# LC notch filter not working!

Discussion in 'Electronic Design' started by [email protected], Apr 26, 2007.

1. ### Guest

Hi,

I am trying to filter out a 13.56 MHz signal (and if possible I would
want to filter some
of its harmonics in succeeding circuits).

I have tried a LC parallel resonance circuit put in series with the
the impedance of the LC parallel circuit becomes infinite at resonance
frequency, i.e.
the circuit becomes open.

I used a fixed inductance L=10uH and a variable C, i.e. a trimmer to
get the product
L*C = 1 / (2*pi*13.56MHz)^2 right. C should be approx 14 pF, however,
due to +/-20%
tolerances in L, I use a trimmer.

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).

Thank you!!

2. ### Jan PanteltjeGuest

Well, 'infinite'.... it depends how much is your termination (the impedance
you drive _from_ and drive _into_?

L
------------ ----------------
Zin C Zout
---------------------------------
But Zin _and_ Zout need to be a lot lower then the high impedance of the parallel LC
in resonance for anything significant to happen.

Often something like this is easier:

Zi --------------------------Zo
|
L
|
C
|
///

Now Zi and Zo can be a few kOhm, and will be practially shorted at resonance.

3. ### Tom BruhnsGuest

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).

There are better circuits for implementing RF notches. With only
passive parts (and no superconductors), you can't get a infinite Q
peak, but you can get an infinitely deep notch. Google 'bridged T
notch circuit.' The key concept is that if you give the RF two paths
to follow that cause phase shifts that are exactly 180 degrees out of
phase at the output at one frequency, and you can adjust the
amplitudes while summing them back together, you can get them to
cancel perfectly. The bridged T notch should work quite well for you
at 13MHz.

Cheers,
Tom

Say Tom,

Any suggestions for building electronically adjustable notch filters where
there's a ballpark of a watt flowing around (i.e., 30dBm in a 50 volt
system -> 10V peak voltage)? Such high voltage seem to rule out using a
varactor diode as the capacitor or using a DC bias current in an inductor to
push it towards saturation. For UHF and above the obvious answer seems to be
YIG filters, but how about for HF and VHF? A single octave of tunability
would do wonders for me at times...

Thanks,
---Joel

5. ### Phil HobbsGuest

One approach would be to use a coaxial stub filter with a bunch of PIN
diode shunt switches arrayed along its length. It would tune in steps,
of course, but depending on the notch width you want, that might work.

I often use a coax patch cord and a thumbtack for this sort of thing.

Cheers,

Phil Hobbs

Hi Phil,

Yeah, that's similar to what I typically do now -- albeit with lumped L-C's by
the time I'm down at HF. By the time you get, e.g., 5% tuning steps though,
that's 15 sections to make an octave. Not bad, I'm just hoping someone knows
some magic that will work even better.

I have seen some commercial notch filters that were electro-mechanical in
nature: something like a motor-driven roller-inductor. That and a handful of
capacitors gets you a very wide tuning range, tons of power, etc. with the
only drawback being that the time to change frequencies is going to be
measured in seconds. It almost seems like the most elegant approach
sometimes...

---Joel

7. ### Michael A. TerrellGuest

In the '60s it was done with motor driven roller inductors and motor
driven variable capacitors.

--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

8. ### mpmGuest

You could reverse feed an isolator into a selective load....
The beauty of this approach is low (even very low) insertion loss.
The drawbacks or course are deep notches, and maybe only 20MHz BW at
UHF.
The latter being primarily a function of the isolator response.

You would still have to make the load (cavity, line section, etc...)
criteria, but it should work.

-mpm

9. ### ehsjrGuest

Can you sweep it to determine where it is resonant?
Sounds like it is either off frequency or has miserable
Q or maybe both. 'Course as others have indicated
the meaurements set up may be part of the problem.

Ed

10. ### jasenGuest

an LC won't touch harmonics.
depends on Q
parasitic capacitance, parasitic inductance.....
if you have room you could maybe try a stub filter.

Bye.
Jasen

11. ### AF6AYGuest

From: on 26 Apr 2007 02:36:19 -0700
Only if you have the theoretical zero-loss components.
OK. At 13.56 MHz, the reactance of the inductor is ~ 852 Ohms. The
impedance
magnitude of a parallel-tuned circuit at resonance is very close to Q
* X where
Q is the total quality factor of both inductance and capacitance and X
is the
resonance reactance of either L or C.

Using a toroid inductor with a Q=150 will get you a resonance
magnitude of
about 127.8 KOhms. Using a solenoidal form inductor the Q is closer
to
50 and the resonance magnitude would be about 42.6 KOhms.

A series impedance between source and load, the load being a finite
resistance of some sort, makes a simple voltage divider...the
impedance
of the parallel-resonant circuit being resistive at resonance. If the
on the order of 100 KOhms or more, the voltage drop will be small; if
it is
on the order of 100 Ohms, the voltage drop at resonance is great.
But,
with a low load resistance, there will be a decided loss of voltage on
either
side of resonance due to finite impedance magnitude of the tuned
circuit.
That depends on just where you are observing. Putting a scope probe
on
the load end will detune the L-C since the probe itself has a
capacitance
which is very close to what you've chosen. That can be calculated and
proven but the impedance math gets more complicated. Note: The load
end also has some capacitance at this frequency and that will detune
the resonance as well.

Someone else suggested a shunting trap of a series L-C rather than a
parallel L-C. That wouldn't be much better since the off-resonance
impedance of a series trap will affect the source end's impedance and
thus its gain. Such an application needs to take into account the
entire
circuit's impedances including circuit capacitance of both source and
as with the parallel L-C that needs to include pass frequency as well
as
notch frequency..
In general, the "trap" circuits used in the past (early TV receivers
of 50
years ago) were only partially-successful, primarily concerned with
bandpass shaping without assuming anything close to high attenuation
at a single frequency. They worked fine at the high source and load
impedances for tubes but not at all optimum for solid-state active
devices.

There are some bridge circuits that might work at a specific frequency
for
attenuation, but those would need to be analyzed for their response
you want to pass. A better bet for attenuating both a specific
frequency - and -
harmonics is to use a lowpass L-C. If your desired bandpass frequency
is
only about a third of the "trap" desired, an Elliptic (aka Cauer)
lowpass
with one of its maximum attenuation frequencies at 13.56 MHz could do
that and attenuate the higher harmonics. The Elliptic function
filters have definite attenuation frequencies in their stopbands.

I'd like to suggest an easy way out, but there really isn't
any...without going to a more elaborate circuit than first realized.

If you wish to pass a rather narrow band of frequencies but attenuate
a
specific frequency well away from those, an ordinary tuned circuit
might
be better. Depending on the frequency desired and Q of the L and C,
the
impedance magnitude drop-off away from resonance might be enough to
do whatever it is you want to do.

73, Len AF6AY

12. ### Jan PanteltjeGuest

Drive from a low impedance with a series resistor.
One big advantage of series LC to ground is that you can connect the trimmer cap
to ground, so it does not detune when you stick a normal screwdriver in it
Somebody may remember the 'tol-trimmer'.

Else I agree with yet an other poster that the T filter is likely the way to go.

13. ### oopereGuest

exhibit an even higher impedance at resonance, which is difficult for
real inductors.
To suggest a solution we should know more details on the source and the

Pere

14. ### Tom BruhnsGuest

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. Obviously PIN diodes would be faster
though. Also, I wouldn't count varactors out of the race, for at
least part of the job. You might not use diodes rated specifically
for varactor service, since you'd be biasing them with tens of volts
most likely. But there's a long ways between these embryonic ideas
and a working design, and I leave that to you. ;-)

Cheers,
Tom

15. ### Tom BruhnsGuest

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.

Cheers,
Tom

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.

---Joel

17. ### Phil HobbsGuest

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. Try 5
milliohms. You'd need as many varactors in parallel for that as you'd
need in series-parallel for the other approach.

Cheers,

Phil Hobbs

18. ### Michael BlackGuest

Tom Bruhns () writes:

Who's the idiot who suddenly decides to cross-post this to so many newsgroups:
sci.electronics.design,
alt.engineering.electrical,
sci.electronics.equipment

You bozos think that just because it might be relevant to a newsgroup, it's
acceptable to cross-post.

The reality is that there are virtually no times when crossposting is
necessary, and it's the mark of iditots too lazy to find the right newsgroup,
or too clueless.

ANd when you see fit to add in newsgroups, you're even greater idiots.

Michael

19. ### Jan PanteltjeGuest

Many years ago I made a digital system, it consisted of 4 coils
10, 20, 40, and 80 uH, and 4 relais to put these in series as needed
(this was a rather high power system).
So the the tuning software switched the relais, giving 15 presets.
That was accurate enough for antenne matching.
And there were 12 of these in a rack....

20. ### ehsjrGuest

Why are you asking me? I have no idea what the load
is - I'm not the OP.

Ed