From:

[email protected] on 26 Apr 2007 02:36:19 -0700

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

load. In theory the impedance of the LC parallel circuit becomes infinite at resonance

frequency, i.e. the circuit becomes open.

Only if you have the theoretical zero-loss components.

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.

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

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.

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.

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..

Any hints ? What am I missing ?

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