From: gary@ke4zv.uucp (Gary Coffman)
Newsgroups: rec.radio.cb,rec.radio.amateur.misc
Subject: Re: But wait! There's More!!!
Date: 19 Aug 92 14:00:30 GMT
Reply-To: gary@ke4zv.UUCP (Gary Coffman)
Organization: Gannett Technologies Group

In article <436@arrl.org> zlau@arrl.org (Zack Lau) writes:
>In rec.radio.amateur.misc, gary@ke4zv.uucp (Gary Coffman) writes:
>
>lots deleted
>>
>>Note in case it isn't clear that the output network of a transmitter is
>>itself a transformer made up of inductors and capacitors. It's job is to 
>>transform a nominal 50 ohm load impedance to the output impedance of the 
>>final amplifier's active device.  In the case of a transistor final, this 
>>is usually only a few ohms, or even a fractional ohm in higher powered 
>>amplifiers. In the case of tube amplifiers, the active device output 
>>impedance is usually a few thousand ohms. The output network transforms 
>>this impedance to an idealized 50 ohms. 50 ohms is chosen because it's a 
>>convienent value of impedance for a coaxial cable. There's nothing magic 
>>about this number. Another impedance could equally well have been chosen.
> more deleted>
>>Gary KE4ZV
>>
>Gary is assuming that amplifier's active device is truly linear. I don't
>believe this is normally the case.  Warren Bruene took the time to write up
>an experiment he did in the November 1991 QST, with a further clarification
>in a Technical correspondence item (May 1992 QST).  

Linear devices have nothing to do with it. A class C stage, or even a
class D stage place the same requirement on an output network. That is
to transform the load impedance to the output impedance of the active
device.

>It is not true that the impedance presented by the transmitter 
>is necessarily 50 ohms when the system is operating as designed (operating
>with a 1 to 1 SWR into a 50 ohm load).  

This is certainly true in many cases, but is a different thing altogether.
The impedance seen when looking backwards *into* an *output* network isn't
necessarily the same impedance seen looking *out* through an output network.
In fact it almost never is. I shouldn't have phrased the sentence "The 
output network transforms this impedance to an idealized 50 ohms." the
way that I did. As the rest of the paragraph shows, I should have said,
"The output network is designed to transform an idealized 50 ohm load
to the active device's impedance." With tunable networks, even this
restriction doesn't apply since load impedances other than 50 ohms
can be matched to the active device's output impedance.

The output impedance of *any* active device is set by the load line
of the device. In most real devices, this load line crosses the Y
axis at a point other than the origin. Class A devices are never
biased to allow signal to reach this point. Class B devices reach
this point at the zero crossing of the waveform. And class C devices
reach this point at less than 180 degrees of waveform swing. Below
this point, the output impedance of the device is effectively
infinite, but can be ignored in real circuits because no power is
being delivered. The output network is designed to match the 
*average* point on the load line over the entire cycle. In RF circuits,
the circulating currents in the output tuned network complete the
parts of the waveform where the active device is biased into cutoff.
This *averaging* action is why the network is designed to match the
*average* impedance of the active device.

Now when the stage isn't *transmitting*, the impedance seen looking
into the output port can be any arbitrary value depending on the
static bias of the circuit. When a transmitter is transmitting, the
impedance seen looking into the output port is the vector sum of the
instantaneous output wave and the impinging incoming wave. This can
be any value from zero to twice the active device's transformed 
impedance depending on relative phase.

>This is a case of using the wrong equations/theory.  Linear theory is fine
>for approximations, but can and does fall apart when carelessly applied to
>nonlinear circuits.  

Sorry, this has nothing to do with linear versus nonlinear circuitry.
It has to do with the fact that a transmitter is a *source* not a *load*.
Naively assuming otherwise leads to false conclusions.

Gary KE4ZV
Newsgroups: rec.radio.cb