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In the following circuit: Schmitt trigger crystal oscillator circuit taken from: https://circuitdigest.com/tutorial/quartz-crystal-oscillator

How is the crystal powered to begin oscillation?

I understand the amplifier is required to prevent the oscillations decaying and that it is required to be inverting to counteract the 180 degree phase shift of the oscillator, however, to first get the oscillator oscillating a voltage must be applied to the crystal. If the initial state of the schmitt trigger output is low then how can the crystal start-up?

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The first problem with the circuit shown is that it shows a schmitt trigger inverter and that is incorrect; it should be a regular inverter. The next problem is that I don't believe the 74HC19 is a recognized part number so, going to the site you linked they specify a 4049 device and that is indeed a non-schmitt trigger device. So, assuming an inverter of the type normally used in these circuits...

Resistor R1 ensures that the inverter is forced into linear amplification mode i.e. R1 applies negative feedback and biases the input at the mid-point where gain is maximum. Then it's a simple case of a little bit of noise causing the output to start to wobble a bit and the rest is history; the oscillations build on the effects of noise.

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  • \$\begingroup\$ Thanks for your answer, that seems to make sense. So no power is directly applied to the oscillator circuit other than the supply rails of the amplifiers. From that the noise generated in the feedback loop is enough to induce oscillation? \$\endgroup\$ – TheAndyEngineer Nov 27 '18 at 12:55
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    \$\begingroup\$ An excellent reference on the theory of these oscillators is ti.com/lit/an/szza043/szza043.pdf \$\endgroup\$ – henros Nov 27 '18 at 12:58
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    \$\begingroup\$ Noise is naturally present at the input (as in any amplifier) and this starts things off. The feedback resistor is just to ensure the inverter is running in a linear part of its characteristic. \$\endgroup\$ – Andy aka Nov 27 '18 at 12:59

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