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I want to measure quartz crystal motional parameters with a VNA, unfortunately all Google searches I've done have lead me to nothing in concrete. I know how to get an impedance curve and measure the parallel and series resonant frequencies but I dont know how to extract the motional parameters from it. Im looking for a relatively easy way to do it, rather than a lot of tweaking in computer software. I found this YouTube video Crystal Filters & Crystals, Part 1 (Adv. 13) which finds the motional parameters but it is an overly complicated process and you need extra software.

I am using a miniVna Pro. Is there a way to do this?

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  • $\begingroup$ I've been thinking about asking this question too. Any good answer will be highly appreciated. $\endgroup$ – zrnzvxxy Jul 7 at 12:50
  • $\begingroup$ How is that process "overly complicated"? You've got a system with quite a few unknown parameters... $\endgroup$ – Marcus Müller Jul 7 at 12:57
  • $\begingroup$ @MarcusMüller you need to fumble around with the simulator to get the parameters, perhaps for one crystal is ok, not very practical if you need to measure several of them. Also, I am used to a crystal impedance meter in which you can get the parameters relatively easy. I'm asking if there is a straight forward way to do it. $\endgroup$ – S.s. Jul 8 at 0:38
  • $\begingroup$ That sounds like the straightforwardest way to me. $\endgroup$ – Marcus Müller Jul 8 at 6:53
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I don't know if this counts as "easy", but:

  1. Find the series resonant frequency. Multiply this frequency by $2\pi$ to convert it to an angular frequency, and call that $\omega_s$.
  2. Note the resistance at this frequency. That's $R$.
  3. Find the parallel resonant frequency, multiply by $2\pi$, and call it $\omega_p$.
  4. Halfway between these resonant frequencies is $\omega_t$. Measure the impedance there and call it $Z_t$.
  5. Calculate the remaining values:

$$ C_p = \mathrm{Re} \left( i (\omega_s^2 - \omega_t^2) \over \omega_t Z_t (\omega_t^2 - \omega_p^2) \right) $$

$$ C_s = {C_p (\omega_p^2 -\omega_s^2) \over \omega_s^2} $$

$$ L = {1 \over C_s \omega_s^2} $$

These are derived below as equations 13, 7, and 2.

If you can measure $C_p$ some other way, then you can skip the measurement at $\omega_t$ and just use the latter 2 of these equations and the resonant frequencies.

Another method is to measure the series resonant frequency with some variable capacitance in series. This might result in some simpler math, and it doesn't require a VNA: only a sweep generator and a power detector.


Explanation:

schematic

simulate this circuit – Schematic created using CircuitLab

The impedance of this circuit is:

$$ Z(\omega) = \left({1 \over -i/(C_s\omega) + i L \omega + R} + i C_p \omega \right)^{-1} \tag 0 $$

When $L$ and $C_s$ have reactance equal in magnitude but opposite in sign, we are very close to series resonance. I say close because $C_p$ has some effect, but it's small because the impedance of the lower components is very much lower. The error is about 0.25 Hz for the 14 MHz crystal in the video. If we neglect that error, the math is simpler.

Let's define $\omega_s$ as the series resonant angular frequency. We can then solve this equation for $C_s$ or $L$.

$$ i \omega_s L = -{1 \over i \omega_s C_s} \tag 1 $$

$$ L = {1 \over C_s \omega_s^2} \tag 2 $$

$$ C_s = {1 \over L \omega_s^2} \tag 3 $$

The series resonance can be found by the VNA by looking for a frequency where reactance is zero and resistance is on the order of 10 ohms. At this frequency, $R$ is the only significant impedance, so:

$$ Z(\omega_s) = R \tag 4 $$

Parallel resonance occurs when the admittance of the two parallel branches of the circuit are equal. Again we're going to accept a little bit of error to simplify the math by neglecting the influence of $R$. Let's call the parallel resonance angular frequency $\omega_p$:

$$ i\omega_p C_p = - \left( i\omega_p L + {1 \over i\omega_p C_s} \right)^{-1} \tag 5 $$

Substitute equation 2 for $L$ and simplify:

$$ i\omega_p C_p = - \left( {i\omega_p \over C_s \omega_s^2} + {1 \over i\omega_p C_s} \right)^{-1} $$

$$ i\omega_p C_p = - \left( {i^2 \omega_p^2 \over i\omega_p C_s \omega_s^2} + {\omega_s^2 \over i\omega_p C_s \omega_s^2} \right)^{-1} $$

$$ i\omega_p C_p = - \left( {i^2 \omega_p^2 + \omega_s^2 \over i\omega_p C_s \omega_s^2} \right)^{-1} $$

$$ i\omega_p C_p = - \left( {i\omega_p C_s \omega_s^2 \over \omega_s^2 - \omega_p^2 } \right) $$

$$ i\omega_p C_p = {i\omega_p C_s \omega_s^2 \over \omega_p^2 - \omega_s^2 } $$

$$ C_p = {C_s \omega_s^2 \over \omega_p^2 -\omega_s^2 } \tag 6 $$

$$ C_s = {C_p (\omega_p^2 -\omega_s^2) \over \omega_s^2} \tag 7 $$

Just one more degree of freedom to solve for. Pick some angular frequency that isn't resonant, call it $\omega_t$. The impedance measured at this frequency is $Z_t$. From equation 0, we can write:

$$ Z_t = \left({1 \over -i/(C_s\omega_t) + i L \omega_t + R} + i C_p \omega_t \right)^{-1} $$

Substitute equations 2 and 6 for $L$ and $C_p$:

$$ Z_t = \left( {1 \over -i/(C_s\omega_t) + i {1 \over C_s \omega_s^2} \omega_t + R} + i {C_s \omega_s^2 \over \omega_p^2 -\omega_s^2 } \omega_t \right)^{-1} \tag 8 $$

Now there is only one variable that can't be measured directly by the VNA: $C_s$. If we can solve for $C_s$ we're golden.

Unfortunately the solution is very hairy. But it gets substantially simpler if we ignore $R$:

$$ C_s = { i(\omega_p^2 - \omega_s^2)(\omega_s^2 - \omega_t^2) \over \omega_s^2 \omega_t Z_t (\omega_t^2-\omega_p^2) } \tag 9 $$

Of course, this is going to give you a complex number, and you can't really have a complex-valued capacitor. But we can gloss over that! Just ignore the complex part. As long as we pick a frequency where $R$ isn't too significant, the error will be small.

Halfway between the series and parallel resonant frequencies seems to work pretty well.


Addendum: it's also possible to start with equation 5 and substitute equation 3 for $C_s$ instead. I wonder if that leads to a simpler solution:

$$ i\omega_p C_p = - \left( i\omega_p L + {1 \over i\omega_p {1 \over L \omega_s^2}} \right)^{-1} $$

$$ i\omega_p C_p = - \left( i\omega_p L + {L \omega_s^2 \over i\omega_p} \right)^{-1} $$

$$ i\omega_p C_p = - \left( {i^2\omega_p^2 L + L \omega_s^2 \over i\omega_p} \right)^{-1} $$

$$ i\omega_p C_p = - {i\omega_p \over i^2\omega_p^2 L + L \omega_s^2} $$

$$ i\omega_p C_p = - {i\omega_p \over L (\omega_s^2 - \omega_p^2)} $$

$$ C_p = {1 \over L (\omega_p^2 - \omega_s^2)} \tag{10} $$

$$ L = {1 \over C_p (\omega_p^2 - \omega_s^2)} \tag{11} $$

Now we can express the impedance in terms of $L$ with substitutions from equations 10 and 3:

$$ Z_t = \left( {1 \over -i L \omega_s^2 / \omega_t + i L \omega_t + R} + {i \omega_t \over L (\omega_p^2 - \omega_s^2)} \right)^{-1} $$

Which is still pretty bad unless $R$ is dropped:

$$ L = { i \omega_t Z_t (\omega_p^2 - \omega_t^2) \over (\omega_p^2 - \omega_s^2)(\omega_s^2 - \omega_t^2) } \tag{12} $$

Or, we can do the same thing for $C_p$ with equations 11 and 7:

$$ Z_t = \left( { 1 \over -i/\left({C_p (\omega_p^2 -\omega_s^2) \over \omega_s^2}\omega_t\right) + {i \omega_t \over C_p (\omega_p^2 - \omega_s^2)} + R } + i C_p \omega_t \right)^{-1} $$

$$ Z_t = \left( { 1 \over {-i \omega_s^2 \over \omega_t C_p (\omega_p^2 -\omega_s^2)} + {i \omega_t \over C_p (\omega_p^2 - \omega_s^2)} + R } + i C_p \omega_t \right)^{-1} $$

$$ Z_t = \left( { 1 \over {-i \omega_s^2 + i \omega_t^2 \over \omega_t C_p (\omega_p^2 - \omega_s^2)} + R } + i C_p \omega_t \right)^{-1} $$

$$ Z_t = \left( { 1 \over {i (\omega_t^2-\omega_s^2) \over \omega_t C_p (\omega_p^2 - \omega_s^2)} + R } + i C_p \omega_t \right)^{-1} $$

Still hairy, unless again removing $R$:

$$ Z_t = \left( {\omega_t C_p (\omega_p^2 - \omega_s^2) \over i (\omega_t^2-\omega_s^2)} + i C_p \omega_t \right)^{-1} $$

$$ C_p Z_t = \left( {\omega_t (\omega_p^2 - \omega_s^2) \over i (\omega_t^2-\omega_s^2)} + i \omega_t \right)^{-1} $$

$$ C_p Z_t = \left( {\omega_t (\omega_p^2 - \omega_s^2) + i^2 \omega_t (\omega_t^2-\omega_s^2) \over i (\omega_t^2-\omega_s^2)} \right)^{-1} $$

$$ C_p Z_t = { i (\omega_t^2-\omega_s^2) \over \omega_t (\omega_p^2 - \omega_s^2) - \omega_t (\omega_t^2-\omega_s^2) } $$

$$ C_p Z_t = { i (\omega_t^2-\omega_s^2) \over \omega_t (\omega_p^2 - \omega_s^2 - (\omega_t^2-\omega_s^2)) } $$

$$ C_p = { i (\omega_s^2 - \omega_t^2) \over \omega_t Z_t (\omega_t^2 - \omega_p^2)} \tag {13} $$

This is a little better!

I threw together an ugly script to check the math, using the values from W0QE's video, and the numbers seem to add up.

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  • $\begingroup$ You are really on to something, thank you for the effort, I will try your procedure and report back. Maybe we can take into account Cp, as it is easily found with a simple LCR meter. $\endgroup$ – S.s. Jul 15 at 2:54
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    $\begingroup$ @S.s. is it? I would think the parallel circuit would confuse the meter. Anyway, I just did a big update! And I'm fairly confident in these equations, at least I wrote a little program to check the math, and it seems to add up. $\endgroup$ – Phil Frost - W8II Jul 15 at 17:00
  • $\begingroup$ Phil, that seems simple enough for me, I will try it out, thanks again. $\endgroup$ – S.s. Jul 15 at 21:01
  • $\begingroup$ Measuring the Cp with a standard LCR meter seems to work perfectly, because the motional parameters only appear when the circuit is oscillating, however Cp is an actual electrical parameter, so maybe using equation 7 to find Cs by using wp, ws and Cp is more than enough. $\endgroup$ – S.s. Jul 16 at 5:46
  • $\begingroup$ @S.s. I think most LCR meters work by applying an AC current and measuring the voltage (or the other way around), and so will oscillate the crystal just the same. $\endgroup$ – Phil Frost - W8II Jul 16 at 13:27

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