Change in resonant frequency of quartz crystal with atmospheric pressure?

Question

We have a bare quartz crystal and we are measuring its resonance frequency to a very high accuracy (1 ppb). As it cycles between atmospheric pressure and vacuum there appears to be a change in frequency. Could this be because the crystal is being compressed? How can I calculate change in frequency if this is so?

Unexpected change, in temperature controlled environment, is about 400 ppb

Answer 1

Remember that a crystal works on mechanical motion. When something vibrates in air, some power is transferred to the air. Loudspeakers rely on this, for example.

Anything vibrating in air will make sound, which means some power from the vibrating thing gets transferred to the air. With air around the crystal, some the energy stored at resonance is lost to the air each cycle. Effectively this lowers the Q of the crystal. This effect must be quite small, but it doesn't seem like a stretch to be able to measure it at the PPB level.

Answer 2

It's a base crystal open to the vacuum

Then the damping on the mechanical movement is reliant on pressure and this can slightly alter the resonant peak (series and parallel). It will generate sound waves that represent a loss to the resonant circuit and in a vacuum this loss would be less and it's likely the resonant frequency will rise slightly.

Answer 3

The resonant frequency of a \$x\$-cut quartz crystal (i.e. of a quartz crystal cut along the \$x\$ crystallographic axis) is related to its thickness \$d\$, according to what writes H.B. Huntington in [1], chapter 7, p. 219, by means of the following formula $$ f=\frac{2.86}{d}\quad\text{($\mathrm{MHz}$)}\tag{1}\label{1} $$ with \$d\$ measured in millimeters. Assuming we are in the realm of linear elasticity, we can then relate \$d\$ to the pressure applied to the crystal plate by Hooke's law $$ \varepsilon_{xx}=-\frac{1}{K_x}\sigma_{xx}\tag{2}\label{2} $$ where

• \$\varepsilon_{xx}\equiv \frac{\Delta d}{d}\$ is the fractional change in the thickness of the crystal plate,
• \$\sigma_{xx}\equiv p_x\$ iss the component of the pressure vector applied along the \$x\$ crystallographic axis,
• \$K_x\$ is the \$x\$ axis component of the elastic tensor (the "modulus of elasticity" along that axis)

In sum, pressure surely influences the resonant frequency of a Quartz crystal: by a careful use of the above formulas and of the (known?) characteristics of your quartz crystal, you can try to evaluate if it is really it that gives you a "so big" change in the resonant frequency you measure. Finally, let me share with you a few notes:

• Huntington gives formula \eqref{1} "as is", without any formal derivation: however, in the book of W. G. Cady ((1946)[1939], Piezoelectricity, 2nd. ed., New-York: Dover Publications: a new 2018 reprint is coming), and perhaps also in the one of W. P. Mason ((1950), Piezoelectric Crystals and Their Application to Ultrasonics, New-York: Van Nostrand) you'll find the precise deduction and also its applicability limit. Ther are also more modern treatises on the topics, but have a look at the following point.
• Beware! I introduced Hooke's Law in its (though simplified) tensor form \eqref{2} just to give you the feel of the mathematics required (even if I believe that a team who has the ability of measuring frequency variation down to one ppb is not too much impressed by such things ;)). If you want to dive deep in such developments, the book of H.F Tiersten ((1969), Linear Piezoelectric Plate Vibrations: Elements of the Linear Theory of Piezoelectricity and the Vibrations Piezoelectric Plates, New-York: Plenum Press, currently reprinted by Springer Verlag) is perhaps the absolute reference.

[1] Blackburn, J. F. (1949), Components Handbook, MIT Radiation Laboratory Series 17, New-York, Toronto and London: McGraw-Hill Book Company, Inc.

Answer 4

Another way to look at the effect (although just an approximation) is that as pressure increases, more atmosphere moves along with the crystal as it vibrates (the skin depth) and in a sense, increases its mass, thus slowing down its vibration. Of course, this model falls apart if the vibration rate puts the crystal motion above the speed of sound....

Answer 5

In addition to what others have written let me say, that’s the error frequency depends on the effective load capacitance ratio on the motional capacitance In addition to the series inductance which results in a resonant Q-value. I have worked with many different types of crystals from 5° X-cut for VLF to the family of curves of your standard AT cut, which has a third order temperature response and a Q>10,000 and the very high Q of 100,000 or more for the SC cut crystals typically found in all OCXO’s.

The pole ability of centre frequency of any Crystal depends only on the Q and the max/min capacitor ratio applied . I am assuming this is for parallel resonance. Considering your results of 400 ppb or 0.4 ppm , I expect this is a standard AT-cut crystal. These can be expected to be pulled by at least +/-200 ppm. I might also assume that you have chosen an angle cut that produces zero sensitivity to temperature at your other T setpoint or null slope point at some temperature.

Therefore a ratio of 0.4/ 200 [ppm/ppm] is only 0.2% but apparently excessive. A ruggedized SC cut crystal should be 1000x smalller.

I hope this insight helps towards your error correction.

At one time in my career I could test any AT crystal and extrapolate the 3rd order equation of f vs T to < 100 ppb by only two f measurements at 40C , 70C from an equation that derived by polynomial curve fitting. This made it possible to make a 25 cent 1ppm TCXO in production.