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I've read interesting answers on this site of how crystals generate oscillations. I would like to know how the crystal goes to its natural oscillation from a voltage applied at its terminals of, for example, 5 V, 60 Hz. I don't understand how the crystal manages to generate its natural frequency from 60 Hz onwards (for example 10 kHz) using a feedback that does not have, in principle, the natural resonance signal of the crystal.

How is the transition from 60 Hz to 10 kHz? Can I apply whatever voltage and frequency and the crystal will end up providing its natural frequency of oscillation at all times?

JRE
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Osmar Idalercio
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    Why do you think that crystals are excited with 50 or 60 or some other than the natural frequency? – Justme Dec 29 '22 at 16:24
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    Osmar, it seems that you are mistaken in your assertion. Maybe you can link one of the interesting answers on this site that allowed you to develop that idea? Crystal oscillators don't work in the way that you appear to believe they do. – Andy aka Dec 29 '22 at 16:26

3 Answers3

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A crystal, on its own, is nothing more than a passive filter with a very particular frequency response that makes it very selective to frequency; it has two so-called resonant frequencies:

  • the "series" resonant frequency is the one where the crystal has very low, purely-real impedance and acts nearly like a short circuit.
  • the "parallel" resonant frequency is the one where the crystal has a very high impedance, acting nearly like an open circuit.

If you apply a frequency away from either of these two resonant frequencies, the crystal will have a behavior similar to an inductor or capacitor.

Oscillation and an output at the resonant frequency occurs as a result of a few possible mechanisms:

  1. The incoming signal is not a pure sine wave (e.g. it includes a transient or harmonics) and contains a component at the crystal's resonant frequency that resonates (if a transient, it may die out over time, because of energy dissipation in the system)

  2. You are talking about a crystal amplifier and not just a crystal. In this case, a circuit such as a pierce oscillator amplifier provides frequency independent gain. In this case, the incoming power is used to power the amplifier itself, typically with DC (a device that takes in 50/60 Hz AC will first rectify and regulate that supply to DC).

    When combined with the crystal, which acts as a frequency-selective filter, a positive feedback loop is established. The combination of crystal and amplifier is now an oscillator. If there is an excitation at the resonant frequency of the crystal, it will be amplified and become self-sustaining.

    This however raises the question of where the excitation comes from, because we are only applying DC, 50 Hz, or 60 Hz to the system. In practice, it comes either from an impulse when the oscillator is powered on, or from thermal noise (which is always present, and present at all frequencies). As soon as the tiniest wisp of energy at the resonant frequency enters the amplifier, it'll get a bit of amplification, pass through the crystal's resonant filtering, and get amplified even more, thus building up and sustaining the oscillation.

Adding some remarks from comments:

The crystal only has two pins, which connect right to the amplifier. The crystal isn't "powered" with the DC source; the amplifier is powered with the DC source. While the amplifier is powered, it's able to provide amplification, while the crystal provides tuning and frequency selection. Together, the two make an oscillator - the amplifier amplifies whatever signal it sees, while the crystal passes only the resonant frequency back to the amplifier.

The signal coming out of the oscillator may be sinusoidal, but it may get "squashed" into a square wave because of saturation and other effects. The exact shape depends on the gain, saturation behavior, and other aspects of the amplifier itself.

nanofarad
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  • Your answer is very good. It helped me a lot to understand a little more about crystals. Sorry if I ask any inappropriate questions, I'm a beginner and the operation of a crystal is not very easy to understand. I take the opportunity for one more question: Is it possible in an oscillator to excite the crystal with 5v DC or AC through source 1 and the transistor at the output with another DC voltage source (15v for example)? Or do the transistor and the crystal always have to be powered by the same source? – Osmar Idalercio Dec 29 '22 at 19:04
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    @OsmarIdalercio The crystal only has two pins, which connect right to the amplifier. The crystal isn't "powered" with the DC source; the amplifier is powered with the DC source. While the amplifier is powered, it's able to provide amplification, while the crystal provides tuning and frequency selection. Together, the two make an oscillator - the amplifier amplifies whatever signal it sees, while the crystal passes only the resonant frequency back to the amplifier. – nanofarad Dec 29 '22 at 19:09
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    Further - if you forcibly inject the wrong frequency into the crystal in an oscillator (e.g. 50 Hz), that signal will die down because the crystal will not pass it back to the amplifier, so there will not be positive feedback. At the same time, the most feeble signal at the correct resonant frequency will pass and will get stronger and stronger until it reaches the maximum voltage that the amplifier can output (then it will stop getting stronger and remain stable) – nanofarad Dec 29 '22 at 19:11
  • GREAT!!! And what kind of wave will the signal that will come out in front of the oscillator have? Always square? Thank you very much. – Osmar Idalercio Dec 29 '22 at 19:17
  • I'm new to this site too. I don't really know how it works. I would like not to lose contact with you. You explain things very well. Any way to reach you when I need to ask a question about electronics? I would like not to lose contact with you. Big hug! – Osmar Idalercio Dec 29 '22 at 19:21
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    @OsmarIdalercio The signal coming out of the oscillator will usually be sinusoidal, but it may get "squashed" into a square wave because of saturation and other effects. My profile has some contact info, but I browse this site often, so just asking a new question is a good idea for a number of reasons (others can correct my mistakes, the answer is available to other readers, etc) – nanofarad Dec 29 '22 at 19:25
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If we hit a crystal with an impulse, it will 'ring' at its natural frequency. Rather as a bell, struck with a hammer. This is the transient behaviour, of crystal, or bell.

If we apply a 50 Hz squarewave, then 100 times a second, we are giving the crystal a kick, and it will ring, rather like a telephone bell, with 100 'pings' of its natural frequency per second.

If instead we apply a 50 Hz sinewave, and wait long enough for the initial transient to die down (as it will, any circuit has finite Q), then the output will be 50 Hz. It cannot be anything else, as that is the only frequency at which energy is being supplied.

If you provide DC power, to power an amplifier that makes up for the losses of the crystal, then you effectively give the amplifier/feedback/crystal combination infinite Q, and it will ring forever. This is called an oscillator.

Neil_UK
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  • Your answer is masterful. Well against my doubts. I just don't understand why if I apply 50 hertz and wait for initial transient die I will get the same 50 hertz. Sorry I'm a beginner. I would be grateful if you could talk more about this and clarify what "initial transient" is. I conclude from your explanation that no matter what frequency I excite the crystal, it will respond at its natural frequency. Am I right? Another question: If in an oscillator I initially excite the crystal with a sine wave, what kind of wave will I get at the output? Or is what I'm talking about nonsense? – Osmar Idalercio Dec 29 '22 at 18:55
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    Take a 1 kHz bell, and hit it with a hammer. it rings at 1 kHz, and then eventually this initial transient dies away. Take a 1 kHz bell, and move it sinuosoidally at 1 Hz. Once any initial ping from the sudden starting to move it has died away, it will move at 1 Hz. There's no way it could move any other way, there's no energy at any other frequency. Similary a crystal of whatever natural frequency, excited with 50 Hz sine wave, will eventually only output 50 Hz. – Neil_UK Dec 29 '22 at 19:14
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    @Neil_UK in practice, a bell probably has a bit of non-linearity that can shift low frequency energy up to higher ones, I know piano's do. I'd guess a crystal is the same, but how much, and is it practically relevant, I'm not sure... – mbrig Dec 30 '22 at 06:51
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how the crystal goes to its natural oscillation from a voltage applied at its terminals of, for example, 5 V, 60 Hz

Crystals are not fed with 60Hz or anything like it. They are in a circuit that naturally oscillates if you close the feedback loop with a simple impedance. The crystal selects which frequency the oscillation happens at. The circuit itself may oscillate at any one of a range of frequencies, only determined by the characteristic of the feedback impedance in the circuit, and by the startup conditions. But, when there's a crystal in the feedback loop, it acts as a very narrow bandpass filter. Since the oscillator gets feedback only at that one frequency, that's the frequency it oscillates at (this is a simplification!).

Another important aspect of applying crystal resonators in oscillating circuits is the maximum power delivered to the crystal. Crystal is a transducer between electrical and mechanical domains. When the drive signal is too strong, it will mechanically overstress the crystal and cause premature aging and eventually mechanical failure.

Crystals are also nonlinear, like a pendulum would be: their mechanical oscillation frequency is only "on the spot" when the amplitude of the oscillation is small. As the amplitude of the oscillation grows, the frequency changes!

Overdriving the crystal introduces immediate frequency offset and other deviations in performance that can be quite substantial relative to the specifications in the datasheet (nominal frequency, aging rate, temperature coefficient, etc.).

When validating crystal oscillators, it is important to measure the AC current flowing through the crystal. Those currents are tiny - in the tens-hundreds of nanoamps. For more on this topic, see this excellent EDN Video Design Notice by Jim Williams.

Kuba hasn't forgotten Monica
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