Impedance calculation rule crystal filter

Hello,

?

In the Ubitx V6 documentation, it is noted that the 45 Mhz filter must see an impedance of 500 Ohm on each side. See on?
--->

"The 45 MHz filter needs 500 ohms termination impedance on both ports.

?We use simple L network to match the filter to either ends of the front-end and the 2nd IF mixer."

Knowing for example that the 45MHZ filter model 45M15F has a characteristic 650 ohm/ 3 PF, what is the calculation rule for the capacitor value and inductance on each side of the filter. The capacitor 0.1?F? being considered began in short circuit at this frequency.

Please complete the calculation? rule of these 2 components, if the filter is different with value 1.5KOhm and 2PF.
The rule is:....so
Values of C212, C214? and L5, L7 are....if filter 1.5KOhm / 2PF

Thanks

#110086

Ugg...memories of circuit analysis in college? have assaulted my brain! I bet a computer can get the answers faster than me. Gotta get back to changing my tractor tire...enjoy.

Scotty WD4PYT?

Show quoted text

On Mon, Mar 25, 2024, 2:55 PM Gerard <kabupos@...> wrote:

Hello,

?

In the Ubitx V6 documentation, it is noted that the 45 Mhz filter must see an impedance of 500 Ohm on each side. See on?
--->

"The 45 MHz filter needs 500 ohms termination impedance on both ports.

?We use simple L network to match the filter to either ends of the front-end and the 2nd IF mixer."

Knowing for example that the 45MHZ filter model 45M15F has a characteristic 650 ohm/ 3 PF, what is the calculation rule for the capacitor value and inductance on each side of the filter. The capacitor 0.1?F? being considered began in short circuit at this frequency.

Please complete the calculation? rule of these 2 components, if the filter is different with value 1.5KOhm and 2PF.
The rule is:....so
Values of C212, C214? and L5, L7 are....if filter 1.5KOhm / 2PF

Thanks

#110088

Re,

It’s never too late to learn...

I think this site??
must answer my question if we consider the crystal filter seen as a resistance of 650 ohm/ 3PF

Case A looks good for filter input (500 --> 650 ohm) and case B for output (650 -->500Ohm)

So capacitor and inductance well placed
Some calculations to do, but tomorrow, because the brain is too hot already LOL

cdt

?

#110089

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.

--
Allison
------------------
Please use the forum, offline and private will go to bit bucket.

#110090

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:

Show quoted text

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.

#110091

In the general case

Show quoted text

On Mon, Mar 25, 2024 at 05:52 PM, Jerry Gaffke wrote:

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.

#110092

The webpage Gerard pointed to keeps things simple by ignoring phase.
The L network example on that webpage assumes purely resistive input and output impedances.
Gerard's filter module of 1500 Ohms resistance in parallel with 2pF of capacitance is not purely resistive.

#110099

I use on-line calculators to figure out matching.? One of my favorites is here:

Using 45 MHz

In the configuration on the schematic, 50 j0 ohms into a series C211 of 0.1uF (X= 0.04 ohms @ 45 MHz) into series L7 of 720 nH (X = 204) into shunt C212 of 22 pF (X = -161) does not appear to work at 45 MHz.? At 38.3 MHz the solutioin comes out almost exactly a series 720 nH to shunt 22 pF.

At 45 MHz, the solve for 50 j0 to 650 j0 is a series 613 nH to shunt 19 pF.

The practical issue of knowing the characteristic impedance of Y1 is clearly a factor.??

Ford N0FP

#110100

The calculated solution for 650 to 500 is 969 nH and 3 pF.

It seems to me that the only way to actually find the correct solution is to lift the filter out of the circuit.? Terminate it using a 500 ohm resistor, and put a NanoVNA on it to measure the R +/-j.? Even then, the NanoVNA is itself a 50 ohm device so an approximation may be to use a 9:1 transformer on the VNA to filter to at least match it to 450 ohms.? Even better would be to use the 9:1 and a series 50 ohm resistor.??

Y1 is known to be extremely reactive over frequency.? The model of Y1 is a complex collection of L/C/R configuration that is highly reactive.? The 650 to 500 solution is an over-simplified description of the actual problem to be solved, which is to match at 45 MHz.

Ford-N0FP

#110104

Hello,

I just made calculations following the instructions of the site I quoted

If we consider that the impedance is 50 Ohm, not 500Ohm? ?(As said Allison) , I find exactly the same values as Ford calculed
L=613nH, C=19PF (c+3pf crystal filter)
so L5 L7 =613nH? and C212 C214 = 16Pf (19-3)

if impedance is really 500 ohm, i found
L5 L7 970nH and C 3pF, so C212 C214 near 0

So is there a mistake of what is indicated.

This seems to correspond much more to 50 Ohm than 500.

Will this statement be corrected?

"The 45 MHz filter needs 500 ohms termination impedance on both ports.

?We use simple L network to match the filter to either ends of the front-end and the 2nd IF mixer."

Can Ashan confirm what we have done?

I want to be sure.

It’s cerebral gesticulation, I agree. LOL

cdt

#110105

What I find interesting is uBitx V3 though 6 the terminations for the filter that is
the crystal filter plus matching network is between a Mixer (DBM) and TIA amp.
Both of which are NOT 500 ohms and are optimized for 50 ohms.? So the matching
network has that as its?input and output? terminations.

The filter is characterized as 1500ohms 2pf That is not what it presents,
its what it expects as a load to produce the desired and expected filter shape.
However... I have parts tha texpect 470ohms to as high as 2000ohms resistive
for the termination. So you do need the data sheet for the filter in use.

The above is critical to the filter used as when done tight the loss though the
filter will match the data sheet and also the passband frequency and shape will
be as specified on the datasheet.? Crystal filters are not an exact match to LCR
filters, most LCR fitlers are designed for restive termination where most crystal
filters are designed to expect some reactance as part of the termination.? Altering
that alters the filters behavior (pass band and shape).

Also I have the experience with V3 and a lot of testing.? I found with the supplied
filter and the matching network it performed only marginally well.? Testing using VNA
(Agilent) the matching network values used both skewed the filter response and
created a large amount of ripple in the pass band. causing RX degradation and
also degrading RX and TX audio (for mine).? The values I used were close to
calculated once I had filter data.? I did have to adjust for the trace to ground
capacitance.? The results when properly matched improved.? The filter
pass band was properly centered at 45.000mhz and the loss was lower.?
FYI the filter used wanted a termination of 1500ohms and parallel of 3pf
as a match.

And like Gerry said its not a pure resistive match.

--
Allison
------------------
Please use the forum, offline and private will go to bit bucket.

#110107

Gerard.?

I believe that statement refers to the crystal filter module, without L7 and C212.
There are such modules that are 500 ohms resistive, though they also have
a significant capacitive reactance in the impedance they present.

As Allison says, it is the input to the complete network (including L7 and C212) that should be 50 ohms resistive.

Jerry, KE7ER

Show quoted text

On Tue, Mar 26, 2024 at 09:39 AM, Gerard wrote:

Will this statement be corrected?

"The 45 MHz filter needs 500 ohms termination impedance on both ports.

#110108

Ford said::? "The calculated solution for 650 to 500 is 969 nH and 3 pF."
As Allison pointed out, we are targeting an input impedance for the network of 50 ohms, not 500 ohms.

He also said:
"It seems to me that the only way to actually find the correct solution is to lift the filter out of the circuit.? Terminate it using a 500 ohm resistor, and put a NanoVNA on it to measure the R +/-j.? Even then, the NanoVNA is itself a 50 ohm device so an approximation may be to use a 9:1 transformer on the VNA to filter to at least match it to 450 ohms.? Even better would be to use the 9:1 and a series 50 ohm resistor."

A nanoVNA is indeed most accurate around 50 ohms, but a few hundred ohms should be well within its range to give good results.
Adding a transformer introduces a bunch more sources of error, and I'd guess would make the result less accurate than a direct measurement.
When measuring the input impedance with a nanoVNA, terminate the far end of the filter module with exactly what the datasheet claims it should be, including any reactance.

However, Allison suggests that you will not measure the datasheet value, as the datasheet specs the
source and load impedance that you should give the filter module, not what the filter gives you.
I defer to Allison on this matter, as she is most often right.

Here's a selection of 45mhz crystal filters:??
I'm guessing Gerard is looking at something like the 45R15A1 at 7 lines down,
which presents an impedance of 650 ohms in parallel with 3pf.
The one immediately below that has an extra spec of Cc = 9pf,
which I assume is capacitance to a metal shell, I'll ignore those parts.

Farhan says the ones he buys are 500 ohms, but I'd guess they are more like the 45R12A1,
which is 500 ohms in parallel with 5pf.
The 5pf is not insignificant at 45mhz, having a reactance of 1/(2*pi*f * c) = 1/(j * 2* 3.1416 * 45e6 * 5e-12) = j707 ohms.
Clearly, we can't use the formulas Gerard found that assume the impedance of the filter module is purely resistive.

Let's go with the one filter module similar to Farhan's.
The impedance of 500 || 5pf is the parallel combiniation of 500 ohms and -707j ohms.
Using the python interpreter, we find that:? ? (500 * -707j) / (500 - 707j)? =??333.3 - 235.7j ohms
We perform that calculation the same as we would for two parallel resistors.
(Python insists that the j follows the number, most other sources have j preceeding the number.)

Plug it all into the calculator on the webpage that Ford found at? https://leleivre.com/rf_lcmatch.html
Our inputs to the calculator:? Freq=45 mhz,? Zs=50+j0, Zl= 333.3-j235.7 ohms,
The top network agrees with the topology Farhan chose, and gives values of L7=530.5nH and C212=16.22pf

Now plug all that into my python script as a double check:

zL = 1j * 2 * 3.1416 * 45e6* 530.5e-9

zC = 1 / (1j * 2 * 3.1416 * 45e6 * 16.22e-12)

zP = (zC * (333.3-j235.7)) / (zC + (333.3-j235.7))

print(zL + zP)

That gives a result of?49.99+0.007j, which is extremely close to 50 ohms resistive.
There is a slight error because Ford's webpage calculator only give 4 significant digits in the answer.

I previously wrote:
"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."

Finding C212 is a bit of a trick.?
It's not obvious how to write an equation such that the real part of that parallel combination is 50 ohms resistive.
I would just make a guess for the value of C212, and then iterate with further guesses until my python code got close enough.
Perhaps that is how the calculator Ford found does it, or perhaps it calculates the value directly using what I assume is a rather complicated formula.
As demonstrated above, even a small capacitance across the module of 2 or 3 pF can have a reactance
greater than the 500 ohms resistance, and board routing will add further reactance to that.
So calculate as best you can, but then measure the results carefully and trim those values for best performance.

Here's a webpage that describes the various L network topologies:

"Class notes Set #5" is about matching networks.
He gives exact formulas involving the Q (quality factor), but only if source and load are both pure resistances.

Here's one that solves the general case of loads with reactive components,?
but only does so graphically using a Smith chart:

That website with the selection of 45mhz filter modules:?
has some tutorials you can click on at the left side of the page that might be of interest.

I always come back to my python scripts, as they keep me grounded in the basics of what is really going on with complex impedances.

You have probably noticed that many formulas include the term 2*pi*freq, with frequency in Hz.

This gives the angular velocity in radians per second, often represented by the greek lowercase?

symbol omega, which looks like a script "w". Since this is often found in the literature,

I tend use it to simplify my python scripts.? ?

Here's a script that iterates to about the same solution of 16.22pF as the website that Ford found.
Actually, the script result is 17.69pF, the error is due tot he fact that we are only trying integer values of reactance in ohms.
If the granularity is improved to 0.01 ohms, the script then gives a result of 16.29pF

###########################

# w is the frequency expressed as an angular velocity in? radians per second

f=45e6;? ?w=2*3.1415926535*f;? ??

?

# Find the impedance of a capactitor, c in Farads

def zC(c):

? return(1 / (1j * w * c))

?

# Find the impedance of an inductor, h in Henries

def zL(h):

? return(1j * w * h)

?

# Find impedance of two impedances in parallel, same formula as for resistors

def zP(zA, zB):

? ? return(zA*zB/(zA+zB))

??

# Find impedance for parallel combination of 500 ohms? and 5pF at 45mhz

zLoad =? zP(500,? zC(5e-12))

? ??

# Try this loop for all values of x between -2000 and 2000 ohms of reactance

# where a negative value is capacitive and a positive value is inductive.

# Find the value of x which in parallel with zLoad has a real part closest to 50 ohms

bestErr = 999999999

for x in range(-2000, 2000, 1):? ? ?

? ? zX = zP(zLoad, 1j*x)

? ? error = abs(zX.real - 50)?

? ? if error < bestErr:??

? ? ? ? bestErr = error

? ? ? ? bestzX = zX

? ? ? ? bestX = x

?

# if reactance of bestX is negative, need a parallel capactitor

# Given a reactance of x ohms, what is the inductance or capacitance at 45mhz?

# Since x = 1/(w * c),? c = 1/(w * x);? ?x = w * h;? h = x/w?

if bestX < 0:

? farads = 1/(w*x)

? print("zLoad:", zLoad, "bestX:", bestX, "bestErr:", bestErr, "farads:", farads)

else:

? henries = x/w

? print("zLoad:", zLoad, "bestX:", bestX, "bestErr:", bestErr, "henries:", henries)

#############################################

As I said previously,? the calculation is simple, but the concept of impedance is not always easy.

Jerry, KE7ER

#110111

Merci à tous pour vos réponses très technique.

but...

It’s a bit jungle with 45mhz filters

On the diagram of the Ubitx V6, it is noted that it is a 45M15F.

I never found the datasheet on the net.

I got closer to the 45M15AF, so 650 Ohm/3pF

Another Ubitx user , said his was a WTL69 45Mhz on his PCB. I never found the datasheet.

To conclude, I bought on Ebay Motorola 20J26 45Mhz and there was also impossible to find the specifications.

On? A...x, I got? filters 45R15AZ. Same, couldn’t find the specs.

It seems that it is similar to a 45M15A, so 1.5Kohm/ 2pf...?

To add to the confusion, it is noted 720nH on the diagram.

it is also indicated on the technical description: Coil Details

L5,? L7: 12 turns on T30-6

but tore calculator give 14 turns for 720nH

With 12 turns, you are to 560nH.

others things:

As my design allows to connect the Si5351 (It is a plug-in module) and as the whole of my 45Mhz filter is also pluggable, I wonder if the best measurement is:

disconnect the Si 5351

Put a generator in TP13 and look at output TP16 or Tp17 (the SSB filter is also removable).?

Sweep between 44 and 46 Mhz and see.

My filter has adjustable capacitors, I would put a coil 14 turns or really 720nH about to have the best curve.

What do you think? you will tell me it is a measure in tension.

?

Moreover the 45R15Az is given as 3rd harmonic, so important detail or not?

?

and to conclude, if my curve gives a central point of the bandwidth at, for example, 44988000 htz, I would modify the pgm of K8CeC accordingly.?

Maybe it’s too fussy? LOL

cdt

#110113

Gerard,

Injecting a signal at TP13 sounds good, the amp at Q11,12,13 should drive the filter
with a proper 50 ohm source impedance if it was built as per the uBitx schematics.

A filter using the third harmonic of the quartz crystal should work.

There's a lot going on in the mixer at? T3,4 so I'd probably cut the trace to the mixer
and put a 50 ohm resistor across the unstuffed C210.

Then sweep the oscillator into TP13 across the 45mhz passband and record the AC voltage
seen across C210 at each step of frequency somehow.? Just a scope would work.

If C212 and C214 are variable, then try adjusting them for a better response curve, keeping
both caps at about the same setting.

Since you don't have a datasheet for the ebay filter, and since the ebay filter might be a cheap fake anyway,
you may need to try several different values for the inductors at L5 and L7, keeping them them equal.
If you don't have a datasheet, the best you can do is try various cap and coil values
and see what works best.

Having everything pluggable and having variable caps in the circuit would add stray capacitance
and inductance.? You may get better results with the high impedance nodes around the filter module
kept tight and clean, only socket things where the impedance is down at 50 ohms.

RF design can look simple going in, it's just a handful of parts.
But understanding how it should work and getting things to work properly can take months.

Jerry, KE7ER? ?

#110114

Gerard,

A nanoVNA in 2 port mode would be the ideal tool to plot the response of your filter,
less tedious than using a signal generator and a scope.

The nanoVNA is designed for a 50 ohm environment, so can be connected
directly to your L networks on input and output.
See page 20 of? ?

Jerry, KE7ER

#110125

Thank you, Jerry. It’s still great.

Yes, I also have a nanoVna.

In fact if you look at the Iso HBTE forum, you will see that I also completely redone the PCB of the Simple SA, a Wifi spectrum analyzer. I have to put it in the case, it’s the longest to do. (I will include the quartz tester)

I also made a variable 0.30V 3A power supply for the Ubitx and other.

I also fixed an oscilloscope 4*200Mhz Yokogawa (see forum Eevblog)

I also just spent some time to restore a Rigol 1052E. (See the same forum)

All this is cheap material found on Ebay.

But it is necessary to invest, I spent several treatments over several months, to understand the functioning of the Jtag and find in fact that there is only the Wriggler which works well despite its slowness (9H00 to program a flash of 8Mo) Testing UsbBlaster clone, HW of lattice...

What matters is the result and to spin the brain.

Well, it’s going to be a beautiful? season, so I’ll be busy with the outside.

I’m not worried about the Ubitx, but it’s true that the higher the frequency, the more a piece of wire becomes something else.

The main thing is that a little theory will help us understand how things work and that it will serve as information for everyone.

?

Have a good day
cdt

#110134

Gerard,

You have your hands full there!
We learn a lot just jumping in and trying stuff.
As you know this hobby is a bottomless pit,
there is always more to learn.

Complex impedance is one of the key mysteries of RF design,
difficult to master but very powerful.

A correction to a script in post? ?/g/BITX20/message/110108
In the two places where I wrote "j235.7", this should be changed to "235.7j"
There was also an extra left paren. The corrected passage is shown below
This can be executed by pasting it into the interpreter at??https://trinket.io/console

#######################
# Now plug all that into my python script as a double check:

zL = 1j * 2 * 3.1416 * 45e6 * 530.5e-9

zC = 1 / (1j * 2 * 3.1416 * 45e6 * 16.22e-12)

zP = zC * (333.3-235.7j) / (zC + (333.3-235.7j))

print(zL + zP)

#######################

Jerry, KE7ER?

#110136

Thanks for the feedback Jerry

Interactive Python Console
>>> zL = 1j * 2 * 3.1416 * 45e6 * 530.5e-9 
... zC = 1 / (1j * 2 * 3.1416 * 45e6 * 16.22e-12) 
... zP = zC * (333.3-235.7j) / (zC + (333.3-235.7j)) 
... print(zL + zP) 
... #######################
(49.99401015606562+0.0069889380417578195j)
>>>

In fact if I understood correctly it's give in the theory : 530nH (so 12 turns give 560nH in the tore calculator), and total of 16pF--> remove 3pf of the crystal filter--> so 13pF for capacitor

ok we are not very far from the schematic

In any case it was very useful to understand the impedance adaptation

So I would see how to calculate the values if my filter is other

I hope it was useful to all

?

thanks

?

#110139

Gerard,

It would be wise to check the inductance of your torroid using your nanoVNA.
Could well be that there is enough capacitance between windings to give poor results.
Surface mount inductors might work better.

If you don't have a datasheet for the 45mhz module, you will have to just try different caps and inductors
to see what's best, measuring the response with your nanoVNA.??
Figuring out how to do the calculations is a hobby in itself, but for a properly working filter
it all comes down to proper testing.

Jerry, KE7ER

Show quoted text

On Wed, Mar 27, 2024 at 07:13 AM, Gerard wrote:

In fact if I understood correctly it's give in the theory : 530nH (so 12 turns give 560nH in the tore calculator), and total of 16pF--> remove 3pf of the crystal filter--> so 13pF for capacitor

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Impedance calculation rule crystal filter