3
\$\begingroup\$

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?

\$\endgroup\$
  • 1
    \$\begingroup\$ One thing missing in your explanation is that the crystal IS the filter. It selects the frequency. Another is that, without powering the amplifier, there is no source of noise power. \$\endgroup\$ – Brian Drummond Feb 2 at 12:06
  • \$\begingroup\$ Because the idle crystal is warm, thermal noise excites its 8 MHz resonance. But it's thermal power is so very tiny, you have no hope of seeing it with your oscilloscope. \$\endgroup\$ – glen_geek Feb 2 at 12:25
  • \$\begingroup\$ a crystal is like a swing .... if nothing pushes on the swing, then it will not move .... if someone gives the swing a push then it will start to oscillate at a certain frequency, its resonant frequency .... small pushes at the right time will build up to the point that the swing reaches its limits ..... crystals work in a similar way ... they have a natural resonant frequency and only a small amount of voltage at the right time will cause it to build and sustain its oscillation \$\endgroup\$ – jsotola Feb 3 at 1:20
  • 1
    \$\begingroup\$ So, crystal is connected to the MCU. MCU is powered down. Now, the crystal picks up thermal noise. Since, the signals are very low in amplitude, it is very difficult to see it in the scope. But at this point, we do have signals from the crystal pin outs during the MCU powered down state. So, when the MCU is powered on, the crystal selects the 8MHz noise from the surrounding and gives it to the amplifier circuit inside the MCU. The MCU, amplifies this 8MHz signal and then satisfies the barkhausen criterion with the additional capacitors and sends it to the other pin of the crystal. Am I right? \$\endgroup\$ – Electronic_Maniac Feb 3 at 5:20
  • \$\begingroup\$ Yes. But different amplifier topologies will select either the series-resonance or the parallel-resonance frequencies. \$\endgroup\$ – analogsystemsrf Feb 3 at 8:58
10
\$\begingroup\$

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:

schematic

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?

enter image description here

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.

==============================

Response to the first comment "How does adding resistors increase the phase shift??"

schematic

simulate this circuit

Here are one more methods to provide phaseshift

schematic

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.

\$\endgroup\$
  • \$\begingroup\$ Thank you very much for the detailed explanation. Forgive me for the basic doubt but how does adding resistors increase the phase shift in you 1st point? We usually add Capacitors and Inductors for this phase shifting right? How does a resistor help? \$\endgroup\$ – Electronic_Maniac Feb 3 at 7:22
  • 1
    \$\begingroup\$ Two comments: (1) Regarding oscillators start - forget about noise! It is always the switch-on transient which allows the oscillator to start. (2) In many crystal oscillators - also in the one shown in the answer - the crystal dies NOT work as a resonant circuit. Instead, it works as a high-Q inductor. Hence, the feedback circuit is nothing else than a 3rd-order R-C-L-C lowpass providing the nececcary 180 deg phase shift when connected in a loop with an inverting amplifier. \$\endgroup\$ – LvW Feb 7 at 14:22
  • \$\begingroup\$ Given the two PI 10_pF caps are not precise nor temperature stable, I consider those two caps to merely provide the phase_inversion I've illustrated, and affect the loop gain (by setting the stepup or stepdown ratio of the PI). I'll add a later simulation, to show how the 10pF affect the frequency; the tool allows zooming in, down to the microHertz resolution. \$\endgroup\$ – analogsystemsrf Feb 7 at 20:19
2
\$\begingroup\$

No, a crystal does not do anything by itself, it is a piece of piezoelectric quartz material with contacts for electrical connections. It needs to be connected to a suitable oscillator circuitry. The crystal is used as a high Q band pass filter which sets the oscillation frequency.

The microcontroller contains the necessary oscillator circuitry (including the amplifier) to make the complete system, so when there is no supply voltage on microcontroller, the oscillator is unpowered and it does not oscillate.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.