What is the relationship between crystal oscillator frequency Vs Die size?
Why it is smaller Die size over the high frequency ranges and bigger Die size in lower frequency ranges?
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Broadly speaking, the resonant frequency (fr) depends on size, shape, elasticity of the material. The heart of Crystal Oscillator is a Quartz Crystal, which operates on the principle of inverse piezoelectricity.
The book "Principles of Electronics" by Mr. V. K. Mehta says,
A Quarts Crystal can be made to distort in an electric field by applying a voltage to an electrode near or on the crystal. It needs to be properly cut and mounted in order to have stabilized Resonant Frequencies.
The resonating frequency nothing but the rate of expansion and contraction of the quartz. The geometry of the crystal changes depending on the polarity of an applied voltage, which is more like a mechanical deformation. The deformation of the crystal is proportional to square of electric field strength applied. And, conversely, when it is subjected to mechanical pressure, it develops charges on its surface.
Refer the equivalent circuit shown below:
(I don't know how to add a formula in the editor so I opted to take a screenshot of the formulae and upload it as an image, hope that its not a problem)
The inductance(L) depends size of the cut and firing pin. I am considering only the inductance here since it would vary in a comparatively higher variation range. Ultimately the resonant frequency depends on the size of the Cut.
Though I didn't comprehend it completely, both, due to lack of interest in minute detailing and lack of wisdom, this radio-electronics article and an article on crystal techonolgy should be a good read if you are looking to dig deeper.