From: zlau@arrl.org (Zack Lau)
Newsgroups: rec.radio.cb,rec.radio.amateur.misc
Subject: Re: But wait! There's More!!!
Date: 19 Aug 92 19:20:43 GMT
Article-I.D.: arrl.444
Organization: American Radio Relay League

In rec.radio.amateur.misc, gary@ke4zv.uucp (Gary Coffman) writes:
>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.

This is known as conjugate matching.  However, Warren showed that this does
not occur, "at least with tube amplifiers operating in the HF range up through
30 MHz."
>
>>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.
>
Its not easy, but it is possible to measure the impedance of an amplifier's
active device as transformed by the output network by injecting a signal
of a slightly different frequency than that used by the amplifier and
looking at the forward and reflected powers (directional couplers) with
a spectrum analyzer.  And surprise, if you do you will probably discover
that the impedance is not what you think it is suppose to be.

>>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.

The theory of conjugate matching works just fine for linear circuits.  And,
if you are studying calculus, its nice to prove it to yourself as a basic
max/min problem.

And, I suppose there isn't a problem if you are just concerned with
the impedance of loads.

Zack Lau  KH6CP/1

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>
>Gary KE4ZV
>