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