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. From the
measurements we have
all seen when loading a tube amplfier the output power will increase as the
load
resistance is reduced up to a point when the output resistance of the tubes
becomes the
limiting factor - with the inductance and capacitance variation provided for in
the tank
circuit.
Possibly making changes to the inductor and the capacitor in the tank circuit
may allow a little more power to be extracted, but this may be because the new
inductor
and capacitor have lower losses, and not due to a re-matching situation. One
cannot be
certain which is the case without a comparison of the two sets of componets.
Essentially the output power from a particular tube is limited in the first
order by the
emissive capabilities of that tube. To suggest that a 4CXxxx could give more
the power
is a false assumption, as it cannot support any extra cathode emission for a
given plate
or screen voltage. That is limited by the physics of the cathode design.
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.
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.
John ZS5JF
----------
> From: Tom Rauch <w8ji@contesting.com>
> To: amps@contesting.com; John Fielding <johnf@futurenet.co.za>
> Subject: Re: [AMPS] Impedance of amplifiers
To: <amps@contesting.com>
> Date: 17 July 1999 03:02
>
> Hi John,
> > Absolutely correct Tom. We choose to use 50 ohms as a
> convenience because
> > transmission lines are usually 50 ohms today.
> >
> > However consider a high power audio amplifier driving a load of 4 ohms.
> > The output Z of the amp isn't 4 ohms - more like approaching zero ohms to
> > be able to deliver the required current!
>
> That's right John. The design criteria is different.
>
> > I think where some folks are going wrong is that they have lost sight of
> > what Z actually is. You can (within reason) make a 50 ohm Z from a myriad
> > of combinations of L, C and R but in only case - the series resonant
> > condition - the resistive portion is actually 50 ohms and the reactance
> > of the L & C cancel out. For lots of others the value of the apparent R
> > is very low - like the audio amplifier example above. Power cannot be
> > dissipated in a reactance, only in a resistance. So the "dissipative Z"
> > theory immediately falls flat on its face because the R portion isn't 50
> > ohms! Clearly in a practical world ALL components have some value of R -
> > due to loss mechanisms. So defining the efficiency of a rf PA isn't easy.
> > I am going through that scenario at work at the moment!
>
> It is easy. Just use a high power 50 ohm attenuator and tweek the
> PA with a little reverse RF near the operating frequency (so the
> tank or harmonic filter is not in cut-off and shows the designed
> impedance) and measure standing wave voltage of the reverse
> generator on the line between the attenuator and the PA at several
> points.
>
> You'll be able to calculate SWR and impedance by measuring that
> voltage while the PA is operating.
>
> Or you could do a load pull.
>
> > I had this exact discussion with Prof Abrey of Pretoria University some
> > years ago and he explained that depending on the configuration of the
> > amplifier the output Z could either very low or very high. Transistor
> > amplifiers are predominately low Z and have very low collector impedances.
> > Tube amplifiers on the other hand usually have highish impedances - as
> > witnessed by the load-line resistance.
>
> I have no idea what he is talking about, unless it is the plate
> resistance which is something you don't match into anyway.
>
> If indeed the source impedance is very low, power would increase
> steadily when load impedance is reduced. It does not do that in
> conventional PA's.
>
> If the source impedance were very high, power in the load would
> steadily increase in power if load impedance was raised. It does
> not do that either.
>
> A load pull is a valid way to determine source impedance, and it
> certainly fails to confirm the fact high efficiency PA's are greatly
> mismatched. As do reverse power voltage measurements along a
> line.
>
> One shouldn't lose sight of all the stray reactances in a tube, the grid
> > to kathode and plate to grid capacitances can be fairly high and at high
> > frequencies these come into the equations. Similarly the inductive
> > reactances, although not often considered at hf, are still there.
>
> One shouldn't loose sight of the fact the tube is a time-varying
> resistance that is out of circuit for much of the cycle in a non-class
> "A" PA, and that it is power limited by the maximum available
> power in the energy conversion process. That power limitation does
> NOT need to be from a dissipative resistance in series with a
> limitless voltage source, or a resistor in parallel with a limitless
> current source.
>
> The system is non-linear at the tube end of the tank, and that rules
> out using any theorems on the tube side of the tank. Class "A"
> PA's would be an exception, except the model still won't give
> dissipation because it makes no allowance for quiescent current.
>
> All this stuff about something being 50% efficient when conjugately
> matched, or adjusted for maximum power transfer, is nonsense.
>
>
> 73, Tom W8JI
> w8ji@contesting.com
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