Resonator Calculations

[redhat]

how to calculate the total capacitance of this resonator?
http://www.geocities.com/aezzat3/resonator2.jpg
should i first terminate the 2 ends with a 50ohm resistors or i can
calculate it directly?

 

[Larry Brasfield]

how to calculate the total capacitance of this resonator?
http://www.geocities.com/aezzat3/resonator2.jpg
should i first terminate the 2 ends with a 50ohm resistors or i can
calculate it directly?

The natural resonance frequency is simply
(1/2 Pi) * sqrt( (1/C7 + 1/C7A + 1/C7B) / L )
The capacitors are just in series w.r.t. the resonance.
The actual peak frequency will depend on the source
and load resistances and be somewhat lower, especially
with 50 Ohms at each port.

 

[Roger Lascelles]

how to calculate the total capacitance of this resonator?
http://www.geocities.com/aezzat3/resonator2.jpg
should i first terminate the 2 ends with a 50ohm resistors or i can
calculate it directly?

At the 94MHz or so frequency of interest, you convert the ends from parallel
to series form.

33pF in parallel with 50R at 94 MHz transforms to 76pF in series with 30
ohsm.

Now your capacitance is 76 pF in series with 76 pF in series with 5p5. The
5p5 dominates.

Q at resonance is 1 / (2 Pi f C R ) = about 5. With such a low Q, the ends
won't have that much effect on tuning. Especially if L7 is not adjustable,
you would want a low Q or else component and construction variations would
move your frequency around too much.

The end capacitors are there to set the Q and hence bandwidth and losses in
the filter.

Roger Lascelles

 

[redhat]

Hi Larry,
how could the capacitors be in series with each other, you haven't
considered input and output ports?
Hi Roger,
you have assumed a frequency of 94MHz to convert the capacitor to
series, what if i don't know the resonant frequency?

regards

 

[Roger Lascelles]

Hi Larry,
how could the capacitors be in series with each other, you haven't
considered input and output ports?
Hi Roger,
you have assumed a frequency of 94MHz to convert the capacitor to
series, what if i don't know the resonant frequency?

I just took the L7 C7 series resonant frequency to get in the ballpark.
Using the transformed ends I could calculate a new resonant frequency, then
recalculate the ends again etc.

Its quite common to do the transformation at one frequency then use the
result over a narrow band of frequencies.

Easiest to just whack it into SPICE, 50 ohms source and load and look at the
response.

The schematic looks as though non of the components is adjustable. In a
commercial piece of equipment, you just might get away with selecting your
components to give the pass frequency you want and always fitting the same
brand and value parts onto the same PCB. Otherwise, you'd make L7 or C7
adjustable and tune it up.


Roger Lascelles

 

[Andrew Holme]

how to calculate the total capacitance of this resonator?
http://www.geocities.com/aezzat3/resonator2.jpg
should i first terminate the 2 ends with a 50ohm resistors or i can
calculate it directly?

Download SCILAB from http://scilabsoft.inria.fr/
You only need a "Compact" installation
Run this script (paste into editor window, then execute):

s = poly(0,'s');

xL7 = s*540e-9;
xC7 = 1/s/5.5e-12;
xC7A = 1/s/33e-12;
xC7B = 1/s/33e-12;

Ri = 50;
Ro = 50;

z1 = 1/(1/Ro+1/xC7B);
z2 = xC7 + xL7;
z3 = 1/(1/xC7A + 1/(z1+z2));

f = z3/(Ri+z3) * z1/(z1+z2);

xbasc(0);
bode(syslin('c', f), 1e7, 9e9, .01);


Voila! A beautiful Bode plot. Even better if you do "set old_style
on" first. With 50 ohm terminations, the resonant frequency is close
to 100 MHz.

 

[Andrew Holme]

With 50 ohm terminations, the resonant frequency is close
to 100 MHz.

Phase passes through 180 degrees around 125 MHz
bode(syslin('c', f), 1.24e8, 1.26e8, .01);

 

[Kevin Doyle]

how to calculate the total capacitance of this resonator?
http://www.geocities.com/aezzat3/resonator2.jpg
should i first terminate the 2 ends with a 50ohm resistors or i can
calculate it directly?

Is there a formula for calculating the phase shift of the resonator circuit
shown.
That would be interesting to compare it with a computer calculated model.

Cheers,
Kevin.