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Matching a monolithic xtal filter

T

Thomas Magma

Jan 1, 1970
0
I am attempting to implement a monolithic crystal filter (ECS-96SMF) into a
50 ohm IF. It will sit between a 50 ohm active mixer and a 50 gain block. I
was really surprised that the manufacture doesn't tell you how match it into
a 50 ohm system or have any app notes. It says the insertion loss should be
around 3 dB but is that relative to a 50 ohm system? and what matching
components do you use? If this were a saw filter I would have the matching
components values and even typical layout of the circuit. All I get from the
crystal filter manufactures is a one page datasheet with no matching info.

Any thoughts on the matter?

Thomas
 
J

Joel Kolstad

Jan 1, 1970
0
Thomas Magma said:
I am attempting to implement a monolithic crystal filter (ECS-96SMF) into a
50 ohm IF. It will sit between a 50 ohm active mixer and a 50 gain block. I
was really surprised that the manufacture doesn't tell you how match it into
a 50 ohm system or have any app notes.

Crystal manufacturers are still rather "old school" and don't provide as much
assistance (at least on data sheets/app notes) as you get these days from,
e.g., the IC manufacturers.
It says the insertion loss should be around 3 dB but is that relative to a
50 ohm system?

No, it's 3dB in an "ideally matched" system. That is, if you terminate the
crystal in the termination impedances specified on the data sheet (keep
reading...).
and what matching components do you use?

The data sheet there tells you the termination impedance that the crystal
wants to see. If you compute the impedance at the specified frequency (e.g.,
for the ECS-96SMF45A30, termination impedance is listed as 1200 ohms in
parallel with 1.8pF = 1200 ohms in parallel with 1.8pF @ 45MHz = -j1965 ohms =
874 - j533.8 ohms), the complex conjugate (874 + j533.8) will be the input
impedance of the crystal (give or take -- often the "termination impedances"
specified have been rounded to "nice numbers"). You now have the standard
problem of, "I have a 50 ohm source to match to an 874 + j533.8 ohm load --
how do I do that?" and this can be solved in many different ways. See, e.g.,
"RF Circuit Design" by Chris Bowick.

---Joel
 
T

Tom Bruhns

Jan 1, 1970
0
Crystal manufacturers are still rather "old school" and don't provide as much
assistance (at least on data sheets/app notes) as you get these days from,
e.g., the IC manufacturers.


No, it's 3dB in an "ideally matched" system. That is, if you terminate the
crystal in the termination impedances specified on the data sheet (keep
reading...).


The data sheet there tells you the termination impedance that the crystal
wants to see. If you compute the impedance at the specified frequency (e.g.,
for the ECS-96SMF45A30, termination impedance is listed as 1200 ohms in
parallel with 1.8pF = 1200 ohms in parallel with 1.8pF @ 45MHz = -j1965 ohms =
874 - j533.8 ohms), the complex conjugate (874 + j533.8) will be the input
impedance of the crystal (give or take -- often the "termination impedances"
specified have been rounded to "nice numbers"). You now have the standard
problem of, "I have a 50 ohm source to match to an 874 + j533.8 ohm load --
how do I do that?" and this can be solved in many different ways. See, e.g.,
"RF Circuit Design" by Chris Bowick.

---Joel

Point of clarification: do you really mean that the filter,
terminated in 874-j534, will look like 874+j534 at its input (in the
passband...), or that you want the driving source to look like
874+j534 to get the maximum power transfer?

I'd have assumed from the data sheet that I want to terminate both the
input and output with 1200 ohms in parallel with 1.8pF to get the
correct frequency response from the filter, but I would not have
assumed that the input to the filter, when terminated at the output in
1200||1.8pF, would necessarily look anything like 874+j534. Of course
it certainly could not outside the passband, if it's a non-dissipative
filter.

Cheers,
Tom
 
T

Thomas Magma

Jan 1, 1970
0
The data sheet there tells you the termination impedance that the crystal
wants to see. If you compute the impedance at the specified frequency
(e.g., for the ECS-96SMF45A30, termination impedance is listed as 1200
ohms in parallel with 1.8pF = 1200 ohms in parallel with 1.8pF @ 45MHz
= -j1965 ohms = 874 - j533.8 ohms), the complex conjugate (874 + j533.8)
will be the input impedance of the crystal (give or take -- often the
"termination impedances" specified have been rounded to "nice numbers").
You now have the standard problem of, "I have a 50 ohm source to match to
an 874 + j533.8 ohm load -- how do I do that?" and this can be solved in
many different ways. See, e.g., "RF Circuit Design" by Chris Bowick.

Thanks Joel, we were working on coming to that same conclusion ourselves,
but it's nice to have it clarified. It should be relatively easy to match
with a Smith Chart program and a network analyzer.
Crystal manufacturers are still rather "old school" and don't provide as
much assistance (at least on data sheets/app notes) as you get these days
from, e.g., the IC manufacturers.

I'll say old school!! I don't see why engineers have to go through the
interpretation and then the matching process each and every time when the
crystal manufactures could just put a 'typical matching circuit' right on
the datasheet. They're all the same too. Sheesh!!

Thomas
 
J

Joerg

Jan 1, 1970
0
Thomas said:
I am attempting to implement a monolithic crystal filter (ECS-96SMF) into a
50 ohm IF. It will sit between a 50 ohm active mixer and a 50 gain block. I
was really surprised that the manufacture doesn't tell you how match it into
a 50 ohm system or have any app notes. It says the insertion loss should be
around 3 dB but is that relative to a 50 ohm system? and what matching
components do you use? If this were a saw filter I would have the matching
components values and even typical layout of the circuit. All I get from the
crystal filter manufactures is a one page datasheet with no matching info.

Any thoughts on the matter?

Usually like this: Series resistor from 50ohm source into filter input,
matching the spec'd input impedance minus 50ohms. Resistor of spec'd
output impedance value to ground on output, then to base of an ordinary
emitter follower which feeds the next 50ohm stage. Plus a wee capacitive
load if they spec it.
 
J

Joerg

Jan 1, 1970
0
Thomas said:
Thanks Joel, we were working on coming to that same conclusion ourselves,
but it's nice to have it clarified. It should be relatively easy to match
with a Smith Chart program and a network analyzer.




I'll say old school!! I don't see why engineers have to go through the
interpretation and then the matching process each and every time when the
crystal manufactures could just put a 'typical matching circuit' right on
the datasheet. They're all the same too. Sheesh!!

Wait until you need a detailed spec for a piezo. That can really drive
you up the wall.
 
T

Thomas Magma

Jan 1, 1970
0
Thanks Joel, we were working on coming to that same conclusion ourselves,
but it's nice to have it clarified. It should be relatively easy to match
with a Smith Chart program and a network analyzer.

I spoke to soon. When trying to model the matching network from 50 ohms to
(874 + j533.8) I get unrealistic values of components. For instance (from 50
ohm to xtal), a series capacitor of 15 pF and a shunt inductor of 770 nH.
That's one huge inductor. So how do I do this then? What are the thousands
of other engineers doing to get decent insertion loss?

Thomas
 
J

Joel Kolstad

Jan 1, 1970
0
Hi Tom,

Tom Bruhns said:
Point of clarification: do you really mean that the filter,
terminated in 874-j534, will look like 874+j534 at its input (in the
passband...), or that you want the driving source to look like
874+j534 to get the maximum power transfer?

Now that I think about it, the later ("for maximum power transfer terminate in
874-j534 ohms") makes sense whereas the former is not necessarily true. I
suspect you have more experience with this than I do anyway. :)

---Joel
 
T

Tam/WB2TT

Jan 1, 1970
0
Thomas Magma said:
I spoke to soon. When trying to model the matching network from 50 ohms to
(874 + j533.8) I get unrealistic values of components. For instance (from
50 ohm to xtal), a series capacitor of 15 pF and a shunt inductor of 770
nH. That's one huge inductor.

What's wrong with 770 nH. Looks good to me. Or did you mean uH? Motorola ap
note AN267 is full of good stuff on matching. I think you can still download
that from the web. Also, why are you matching the filter to 50 Ohms, instead
of the input impedance of the IF amp?

Tam
 
J

John Larkin

Jan 1, 1970
0
I am attempting to implement a monolithic crystal filter (ECS-96SMF) into a
50 ohm IF. It will sit between a 50 ohm active mixer and a 50 gain block. I
was really surprised that the manufacture doesn't tell you how match it into
a 50 ohm system or have any app notes. It says the insertion loss should be
around 3 dB but is that relative to a 50 ohm system? and what matching
components do you use? If this were a saw filter I would have the matching
components values and even typical layout of the circuit. All I get from the
crystal filter manufactures is a one page datasheet with no matching info.

Any thoughts on the matter?

Thomas

I'd be surprised if the mixer output impedance, or the gain block
input, were actually 50 ohms. So the filter response may not match
what the mfr designed it to be, since they no doubt assumed perfect 50
ohm resistive source and load.

What's the gain block? I've measured a bunch of MMICS and they're
usually low, as low as 30-ish.

John
 
T

Tom Bruhns

Jan 1, 1970
0
I spoke to soon. When trying to model the matching network from 50 ohms to
(874 + j533.8) I get unrealistic values of components. For instance (from 50
ohm to xtal), a series capacitor of 15 pF and a shunt inductor of 770 nH.
That's one huge inductor. So how do I do this then? What are the thousands
of other engineers doing to get decent insertion loss?

Thomas

Huh?? http://www.coilcraft.com/0805cs.cfm. If that's "huge," just
what are you working on?? It's more of a problem that you need to use
fairly high Qu parts in a matching network that goes between such
different impedances. Instead, use a transformer such as MiniCircuits
TC16-1T. That and a 6.8pF series cap on the filter side gets you from
50 ohms to about 800-j520, pretty close to the recommended load for
the filter. It's pretty unlikely you'd need to match better than
that, for either frequency response or signal loss reasons. (I have a
feeling that you are trying to match to the wrong polarity of
reactance....or maybe I got it wrong...).

Often these filters are used in circuits that aren't implementing 50
ohm signal paths at that point...

Cheers,
Tom
 
F

Fred Bloggs

Jan 1, 1970
0
Thomas said:
I spoke to soon. When trying to model the matching network from 50 ohms to
(874 + j533.8) I get unrealistic values of components. For instance (from 50
ohm to xtal), a series capacitor of 15 pF and a shunt inductor of 770 nH.
That's one huge inductor. So how do I do this then? What are the thousands
of other engineers doing to get decent insertion loss?

Thomas

You use either coupled inductor or capacitor transformers. Your step-up
is only a factor of 18 or so.
 
J

John E. Perry

Jan 1, 1970
0
Why not a common-source input amplifier? An rf jfet and a drain
resistor, with maybe a source resistor for matching.

I'm not an rf type (for me, 45MHz is rf), so please be gentle :).

john perry
 
J

John E. Perry

Jan 1, 1970
0
John said:
...
Why not a common-source input amplifier? An rf jfet and a drain
resistor, with maybe a source resistor for matching.

....duh, I meant, of course, a common-gate amp.

jp
 
T

Tony Williams

Jan 1, 1970
0
Tom Bruhns said:
...............Instead, use a transformer such as MiniCircuits
TC16-1T. That and a 6.8pF series cap on the filter side gets you
from 50 ohms to about 800-j520, pretty close to the recommended
load for the filter.

Tom. Interested question..............

That 1:4 stepup transformer and 6.8pF raises the 50 Ohm
source to nearly the 874-j534 series-equivalent (at
45MHz only) of 1200//1.8pF.

What is the disadvantage of using a 1:5 stepup to get
the 50 Ohm up to 1250, plus a parallel 1.8pF, in order
to get a direct (wideband) output Z of 1200//1.8pF?
 
R

RST Engineering \(jw\)

Jan 1, 1970
0
Except that method loses you 6 dB right off the crack of the bat in a place
where loss adds directly to noise figure.

Most monolithic crystal manufacturers presume that people buying a crystal
filter have done a little thoughty work up front and know how to match from
a mixer's collector load into the front end of a gain block that has been
designed for a conjugate match. It's like buying a Ford and expecting the
owner's manual will teach you how to drive. They do expect a bit of a
priori experience, right or wrong.

Me? I'd see if an L-network gives reasonable answers for real components,
both in and out, with a tunable inductor that can give you a little variable
X to take care of the imaginary component. Mouser has some real cheap
variable inductors these days.

I'm just presuming that you are using a 10.7 or 21.5 MHz. monolithic?

Jim
 
F

Fred Bloggs

Jan 1, 1970
0
Except that method loses you 6 dB right off the crack of the bat in a
place where loss adds directly to noise figure.

That would be 3dB in power and who says it has any effect on total NF at
this stage of IF processing anyway. The mixer is always a bad actor when
it comes to NF degradation and harmonic distortion. If it's a narrowband
communication application, it's always better to lose a little signal
strength and buy spec'd harmonic distortion performance by isolating all
the mixer ports with attenuation.
Most monolithic crystal manufacturers presume that people buying a
crystal filter have done a little thoughty work up front and know how
to match from a mixer's collector load into the front end of a gain
block that has been designed for a conjugate match. It's like buying
a Ford and expecting the owner's manual will teach you how to drive.
They do expect a bit of a priori experience, right or wrong.

Me? I'd see if an L-network gives reasonable answers for real
components, both in and out, with a tunable inductor that can give
you a little variable X to take care of the imaginary component.
Mouser has some real cheap variable inductors these days.

I'm just presuming that you are using a 10.7 or 21.5 MHz. monolithic?

If he's way down there at that frequency it's almost not RF anymore. He
should use a stock step-up transformer from MCL. Joerg's suggestion gets
him some ridiculous refection coefficient like 10LOG(1-rho^2) loss at
the filter, and that really would degrade NF.
 
J

Joerg

Jan 1, 1970
0
RST said:
Except that method loses you 6 dB right off the crack of the bat in a place
where loss adds directly to noise figure.

Most monolithic crystal manufacturers presume that people buying a crystal
filter have done a little thoughty work up front and know how to match from
a mixer's collector load into the front end of a gain block that has been
designed for a conjugate match. It's like buying a Ford and expecting the
owner's manual will teach you how to drive. They do expect a bit of a
priori experience, right or wrong.

Me? I'd see if an L-network gives reasonable answers for real components,
both in and out, with a tunable inductor that can give you a little variable
X to take care of the imaginary component. Mouser has some real cheap
variable inductors these days.

I'm just presuming that you are using a 10.7 or 21.5 MHz. monolithic?

I'm not using one, it's Thomas. Anyhow, if losses are a concern you can
also rig up a common gate stage up front between mixer and filter. I
think someone had already suggested that. These have tons of dynamic
range, you can taylor the output impedance, their input impedance
typically suits a mixer nicely, they give you the needed gain and cost
next to nothing. In commercial designs I often employed switcher FETs
like the BSS123 for that because they are just a few cents. When I had
enough dynamic range I even skimped on the output inductor because it
would have added 3-4 cents.

Oh, did I mention that a professor at my alma mater taught us that
common-gate stages are stupid? Well, that is until I brought him a stack
of schematics. Not just mine but also some from reputable companies such
as Collins and Atlas.
 
T

Tom Bruhns

Jan 1, 1970
0
Tom. Interested question..............

That 1:4 stepup transformer and 6.8pF raises the 50 Ohm
source to nearly the 874-j534 series-equivalent (at
45MHz only) of 1200//1.8pF.

What is the disadvantage of using a 1:5 stepup to get
the 50 Ohm up to 1250, plus a parallel 1.8pF, in order
to get a direct (wideband) output Z of 1200//1.8pF?

Hi Tony,

That would be just fine, too, 'cept that in my quick look, it looked
like the 4:1 turns ratio from MiniCircuits was much cheaper than any
suitable 5:1 in their catalog. M/A-com has some similar small
transformers and may have something in a 5:1 turns ratio that would be
appropriate.

I'm not sure how sensitive these particular filters are to load
impedance. With many filters, if you're not trying for the absolute
best conformance to the specified filter shape, a modest mismatch from
the recommended load and source impedances doesn't really matter that
much. It's one of those "YMMV" things--test to be sure you get what
you want.

Cheers,
Tom
 
T

Tom Bruhns

Jan 1, 1970
0
Hi Tom,




Now that I think about it, the later ("for maximum power transfer terminate in
874-j534 ohms") makes sense whereas the former is not necessarily true. I
suspect you have more experience with this than I do anyway. :)

---Joel

Well, actually, I have a fair amount of experience with LC filters of
various types, but not so much with crystal filters and especially not
that much with monolithic crystal filters. However, what I know of
filters suggests to me that the recommended load is probably more to
insure proper frequency response than it is for the absolute maximum
power transfer. The monolithic filters I do know about have nothing
besides plated quartz blanks inside, AFAIK, so capacitance in the load
at each end would be important for response shaping.

Cheers,
Tom
 
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