF energy. It's
everywhere. Bet you weren't aware of the PFC and power grid matching?
Dave - W?LEV


Re: O S L on antenna side of a balun / choke with stud terminals?

 

Yes, PFC is an issue at 60hz.
Just one nit to pick:

Harmonic content is a signature of a reactive load - a
purely resistive load can not generate harmonics. here.
An inductive or capacitive load is still linear, it does not generate harmonics.
All it does is cause the current through the device to be out of phase with the voltage.

Small switching power supplies used to be just a bridge rectifier on the AC
feeding a big cap, the high voltage from that was then brought down
to 12vdc or whatever using some switch mode topology, perhaps a flyback converter.
But the bridge rectifier only has current flowing in spikes at the positive and negative
voltage peaks of the incoming 60hz voltage sine wave, and that definitely does
create harmonics. A number of power factor correction schemes have been
developed to smooth out the current draw from the mains for this application.

In some other thread, we spoke of using a common mode choke in the antenna
feed line to keep household noise from sneaking into the receiver,
assuming the antenna itself is far from those noise sources. Those nasty
little wall warts without PFC would be a prime example of why this might be needed.

I don't know a lot about the various kinds of electric motors, but many
have a "run capacitor" in there to balance out the inductive load for
some crude PFC. Not just to be nice to the utility company,
but to be able to get some torque out of the motor.

Jerry, KE7ER


On Thu, Aug 20, 2020 at 02:24 PM, David Eckhardt wrote:

Some excellent points, Jim!

Let me take this concept of reflected power to something we are all more
familiar with: The AC power sockets in our walls. Whether its 120 vrms,
220 vrms, 308 vrms, 408 vrms, or 480 vrms, at 60 Hz, the power source
assumes a purely resistive load. Where else have we run into this
concept? Definition of resonance: no reactive component to the load (or
source). Why does the power provider prefer resistive-only loads? Their
generators run cooler and the whole grid distribution system operates more
efficiently. They are a resonant system. Now what happens if your
appliance, not a filamented light bulb, your toaster, your electric oven
which are purely resistive, happens to not resonate at 60 Hz. Heavy
inductive motor start-up is a good example. Note how your lights dim when
these items are first started? Reactive loads reflect power back onto the
power grid and ultimately the generators. Those non-resistive appliances
reflect some complex portion of their load onto the power grid and
ultimately the generators. These reactive loads many times require what is
called power factor correction (PFC). This is nothing more than a matching
network to cancel the complex portion of the load which those reactive
loads present to the power grid and generators to keep the whole system
cooler.

By law, appliances less than 70-watts power consumption are exempted from
PFC (Power Factor Correction). Above that, laws driven into the systems by
the power utilities, the FCC, and the EU require PFC. The EU requires (by
law) limits on harmonic output from appliances and power consuming
installations. Harmonic content is a signature of a reactive load - a
purely resistive load can not generate harmonics. Germany is a stickler
for this. However, large consumers of power like large factories (think
Detroit auto manufacturers) are exempted from PFC. If California required
PFC on all power consuming installations, I'm willing to bet there might be
no need for rolling power outages. I can't prove that statement, but I'm
sure requiring PFC would alleviate much of the problem Cal. has every
summer. Of course, the power providers make more money in supplying more
power to deal with the reflected waves from complex loads, so, they may
prefer no PFC (California power lobby?). Money talks.

These PFC circuits are nothing more than matching circuits. They are
designed to provide to the power grid as much as possible a purely
resistive load from a complex power consuming appliance. Since the power
grid is pretty much zero-ohms, non-reactive, any resistive load can be
handled with grace and no need for generating extra power to compensate for
reflections caused by reactive loads to the grid.

Matching loads to sources doesn't apply just to RF energy. It's
everywhere. Bet you weren't aware of the PFC and power grid matching?


Re: O S L on antenna side of a balun / choke with stud terminals?

 

Some excellent points, Jim!

Let me take this concept of reflected power to something we are all more
familiar with: The AC power sockets in our walls. Whether its 120 vrms,
220 vrms, 308 vrms, 408 vrms, or 480 vrms, at 60 Hz, the power source
assumes a purely resistive load. Where else have we run into this
concept? Definition of resonance: no reactive component to the load (or
source). Why does the power provider prefer resistive-only loads? Their
generators run cooler and the whole grid distribution system operates more
efficiently. They are a resonant system. Now what happens if your
appliance, not a filamented light bulb, your toaster, your electric oven
which are purely resistive, happens to not resonate at 60 Hz. Heavy
inductive motor start-up is a good example. Note how your lights dim when
these items are first started? Reactive loads reflect power back onto the
power grid and ultimately the generators. Those non-resistive appliances
reflect some complex portion of their load onto the power grid and
ultimately the generators. These reactive loads many times require what is
called power factor correction (PFC). This is nothing more than a matching
network to cancel the complex portion of the load which those reactive
loads present to the power grid and generators to keep the whole system
cooler.

By law, appliances less than 70-watts power consumption are exempted from
PFC (Power Factor Correction). Above that, laws driven into the systems by
the power utilities, the FCC, and the EU require PFC. The EU requires (by
law) limits on harmonic output from appliances and power consuming
installations. Harmonic content is a signature of a reactive load - a
purely resistive load can not generate harmonics. Germany is a stickler
for this. However, large consumers of power like large factories (think
Detroit auto manufacturers) are exempted from PFC. If California required
PFC on all power consuming installations, I'm willing to bet there might be
no need for rolling power outages. I can't prove that statement, but I'm
sure requiring PFC would alleviate much of the problem Cal. has every
summer. Of course, the power providers make more money in supplying more
power to deal with the reflected waves from complex loads, so, they may
prefer no PFC (California power lobby?). Money talks.

These PFC circuits are nothing more than matching circuits. They are
designed to provide to the power grid as much as possible a purely
resistive load from a complex power consuming appliance. Since the power
grid is pretty much zero-ohms, non-reactive, any resistive load can be
handled with grace and no need for generating extra power to compensate for
reflections caused by reactive loads to the grid.

Matching loads to sources doesn't apply just to RF energy. It's
everywhere. Bet you weren't aware of the PFC and power grid matching?

Dave - W?LEV

On Thu, Aug 20, 2020 at 7:38 PM Jim Lux <jimlux@...> wrote:

On 8/20/20 7:15 AM, Jerry Gaffke via groups.io wrote:
Jim,

Then Miro's question remains: What happens to power reflected back from
the antenna feedpoint
if there is no antenna tuner? Without a tuner, it is not all getting
sent back out to the antenna.
Conservation of energy says the transmitter has to put it somewhere.
It could reflect back from the transmitter toward the antenna (and some
be lost in the transmission line) - bouncing back and forth.

Don't forget that most transmitters do not have a 50 ohm resistive
output impedance (particularly an amplifier with a tuned circuit in the
output).

What happens with a amplifier with a design output impedance connected
to a 25 ohm load (no transmission line). Is it "reflected power" that
is increasing the dissipation of the transmitter (if it is increased..
some amplifiers will just put out a different amount of power), or is it
a change in the operating point of the transmitter.

I find it helpful in these cases to consider some limiting cases - a
stiff DC power supply has a 0 ohm output impedance, so it's "mismatched"
into pretty much any load. Consider a AC source with zero output
impedance - it can launch a wave into a transmission line or towards a
load, but does that mean that there's a reflection and that some power
has to be absorbed? Not necessarily.

So what's "bad" about reflected power (or mismatch) - they're really the
same thing, just one is in terms of a traveling wave and the other isn't.

In terms of practical amplifier design, you wind up with either a
voltage or a current that is higher than in the nominal case - and
typically, one doesn't want to overdesign the amplifier - if you're
putting out 70V for 100W into a 50 ohm load, do you want to go buy 200V
transistors (to cover the worst case), or 100V transistors.



If I have a source with 50 ohm output impedance, designed to put 1 volt
into a 50 ohm load - what I've really got is a 2 volt source with zero
output impedance and a 50 ohm series resistor. And in the "theoretical"
sense, that 50 ohm series resistor would dissipate the same amount of
power as the load. If I feed an open circuit, now the voltage at the
output terminal (keeping the 2V on the AC source) is now 2V. If I feed
a short, the voltage is zero, but now the current is twice.

In reality, of course, you don't have a zero ohm source or a 50 ohm
resistor in the transmitter. You've got something else which has a
fairly complex relationship between voltages, currents, etc. And that
something has practical limits (or cost constraints).

You've no doubt seen the ads for 1kW RF FETS that say "tolerates any
mismatch" with the video of the guy putting a screwdriver intermittently
across the output with sparks and the watt meter banging back and forth.

Well, it turns out this is easy - if your bus voltage is 100V and you've
got a 300V transistor in there, no matter what sort of resistive load
you put on it, you're NEVER going to go past 200V. Same sort of argument
for the output current - even into a short, you're not going to melt the
part, because the power supply (or series resistance in the transistor)
will limit the current and the dissipation.

In none of these situations, though, are you anywhere close to this idea
of "reflected power being absorbed in the transmitter"

For some specific cases (Thevenin equivalent sources with resistive
output impedances) the math works out right as if it were being
"absorbed" in some way, but that's not a real situation:

2V source 50 ohm series resistance

Load: open - perfect reflection - where's the reflected power? it's
zero, because there's no forward power.
Load: 50 ohms - perfect match - 20mW dissipated in source, 20mW
dissipated in load - 40mW consumed from source
Load: Short - perfect reflection - 80mW consumed from 0 ohm source, 80mW
dissipated in internal resistance

Load: 25 ohms - 2:1 VSWR - voltage at load is 0.667 V, about 9mW.
Dissipation in source resistance is 36 mW, so total source power is 45
mW - But let's look at the "reflected power" 2:1 is a reflection
coefficient of 0.333, or a return loss of 9.5 dB - that is, about 1/10th
of the forward power is reflected. The forward power toward the load is
9 * 1.1 or around 10mW, so the reflected power is 1 mW.

Hey, but my ideal voltage source is putting out 45 mW instead of 40mW -
that's 5 mW more, not 1 mW more. And the series source resistor is
dissipating 36 mW instead of 20mW. That's a WHOLE lot more than 1mW.


All the whole "matching theorem" says is that for maximum power transfer
the source and load should be conjugates. That is, if you plot "DC power
into perfectly efficient 0 ohm source" vs "power dissipated in load" the
ratio will be lowest at a conjugate match, and will be 50%

it doesn't say anything about "where the reflected power is absorbed" or
amplifier efficiency or heat dissipation. It is *entirely* a convenient
analytical model: nobody in their right mind would build a high power
source consisting of a zero ohm source with a series resistor.

However, we DO use that kind of design all the time in digital circuitry
- a digital gate has a very low output impedance, so you put a 50 ohm
resistor in series, before driving a long coax run to a (hopefully) 50
ohm load. It's not high power, so we don't are about efficiency or
power transfer - what we're concerned about here is not having a
discontinuity which leads to a reflection in the time domain.

And, such sources are used in test equipment - The standard lab function
generator (like a Keysight 33620 series) is basically a power DAC with
zero ohm output impedance driving a 50 ohm series resistor - again,
we're not concerned with efficiency or power transfer, just having a
stable and known impedance that closely approximates a simple
theoretical model.

But when it comes to high power amplifiers, most designers spend a fair
amount of time trying to get their design "aligned" to the load (to
avoid the word "match") so that it has decent overall efficiency - if
your power amplifier devices aren't zero ohms (generally the case) then
you try and design a *lossless* network that transforms whatever it is
to what the intended load is (which might be 50 ohms resistive, 75 ohms
resistive or something else entirely..) And you hope that this works
over the frequency range of interest (since low loss matching tends to
be narrow band, or the devices themselves tend to have varying output Z)

Sometimes this isn't possible - too wide a frequency range for the
devices you've got. Then you rely on two or three time tested approaches:
1) Design enough margin into the circuit that it can tolerate 2x Voltage
and 2x current.
2) put a nonlinear device in the circuit (a circulator) that an send the
reflected power to a load.
3) design the circuit to foldback or limit, when it detects the voltage
getting too high or current getting to high.
4) Way over build in terms of power and put a pad in series.

Your choice depends on your design constraints and budgets.

In the amateur radio world, there's a "RF output power" constraint (it
used to be DC power to the final stage, and before that it was "HP to
the spark gap") - and a dollar constraint - so you design amps that are
efficient but require good load management. (it's funny, if you didn't
care about DC power requirements, you could sell a ham transmitter that
could put 1kW into "any load" )


In the commercial communications world, they're not "transmitter power"
limited, but rather "radiated power" limited - so you see things like
kilowatt amplifiers into a terminated folded dipole - or, the
equivalent, a untuned dipole with a 6dB pad.

In space communications, we're obsessed with DC power efficiency and
mass, and the selection of parts is limited - so you see circulators and
loads, more for protection than out of need.

It's interesting - the desire to have more output power for solid state
amps in space has led to interesting designs with combiners (since the
individual amplifier dice only put out a few watts) that cleverly
reflect harmonic power back into a optimized resonant amplifier (Class
E/F1).





I can't think of anything the transmitter could do with this energy
other than turn it into heat.
Either by design with your isolator and load, or just heating up
whatever's handy.

mmmm, not really - most amplifiers don't have loads in them - like I
said, it's more about amplifier efficiency changes.




I don't know for sure exactly where it goes in a typical amplifier.
Hence my weasel words: "What happens then depends on the transmitter
design."

Jerry, KE7ER



On Thu, Aug 20, 2020 at 06:55 AM, Jim Lux wrote:
On 8/19/20 10:09 PM, Jerry Gaffke via groups.io wrote:
Without the antenna tuner, that reflected power comes back to the
transmitter.
What happens then depends on the transmitter design.
At least some of it will likely be getting dissipated as heat in the
transmitter.
Only if the transmitter has some sort of isolator (in a logical sense)
to push
the reflected power to a load, and that's pretty rare in a HF
transmitter.

What really happens is that the transmitter is operating less
efficiently with
a load other than its optimum. And that depends a lot on whether it's
something like a tube amplifier with a tuned tank, or a broadband solid
state
design.




--
*Dave - W?LEV*
*Just Let Darwin Work*


Re: What is the power rating of the 50 Ohm dummy load provided with the nanoVNA?

 

On 8/20/20 11:39 AM, rahul@... wrote:
Can I use it for a 5 W radio to calibrate it?
< 1/2 Watt is typical.

the ones with a "hat" might be rated higher (1 or 2W) but probably not.

You really don't want to overheat a cal standard.




Re: What is the power rating of the 50 Ohm dummy load provided with the nanoVNA?

 

Heed the cautions, or you will be without cal. standards!!!

2-watt carbon comp. or thick film resistors are pretty good for the purpose
through 50 MHz (I just checked with the VNA). They are available from
Digikey and Mouser, and likely other suppliers. Likely 5-watters are also
available. The 2-watt units will easily take 5-watts for a short time.

Dave - W?LEV

On Thu, Aug 20, 2020 at 6:43 PM <rahul@...> wrote:

Can I use it for a 5 W radio to calibrate it?



--
*Dave - W?LEV*
*Just Let Darwin Work*


Re: O S L on antenna side of a balun / choke with stud terminals?

 

On 8/20/20 7:15 AM, Jerry Gaffke via groups.io wrote:
Jim,
Then Miro's question remains: What happens to power reflected back from the antenna feedpoint
if there is no antenna tuner? Without a tuner, it is not all getting sent back out to the antenna.
Conservation of energy says the transmitter has to put it somewhere.
It could reflect back from the transmitter toward the antenna (and some be lost in the transmission line) - bouncing back and forth.

Don't forget that most transmitters do not have a 50 ohm resistive output impedance (particularly an amplifier with a tuned circuit in the output).

What happens with a amplifier with a design output impedance connected to a 25 ohm load (no transmission line). Is it "reflected power" that is increasing the dissipation of the transmitter (if it is increased.. some amplifiers will just put out a different amount of power), or is it a change in the operating point of the transmitter.

I find it helpful in these cases to consider some limiting cases - a stiff DC power supply has a 0 ohm output impedance, so it's "mismatched" into pretty much any load. Consider a AC source with zero output impedance - it can launch a wave into a transmission line or towards a load, but does that mean that there's a reflection and that some power has to be absorbed? Not necessarily.

So what's "bad" about reflected power (or mismatch) - they're really the same thing, just one is in terms of a traveling wave and the other isn't.

In terms of practical amplifier design, you wind up with either a voltage or a current that is higher than in the nominal case - and typically, one doesn't want to overdesign the amplifier - if you're putting out 70V for 100W into a 50 ohm load, do you want to go buy 200V transistors (to cover the worst case), or 100V transistors.



If I have a source with 50 ohm output impedance, designed to put 1 volt into a 50 ohm load - what I've really got is a 2 volt source with zero output impedance and a 50 ohm series resistor. And in the "theoretical" sense, that 50 ohm series resistor would dissipate the same amount of power as the load. If I feed an open circuit, now the voltage at the output terminal (keeping the 2V on the AC source) is now 2V. If I feed a short, the voltage is zero, but now the current is twice.

In reality, of course, you don't have a zero ohm source or a 50 ohm resistor in the transmitter. You've got something else which has a fairly complex relationship between voltages, currents, etc. And that something has practical limits (or cost constraints).

You've no doubt seen the ads for 1kW RF FETS that say "tolerates any mismatch" with the video of the guy putting a screwdriver intermittently across the output with sparks and the watt meter banging back and forth.

Well, it turns out this is easy - if your bus voltage is 100V and you've got a 300V transistor in there, no matter what sort of resistive load you put on it, you're NEVER going to go past 200V. Same sort of argument for the output current - even into a short, you're not going to melt the part, because the power supply (or series resistance in the transistor) will limit the current and the dissipation.

In none of these situations, though, are you anywhere close to this idea of "reflected power being absorbed in the transmitter"

For some specific cases (Thevenin equivalent sources with resistive output impedances) the math works out right as if it were being "absorbed" in some way, but that's not a real situation:

2V source 50 ohm series resistance

Load: open - perfect reflection - where's the reflected power? it's zero, because there's no forward power.
Load: 50 ohms - perfect match - 20mW dissipated in source, 20mW dissipated in load - 40mW consumed from source
Load: Short - perfect reflection - 80mW consumed from 0 ohm source, 80mW dissipated in internal resistance

Load: 25 ohms - 2:1 VSWR - voltage at load is 0.667 V, about 9mW. Dissipation in source resistance is 36 mW, so total source power is 45 mW - But let's look at the "reflected power" 2:1 is a reflection coefficient of 0.333, or a return loss of 9.5 dB - that is, about 1/10th of the forward power is reflected. The forward power toward the load is 9 * 1.1 or around 10mW, so the reflected power is 1 mW.

Hey, but my ideal voltage source is putting out 45 mW instead of 40mW - that's 5 mW more, not 1 mW more. And the series source resistor is dissipating 36 mW instead of 20mW. That's a WHOLE lot more than 1mW.


All the whole "matching theorem" says is that for maximum power transfer the source and load should be conjugates. That is, if you plot "DC power into perfectly efficient 0 ohm source" vs "power dissipated in load" the ratio will be lowest at a conjugate match, and will be 50%

it doesn't say anything about "where the reflected power is absorbed" or amplifier efficiency or heat dissipation. It is *entirely* a convenient analytical model: nobody in their right mind would build a high power source consisting of a zero ohm source with a series resistor.

However, we DO use that kind of design all the time in digital circuitry - a digital gate has a very low output impedance, so you put a 50 ohm resistor in series, before driving a long coax run to a (hopefully) 50 ohm load. It's not high power, so we don't are about efficiency or power transfer - what we're concerned about here is not having a discontinuity which leads to a reflection in the time domain.

And, such sources are used in test equipment - The standard lab function generator (like a Keysight 33620 series) is basically a power DAC with zero ohm output impedance driving a 50 ohm series resistor - again, we're not concerned with efficiency or power transfer, just having a stable and known impedance that closely approximates a simple theoretical model.

But when it comes to high power amplifiers, most designers spend a fair amount of time trying to get their design "aligned" to the load (to avoid the word "match") so that it has decent overall efficiency - if your power amplifier devices aren't zero ohms (generally the case) then you try and design a *lossless* network that transforms whatever it is to what the intended load is (which might be 50 ohms resistive, 75 ohms resistive or something else entirely..) And you hope that this works over the frequency range of interest (since low loss matching tends to be narrow band, or the devices themselves tend to have varying output Z)

Sometimes this isn't possible - too wide a frequency range for the devices you've got. Then you rely on two or three time tested approaches:
1) Design enough margin into the circuit that it can tolerate 2x Voltage and 2x current.
2) put a nonlinear device in the circuit (a circulator) that an send the reflected power to a load.
3) design the circuit to foldback or limit, when it detects the voltage getting too high or current getting to high.
4) Way over build in terms of power and put a pad in series.

Your choice depends on your design constraints and budgets.

In the amateur radio world, there's a "RF output power" constraint (it used to be DC power to the final stage, and before that it was "HP to the spark gap") - and a dollar constraint - so you design amps that are efficient but require good load management. (it's funny, if you didn't care about DC power requirements, you could sell a ham transmitter that could put 1kW into "any load" )


In the commercial communications world, they're not "transmitter power" limited, but rather "radiated power" limited - so you see things like kilowatt amplifiers into a terminated folded dipole - or, the equivalent, a untuned dipole with a 6dB pad.

In space communications, we're obsessed with DC power efficiency and mass, and the selection of parts is limited - so you see circulators and loads, more for protection than out of need.

It's interesting - the desire to have more output power for solid state amps in space has led to interesting designs with combiners (since the individual amplifier dice only put out a few watts) that cleverly reflect harmonic power back into a optimized resonant amplifier (Class E/F1).




I can't think of anything the transmitter could do with this energy other than turn it into heat.
Either by design with your isolator and load, or just heating up whatever's handy.
mmmm, not really - most amplifiers don't have loads in them - like I said, it's more about amplifier efficiency changes.



I don't know for sure exactly where it goes in a typical amplifier.
Hence my weasel words: "What happens then depends on the transmitter design."
Jerry, KE7ER
On Thu, Aug 20, 2020 at 06:55 AM, Jim Lux wrote:
On 8/19/20 10:09 PM, Jerry Gaffke via groups.io wrote:
Without the antenna tuner, that reflected power comes back to the
transmitter.
What happens then depends on the transmitter design.
At least some of it will likely be getting dissipated as heat in the
transmitter.
Only if the transmitter has some sort of isolator (in a logical sense) to push
the reflected power to a load, and that's pretty rare in a HF transmitter.

What really happens is that the transmitter is operating less efficiently with
a load other than its optimum. And that depends a lot on whether it's
something like a tube amplifier with a tuned tank, or a broadband solid state
design.


Re: O S L on antenna side of a balun / choke with stud terminals?

 

Hi Jerry,

There is matching the feedline to the antenna and matching the feedline to the transmitter. Only if the antenna comes all the way to the transmitter does the box in the shack match the antenna. Nothing is "tuned". However we know what people really mean when they abuse that label. Ideally you would math the feedline to the antenna AND if necessary match the feedline to the transmitter at the other end.

SWR does NOT indicate a resonant antenna either.

73,

Bill KU8H

bark less - wag more

On 8/20/20 1:43 PM, Jerry Gaffke via groups.io wrote:
Dave,
I agree with you that the antenna itself need not be resonant.
Though I'd say that when an antenna tuner creates a suitable
matching network for that non-resonant antenna, we can properly say
that the entire antenna system of tuner plus feedline plus antenna
has been "tuned".
I didn't see anything at to convince me otherwise.
DIgging around some on his website, I find this on
IT DOES NOT 'TUNE' THE ANTENNA! NEVER! (with one exception)
To do that, the matchbox would have to be located at and be a physical part of the antenna.
AND IF IT WERE PART OF THE ANTENNA, IT WOULD NOT NEED A SEPARATE NAME AND DESCRIPTION!
But it does match the antenna and its transmission line (the antenna system) to the transmitter.
Therefore I prefer to say "match" and not "tune".
So he figures something that "tunes the antenna" must be physically part of the antenna.
OK, fine with me, but that strikes me as a rather silly distinction.
Perhaps we can all agree that an antenna tuner tunes the "antenna system",
where the antenna system includes the tuner, the feedline, and the antenna.
Also, on that same page he says:
NOTE 1: [very important!] Just because the matchbox is able to match the antenna and transmission line to the transmitter with a perfect match (SWR = 1.0:1), that does not necessarily mean that all of the power being passed through the matchbox to the antenna.
THERE IS ALWAYS SOME POWER LOSS WHEN USING A MATCHBOX.
THIS IS VERY IMPORTANT TO UNDERSTAND.
NOTE 2: [and also very important!] Just because power is efficiently transferred into the transmission line (with very little loss inside the matchbox), do NOT assume the it effeciently arrives at the antenna! Don't make that mistake! ? Transmission lines, ESPECIALLY COAX, can add significant loss to the antenna system.
That strikes me as a bit over the top.
If the antenna is somewhat close to a match to begin with, perhaps a 3:1 or 4:1 SWR
so the reflections aren't just too severe, and if the coax is not terribly lossy,
then at HF you probably don't lose enough power in the coax for the guy at the far end to even notice.
Maxwell's "Reflections III" makes the case in section 4.4,
Later in the book he gave a more definitive formula for the loss, but I'm not finding it just now.
The organization of that book is more than a little bit scattered.
It is legitimately available on the web as a pdf.
And if you are worried about those coax losses, use ladder line.
Then even with some really drastic reflections between antenna tuner and antenna feedpoint,
the antenna system will still radiate power efficiently.
Jerry, KE7ER
On Thu, Aug 20, 2020 at 09:43 AM, David Eckhardt wrote:

Other than the name, "ANTENNA TUNER", why is the 'conventional knowledge'
so prevalent that the "ANTENNA TUNER" actually "tunes" the antenna. *IT
DOES N O T !!!!!* It functions ONLY AS A MATCHING NETWORK, nothing more,
nothing less!!!!!!
.


Re: O S L on antenna side of a balun / choke with stud terminals?

 

Matching circuit as part of the *system* consisting of the antenna,
transmission, and matching network is quite acceptable so long as we
realize there is no antenna "tuning" involved. With that "system" concept,
yes, the matching network, a.k.a. antenna 'tuner', functions to resonate
the *system*. However, again, it does NOT 'tune' the antenna. It only
brings the *system* into resonance at a chosen frequency. It's an embedded
matching network just like the broadband transformers in our solid state
PA's.

If one chooses the proper matching network topology, the losses are
extremely minimal. The L-Network is the only network where a single match
solution is available between the two adjustments, inductance and
capacitance. Also, it is the only network where this unique matching set
of adjustable parameters yields highest efficiency. So you get guaranteed
maximum efficiency at the unique matched condition. Many matching networks
marketed to the amateur ignore this fact. I use such a system as
described. Even after 5-minutes of key-down conditions at 1.5 kW on any
band (I don't do 10 or 12-meters- too much like CB), there is only a very
minor rise in temperature within *any* of the two components. Remember,
generated heat indicates losses within the network. I have very little
loss in the matching network.

Yes, mismatches in a coaxial system can and do lead to additional losses,
especially at the higher frequencies. We should all know that by now.
That's why I do not use coax in my HF system (for reasons other than
costing more than alternatives). There are only 5-feet or so of coax in my
entire feed system and all of that is at 50-ohms (non-reactive). I use
parallel wire, open wire, window line, balanced line (pick your
terminology, they all refer to the same type of transmission line). And so
long as balanced feed and load are assured, there is minimal interaction
with my balanced antenna, a 450-foot long center-fed doublet. For that
purpose, I use multiple common mode chokes (verified on the HP 8753C VNA)
taylored for the operating frequency at the output of the matching network.

Addressing open wire feeders and commercially available 'tuners': If
currents are properly balanced (equal amplitude and opposite phase), no
radiation to or from the feedline occurs. This is generally NOT provided
for in commercially available matching networks marketed to the amateur
community. Transformers have no way of assuring a balanced source/load
to/from the feedline. Most claim a 4:1 balun at the output of the
'tuner'. It's a very poor balun, at best, which does not function well at
all, especially with a complex portion of the impedance to be matched.
This introduces even more (undocumented) losses in the commercially
available system. This time I'll get it right: refer to:



for an excellent treatment of the whole balun subject. And read and digest
all the embedded links on the subject within that site. It's good reading
and very enlightening!!!

Dave - W?LEV

On Thu, Aug 20, 2020 at 5:43 PM Jerry Gaffke via groups.io <jgaffke=
[email protected]> wrote:

Dave,

I agree with you that the antenna itself need not be resonant.

Though I'd say that when an antenna tuner creates a suitable
matching network for that non-resonant antenna, we can properly say
that the entire antenna system of tuner plus feedline plus antenna
has been "tuned".

I didn't see anything at to convince me
otherwise.
DIgging around some on his website, I find this on


IT DOES NOT 'TUNE' THE ANTENNA! NEVER! (with one exception)
To do that, the matchbox would have to be located at and be a physical
part of the antenna.
AND IF IT WERE PART OF THE ANTENNA, IT WOULD NOT NEED A SEPARATE NAME AND
DESCRIPTION!
But it does match the antenna and its transmission line (the antenna
system) to the transmitter.
Therefore I prefer to say "match" and not "tune".


So he figures something that "tunes the antenna" must be physically part
of the antenna.
OK, fine with me, but that strikes me as a rather silly distinction.

Perhaps we can all agree that an antenna tuner tunes the "antenna system",
where the antenna system includes the tuner, the feedline, and the antenna.

Also, on that same page he says:

NOTE 1: [very important!] Just because the matchbox is able to match the
antenna and transmission line to the transmitter with a perfect match (SWR
= 1.0:1), that does not necessarily mean that all of the power being passed
through the matchbox to the antenna.
THERE IS ALWAYS SOME POWER LOSS WHEN USING A MATCHBOX.
THIS IS VERY IMPORTANT TO UNDERSTAND.
NOTE 2: [and also very important!] Just because power is efficiently
transferred into the transmission line (with very little loss inside the
matchbox), do NOT assume the it effeciently arrives at the antenna! Don't
make that mistake! ? Transmission lines, ESPECIALLY COAX, can add
significant loss to the antenna system.


That strikes me as a bit over the top.
If the antenna is somewhat close to a match to begin with, perhaps a 3:1
or 4:1 SWR
so the reflections aren't just too severe, and if the coax is not terribly
lossy,
then at HF you probably don't lose enough power in the coax for the guy at
the far end to even notice.

Maxwell's "Reflections III" makes the case in section 4.4,
Later in the book he gave a more definitive formula for the loss, but I'm
not finding it just now.
The organization of that book is more than a little bit scattered.
It is legitimately available on the web as a pdf.

And if you are worried about those coax losses, use ladder line.
Then even with some really drastic reflections between antenna tuner and
antenna feedpoint,
the antenna system will still radiate power efficiently.

Jerry, KE7ER



On Thu, Aug 20, 2020 at 09:43 AM, David Eckhardt wrote:

Other than the name, "ANTENNA TUNER", why is the 'conventional knowledge'
so prevalent that the "ANTENNA TUNER" actually "tunes" the antenna. *IT
DOES N O T !!!!!* It functions ONLY AS A MATCHING NETWORK, nothing more,
nothing less!!!!!!


--
*Dave - W?LEV*
*Just Let Darwin Work*


Re: O S L on antenna side of a balun / choke with stud terminals?

 

Jerry, I second your statements!

I agree that "match" is more appropriate in this context then "tune", but "antenna tuner" rolls off a tong easier then an "antenna matcher" :)

To some extent, "tuner" does make sense, under some assumptions - I guess that most of us would agree that "resonant antenna" (or antenna on resonant frequency) only means that that, at the feed point, antenna appears to have restive component of the impedance only. It's not quite a stretch to then assume that when someone is "tuning" an antenna (adjusting the length of a dipole for example), the goal is to bring that antenna into resonance (well, not always, but let's say most of the time). If we agree so far, then a device that can make a non resonant antenna appear as restive load can be called a "tuning device", right?

I still agree that "device that matches one complex impedance to another" is much more accurate, but try to say that 3 times. Antenna tuner still "sounds" better :)


Dave,

It was not my point to condemn the need/use of matching networks (either wide band transformers, LC, t-line based or any other), my point was that if you have your TX properly designed to present 50 ohms, using antenna tuner to connect it to some crazy impedance that comes as a result of your "random" antenna and even more random t-line length and impedance and "see" SWR 1:1 does not mean that you maxed out radiated power.

If that "antenna tuner" (sic) hides the mess of various mismatches behind, it is still possible that significant part of your power is not "hitting the air".

And yes, I'm not concerned with SWR 1:3, it's only affecting half the S unit on the receiving side, all else being the same. Still can't believe that you need 6db difference for full S unit on the reception side, but thats what smart people say :)


Re: What is the power rating of the 50 Ohm dummy load provided with the nanoVNA?

 

Yes, it doesn't matter the power used for the calibration. You just have
to be sure of the resistance value.
Un saludo

El jue., 20 ago. 2020 21:04, Larry Rothman <nlroth@...> escribi¨®:

Rahul,
NEVER use the calibration parts that came with the nanovna for anything
other than to calibrate the nanovna with!!
The 50 ohm calibration load has the equivalent of about a 1/16W precision
resistor in it.
Also - NEVER push any power level more than 10dbm (10mW) into the nanovna
- it's front end bridge is also made of surface mount resistors that cannot
dissipate any higher power.

The nanovna is not like a standard through-line VSWR/watt meter - it's a
self-contained measuring instrument.
Read up about VNAs in the Wiki and Files areas - lots of beginner info
there to go through.
...Larry

On Thursday, August 20, 2020, 2:43:38 p.m. EDT, rahul@... <
rahul@...> wrote:

Can I use it for a 5 W radio to calibrate it?







Re: What is the power rating of the 50 Ohm dummy load provided with the nanoVNA?

 

As Larry says, 5W into your calibration standard would be a really bad idea.
If in the US, buy a couple Xicon 283-100-RC from Mouser, use two in parallel
for a 50 ohm load, will dissipate 6W all day long.

Or use twenty in parallel of 283-1.0K-RC in parallel,
good for 60W, 100W for short periods.
Total cost of US$2. (Plus $10 shipping)

Jerry, KE7ER

On Thu, Aug 20, 2020 at 11:43 AM, <rahul@...> wrote:

Can I use it for a 5 W radio to calibrate it?


Re: What is the power rating of the 50 Ohm dummy load provided with the nanoVNA?

 

Rahul,
NEVER use the calibration parts that came with the nanovna for anything other than to calibrate the nanovna with!!
The 50 ohm calibration load has the equivalent of about a 1/16W precision resistor in it.
Also - NEVER push any power level more than 10dbm (10mW) into the nanovna - it's front end bridge is also made of surface mount resistors that cannot dissipate any higher power.

The nanovna is not like a standard through-line VSWR/watt meter - it's a self-contained measuring instrument.
Read up about VNAs in the Wiki and Files areas - lots of beginner info there to go through.
...Larry

On Thursday, August 20, 2020, 2:43:38 p.m. EDT, rahul@... <rahul@...> wrote:

Can I use it for a 5 W radio to calibrate it?


What is the power rating of the 50 Ohm dummy load provided with the nanoVNA?

 

Can I use it for a 5 W radio to calibrate it?


Re: O S L on antenna side of a balun / choke with stud terminals?

 

Dave,

I agree with you that the antenna itself need not be resonant.

Though I'd say that when an antenna tuner creates a suitable
matching network for that non-resonant antenna, we can properly say
that the entire antenna system of tuner plus feedline plus antenna
has been "tuned".

I didn't see anything at to convince me otherwise.
DIgging around some on his website, I find this on

IT DOES NOT 'TUNE' THE ANTENNA! NEVER! (with one exception)
To do that, the matchbox would have to be located at and be a physical part of the antenna.
AND IF IT WERE PART OF THE ANTENNA, IT WOULD NOT NEED A SEPARATE NAME AND DESCRIPTION!
But it does match the antenna and its transmission line (the antenna system) to the transmitter.
Therefore I prefer to say "match" and not "tune".


So he figures something that "tunes the antenna" must be physically part of the antenna.
OK, fine with me, but that strikes me as a rather silly distinction.

Perhaps we can all agree that an antenna tuner tunes the "antenna system",
where the antenna system includes the tuner, the feedline, and the antenna.

Also, on that same page he says:

NOTE 1: [very important!] Just because the matchbox is able to match the antenna and transmission line to the transmitter with a perfect match (SWR = 1.0:1), that does not necessarily mean that all of the power being passed through the matchbox to the antenna.
THERE IS ALWAYS SOME POWER LOSS WHEN USING A MATCHBOX.
THIS IS VERY IMPORTANT TO UNDERSTAND.
NOTE 2: [and also very important!] Just because power is efficiently transferred into the transmission line (with very little loss inside the matchbox), do NOT assume the it effeciently arrives at the antenna! Don't make that mistake! ? Transmission lines, ESPECIALLY COAX, can add significant loss to the antenna system.


That strikes me as a bit over the top.
If the antenna is somewhat close to a match to begin with, perhaps a 3:1 or 4:1 SWR
so the reflections aren't just too severe, and if the coax is not terribly lossy,
then at HF you probably don't lose enough power in the coax for the guy at the far end to even notice.

Maxwell's "Reflections III" makes the case in section 4.4,
Later in the book he gave a more definitive formula for the loss, but I'm not finding it just now.
The organization of that book is more than a little bit scattered.
It is legitimately available on the web as a pdf.

And if you are worried about those coax losses, use ladder line.
Then even with some really drastic reflections between antenna tuner and antenna feedpoint,
the antenna system will still radiate power efficiently.

Jerry, KE7ER

On Thu, Aug 20, 2020 at 09:43 AM, David Eckhardt wrote:

Other than the name, "ANTENNA TUNER", why is the 'conventional knowledge'
so prevalent that the "ANTENNA TUNER" actually "tunes" the antenna. *IT
DOES N O T !!!!!* It functions ONLY AS A MATCHING NETWORK, nothing more,
nothing less!!!!!!


Re: O S L on antenna side of a balun / choke with stud terminals?

 

I have both the first and second editions of J. D. Kraus's book, "*ANTENNAS*".
Either edition is likely the best book on antennas that isn't 'big time'
heavy in math, although a working knowledge of calculus should be under
your belt. Others are FAR heavier into math and not recommended for the
amateur. The second edition:

"*ANTENNAS*", by J. D. Kraus, publisher: McGraw Hill, ISBN:
0-07-0354 22-7

The first edition:

"*ANTENNAS*", by J. D. Kraus, publisher: McGraw Hill, ISBN:
None given

Either is an excellent and rigorous treatment of antennas and theory. I
highly recommend either edition to anyone who desires a starting reference
on the subject.

Another "reasonable" but brief reference I stumbled on while at HRO, Denver
follows a couple of years ago:

"*ANTENNAS AND TRANSMISSION LINES*", by: John A. Kuecken,
Publisher: MFJ, ISBN: None given

Dave - W?LEV

On Thu, Aug 20, 2020 at 4:27 PM Jerry Gaffke via groups.io <jgaffke=
[email protected]> wrote:

I assume the book you are referring to is "Antennas" by John D Kraus.
Jim mentioned it here: /g/nanovna-users/message/16645

FIrst edition printed in 1950, second edition (with Ronald Marhefka) in
1988.

I poked around on the web.
Lots of pdf downloads out there, they all look kind of dodgy.
For such a thick read, I'd probably want a paper copy anyway.

Amazon sells the 2001 third edition for $64.41, the fourth (2010) for
$847.00, and the fifth (2018) for $32.23.


Or maybe Amazons price for the third in paperback is $54.41, hardcover for
$139.65.



Here's a 5'th edition, $11.95 plus $13.03 to ship it from India:

"Territorial restriction maybe printed on the book. This is an Int'l
edition, ISBN and cover may differ from US edition ..."
Apparently the same one that Amazon sells for $32.23.

So kind of confusing.
I may just give Bezos his $32.23.
If somebody thinks there's a legit pdf out there, I'd probably sample that
first.
More than likely, I'll get sidetracked after a few pages anyway, trying to
figure out
Maxwell's equations again or some long forgotten math.

Jerry, KE7ER



On Thu, Aug 20, 2020 at 07:26 AM, Dallas wrote:


Thanks for the book link. I just downloaded it and read the first few
pages,
it will be a good read.

Dallas
N5fee


--
*Dave - W?LEV*
*Just Let Darwin Work*


Re: Can the NanoVNA be used on 75 ohm cables/ antennas --- Part 2 #75 ohm measurements

 

For those of you who would like to use the resistive pad to match 75 ohms to the 50 ohm nanovna, the series resistor is 25 x sqrt of 3 (about 43.3 ohms) and the shunt resistor on the 50 ohm side is 50 x sqrt 3 (about 86.6 ohms) . With that you can use 1% and 1/2% resistors and get as close as you want! Or you can distribute the resistance around the connectors to minimize inductance and capacitance.


Re: O S L on antenna side of a balun / choke with stud terminals?

 

Other than the name, "ANTENNA TUNER", why is the 'conventional knowledge'
so prevalent that the "ANTENNA TUNER" actually "tunes" the antenna. *IT
DOES N O T !!!!!* It functions ONLY AS A MATCHING NETWORK, nothing more,
nothing less!!!!!!

Please consult and digest the information presented at the following URL:



If you are one of the many that believe (for better or worse) that matching
networks (yes, the proper name for the function) are evil, consider the
following. At the collector or drain of the output stage of your modern
transceiver, the impedance presented to the 'load' is typically 0.5 to a
couple of ohms real and a bit of reactance. That plots to the extreme left
side of the Smith Chart which is the 'shorted' side of the chart. It may
be slightly above or below the horizontal line which represents only real
resistance with no reactance. Above or below that horizontal line
represents reactance of the impedance, only.

OK, that's a terrible mismatch to our communications standard of 50-ohms
(real, no reactance): a 50:1 SWR!! How do the designers of that PA get
from such a low impedance to 50 ohms, real??? Guess what?? A broadband
matching transformer is used between the collectors or drains and the
inputs to the various filters to keep FCC happy. That broadband matching
transformer can be viewed as a "matching network" that transforms a 50:1
SWR to a 1:1 SWR. Is that evil? No. Is there loss? Yes, and a bit more
that the typical "antenna tuner" introduces to the overall system. Is it
necessary? Yes, if we are to utilize a 50-ohm non-reactive 'load'. The
option is to drive a 1 to 2-ohm transmission line supported by a 1 to 2-ohm
load at the antenna.

Matching networks are necessary and an integral part of any RF design, be
it for high power or small-signal low noise amplification. We wouldn't be
able to utilize the 'wonders' of the RF world without matching networks.

Dave - W?LEV

On Thu, Aug 20, 2020 at 4:07 AM Miro, N9LR via groups.io <m_kisacanin=
[email protected]> wrote:

So, based on what Dave and Jerry are pointing out, I have two VALID
conclusions

1) Antenna tuners suck! They make t-line/antenna go gently on TX, but they
hide what's behind them. Strive to have (at least) antenna matched to
t-line, then antenna tuner (might) make more sense, or even better, match
TX to t-line to antenna and sell antenna tuner on ebay

2) I still have no idea what is the answer to my question :)
* short run of low loss t-line
* no match between TX, t-line, antenna impedances
* antenna tuner makes t-line/antenna appear to TX as TX's impedance (SWR
at that point is 1:1)





--
*Dave - W?LEV*
*Just Let Darwin Work*