On 14 mayo, 18:06, Cecil Moore wrote:
On May 13, 5:51*pm, Wimpie wrote:

When a source behaves like a 50 Ohms source, it will not rereflect the
reflected power back to the load. In other words the forward power
generated by the source will not change when changing the load. The
reflected power that is absorbed by the source may result in increase
or reduction of power dissipation of the active device. It all depends
on the change of the integral of V*I (for the active device).

Where is the reflected power that is "absorbed by the source"
dissipated? How can a "reduction of power dissipation" be the result?

Hello Cecil,

Ever looked at the plate current indicator during tuning? I think you
did. Here you see that plate loading affects plate current, hence DC
input power to the PA.

Just convert the voltage reflection coefficient (as seen by the PA) to
an impedance and work this back through the Pi-filter to the plate.
Now you can forget the whole reflected power story.

What happens now depends on what the plate sees. Due to internal
feedback (grid capacitances) and/or voltage saturation issues, both
plate dissipation and average plate current (so PA input power) may
change. Same is valid for solid state PAs also. Changing the load
changes device dissipation and DC input power.

I mentioned this in this thread and a similar thread about a year
ago.

your question directly: reduction in power dissipation can occur when
the impedance as seen from the active devices increases. It will go
into voltage saturation, and this leads to a reduction in average (DC)
current and integral of V*I product. V can be plate voltage, I can be
plate current.


The answer is that if the power dissipation of the source decreases
when the reflected power is incident, then destructive interference
energy is being redistributed back toward the load. That destructive
interference energy includes some (or all) of the reflected energy.

If some of the reflected wave re-reflects at the cable-PA transition,
you will notice this very likely at the forward power indicator. It
also shows that in that case the output impedance of the PA isn't 50
Ohms (as is the case with many PAs).


The forward wave is associated with EfwdxHfwd power. The reflected
wave is associated with ErefxHref power. The energy in those two waves
traveling in opposite directions at the speed of light in the medium
must be conserved.

For instance, when the load at the end of an ideal 1/4WL feedline is
changed to a short, the forward power is still the same and the
reflected power is equal to the forward power.

Fully agree, if the source remains a 50 Ohms source under this massive
mismatch.

Yet, the source
dissipation has been reduced to zero even though the reflected power
"is absorbed by the source".

I do generally not agree on this one, please convert your Thevenin
source to a Norton source. Though the source doesn't deliver any power
to the shorted quarter wave line (as its input impedance is infinite),
there is still dissipation in the source.

Practical issue (very related to PAs): when you put your shorted
quarter wave line directly to a PA with a type of L match (as is used
in low voltage solid state PAs), the L match will "convert" this
impedance to a very low value, resulting in an near AC short at the
drain or collector. This will greatly increase the power dissipation
in the BJT or MOSFET.

In the source resistor, the forward
current is equal in magnitude and 180 degrees out of phase with the
reflected current and the net current superposes to zero. That's the
definition of "total destructive interference" and according to the
conservation of energy principle, the energy involved has to go
somewhere - so it is redistributed in the only possible direction -
back toward the load associated with the "total destructive
interference" that occurs in the 50 ohm source resistor.

The main point to this part of the discussion is to realize that the
forward and reflected wave energy is flowing through the 50 ohm source
resistor because the physical reflection coefficient looking back into
the 50 ohm source from a 50 ohm feedline is zero.

--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

For the experimenters:
Make this setup: PA – forward/reflected power indicator – tuner – 50
Ohms load. Have a means to measure the PA input current.

Notice DC input current and forward power when the tuner is not used
(bypassed) and don't touch PA settings anymore. Now activate the tuner
and show slight mismatch to your PA (with your tuner). Notice the PA
input current and forward power reading. You may need to turn of ALC
or other means that automatically change the drive level of the actual
power device (valve or semiconductor).

Very likely both PA input current and forward power reading will
change. You may do this at various power levels and not the
difference.

With kind regards,

Wim
PA3DJS
www.tetech.nl