Peter Chadwick wrote:
>
>Something I don't understand - maybe Carl or Rich or Jon can explain.
>
>If I take an inductor (start with a 'perfect' inductor) and put a resistor
>in series with it, at LF it looks like the inductor isn't there. If I make
>the inductor out of resistance wire, it still looks like a resistor.
>
>Now increase the frequency. Inductive reactance rises, and the Q, given by
>2*pi*F*L /R increases, eventually becoming infinite at infinite frequency
>(OK,
>3-500s don't have any gain there!) So I have a series inductive reactance
>with Q increasing with frequency.
>
>Now take the coil, wind it out of silver tape if you like, and put a 33 ohm
>resistor across it. At this stage, it's a perfect resistor. At LF, the low
>reactance of the coil shorts out the resistor, but as the frequency rises,
>so does the coil reactance. Because it's a parallel circuit, the impedance
>tends asymptotically to 33 + j0. Even a practical resistor will tend that
>way. As Q is Rp/1*pi*F*L, the Q is falling with frequency - in fact, it will
>have a maximum value at VLF.
>
>If I understand Rich's argument, the Q of a paraistic suppressor should fall
>with increasing frequency. So there seems to me to be an anomaly - from the
>above argument, an L-R shyunt circuit is indicated.
>
>can someone explain?
Peter, what have you done? You've awoken the undead!
Prepare for at least six weeks of mayhem and misquotation...
I used to think that the Q of a parasitic suppressor had some meaning.
having looked at the problem in detail, I don't believe that any more.
The basic question is: will my HF amplifier oscillate at VHF?
That depends on three things:
1. Existence of a parasitic VHF resonance - we know that does exist,
due to unavoidable stray inductance between the tube and C1 of the
Pi(L) tank.
2. Existence of a feedback path from output to input of the tube.
3. Sufficient gain at the VHF resonance to permit oscillation.
Note: ALL THREE of those requirements must be met, or else the amp will
not oscillate.
The parasitic suppressor operates only on #3 and #1. But in fact the big
unknown is #2, because it depends on the individual design and
construction. That's why you always have to develop the suppressor by
cut-and-try methods.
The suppressor is wired in series with the tube, but the way it works is
that its resistive part creates a permanent PARALLEL load impedance for
the tube at VHF, low enough to make its gain drop below the oscillation
threshold. If you analyse the circuit in detail, you find that the
series-to-parallel impedance transformation depends on the stray
inductance that is causing the VHF resonance. There's the problem: the
load impedance seen by the tube depends partly on the suppressor
characteristics BUT ALSO on the stray inductance in the amplifier.
So, however much you know about a parasitic suppressor, it's of no real
use unless you also know about the strays in the individual amplifier in
which it will be used.
Finally, the characteristics of the suppressor consist of TWO numbers, R
and X (both of which will vary with frequency... yes, even R). To work
out what the suppressor will do for an amplifier, you need BOTH of those
numbers separately. If you roll the two numbers together and talk only
about a single value of Q, you have just thrown away any chance of truly
understanding what is going on!
That's why any debate about "the Q of a parasitic suppressor" is doomed
to failure - it's only taking about one small corner of the whole
problem.
This has already happened at least three times. The bodies are buried in
DejaNews and the web archive... buried, but not dead.
73 from Ian G3SEK Editor, 'The VHF/UHF DX Book'
'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.demon.co.uk/g3sek
--
FAQ on WWW: http://www.contesting.com/ampfaq.html
Submissions: amps@contesting.com
Administrative requests: amps-REQUEST@contesting.com
Problems: owner-amps@contesting.com
Search: http://www.contesting.com/km9p/search.htm
|