I have read the answers to the question,"How do crystal oscillator start to work" in this forum. And the common answer which I got is that, there are many frequency around the crystal. The crystal will pick up these all noise signal frequencies and send it to the internal amplifier and filter which will select the frequency and send it back to the crystal again which the barkhausan criterion being satisfied? Am I right? Please correct if I am wrong.
My doubt is that, I have a quartz crystal of 8MHz connected to an MCU. When I don't power the MCU, I am not getting any signal from the ends of the crystal (I probed using an Oscilloscope). I should get the noise frequencies at one end of the crystal,right? Since, the MCU (system) is powered down, I may not get the desired 8MHz with amplification. But I should get some signals at the ends of crystal,right? Or am I wrong?
Best Answer
The noise level of an unpowered XTAL oscillator is approximately computable; in this circuit, even with NO power, the 1Kohm resistor provides broadband electrical random noise:
simulate this circuit – Schematic created using CircuitLab
Assume 1Kohm Rnoise, which produces 4 nanoVolt RMS voltage per root_Hertz of bandwidth. The noise power scales up linearly with bandwidth; thus the noise voltage increases by the square_root of bandwidth. A 10 Hz bandwidth produces 4nV * sqrt(10) = 4 * 3.1 = 12.4 nanovolts RMS. That 8MHz crystal likely has Q of 80,000 or a bandwidth of 100 Hz. The noise voltage will be 4nV * sqrt(100) = 4nV * 10 = 40 nanoVolts.
Can we simulate that?
What is going on, in this simulation?
1) examine the lower-left plot "overall gain/phase response" and notice the RED-colored curve has a dip in the phase, but that phase does NOT REACH -180 degrees, thus Barkhausen is not satisfied; another 5 (FIVE) degrees phaseshift is needed. Hence the insertion of resistors into Crystal oscillators, to provide more phase shift.
2) what is the Q? I assumed 80,000. What is the computable Q, ignoring the 1,000 ohm in the "Sensor" voltage source? Examine the central "impedance stage definition" values, and notice the crystal circuit and component values. The inductor is 0.01 Henry. Reactance at 8MHz is 2 * 3.14 * 8,000,000 * 0.01 (2 * Pi * F * L) = 500,000 ohms for XL. The lossy element is (the motional quartz loss model) 100 ohms.
Thus the Q is only 500,000 / 100, or 5,000.
3) Now for the output noise (normally this will be the startup random noise used by the active-element amplifer/transistor/MOSFET). There are two voltages show by the two curves --- orange and red --- in the lower right plot "thermal noise (rms)". The red curve, noise from 1Kohm in the stimulus (the left-most stage "S0" in the simulation), peaks at 14 nanoVolts RMS. The orange curve, from Stage1 "S1", noise from 100 Ohms in the Crystal "impedance stage definition" menu, peaks at 22 nanoVolts RMS.
The total modeled random noise, at 8Mhz, is RSS of 14nV and 22 NV, or about 30 nanoVolts RMS.
What is your oscilloscope noise floor? The scope is BROADBAND. Assume the digitizer uses 5 pF Csample; using Vnoise = sqrt( K * T /C), and knowing 10pF produces exactly 20 microvolts rms noise at 290 degrees Kelvin (+17C), the 5pF contributes 20 * 1.414 = 28 microVolts RMS noise.
Give the scope floor is 28 microVolts, the 30 nanoVolts of the UNPOWERED crystal cannot be measured.
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Response to the first comment "How does adding resistors increase the phase shift??"
simulate this circuit
Here are one more methods to provide phaseshift
simulate this circuit
THOUGHTS on simulation: include a lossy model of your ESD structures. At 40MHz, the reactance of 10pF is -j400 ohms; reactance of 3pF is -j1200 ohms. Lots of your PI_network circulating energy will flow thru the ESD structures and turn into heat. Additionally, the ESD diode reverse bias capacitance is NONLINEAR.