Basic question (at least for me) for a very poor antenna matching :
-100 w reach the antenna and 50 w are radiated.
- 50 w are "reflected", what is their fate ?
Are they definitely lost for radiation and just heat the line, the final....
JC

On 6 jun, 16:22, "JC" wrote:
Basic question (at least for me) for a very poor antenna matching :
-100 w reach the antenna and 50 w are radiated.
- 50 w are "reflected", what is their fate ?
Are they definitely lost for radiation and just heat the line, the final....
JC

Hello,

If the source has a linear 50 Ohms output impedance (assuming 50 Ohm
cable, no loss), all power will go back into the source. Partly in the
form of heat, partly in the form of power saving. However a PA is not
a 50 Ohms source, so what happens may vary.

It is very likely that some reverse power reflects back to the antenna
(as the transmitter is not a source with 50 Ohms output impedance). So
the reading on an instrument that measures true forward power is the
sum of the real incident power plus the reflected power from the
transmitter towards the antenna. Changing the load may result in a
reduction or increase of total forward power. In case of a real 50
Ohms source, the forward power will not change, no matter the load.

The bad match seen from the transmitter may result in a reduction of
DC input power (input current reduces), but may also result in an
increase of DC input power.

So when you have reverse power reading of 50W, you cannot just say
that the PA has to dissipate an extra 50W, it can be more, less and
even negative. The negative case is when the input power reduces
significantly and the active element has to dissipate less (with
respect to the best match condition). In particular high efficiency
amplifiers (where the active devices are used as switches), show
strong variation in supply current versus load change. Under certain
load conditions such amplifiers may show a strong decrease in
efficiency resulting in high relative increase of device dissipation.

Best regards,

Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me

On Sun, 6 Jun 2010 07:42:31 -0700 (PDT), Wimpie
wrote:

However a PA is not a 50 Ohms source

Hi Wimpie,

You say this like others, with the air of "knowing." However, when I
ask in response of those who "know" what the PA is NOT, what IS it?

Give me the Z value of your transmitter. Specify all initial
conditions.

We have had lengthy correspondence with Walt Maxwell's very rigorously
measured Kenwood TS830s that demonstrates a Z of 50 Ohms, or nearly
that as is practicable (say +/- 20%); and yet your voice was missing
from this discussion with evidence to the contrary.

73's
Richard Clark, KB7QHC

On 6 jun, 19:00, Richard Clark wrote:
On Sun, 6 Jun 2010 07:42:31 -0700 (PDT), Wimpie
wrote:

However a PA is not a 50 Ohms source

Hi Wimpie,

You say this like others, with the air of "knowing." *However, when I
ask in response of those who "know" what the PA is NOT, what IS it?

Give me the Z value of your transmitter. *Specify all initial
conditions.

We have had lengthy correspondence with Walt Maxwell's very rigorously
measured Kenwood TS830s that demonstrates a Z of 50 Ohms, or nearly
that as is practicable (say +/- 20%); and yet your voice was missing
from this discussion with evidence to the contrary.

73's
Richard Clark, KB7QHC

Hello Richard,

Try the following experiment:

Measure the forward power of your PA at a convenient load. Use a
directional coupler for that, not a voltage meter calibrated for
power. The "SET" or "CAL" position of a VSWR meter can be used as
forward power indicator.

Now make some mismatch (for example VSWR=2 at different phase) and
read the forward power. Did it change? If so, the output impedance is
no (longer) 50 Ohms. If possible, disable automatic protection to
avoid changing drive level. I don't know the value for my FT7B, but I
know forward power changes with load variations (as I use it as
"measuring instrument" sometimes).

Virtually all power amplifiers I designed do not have a large signal
output impedance of 50 ohms under significant load change. For some I
measured it because of a discussion on this between colleges.

Measuring method used: change in resistive load, from voltage change
you can calculate the current change, hence the output impedance. One
note, except two, all where solid state.

All high efficiency designs (class E, D) that I did have output
impedance far from the expected load impedance. With "far" I mean
factor 2 or factor 0.5. I did not measure that (as it is not
important in virtually all cases), but know it from the overload
simulation/measurement and I did the design myself.

The reason for not being 50 Ohms (after matching) is that when you
change the load, the active device will go into voltage or current
saturation. This is not a hard process, so for small load variation
(low VSWR values), forward power will not change much. I think this is
especially true for vacuum triode PA where you have significant tube-
internal feedback. For large variation (for example VSWR = 2.5,
reflected power 18%), you will notice change in forward power for most
power amplifiers.

High efficiency CW amplifiers use saturated switches (for example
half, full bridge or push-pull), so behave (seen at the active device)
as a voltage source. Depending on the total phase shift of the filter
sections this may convert to a current source behavior (seen at the
output). I had to spent much time to avoid destruction of some
circuits in case of mismatch.

If you can drop me a link to the discussion of the TS830s, I would
appreciate that.

Best regards,


Wim
PA3DJS
www.tetech.nl
Remove abc first in case of PM.

On Sun, 6 Jun 2010 15:21:59 -0700 (PDT), Wimpie
wrote:

Try the following experiment:

-sigh-

Hi Wimpie,

This is not what I asked for. You said what the source was NOT, but
you cannot say what the source IS.

The source is not a bipedal reptile. The source is not an elongated
diphthong. The source is not the resurrection of a deity (some may
argue that more than I would care to follow). The source is not....

There are many ways to say what the source is NOT, and that will never
inform us about the source. I see many draw deuces to this question
and try to convince everyone that the card is a pair of winning aces.

Now make some mismatch (for example VSWR=2 at different phase) and
read the forward power. Did it change? If so, the output impedance is
no (longer) 50 Ohms.

What IS it now? If you could measure it once, you should be able to
tell us what it is this time too. I did this for years to methods set
by the National Bureau of Standards. You have drawn a deuce, not two
aces.

Measuring method used: change in resistive load, from voltage change
you can calculate the current change, hence the output impedance. One
note, except two, all where solid state.

Yes, that is called a load pull, but you offer no data - another
deuce. I have done a lot of load pulls.

All high efficiency designs (class E, D) that I did have output
impedance far from the expected load impedance. With "far" I mean
factor 2 or factor 0.5. I did not measure that (as it is not
important in virtually all cases), but know it from the overload
simulation/measurement and I did the design myself.

You didn't measure it, but you can state the value - interesting state
of guessing. So, what did the simulation/non-measurement give you as
a value? What IS the value? Another deuce.

The reason for not being 50 Ohms

Reasons abound. BP is giving us reasons why the Gulf Coast shouldn't
worry. Data has proven that reasons don't work and neither does their
well. That is a Joker draw.

High efficiency CW amplifiers

The TS830S is not such an animal. Another Deuce.

If you can drop me a link to the discussion of the TS830s, I would
appreciate that.

Google "Plate Resistance" in this group.

73's
Richard Clark, KB7QHC

On 7 jun, 03:04, Richard Clark wrote:
On Sun, 6 Jun 2010 15:21:59 -0700 (PDT), Wimpie
wrote:

Try the following experiment:

-sigh-

Did you do the experiment (forward power versus mismatch)?

Hi Wimpie,

This is not what I asked for. *You said what the source was NOT, but
you cannot say what the source IS.

The source is not a bipedal reptile. *The source is not an elongated
diphthong. *The source is not the resurrection of a deity (some may
argue that more than I would care to follow). *The source is not....

There are many ways to say what the source is NOT, and that will never
inform us about the source. *I see many draw deuces to this question
and try to convince everyone that the card is a pair of winning aces.

Probably you consumed something wrong here.

Now make some mismatch (for example VSWR=2 at different phase) and
read the forward power. Did it change? If so, the output impedance is
no (longer) 50 Ohms.

What IS it now? *If you could measure it once, you should be able to
tell us what it is this time too. *I did this for years to methods set
by the National Bureau of Standards. *You have drawn a deuce, not two
aces.

Some values:
9 +/-1 Ohm (real impedance), reference impedance 16 Ohms, 8 MHz
amplifier (ISM). Exact value requried because of additional filtering,
based on several IRF110, push pull.

2 Ohms (small reactance present), 10V/1A driver (sinusoidal), 700
kHz, slightly saturating class C output stage based on single mosfet.
RF feedback present to guarantee value below 2 Ohms. also here,
certain maximum value was required for the application

|RC| 0.80 (actual value depending on type of additional filtering in
between), 5kW pulsed power amplfiier, non 50 Ohm application.
Actual value not important, but is a recent project, so I knew from
memory.

See also EL34 example in posting to Walt.



Measuring method used: change in resistive load, from voltage change
you can calculate the current change, hence the output impedance. One
note, except two, all where solid state.

Yes, that is called a load pull, but you offer no data - another
deuce. *I have done a lot of load pulls. *

All high efficiency designs (class E, D) that I did have output
impedance far from the expected load impedance. With "far" I mean
factor 2 or factor 0.5. I did not measure that (as it is not
important in virtually all cases), but know it from the overload
simulation/measurement and I did the design myself.

You didn't measure it, but you can state the value - interesting state
of guessing. *So, what did the simulation/non-measurement give you as
a value? *What IS the value? *Another deuce.

Please read again last part of sentence above your text.


The reason for not being 50 Ohms

Reasons abound. *BP is giving us reasons why the Gulf Coast shouldn't
worry. *Data has proven that reasons don't work and neither does their
well. *That is a Joker draw.

High efficiency CW amplifiers

The TS830S is not such an animal. *Another Deuce.

Not al hams use TS830S, See more recent reply of Walt to Tom's
posting.

If you can drop me a link to the discussion of the TS830s, I would
appreciate that.

Google "Plate Resistance" in this group.


73's
Richard Clark, KB7QHC

I hope, further replies from you will be constructive,


Wim
PA3DJS

On Jun 6, 3:21*pm, Wimpie wrote:

Measuring method used: change in resistive load, from voltage change
you can calculate the current change, hence the output impedance. One
note, except two, all where solid state.

Of course, since the source impedance may be anywhere in the complex
plane, you need to change more than just the resistive part of the
load, I believe, to get an accurate picture...

All high efficiency designs (class E, D) that I did have output
impedance far from the expected load impedance. With "far" I mean
factor 2 or factor 0.5. I did not measure that (as it is not
important in virtually all cases), but know it from the overload
simulation/measurement and I did the design myself.

What I've seen in similar situations is that the source impedance is
likely to be strongly reactive, depending on the filter network you
use to get sinusoidal output. In any event, the source impedance is
likely to have a reflection coefficient magnitude that is quite close
to unity. That is exactly what you should expect: there's nothing to
absorb reflections. You could (theoretically at least) use feedback
to make the output look like 50 ohms, but just as you say, Wim ...
why?? There's really almost never any point in doing so.

....


Cheers,
Tom

On Jun 6, 10:06*pm, K7ITM wrote:
On Jun 6, 3:21*pm, Wimpie wrote:

Measuring method used: change in resistive load, from voltage change
you can calculate the current change, hence the output impedance. One
note, except two, all where solid state.

Of course, since the source impedance may be anywhere in the complex
plane, you need to change more than just the resistive part of the
load, I believe, to get an accurate picture...



All high efficiency designs (class E, D) that I did have output
impedance far from the expected load impedance. With "far" I mean
factor 2 or factor 0.5. I did not measure that (as it is not
important in virtually all cases), but know it from the overload
simulation/measurement and I did the design myself.

What I've seen in similar situations is that the source impedance is
likely to be strongly reactive, depending on the filter network you
use to get sinusoidal output. *In any event, the source impedance is
likely to have a reflection coefficient magnitude that is quite close
to unity. *That is exactly what you should expect: *there's nothing to
absorb reflections. *You could (theoretically at least) use feedback
to make the output look like 50 ohms, but just as you say, Wim ...
why?? *There's really almost never any point in doing so.

...

Cheers,
Tom

Again Wim, we're not on the same page, so let me quote from your
eariler post:

"Virtually all high efficient amplifiers work in voltage saturated
mode
and are therefore not operated at maximum available power, therefore
their output impedance doens't match the expected load."

I have not been talking about MAXIMUM available power, only the power
available with some reasonable value of grid drive. In the real world
of amateur radio operations, of which I'm talking, when we adjust the
pi-network for that given drive level, we adjust for delivering all
the AVAILABLE power at that drive level. When all the available power
is thus delivered, the output source resistance equals the load
resistance by definition. We're now not talking about changing the
load, phase or SWR--we're talking about the single condition arrived
at after the loading adjustments have been made.

And Tom, why would the source be reactive when the pi-network is tuned
to resonance? And because the source resistance of the (tube) power
amp is non-dissipative, its reflection coefficient is 1.0 by
definition, and so it cannot absorb any reflected energy, and
therefore re-reflects it.

Walt, W2DU

On 6 jun, 19:00, Richard Clark wrote:
On Sun, 6 Jun 2010 07:42:31 -0700 (PDT), Wimpie
wrote:

However a PA is not a 50 Ohms source

Hi Wimpie,

You say this like others, with the air of "knowing." *However, when I
ask in response of those who "know" what the PA is NOT, what IS it?

Give me the Z value of your transmitter. *Specify all initial
conditions.

We have had lengthy correspondence with Walt Maxwell's very rigorously
measured Kenwood TS830s that demonstrates a Z of 50 Ohms, or nearly
that as is practicable (say +/- 20%); and yet your voice was missing
from this discussion with evidence to the contrary.

73's
Richard Clark, KB7QHC

Hello Richard,

Walt did respond and did a solid statement regarding the amplifier
with matching section to obtain maximum output (so I know the
conditions) This is the point where the tube/transistor is at the
edge of current/voltage saturation.

At that operating point the output impedance is 50 Ohms (for small
load variations). When you change the load significantly (or change
drive level), current or voltage saturation will dominate, hence the
output impedance is no longer 50 Ohms. This is also the reason that
PA intermodulation may occur in close spaced transmitters where some
power from transmitter A enters the amplifier of transmitter B and
vice versa. This also proves that there is no linear 50 Ohms output
impedance.

Best regards,

Wim
PA3DJS
www.tetech.nl

On Jun 6, 6:36*pm, Wimpie wrote:
On 6 jun, 19:00, Richard Clark wrote:



On Sun, 6 Jun 2010 07:42:31 -0700 (PDT), Wimpie
wrote:

However a PA is not a 50 Ohms source

Hi Wimpie,

You say this like others, with the air of "knowing." *However, when I
ask in response of those who "know" what the PA is NOT, what IS it?

Give me the Z value of your transmitter. *Specify all initial
conditions.

We have had lengthy correspondence with Walt Maxwell's very rigorously
measured Kenwood TS830s that demonstrates a Z of 50 Ohms, or nearly
that as is practicable (say +/- 20%); and yet your voice was missing
from this discussion with evidence to the contrary.

73's
Richard Clark, KB7QHC

Hello Richard,

Walt did respond and did a solid statement regarding the amplifier
with matching section to obtain maximum output (so I know the
conditions) *This is the point where the tube/transistor is at the
edge of current/voltage saturation.

At that operating point the output impedance is 50 Ohms (for small
load variations). When you change the load significantly (or change
drive level), current or voltage saturation will dominate, hence the
output impedance is no longer 50 Ohms. * This is also the reason that
PA intermodulation may occur in close spaced transmitters where some
power from transmitter A enters the amplifier of transmitter B and
vice versa. This also proves that there is no linear 50 Ohms output
impedance.

Best regards,

Wim
PA3DJSwww.tetech.nl

Sorry Wim, I can't agree with some of your statements in your last
post. Concerning maximum output, I will agree that saturation will
occur when the minimum of the peak AC plate voltage equals the peak AC
grid voltage. This condition occurs when the tube is delivering its
total maximum possible power. However, when the grid drive level is
less than that which brings the plate-voltage minimum down to the grid-
voltage level, saturation does not occur.

In addition, when the pi-network has been adjusted to deliver all the
available power at some drive level less than the maximum possible
power, the source resistance at the output of the pi-network will be
exactly equal to its load resistance, not somewhat higher or lower.
This follows from the Maximum Power Transfer Theorem. This is not
speculation, but proof determined by data from many, many
measurements.

Walt, W2DU