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I've done google search on the working principle of crystals and piezoelectricity.

Nothing explains the actual principle. Is it piezoelectricity or inverse piezoelectricity?

From what I've understood, quartz crystals that are used in oscillators work on the principle of inverse piezoelectricity.

Due to any transients or thermal noise, the amplifier amplifies them and sends the AC voltage noise on all frequencies to the crystal. This voltage hits the crystal and deforms it and the crystal produces mechanical vibrations or oscillations nears its resonant frequency which is then fed back to the input of the amplifier.

So, it is inverse piezoelectric effect. Am I right?

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    \$\begingroup\$ It's both. The conversion of energy between mechanical and electrical happens in both directions. \$\endgroup\$
    – Dave Tweed
    May 26, 2022 at 10:37
  • \$\begingroup\$ @DaveTweed Oh, so first, the voltage causes the crystal to vibrate (Inverse piezoelectric effect) and then the crystal vibrations result in production of voltage (piezoelectric effect) - Am I right? \$\endgroup\$
    – user220456
    May 26, 2022 at 11:00
  • \$\begingroup\$ I'm not sure you actually can separate those out as two distinct physical behaviors. Consider, if you stress a crystal, then a voltage appears. OK, and if you apply a voltage, then it becomes stressed. But either way, the end result is the same, the crystal is stressed and it has a voltage gradient. If you could only examine the crystal itself, and not the apparatus to which it was connected, you would have no way to know how it got that way. \$\endgroup\$ May 26, 2022 at 14:21

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Piezoelectric effect works both ways:

Applying voltage across a piezo material will cause it to change shape (usually expand or bend).

Bending or squishing a piezo material will cause it to generate charge and current.

In the case of a crystal oscillator, the crystal is cut in such a way to resonate at the desired frequency.

If you want to think about it in the time-domain, if you apply a voltage pulse to the crystal, it will bend and then resonate, convert these vibrations back into voltage, and it will output a damped oscillation at its characteristic frequency. It is a bit like a tuning fork: you hit it, then it goes "dunnnnnnnnnn" as it keeps oscillating. Without any extra energy input, the the energy will dissipate, and the oscillations will dampen and eventually stop.

In the frequency domain, this translates into an impedance curve with a high-Q peak and a sharp phase shift at the resonance frequency.

The role of the feedback amplifier is to feed back the signal into the crystal so the oscillations keep going. It's like pushing someone on a swing: to keep it going, the push has to be applied with the correct phase.

As per Nyquist's stability criteria, if a feedback circuit has 180° phase shift and unity gain at a frequency, then it will oscillate at this frequency. The sharp impedance peak and sharp phase transition from the crystal at resonance create this condition, so the amplifier feeds back the signal into the crystal with the correct phase to sustain oscillations.

What you say about noise is more about how the circuit starts. When it is powered up, the crystal is not oscillating, but the circuit will amplify its own noise until there is enough to excite the resonance.

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  • \$\begingroup\$ Thank you for the answer. But the crystal working principle and the operating principle is based on the inverse piezoelectric effect, right? \$\endgroup\$
    – user220456
    May 26, 2022 at 10:58
  • \$\begingroup\$ Oh, so first, the voltage causes the crystal to vibrate (Inverse piezoelectric effect) and then the crystal vibrations result in production of voltage (piezoelectric effect) - Am I right? \$\endgroup\$
    – user220456
    May 26, 2022 at 10:59
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    \$\begingroup\$ Yes, it uses both directions of the piezo effect. Besides that, it looks quite similar to a LC circuit from the electrical side. The reason why we use crystals instead of LC circuits, is because it is possible to cut a crystal to a very stable and accurate frequency for a low price. \$\endgroup\$
    – bobflux
    May 26, 2022 at 11:58
  • \$\begingroup\$ Thank you for the answer and clarification \$\endgroup\$
    – user220456
    May 26, 2022 at 14:05

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