[AMPS] Impedance of amplifiers

[Tom Rauch]

Hi John,

> I think essentially we are talking along the same lines Tom.  I said that
> the output Z of a tube amplifier is "highish" which implies several kohms.

Except the output Z of the tube is anything depending on the 
particular part of the cycle where you look at it.

It is near-infinite for less than 180 degrees, and gets pretty low for a 
few degrees of the RF cycle. It has an "impedance" we can't define 
as one value until the tank comes into play. 

Then the normal criteria in amps driving fixed loads is to match into 
E/I of the RF energy available at that point in the system at the 
maximum expected power.

> Getting back to the 50% efficiency argument.  This came about from the
> understanding of the output power capabilities of signal generators, often
> the analagy is used about the internal resistance of a battery to draw a
> similarity.  In this case because we need the source to always look like a
> true 50 ohms it is common practice to use a power amplifier followed by a
> rather high value attenuator to give a 50 ohm output match. 

That's certainly true. A signal generator like you describe does 
behave like a 50 ohm source no matter what the load or power 
level, except we still can't use the model to determine the 
dissipation or efficiency of the black box making up the generator 
and/or pad. 

But in the case of the generator, maximum power transferred is 
50% or less of that available as you say below. 

I just think the term "efficiency" is misleading because it implies 
the energy conversion is perfect and the generator has a perfect 
output impedance without using an attenuator pad, which we both 
know is impractical.

> In such a case the best the efficeincy can be is 50% because of the
> arguments which have been made before, so I won't waste bandwidth
> repeating them again.  It is a common mistake to think that signal
> generators and high power transmittiers are the same, which of course they
> are not.  Signal generators rarely are capable of more than +20dBm output,
> because of the attenuator in series with the output.  Usually this
> attenuator needs to be about 10dB to ensure a good output match, in the
> 10dB attenuator case the worst the output return loss can be is 20dB (VSWR
> = 1,2:1) which is still poor.  Laboratory grade signal generators often
> have output return losses of  30dB or more, which is a VSWR of 1,06:1
 
> I fear that some of the amateur handbooks may have helped propagate this
> misconception. I don't know for sure if the ARRL Handbook uses this
> argument, but if it does someone ought to tell the publishers of the
> mistake.

The Handbook did repeat the mistake you outlined. When 
Maxwell's excellent text on matching was removed, the pure  
nonsense about 50% efficiency crept in.  

Instead of improving on the Maxwell text by simply pointing out the 
PA didn't HAVE to be conjugately matched and often isn't during 
operation, something much worse and more misleading replaced it.

I think they have removed the 50% efficiency stuff now.



They actually showed a Thevinin model and described PA 
efficiency using it. Not good.

73, Tom W8JI
w8ji@contesting.com

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