J
Joel Kolstad
- Jan 1, 1970
- 0
Phil Hobbs said:Nope, you've got the square root in the wrong place: a 100:1 transformer
changes the impedance by 10,000:1, not 10:1.
Oops, thanks for catching that Phil! Sheesh...
Phil Hobbs said:Nope, you've got the square root in the wrong place: a 100:1 transformer
changes the impedance by 10,000:1, not 10:1.
From: Tom Bruhns said:This particular part is not true.
A parallel-resonant trap placed in
series is not detuned by load capacitance at the output. It's still
parallel-resonant at the same frequency. A simulation shows this
easily.
The same is true of a series resonant shunt trap.
Seconds? You should check out the old Collins 490T antenna tuner. As
I recall, the spec was something like 5 seconds max to tune any new
load within the specified range, any new frequency within range, but
you knew that if it didn't tune in a second and a half, something was
most likely broken. Motors don't have to be slow. The inductor went
from one end to the other in I suppose under half a second. Seems
like it was 8 or 10 turns.
But there's a long ways between these embryonic ideas
and a working design, and I leave that to you. ;-)
HI Tom,
Yeah, I suppose that if you start back-biasing your diodes at 100V, suddenly a
volt starts to look like a small signal again.
Someone else mentioned that a straightforward means to drop the voltage swings
is by dropping the "system" impedance. A 1:100 transformer takes 10V to a
mere 100mV, although now the 50 ohm system is 5 ohms so Q has to be 10 times
better to obtain the same notch depth. Still, probably worth pursuing.
Speaking of Collins radios, here's a rather sad site:http://cgi.ebay.com/Collins-490T-1-Radio_W0QQitemZ120112545170QQihZ00...
---Joel
....Yes, you are correct...caught me with a low level of caffeine on
Thursday.
Yes on the parallel L-C for the trap frequency. But, under low source
impedance and high load impedance, with the approximate L and C given,
there is a voltage increase at a frequency below the trap frequency.
Tom Bruhns said:Anyway, Joel, WHY would you think that the Q needs to be any
different??
...
Trimmed off a couple of the groups and much of the message, though all
was noted. Thanks for the additional info; I trust the OP will finde
it useful, if he's still around. (Pet peeve: posters who don't
bother to get back to say "Hey, that helped," or "Huh?" or give some
indication they are still lurking.)
Yes, to be sure the response depends on the load. In fact, even at
the notch frequency, if you start with a high load and add
capacitance, you significantly affect the depth of the notch with the
added capacitance.
It can be quite useful to add another capacitor (or inductor) to a
series or shunt trap, to get the response at a frequency you
specifically want to pass to be high. You can do the same thing with
transmission line stubs, which becomes practical at higher
frequencies. For example, you can put a shorted stub across a line,
where the stub length is 1/2 wave on the frequency you want to
"kill." But then the response at nearby frequencies will also be
attenuated. You can then view that first stub as a reactance at the
frequency you want to pass, and add another stub of the same reactance
magnitude but opposite polarity. You'll find, of course, that the two
stubs total a wavelength, assuming both are shorted at the ends away
from the point the join the through line. With low loss line, this
can be a very effective way to get rid of a large signal in a fixed-
frequency receiver system. The capacitor-or-inductor-added-to-the-
trap is a lumped equivalent of this idea.
I suppose in an absolutely accurate analysis, the impedance versus
frequency charaterisitic of a load that includes capacitance may be
such that the frequency of the maximum attenuation of a finite-Q notch
is shifted ever so slightly, but for sure it won't be shifted enough
to notice; the proximity of the metal in the probe to the coil is
likely to affect the resonant frequency more.
Joel Kolstad said:Say you're in a 50 ohm system. If, at resonance, your series shunt L-C
exhibits a resistance of 0.5 ohms, that's about a 40dB notch. However, in
a 5 ohm system, it's only a 10dB notch.
What input impedance is your scope? In a very slightly more accurate
theory, and a much more useful one, the impedance does NOT become
infinite, but rather becomes Q times the reactance at resonance. The
reactance in your case is about 850 ohms. The Q I have little idea
about: it could be 10 (pretty easily), it could be 1000 (with quite a
bit of difficulty). You may do much better if you put a lower load
resistance on the output of the filter -- in the RF world, 50 ohms
would be usual, but at least something much lower than a 1 megohm
scope input (as I suspect you're using).
Hi Tom,
The input impedance of the scope is 1 MegOhms for the passive probes
I use, but i can change the coupling to 50 Ohms as well.
|----+
Floyd L. Davidson said:Consider that your circuit more or less looks something like this,
Rtrap
input signal >-----+----/\/\/\/\/----+-----> output
| |
/ /
\ \
Rin / / Rout
\ \
/ /
\ \
| |
| |
----- -----
--- ---
- -
Hmmm, looks just like a plain old RC attenuation pad! Except the
Rtrap is actually a parallel tuned circuit that is a high impedance
at one frequency and lower impedances at other frequencies.
However, I can turn the trimmer (in the range from 10 to 20pF) as much
as I want and I don't see ANY effect at all on my scope.
Any hints ? What am I missing ?
I have also looked at active notch filters, but this seems to be
rather difficult at these high frequencies (see http://focus.ti.com/lit/an/slyt235/slyt235.pdf).