what happens to reflected energy ?
 

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.

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

On 7 jun, 01:51, walt wrote:
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

Hello Walt,

I think the problem is in the initial conditions that we use as
starting point. Your starting point is matched to maximum power output
in all cases, also under bad antenna VSWR. My starting point was a
fixed tuned amplifier, made for 50 Ohms and experiencing large
mismatch (50W reflected, 100W forward).

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 that case the final tube will more or less behave like a current
source (LF appraoch, there will be feedback via capacitance), so the
impedance will change from the value for maximum power.

There are two saturation conditions: voltage saturation, this lowers
the plate impedance, current saturation (below maximum voltage swing),
this increases the plate impedance (especially for pentodes).

We start with a maximim power matched situation (so ZL = Zout) and
don't touch the tuning.
When you now change the load (for example VSWR =2), you will probably
have a strong voltage saturation condition or strong current
saturation condition, so voltage source or current source behavior
dictates, but not both. Therefore the output impedance at the plate
will rise (current saturation) or decrease (voltage saturation). For
triodes the effect is less then for pentodes as the transition for
current saturation to voltage saturation is relatively smooth for
triodes.

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.

Fully agree with this, but is this always the case under practical
circumstance?

When you change one of the paramers (drive level or VSWR, but not the
matching), you will divert from the optimum setting, hence impedance
will change. You can try the VSWR dependence by measuring the forward
power under varying VSWR. I am sure it will change (keeping drive
level and pi filter matching the same).

Most solid state amplifiers do not have the possibility of matching
when you change VSWR, so with bad VSWR, you will be in the current or
voltage saturating range of the active device and your output
impedance will change.

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.

Except for CW, most SSB amplifiers run in the current saturated mode
(so below maximum voltage swing), so they do not work at their maximum
power point given the instantaneous drive level (or you need to
modulate your supply also).

In real world the situation is more complex because of phase changes
in VSWR and parasitic feedback that may change the gain of the active
device (worst case spoken, you may get instability). I discovered
(when I was younger) that with a class C fixed tuned amplifier, small
changes in VSWR resulted in change of forward power, but significant
change in DC supply current (and heatsink temperature). So what
happens with the reflected power depends on many factors.


Walt, W2DU

Wim
PA3DJS
www.tetech.nl

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart
wrote:

The 100W forward and 50W reflected have no relation to actual powers

From HP Journal V.7n.2:

If an incident wave is applied at
the left as shown in the diagram, the
wave passes down the main arm. In
the region where the lines are cou
pled, a wave 20 db below the inci
dent wave will be coupled to the
"Forward" terminal, while a second
wave 20 db below the incident wave
will be coupled to the resistive ter
mination in the "Reverse" arm.
Since the combined power in the
two split-off waves amounts to only
2% of the power in the main wave,
the main wave is essentially unalt
ered and continues to the right-hand
terminal.
A wave applied at the right end
of the coupler is coupled in an analo
gous manner. Waves 20 db below the
left-traveling wave will be coupled
to the "Reverse" terminal and to the
resistive termination in the "For
ward" arm, while the main part of
the wave continues to the left-hand
terminal.
The couplers thus provide equal
fractions of right-traveling and left traveling
waves at separate termin
als. The ratio of these waves will be
equal to p *, the magnitude of the
reflection coefficient of any device
connected to the output of the cou
pler. This ratio can be measured
(Fig. 4) by applying the outputs of
the "Forward" and "Reverse" ter
minals to the -hp- 416A Ratio Meter,
using suitable detectors to demodu
late the amplitude-modulated power
which must be used with the system.
In an ideal directional coupler, no
power from a forward wave would
be received at the reverse terminal
and no power from a reverse wave
would be received at the forward ter
minal. In practice some undesired
power is received at these terminals,
although it has been possible to de
sign the couplers so that this unde
sired power is at least 46 db below
the parent wave, i.e, at least 26 db
below the desired wave at the oppo
site terminal. In other words the di
rectivity of the couplers is at least
26 db (30 db in the lower frequency
couplers) over the complete fre
quency range [Fig. 3(a), (b)].
The coupling mechanism itself
consists of quarter-wavelength sec
tions of the conductors placed suit
ably near one another to achieve the
desired degree of coupling. The
combined effects of electrical and
magnetic coupling impart directiv
ity to the coupled wave. The unused
terminal of each of the auxiliary
arms is terminated in a special widerange
low-reflection resistor to ab
sorb any power coupled to that ter
minal.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart
wrote:

The 100W forward and 50W reflected have no relation to actual powers

From HP Journal V4n5-6:

The multi-hole coupler design has now
been extended to a 3 db coupler, that is, a
coupler in which half the power entering the
main guide is coupled into the auxiliary
guide.
....
The -hp- multi-hole couplers consist of
two sections of wave guide mutually coupled
by two rows of coupling holes (Fig. 2).
Power entering the input arm of the coupler
flows down the main guide and divides at the
coupling mechanism. Part of the power con
tinues down the main guide where it will be
incident on any device connected at the end
of the guide. The other part of the power is
coupled into the auxiliary guide. It is a prop
erty of directional couplers that the power
coupled into the auxiliary guide flows essen
tially in only one direction. In the -hp- coup
ler, this power flow is in the same direction
as the power in the primary guide.
....
R E F L E C T O M E T E R S E T - U P
....
Two directional
couplers are connected back-to-back
as shown.
*******
I will leave it to the readership to imagine why.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Jun 6, 9:34*pm, Richard Clark wrote:
On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart

wrote:
The 100W forward and 50W reflected have no relation to actual powers

From HP Journal V4n5-6:

The multi-hole coupler design has now
been extended to a 3 db coupler, that is, a
coupler in which half the power entering the
main guide is coupled into the auxiliary
guide.
...
The -hp- multi-hole couplers consist of
two sections of wave guide mutually coupled
by two rows of coupling holes (Fig. 2).
Power entering the input arm of the coupler
flows down the main guide and divides at the
coupling mechanism. Part of the power con
tinues down the main guide where it will be
incident on any device connected at the end
of the guide. The other part of the power is
coupled into the auxiliary guide. It is a prop
erty of directional couplers that the power
coupled into the auxiliary guide flows essen
tially in only one direction. In the -hp- coup
ler, this power flow is in the same direction
as the power in the primary guide.
...
R E F L E C T O M E T E R * S E T - U P
...
Two directional
couplers are connected back-to-back
as shown.
*******
I will leave it to the readership to imagine why.

73's
Richard Clark, KB7QHC

Walter was the first to place things in a book and a book is the first
point of reference in this newsgroup
Nothing personal Richard, but it is obvious that engineers are totally
split as to what theorem is correct and the one to be used. There is
no way
you will get agreement if they all insist their position is correct
and thus other positions are not worth consideration.You may be
correct in your position but the idea of change is not in your favor.