> From: Peter Chadwick <Peter.Chadwick@gpsemi.com>
To: <amps@contesting.com>
> Date: Tue, 2 Sep 97 08:50:53 +0000
> >>I must admit I don't understand this. If the input SWR is good
> >>(<1.5:1?), then why should the driving transceiver come into it? Why
> >>should the input impedance be dependent upon the driving transceiver?
> Or
> >>have I missed something?
You have missed NOTHING.
> Hopefully, no tx has a 50 ohm output impedance. If it does, it's limited
> to a 50% efficiency.
Totally untrue. That claim comes from the mistaken assumption a PA's
source resistance is a dissipative resistance. That is true for a
class A pa, but NOT true for a class B pa.
>Desirably, the output impedance of a transmitter is
> zero ohms.
Not unless the PA current source reserve is infinite.
>Although this apparently violates Thevenin. To strictly apply
> Thevenin, you need an infinite power capacity, or at least one high
> enough that the matched power condition can be attained. The fact that
> practical linears tend to come out about 50% efficient gives them an
> effective output Z of around 50 ohms, give or take, but that's
> fortuitous. A Class C or a Class D PA running at 90% efficiency doesn't
> look like a 50 ohm source.
Peter, I'll forward you a copy of a paper that effectively debunks
those myths. I've measured class C PA's here, as well as other
classes, and they ALL put out maximum power and have maximum
efficiency with a conjugate match. Non-dissipative source resistance
is often confused with a dissipative resistance like a resistor.
If you look at the rules for a Thevenin model, you'll see:
1.) The rules EXCLUDE the use of the model in a non-linear
application. Since the conduction angle of the dissipative element
(tube, FET, transistor, or even spark gap) in a PA is non-linear
over the RF cycle, Thevenin OR Norton equivalents can not be used AT
the non-linear point.
2.) Thevenin and Norton equivalents are by definition fully
interchangeable. But if you have a case of 90% efficiency using
the Thevenin model to wrongly estimate source efficiency, and switch
to the Norton equivalent, you'll find the efficiency is TEN
percent!!! The reason they models won't interchange is the basic
rules plainly say the models can NEVER be used to describe what
happens INSIDE the source. That rule EXCLUDES their use in
determining source efficiency!!
3.) Thevenin and Norton models are only used to describe energy
transfer, and by definition tell us nothing about power loss in the
source. With a conjugate match, efficiency can be one percent or 99
percent, the ONLY thing we know is that is the point of maximum
energy transfer in the system.
The ARRL really screwed up when they changed that stuff in the new
Handbooks.
But like you said, the impedance of the source has NOTHING to do with
the system SWR. Rich never did get that right, even though it is
common knowledge that is easily tested.
> Thanks for the comments on the vacuum relays and PINs.
PIN diodes in the 1.5 kW CCS range are rated as high as 100 kW
pulse power. Hot switching either a vacuum relay or a PIN diode will
ruin either one in short order.
73 Tom
73, Tom W8JI
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