Matching a monolithic xtal filter
Thomas Magma
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
Joel Kolstad
news:lWiEi.138028$rX4.48697@pd7urf2no...
>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
Tom Bruhns
wrote:
> "Thomas Magma" <somewh...@overtherainbow.com> wrote in message
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
Thomas Magma
> 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
Joerg
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.
--
Regards, Joerg
Joerg
Wait until you need a detailed spec for a piezo. That can really drive
you up the wall.
Thomas Magma
> 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
Joel Kolstad
"Tom Bruhns" <k7...@msn.com> wrote in message
news:1189202317....@d55g2000hsg.googlegroups.com...
> 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
Tam/WB2TT
>
>
>> 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.
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
John Larkin
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
Tom Bruhns
wrote:
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
Fred Bloggs
You use either coupled inductor or capacitor transformers. Your step-up
is only a factor of 18 or so.
John E. Perry
>
>
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
John E. Perry
> ...
> 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
Tony Williams
Tom Bruhns <k7...@msn.com> wrote:
> ...............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?
--
Tony Williams.
RST Engineering (jw)
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
--
"If you think you can, or think you can't, you're right."
--Henry Ford
"Joerg" <notthis...@removethispacbell.net> wrote in message
news:eVjEi.24330$eY....@newssvr13.news.prodigy.net...
Fred Bloggs
> 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.
Joerg
> 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.
Tom Bruhns
> In article <1189210614.332653.299...@d55g2000hsg.googlegroups.com>,
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
Tom Bruhns
wrote:
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
RST Engineering (jw)
engineering lecture on dB?
Jim
--
"If you think you can, or think you can't, you're right."
--Henry Ford
"Fred Bloggs" <nos...@nospam.com> wrote in message
news:46E43CC...@nospam.com...
Tony Williams
Tom Bruhns <k7...@msn.com> wrote:
> On Sep 9, 1:37 am, Tony Williams wrote:
> > 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?
> 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.
Thanks Tom. As you know rf is not my bag so I'm an
interested lurker on these threads, trying to follow
the sums..... and where the sums produce something
that may not be practical to implement. That was the
reason for the question.
> M/A-com has some similar small transformers and may have
> something in a 5:1 turns ratio that would be appropriate.
A 1:4 transformer could be used as a 1+4 auto of course.
> 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.
AFAIR from the data sheet, that filter is about 75KHz
bandwidth, but they also specify a guaranteed attenuation
out at +/- 1MHz from the centre frequency. So presumably
the source/load impedance has to be somewhere near the
req'd 1200//1.8pF over that range.
--
Tony Williams.
Fred Bloggs
It's like all the other crystal parameters, that impedance derives from
the motional parameters near resonance and in most applications
designers use bandpass impedance matching to avoid exciting spurious
responses and non-linear mixing in the crystal.
Tony Williams
Fred Bloggs <nos...@nospam.com> wrote:
> Tony Williams wrote:
> > AFAIR from the data sheet, that filter is about 75KHz
> > bandwidth, but they also specify a guaranteed attenuation
> > out at +/- 1MHz from the centre frequency. So presumably
> > the source/load impedance has to be somewhere near the
> > req'd 1200//1.8pF over that range.
> >
> It's like all the other crystal parameters, that impedance
> derives from the motional parameters near resonance and in most
> applications designers use bandpass impedance matching to avoid
> exciting spurious responses and non-linear mixing in the crystal.
I take that to mean that the 1200//1.8pF need
only be maintained for not much more than the
75KHz passband. Thanks Fred.
--
Tony Williams.
Tom Bruhns
> In article <46E69E30.3070...@nospam.com>,
A bit more than that, actually, Tony. It's good to not hit a crystal
filter with out-of-band signals that are too high in amplitude,
because crystal filters (a) have spurious responses (should be pretty
low if it's a good design, and well implemented), and (b) are not
strictly linear devices and thus will allow mixing of two out-of-band
signals to produce an in-band signal. Crystal filters are generally
pretty low distortion, but people going for the ultimate in receiver
performance end up paying a lot of attention to their crystal
filters. So anyway, it helps to have some filtering in front of the
filter, to avoid those problems, though there's an obvious limit to
how much you can do. At a 45MHz center frequency, without going
really overboard with the LC filter, you probably will end up with a
3dB bandwidth at least a couple MHz wide. To do much better while
keeping the filter loss low requires coils with high Q, which get
physically large.
On the other hand, in a lot of applications, the crystal filter
distortion and spurious responses are low enough that relatively
broadband coupling is not a problem. You need to look at the whole
system to decide what's appropriate.
Cheers,
Tom
Tom Bruhns
wrote:
Let's see here...if I didn't miss anything, to begin with, the loss
will be quite a bit more than 6dB. The filter load/source is supposed
to be 1200 ohms in parallel with 1.8pF. I believe Joerg left off the
1.8pF in his suggestion of a series 1150 ohm resistor. Then the
output is to be terminated in net 1200 ohms and 1.8pF. For the
purpose of calculation, assume in the passband that the filter looks
like a simple short circuit. So the 50 ohm source now is delivering
power to 2400 ohms in parallel with 0.9pF. How does the power
delivered to the output 1200 ohms || 1.8pF compare with the power the
50 ohm source could deliver to a 50 ohm load? THAT's how much you are
giving up. And a bonus question: what other problems are caused by
not terminating the 50 ohm source in 50 ohms, and how much MORE loss
is there if you do add a resistor there to yield a net 50 ohm load?
(Hint: double-balanced mixers generally perform much worse with
respect to distortion products if they are not properly terminated...)
Cheers,
Tom
RST Engineering (jw)
be a great deal more than 6 dB; my ORIGINAL point said "...right off the
crack of the bat..." with resistive matching. If I had carried it through
to a logical conclusion it would have been much worse.
No, my point to the person who said that if you lose 6 dB in voltage you
have lost 3 dB in power was the ultimate decibel freshman student blunder.
I'm sure you will agree that if you lose 6 dB, you lose 6 dB measured in
voltage (2:1) or power (4:1).
Jim
--
"If you think you can, or think you can't, you're right."
--Henry Ford
"Tom Bruhns" <k7...@msn.com> wrote in message
news:1189527073.5...@r34g2000hsd.googlegroups.com...
Joel Kolstad
news:13edgf0...@news.supernews.com...
>>> Excuse me, you wanna rethink that answer before I give you the sophomore
>>> engineering lecture on dB?
I actually had a homework question that asked you to describe why resistive
matching was generally undesirable for low-level, RF signals. It almost
seemed insulting by then, as this was a senior-level RF class...
Of course I also had a freshman or sophomore calculus question asked you to
painfully describe why, if a cop times you traversing a mile in, say, 40
seconds, he knows you've exceeded the speed limit of 60Mph... aiee....
Joerg
Thomas wrote he had an active mixer in there ;-)
If it's got enough gain like >8dB or so you are usually fine with a
passive scheme plus a follower at the output.
Joerg
> That wasn't the point, Tom. You are absolutely correct, the total loss will
> be a great deal more than 6 dB; my ORIGINAL point said "...right off the
> crack of the bat..." with resistive matching. If I had carried it through
> to a logical conclusion it would have been much worse.
>
> No, my point to the person who said that if you lose 6 dB in voltage you
> have lost 3 dB in power was the ultimate decibel freshman student blunder.
> I'm sure you will agree that if you lose 6 dB, you lose 6 dB measured in
> voltage (2:1) or power (4:1).
>
The old saga of voltage dBs and power dBs. Or the perpetuum mobile :-)
Once when someone "higher up" brought up power dBs in a seminar back in
college I asked him whether power dBs get sick more often than voltage
dBs. I got kicked out of the room ...
Joerg
That's where the old concept of the Q-multiplier comes in. After that it
only boils down to how good you are able to control the CF of a resonant
circuit up front. But shhht, don't tell anyone. The younger lads out
there don't have the foggiest idea what that is.
[For the uninitiated: No, it has nothing to do with Q-Tips or luxury
cars of the Infinity brand ...]
> On the other hand, in a lot of applications, the crystal filter
> distortion and spurious responses are low enough that relatively
> broadband coupling is not a problem. You need to look at the whole
> system to decide what's appropriate.
>
Yep.
Fred Bloggs
He should at least have a trap for the mixer LO feedthrough, that will
be the biggest undesirable in the mix and catches up with most designs.
Of course if it's tunable, he has problems.
Fred Bloggs
RST Engineering (jw) wrote:
> No, my point to the person who said that if you lose 6 dB in voltage you
> have lost 3 dB in power was the ultimate decibel freshman student blunder.
Yeah, well no person said that. I did say that half-power is 3dB. You
have a problem with that?
Tom Bruhns
wrote:
> Tom Bruhns wrote:
...
> > how much you can do. At a 45MHz center frequency, without going
> > really overboard with the LC filter, you probably will end up with a
> > 3dB bandwidth at least a couple MHz wide. To do much better while
> > keeping the filter loss low requires coils with high Q, which get
> > physically large.
>
> That's where the old concept of the Q-multiplier comes in. After that it
> only boils down to how good you are able to control the CF of a resonant
> circuit up front. But shhht, don't tell anyone. The younger lads out
> there don't have the foggiest idea what that is.
Ouch! Not around my designs, thank you. :-(
[For the uninitiated: just stay away from them.]
Cheers,
Tom
Joerg
What made you gun-shy here? Got hurt by them? Nowadays you can create
nice gain controlled amps and in most of my cases this is under full
computer-control. That was way different when I started as a teenage
hobbyist where the price tag of an Apple II would make you cringe. Now
you can buy a uC for a buck.
Tom Bruhns
Prove to me you can add one and maintain +55dBm IIP3 and I might think
about it. Well, heck, prove to me that you even NEED it in front of a
GOOD crystal filter, too.
Cheers,
Tom
RST Engineering (jw)
No, I have no problem with that. What I have a problem with is when you
said...
Jim 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."
Fred said: "That would be 3dB in power and who says it has any effect on
total NF at
this stage of IF processing anyway."
If there is any way of interpreting that NOT to mean that you said a 6 dB
voltage loss is a 3 dB power loss, then I will stand corrected.
Jim
Joerg
Just as an example which you can still occasionally buy but someone
would almost have to die first because they tend not to part with it,
scroll down to the Drake 2B:
http://home.earthlink.net/~kf6gk/
One of the finest receivers ever made, have used it myself. For
something like $40 extra you could upgrade to a Q-multiplier inside the
speaker cabinet (early 60's pricing, gets me drooling). This puppy has a
dynamic range from here to the Klondike.
Tom Bruhns
And a 2b's IIP3 (with no pad at the front end) is how much? Under the
same conditions, the NF is how much?
Cheers,
Tom
Joerg
I don't remember. Let's put it like the horsepower of a Rolls Royce:
Adequate. Even 30 years later that receiver outperformed new gear in
terms of operation in the vicinity of strong transmitters. Ok, it's got
tubes which gives it a head start.
Why do you think a modest Q-multiplication would worsen that, provided
there is plenty of supply voltage? If you just want to reign in the LC
before a filter you don't need to reach Qs of a gazillion.
Joerg
Other question: Where do you have +55dBm right now? And why do you need
that much?
Tom Bruhns
Why: because at that level, a couple 0dBm signals from 7MHz broadcast
transmitters will yield only -110dBm third-order product that will
make unintelligible a microvolt-level signal.
Where: look at what people are doing with H-mode mixers, for example;
but that's not the only place.
Cheers,
Tom
Phil Hobbs
Besides the nonlinearity problem, Q-multipliers aren't the quietest
things on the planet either. That isn't usually a big problem at HF,
but (repeat after me) Negative Resistors Don't Have Imaginary Noise. ;)
Cheers,
Phil Hobbs
Joerg
Sure. However, Q-multipliers are usually placed after there has already
been some gain. Also, often noise is less important than selectivity.
It's like pulling a trailer up a hill where your car won't run very
quiet and not the cleanest but you've got to get up there ;-)
Joerg
Ok, but at some point it's "good enough". At 7MHz you'll rarely deal
with uV level signals. At 20-30MHz that can happen though. Point is, the
old Drake 2B performed marvelously in such situations. In ham radio the
field days and the contests were the major challenges. Not multiband
because you'd be running the 2nd transmitter at another band and the
input filters take care of most of it but it was the guys on the next
hill half a mile away, competing against your group with a huge antenna
and lots of power. Often they were just 20-30kHz away. All the older
generation transistor rigs fell off the rocker in that situation. BTDT,
many times.
In CW mode (morse code) a steep enough mechanical or crystal filter was
not affordable in those days unless you had one of the Rockefellers on
the team. So the Q-multiplier saved the day.
Tom Bruhns
wrote:
...
> Ok, but at some point it's "good enough".
The best currently available is still not quite good enough, I'm
afraid. There are those who have valid needs for better. Though you
sing the praises of the Drake 2B, I've seen other reports on it that
are not nearly so glowing, and we still have no actual measure of its
performance. I seriously doubt that it's anywhere near as good at the
good modern designs that have IIP3-NF > 40dB, and phase noise well
below -100dBc/Hz at 100Hz offset. There's certainly no _need_ for a Q
multiplier in front of a first crystal filter to achieve this
performance, and I doubt that you could add one there while
maintaining the same level of noise and dynamic range performance. I
presume the 2B's Q multiplier was at or near the end of the IF strip,
not at the RF amplifier or even at the mixer output.
Cheers,
Tom
Joerg
After filtering AFAIR. Again, I agree that xtal filters generally do not
need a pre-filter, that was brought up by another poster. In the case of
a large ham radio contest that is very different. While you may have a
nice 2.3kHz filter for SSB and if you splurged maybe another 600Hz
filter that often wasn't enough when the narrow CW portions of a band
where crammed with others who also wanted to win that contest. I now own
a 300Hz filter here but as a young lad back then I certainly did not
have $150 for such luxury. Same for the others. Now, the $100 or so for
the booze during the contest, that was not a problem ;-)
The other thing you could do with a Q-muliplier was often even more
valuable. You could use it as a very sharp notch filter. All in all I
find these things rather useful.