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Re: Impedance calculation rule crystal filter


 

The calculation is simple, but the concept of impedance is not always easy.
Few radio amateurs without an engineering background ever figure this stuff out completely.

Impedance is like resistance, but adds the complication of phase.
To keep the formulas simple we use "complex numbers".


Normal "real" numbers only have one dimension, along the number line.
Complex numbers have a second dimension perpendicular to the real axis.
Weirdly enough, the math all works out beautifully if we assume the perpendicular axis
has units of the square root of -1, often expressed with the symbol j.

Once we express impedances as complex numbers, we can use the same formulas
to combine impedances as we do for resistors.

Ohm's law still works too, divide an AC RMS Voltage by and impedance and?
you get a current in Amps, where all three values are complex numbers.
The phase of the complex number for the current will be shifted with respect to
the phase of the voltage, exactly as?you would see with an oscilloscope.



Lets assume the 45mhz filter does have an input impedance of 650 ohms?
in parallel with 3pF as Gerard shows in his diagram.
The 3pF is in parallel with the 22pF of C212, resulting in 25pF of capacitance.
The formula for the impedance of an inductor is zL = j * 2*pi*f * l)
where f is the frequency in Hz and l is the inductance in Henries.
The formula for the impedance of a capacitor is zC = 1/(j * 2*pi*f * c)
where f is the frequency in Hz, c the capacitance in Farads, and zC the impedance in ohms.

That capacitive impedance is in parallel with the 650 ohms resistance,
which can be calculated as:? ?zP = zC*650/(zC+650)
And that impedance in series with the 720nH inductor L7 is zIN = zL + zP

Python is a simple programming language that knows about complex numbers,
We use exponential notation for the big and small numbers, so 720e-9 is 0.000000720
We can write this up as a python script:
?
zL = 1j * 2 * 3.1416 * 45e6* 720e-9
zC = 1 / (1j * 2 * 3.1416 * 45e6 * (22e-12 + 3e-12))
zP = zC * 650 / (zC + 650)
print(zL + zP)

If you copy and paste that script into the python interpreter at? ?https://trinket.io/console
then it will print the following result:
(29.4 + 68.5j)?
That's 29 ohms resistive plus 68 ohms inductive.

Replacing 650 || 3pf from Gerard with the 500 ohms (and no parallel capacitance)
that Farhan mentions in his notes I get? (46.8 + 57.9j)
The resistive part of 46.8 ohms is close to 50, but we may want a smaller inductor.
Unless my calculations are wrong, an inductance around 515nH might be a better choice.
Perhaps Farhan tried various parts, and 720nH is what worked best.

Using Gerard's figures of 1.5k || 2pF for the filter itself, I get an input impedance of:
(14.339296314084606+57.61908010693662j)

When choosing L7 and C212, first find C212 in parallel with the?
filter module such that the network has a resistive component of 50 ohms
Then find an L7 to cancel out any capactive reactance.

The nanoVNA is a cheap and very educational tool if you wish to understand impedances.
For example, a nanoVNA could measure the input impedance of this filter network,?
and hopefully give a result that agrees with the python script.

Jerry, KE7ER



On Mon, Mar 25, 2024 at 01:02 PM, ajparent1/kb1gmx wrote:

The IF amps and associated DBMs were 50 ohm not 500 ohms.?
The TIA amp is 50 ohms and the DBM is also 50 ohms.

The actual network is L network and the calculation is simple.

Note C250 and 216 are not there (zero value means not installed.).

Also c211 and C251 are large enough to be near zero reactance at 45mhz
and serve as DC blocking.

So the active network is L7 and C212 and the mirror components on the other side.

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