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