# LC notch filter not working!

oopere wrote:

Why are you asking me? I have no idea what the load is - I'm not the OP.
Ed
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
From: snipped-for-privacy@googlemail.com on 26 Apr 2007 02:36:19 -0700

Only if you have the theoretical zero-loss components.

+/-20%
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0 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 load is 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 load, 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
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On a sunny day (26 Apr 2007 17:17:20 -0700) it happened AF6AY

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.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
...

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
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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. [there are four combinations of 3 components for L and C circuits, each with a peak versus dip impedance response, me has to keep reviewing those to avoid confusion] To explain, my (later) analysis model was as follows:
One-Ohm impedance current source. Parallel L-C in series with load, L1 = 10 uHy with Q of 150, C1 = 14 pFd. Load is 1 MOhm in parallel with C2, C2 varying 10, 20, 30 pFd. Capacitors were assumed essentially lossless since their typical Q at these frequencies can be 1000 or more. Minimum voltage response was at a nearly constant frequency regardless of C2 value. Maximum voltage response frequency varied considerably. Using 1.0 V RMS reference for 0 db, the response v. C2 value was:
C2 = 30 pFd, Vout peak +22 db at 7.8 MHz, Vout minimum -35 db. C2 = 20 pFd, Vout peak +26 db at 8.25 MHz, Vout minimum -32 db C2 = 10 pFd, Vout peak +21 db at 10.4 MHz, Vout minimum -26 db
I could have done the above with L1 Q of 50 but that would simply decrease the lower frequency peak voltage, show a lesser voltage minimum at the upper trap frequency, the rest about the same.
* At this point someone will get hot about "ya can't have voltage * gain with no amplifier!" or equivalent. :-) Yes, one can since * a voltage increase only means a current decrease at one * frequency...the only power loss is in the Qs of the components.

Yes, but only for the series resonance frequency. There's a variation in the overall voltage response depending on the load resistance and its parallel load (and probe) capacity. For sure, a series-resonant circuit across the source is going to affect the gain of the driving source from its frequency variation of impedance.
This is one of those seemingly-inocuous circuit applications which can get very tricky to apply with any repeatability. Especially so when the source and load were unspecified. It's safe to say that EVERYTHING interacts over frequency and one cannot just assume anything. That includes scope probes which far too many apply thinking just of their 10 Meg input resistance and forgetting they all have capacity to ground in parallel. :-(
Thanks for reminding me to go back to earlier basics, Tom. A number of years ago I worked the math on impedance of the four basic 3- component combinations and wrote it up for a work application (that would have been a high production failure situation if used as-is) and thought memory "would always be there." Actually it was but my mind gets cluttered with other stuff on a disorganized basis. :-)
BTW, I used my own LINEA (DOS-only) analysis program and LTSpice (free Windows compatible full package from Linear Technology) to run this simple circuit model. Results agreed.
73, Len AF6AY
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>