Roger D. Johnson wrote:
>Poor Jon is going to think I'm picking on him but I must disagree on his
>statements about
>conjugate matching of amplifiers. In reality the PA tube is only matched
>to the extent that
>it produces the desired power output.
Nope, not picking on me! It's a good discussion!
First of all, I do agree with your statement. In reality your right, we
match the tube for our desired power output. But is "desired" power
output really the most efficient or is it the maximum available power
output?
>We can see this by the design
>steps required. First,
>the output power is selected. Then we pick a tube and a reasonable anode
>voltage to
>produce that power. Then we compute the necessary current and multiply
>that by a fudge
>factor for the class of operation. The anode voltage is divided by the
>previous value and
>that gives the impedance the tube must work into to produce the desired
>power output.
OK, You are correct here. The tube needs a specific load impedance to
work into. For my 4-1K if I remember correctly it is on the order of
4700 Ohms real resistance.
>A matching network is then designed to transform that impedance to 50 ohms
>(usually).
Correct.
>Notice that at no time was the actual output impedance or load
>resistance of the tube used!
Huh? It is a fact that maximum power transfer occurs when a device is
operating into a load of the same real impedance that that device has.
Period. Now, we really don't mess with the output impedance of the tube
because the tube is not linearly biased. It acts more as a switch and
therefore that impedance varies over the drive cycle. This is the reason
also that the tank circuit needs to have a fairly decent Q so that it can
constantly deliver energy to the load over the drive cycle of the tube.
I wouldn't want to stake my life on it, but that specified load impedance
is probably pretty close to the average output impedance of the tube over
the drive cycle. That's the way the laws of physics require it to be in
order to maximize power transfer.
>The fact that the tube is not "matched" to the load also explains why
>most of the reflected
>power is again reflected at the amplifier and goes back towards the
>antenna.
No, that's not correct. If the output of the amplifier tank circuit is
conjugately matched to the impedance of the feed line, you won't get any
power reflected back to the antenna. The fact that the tube might not be
matched completely to the tank doesn't affect that. The tube not being
completely matched to the tank means that you won't be able to deliver
all of the available power to the tank in the first place. In other
words, a mismatch between tube and tank decreases gain and efficiency.
I don't have the time right now to go and dig through my college
textbooks and come up with references to this. The bottom line though is
that if you want MAXIMUM efficiency and MAXIMUM power transfer between
tube and antenna, you must have a matching network that transforms the
working impedance of the tube to the impedance of the antenna system. If
you design for something other than maximum power transfer, that is fine
as you may not need the maximum power transfer and efficiency.
73,
Jon
KE9NA
--------------------------------------------------------------------------
The Second Amendment is NOT about duck hunting!
Jon Ogden
jono@enteract.com
www.qsl.net/ke9na
"A life lived in fear is a life half lived."
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