ISEGORIA / MATH ENCYCLOPEDIA
Quartz resonators: a crystal that keeps time
Why a sliver of quartz makes a better timekeeper than any balance wheel. Its tuning-fork shape sets 32,768 Hz through the beam equation; the piezoelectric effect turns it into an electrical circuit with a quality factor near 100,000, whose frequency is trimmed by a few picofarads; and its one real weakness, a parabolic dependence on temperature, is what separates a watch that gains a few seconds a month from one accurate to seconds a year.
Before you begin: Waves, elasticity and AC circuits
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Why 32,768 Hz: the tuning fork as a beam
A watch crystal is a tiny tuning fork cut from a quartz wafer, about 2 mm long. Each tine bends like a cantilever clamped at the base, and Euler–Bernoulli beam theory gives its lowest frequency: proportional to the tine’s width in the direction of bending and inversely to the square of its length. The two tines swing in antiphase, so the base, where the fork is mounted, barely moves and little energy leaks away. The target is 32,768 Hz = 2¹⁵ Hz: above hearing, low enough to draw almost no power, and exactly fifteen halvings from one pulse a second. Drag the gold handle at the tip of a tine to change its length, set the width, and read the frequency and what fifteen flip-flops make of it.
Worked example. For quartz with \(E\approx78.7\ \mathrm{GPa}\) and \(\rho=2649\ \mathrm{kg/m^3}\), \(\sqrt{E/12\rho}=1573\ \mathrm{m/s}\) and \(f_1=880\,t/L^2\) in SI units. A tine 0.20 mm wide needs \(L=\sqrt{880\times0.2\times10^{-3}/32768}=2.32\ \mathrm{mm}\). Making it 1 µm shorter raises the frequency by \(2\times10^{-3}/2.32=0.086\%\), 860 ppm, which would make the watch gain 74 s a day: so each fork carries small gold weights at its tips, and a laser trims them away while the frequency is measured.
Watch out. Quartz is anisotropic, so \(E\) depends on the crystal cut; 78.7 GPa is a representative value for the direction in which watch forks bend, and the model ignores the base, the electrodes and the gold tip weights, which is why real forks with this frequency differ somewhat from the formula. The cantilever mode shape drawn is the exact first mode of a clamped–free beam, exaggerated in amplitude by a factor of thousands; the real tip moves well under a micrometre.
Why does the frequency go as t/L², not as 1/L like a string?
A string’s restoring force comes from tension and involves the second derivative of the displacement; a beam’s comes from its bending stiffness and involves the fourth. For a mode of wavelength proportional to \(L\) the fourth derivative brings \(L^{-4}\) against the inertia’s \(\omega^2\), so \(\omega\propto L^{-2}\). The bending stiffness per unit mass, \(EI/\rho A\), grows as \(t^2\) because \(I=bt^3/12\) and \(A=bt\), which gives the factor \(t\) after the square root.
2. The crystal as a circuit, and trimming it
Squeeze quartz and charge appears on its faces; apply a voltage and it deforms. Through this piezoelectric coupling the vibrating fork looks, to the electronics, exactly like a series R–L–C circuit (the motional arm: mass as inductance, compliance as capacitance, loss as resistance) in parallel with the ordinary capacitance C₀ of its electrodes. That is the Butterworth–Van Dyke model. Its impedance falls to a sharp minimum at the series resonance fₛ and peaks at the parallel resonance fₚ, less than a thousandth higher; in an oscillator the crystal runs between them, at a frequency set by the load capacitance C_L of the circuit. Move C_L and watch the operating point slide: a few picofarads trim the rate by seconds a day, which is how watches used to be adjusted with a trimmer capacitor.
Worked example. A typical watch crystal has \(C_1=2.0\ \mathrm{fF}\), \(C_0=1.3\ \mathrm{pF}\), \(R_1=35\ \mathrm{k\Omega}\). At \(f_s=32{,}768\ \mathrm{Hz}\) the motional inductance is \(L_1=1/((2\pi f_s)^2C_1)=11.8\ \mathrm{kH}\), eleven thousand henries from a sliver of stone, and \(Q=69{,}000\). The parallel resonance is \(C_1/2C_0=769\ \mathrm{ppm}\) above \(f_s\). With \(C_L=12.5\ \mathrm{pF}\) the crystal runs 72.5 ppm above \(f_s\), and each extra picofarad of load lowers it by 5.3 ppm, about 0.45 s a day.
Watch out. The component values are representative of 32 kHz tuning-fork crystals, not a particular part. The load-pulling formula is first order in \(C_1/(C_0+C_L)\), which is below \(10^{-3}\), so it is excellent here. Modern quartz watches are no longer trimmed by capacitance: the crystal is cut slightly fast and the divider periodically skips pulses (inhibition), a digital correction shown in the next experiment. The frequency axis is in parts per million from \(f_s\); the resonance is so sharp that a linear axis in hertz would hide it.
How can a mechanical vibration have an inductance of eleven thousand henries?
Through the piezoelectric coupling, velocity of the tines becomes current and force becomes voltage. Inertia resists changes of velocity, so it appears as something that resists changes of current: an inductance, whose value is the effective mass divided by the square of the coupling factor. The coupling in quartz is weak, so a small mass becomes a huge inductance, and for the same reason the spring appears as a tiny capacitance \(C_1\). Weak coupling is also what makes the resonance so clean: the electrode capacitance \(C_0\) is a thousand times larger than \(C_1\), so the circuit barely loads the mechanics.
3. Temperature: the parabola that sets the monthly error
The tuning-fork cut is chosen so that its frequency is stationary at about 25 °C, the turnover point; away from it the frequency always falls, as a parabola, by 0.034 ppm per degree squared. Ten degrees either side costs 3.4 ppm, a third of a second a day, and it is always a loss, whether the watch is warmer or colder. A watch on the wrist sits near 31 °C by day and perhaps 20 °C on a bedside table at night, and the rate averaged over that cycle is what the owner sees. Set the temperatures and hours, and the digital trim, which removes a number of pulses from the divider each minute to cancel the crystal’s deliberate fast cut, then compare an ordinary movement with a thermocompensated one that measures its own temperature.
Worked example. Worn 16 hours a day at 31 °C and left 8 hours at 20 °C, the temperature term averages \(-(16\times0.034\times36+8\times0.034\times25)/24=-1.10\ \mathrm{ppm}\), a loss of 2.8 s a month. A crystal cut \(+10\ \mathrm{ppm}\) fast with 20 pulses removed per minute sits at \(10-20\times0.509=-0.17\ \mathrm{ppm}\) at 25 °C; one pulse fewer, 19, gives \(+0.33\ \mathrm{ppm}\) at the turnover and a net monthly error of \(-0.77\ \mathrm{ppm}\), about \(-2\ \mathrm{s}\). Trimming steps of 0.509 ppm are 0.044 s a day.
Watch out. The parabola coefficient and turnover temperature vary from crystal to crystal by roughly \(\pm10\%\) and \(\pm5^\circ\mathrm C\). The thermocompensated curve is modelled as removing 95% of the parabola, a stand-in for the digital correction such movements apply from a built-in thermometer; real ones are specified at a few seconds a year, where crystal ageing, typically a few ppm in the first year and slowing after, becomes the larger term. Shock, magnetism and battery voltage are not modelled.
Why does the error depend on how many hours the watch is worn, not just on the average temperature?
Because the frequency is a quadratic function of temperature, the average of the frequency is not the frequency at the average temperature: \(\overline{(T-T_0)^2}=(\bar T-T_0)^2+\operatorname{Var}(T)\). A watch that alternates between 31 °C and 19 °C has an average of exactly 25 °C, the turnover, yet it still loses, by \(\beta\) times the variance of its temperature. Swinging temperatures always cost; only their spread and their offset from the turnover matter.
In practice
Where this mathematics and physics is at work, in explainers that take the real thing apart.