I thought I’d really delve into crystal filter analysis now. I’ve been looking at it off and on for the last few weeks but now I’ve decided it would be good to compile some of my learnings for future reference (for myself and of course maybe helpful to others).
As a disclaimer — as usual, what I post in my blog is not an “official” source of anything. It isn’t a peer-reviewed journal. But I do the best I can to be accurate and I give references to where I find equations or explanations. Please let me know if you see a typo in a formula or other error.
I decided to look at the Bitx-20 ladder filter. The one I’m studying came with the Hendricks kit version (which is the one I’m building). There are 4 crystals, each for 11 MHz. I don’t know the manufacturer, but they look like this:
For me the first step in this process was to analyze the crystals used in the filter. I know I can analyze filters all day long in software programs, but I wanted to be able to build a somewhat accurate model of the crystals I’m actually using, then see if I can model the crystal in software, and finally use that crystal model to model the whole filter in software. I saw this done in the Soldersmoke blog and it seemed like a fun and educational exercise. I also thought it might be good to gather these skills for my future projects.
I first put an individual crystal onto my TAPR VNA for study (FYI — I saw one of these for sale on E-bay recently for about $192. Someone apparently snapped it up already and I don’t see any for sale. I’ve found it to be a GREAT little piece of test gear though and at less than $200 it’s a steal, given the original retail in ~2006 was around $800! And keep in mind “real” pro-grade VNAs are generally many thousands of dollars). The crystal was placed in series between the two test leads. I obtained the textbook plot right away. Very satisfying! You can clearly see the series resonance (the peak) and the parallel resonance (the big dip) on the plot of S21 (the forward transmission coefficient):
To understand these resonances you have to study the equivalent circuit of a crystal. The typical representation of this is shown below (see ARRL Handbooks, EMRFD, or Introduction to Radio Frequency Design by Hayward for just a few of many references):
When series resonance is reached, the reactance of the two motional components cancel one another and you’re left with just the series resistance, Rs, and the Cp capacitance. This shows up as an almost dead short, with only the Rs providing loss. The loss I measured, in dB, was about -0.8. This number represents 20 times the log of Vout/Vin (also called insertion loss or “IL”), Vout being (arbitrarily, because the circuit is symmetrical) the test probe on the right and Vin the one on the left.
There are at least two ways to calculate Rs from this information. One is ridiculously complicated, one is easy (and I’m sure there are many more, but once I found the easy one, I stopped). The hard way involves calculating the actual voltages across the input, output and series resistances and using power calculations. The smart and easy way is just to see this as a simple voltage divider situation (neglecting the reactance of Cp at series resonance, which is assumed to be huge in comparison to Rs):
Here, “RL” is just the terminating impedance to the right of the crystal (in my case with the TAPR VNA, 50 ohms). So, you know RL and you know the ratio of Vout/Vin from the insertion loss (once you solve for it with that log equation). Rs is easily solved for.
Note that this nice little relationship did NOT fall out of the pages of any of the references I looked at! As is often the case, it is “assumed” you know how to figure it out once you know insertion loss. I guess I was a bit rusty because I didn’t see it by inspection. I suspect many others out there might be similarly rusty.
So, one down, three to go.
The trick with determining the motional components is to realize that once you solve for one, you know the other one automatically because you’re in series resonance. You can just use the formula for LC resonance and calculate. Which, stated in another way, you know that the reactance of Lm is equal to the reactance of Cm, wLm = 1/(wCm). Plug and chug.
Usually it seems the methods focus on solving for Lm first, although it doesn’t matter. Chapter 7 of EMRFD (Experimental Methods in RF Design) shows a method using the crystal under test in a Colpitts oscillator and a capacitor that is either in the circuit or shorted, shifting the oscillator frequency by some amount. An equation is shown that will determine Cm given the frequency shift from the shorted capacitor experiment. Lm is then quickly determined, again, from the resonant frequency (when the bonus cap is shorted and not part of the circuit).
Another test circuit uses the crystal like a “trap”, in a shunt configuration, so that at resonance you get a huge dip in signal transmission from input to output instead of a spike like in my setup. I’ve seen a few others in a QEX article.
If I didn’t have the VNA, I’d have used the Colpitts method. It seems very intuitive and simple. Having the VNA though, my method of just “hook it up and run” is VERY simple!
A mistake that we all sometimes make (well, I do anyway, so I’m guessing others may) is to grab for a formula from a book/online that seems to be solving for what we want. We don’t know the origin of the formula, but we cross our fingers and chug through it, hoping for a reasonable result. Sometimes you may mostly know the origin of the formula and you think you’re ok. It seems to me though that usually if you don’t really understand the formula you’re using (from detailed explanations in the source or from your own derivation), you get into trouble! Such was the case for me here. I plugged in numbers for a Q equation in EMRFD that had nothing to do with the measurement I had made (it was for the shunt measurement). The confusion that ensued led me to figuring out all the Q relationships though, so I think it was good it didn’t work!
The next step in this process (assuming you didn’t use the Colpitts method and you still need to get Lm/Cm, as I did) is often to solve for Q. I personally hadn’t remembered or realized that there was a “loaded” and “unloaded” Q, but there is. The unloaded Q in this case is the crystal sort of in isolation without any source or load modifying its behavior. The loaded Q is the Q when the crystal is hooked up to a circuit, such as the Colpitts or my VNA. That’s the one we typically solve for, then we can get at unloaded Q and our reactance we’re trying to find.
For a series LC circuit, Q is defined as:
Q (the “loaded” one) can also be determined from a sweep of the frequency characteristics of the crystal, finding the resonance peak and the 3 dB points to either side:
The next step is to solve for unloaded Q from loaded Q. How, you ask? Well again there is that “plug and chug” option or the “full understanding” option. The plug and chug is to use the formula from Wes’s book (Radio Frequency Design). You use an equation for insertion loss that includes a ratio of QL/QU, and solve for QU. It works. You can then use the formula for QU given above and solve for Lm. Cm follows using res. freq. calcs. You’re done.
But this stuck in my craw. A formula with a mysterious ratio of “Q”s in it? I felt like I might be once again using a formula I didn’t fully understand, and I could once again calculate something that’s way off the mark. So I deciphered its origins.
This took me a surprisingly long time but in the end it is a simple formula.
You first have to look at another equation for the loaded Q, the non-bandwidth version. It’s identical to the unloaded Q formula except now it has the terminating resistance on the bottom:
In my setup, the RL would be the 50 ohms terminating the right side of the crystal (the VNA probe). Input voltage is defined as being on the left side of the crystal (and after the 50 ohms of the source impedance — otherwise you’d have 2RL on the denominator).
If you scroll back up and look at the Vout/Vin equation, you’ll see that they have the same denominator. So that funky ratio of Qs (QL/QU) gives me (almost) Vout/Vin.To get Vout/Vin, you have to subtract it from 1. Voila:
So, you have Vout/Vin from a ratio of Qs (one of which, the unloaded, you don’t know). Why do we care? Because all you have to do is take the log and you’re back at insertion loss, which we also know:So that’s it!! That’s the origin of the formula that popped up in the text. We can now solve for QU.
— Obtain the series resonant frequency and 3dB points of the crystal.
— Solve for Rs using the voltage divider formula and insertion loss from the sweep
— Use the 3dB points to solve for loaded Q.
— Use the Q-version of the insertion loss formula to solve for unloaded Q
— Use unloaded Q formula to solve for Lm
— Use resonant freq formula to solve for Cm
You’re almost done. All that remains now is Cp. I read online that “this can easily be measured by a simple LCR meter”. So, I used my LCR meter to measure mine, and sure enough, I got a value (5 pF) that was in line with what others are reporting. I don’t know precisely why this works (not knowing how the LCR meter working), but my guess is that the impedance of Cp at the LCR meter frequency is much much less than that of the non-resonant Lm/Cm series combination (which, with typical values at ~5 MHz, I’ve calculated the difference being a megohm for the Lm/Cm vs. a few kohms for a Cp of 5 pF). So the meter is seeing primarily Cp and determining its capacitance. Knowing Lm, Cm and Rs, one could also calculate Cp using algebra for parallel resonance (but hey, the LCR meter is a lot easier!). I may still check this.
The values I obtained using the above formulas for my crystal were:
Rs: 4.8 ohms
Lm: 8,675 uH
Cm: 0.0241 pF
Cp: 5 pF
These numbers seem roughly in line with others I’ve seen published or online. The unloaded Q is often over 100,000. The inductance is often well into the mH and the capacitance less than a pF. The only parameter that seems a bit off for me is Rs. Other numbers I’ve seen have this number in the 10s of ohms. I don’t yet know why mine came out so low.
I’ve also realized, writing this up, that there is another way to calculate Lm. Instead of determining the unloaded Q and using that to find Lm, you could save some trouble and just use a loaded Q formula. I tried that and got the same answer! Maybe I’m too deep into this now to see why using the unloaded Q is a better way to go. Or, maybe it’s just good to have the unloaded Q number for some of the software modelling programs out there. I’m not sure right now.
Here’s what the actual filter looked like on my VNA. The faded green trace in the background is the same filter but over a wide span (and was accidentally displayed!):
Kind of lumpy/bumpy in the passband. The overall spread of the humps is about 3 kHz, but many of those undulations go below 3 dB from peak. I hope this is a symptom of a noisy measurement and not the actual filter response! Seems like there would be some notches in desired signal taken out of the middle.
I then took my calculated crystal parameters and moved them into Puff models. This is where I started to realize that a dedicated crystal filter program might be advantageous. While it can all be done via Puff, I found it clumsy to have to edit the crystal model, make sure I swept a proper span, save, then edit the filter model, adjust span, etc etc. I will be migrating to some of the other packages soon!
The first plot shows the filter response, the second the crystal. Puff gave me about 1 kHz bandwidth (too narrow) and a much smoother curve (although I guess I was hoping for a more flat top and not rounded). The BW seems about one third the measurement from the VNA. I suspect that the -0.8 dB IL was overly optimistic from the VNA, and so my Rs is too small on Puff.
The bottom plot clearly shows the series and parallel resonances. But the frequency where they occur was off from the VNA slightly. But I’m not sure I can get better agreement with the resolution of tools at hand.
I hope to take my learnings and move to some of the other tools out there. For now, I feel I’ve gotten very close to modelling the crystals I’m using for the Bitx20 kit and I certainly understand the formulas I’m using, which is always good!
Time to get back to melting some solder again.