On 7 jun, 22:36, Roy Lewallen wrote:
I haven't followed this thread since it's been beaten to death so many
times here before. But there's an interesting fact that might have been
overlooked, and might (or might not) be relevant:

If you put a directional coupler such as a Bruene circuit at the output
of a transmitter, and use its "forward" output to control the
transmitter power to keep the "forward" directional coupler output
constant, you'll find that the power output vs load resistance
characteristic is exactly the same as if the transmitter had 50 ohm
output impedance. This is assuming that the directional coupler is
designed for a 50 ohm system, and that the load is purely resistive. It
also assumes that any load is left in place long enough for the feedback
circuit to stabilize. The effective source impedance to a very rapidly
varying load (that is, one changing so fast that the ALC feedback system
doesn't have time to respond) would be the open-loop output impedance
which could be quite different. I haven't taken the time to analyze how
it behaves with a complex load.

I stumbled across this some time ago when designing a rig using this ALC
method and found it interesting. I believe many if not most modern
solid-state transmitters use this ALC method.

Roy Lewallen, W7EL

Hello Roy,

If you have sufficient headroom and under the conditions you
mentioned, you mimic a 50 Ohms source. I think it also works for any
(complex) loads (I couldn’t find why not).

The difference between a real 50 Ohms circuit may be that the phase of
the belonging EMF may change, in many amplifiers phase shift is
somewhat excitation dependent, but who bothers?

I expect such scheme in combination with VSWR measurement also, as
several PAs have a reverse power indicator present (the other half of
a the Bruene circuit). You need the reverse power indication to avoid
destroying active devices and/or intermodulation distortion due to
voltage saturation.

Imagine that you have full reflection |RC| = 1 and it appears at your
active device as RC=-1. You want to maintain the original forward
power. Your active device has to deliver in that case double the
current at zero collector/plate voltage to maintain same forward power
as under matched condition. The actual power delivered to the load is
zero (as the active device supplies current, but no voltage, RC=-1
means a short circuit). This will result in massive dissipation in the
active device.

In case of RC=+1, it has to provide double the voltage with no
current. In other circumstances you will have a significant phase
shift between current and voltage resulting also in increased device
dissipation and inconvenient combinations of instantaneous voltage and
current. So above some value for VSWR, you may have to reduce the
forward power

I had a discussion recently about the power control scheme for TETRA
terminals, but we couldn't find the answer to what is happening under
high VSWR (so we have measure it). It only states VSWR2.

Best regards,


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