Quartz oscillator - why it has so high Q?

Hi there,

why quartz oscillators has so high Q = 10000 or so?

Thank you! I can't find it anywhere...

Dear kbzium
Hi
Do you know about the equivalent circuit of a crystal quartz ? and do you know basics of circuit analyzing ? if yes , you can find , that why Q is very high .
consider a series RLC circuit that is in parallel with a capacitor . it is a real crystal . you can change this complicated circuit to a simple series circuit and at series circuit Q , is X/R

Good luck
Goldsmith

Let's say that i know "math" . I know that r is low and L/C is high so Q is high too. But... why? Why L is so high and r&C are so low?

Thanks!

An electric field applied to a quartz crystal causes a mechanical displacement and vice versa. Thus it is electromechanical transducer. The electrical equivalent circuit for a quartz crystal contains series LCR circuits that represent resonant modes of the XTAL. And here L, C, R are the electrical equivalents to the inertia, stiffness and internal losses of the mechanical vibrating system. That's why the Q value of the motional arm as compared to the LC circuit is extremely high.

In physics and engineering the quality factor or Q factor is a dimensionless parameter that describes how under-damped an oscillator or resonator is, or equivalently, characterizes a resonator's bandwidth relative to its center frequency. Higher Q indicates a lower rate of energy loss relative to the stored energy of the oscillator;

Click to expand...

Q factor - Wikipedia, the free encyclopedia

Crystal oscillator - Wikipedia, the free encyclopedia

So the thing is that there are no real L,C,R but an QO just acts like they were there? And real inductance (such high with small loses) simply doesn't exist so it can't be anywhere near to QO in terms of quality, yeah? Is it that simple ?

Another thing:
There is a LC circuit with no L,C given. We have only f0 = 1MHz and Q0=20. That would be easy, but there is a load connected - RL = 50kOHM. We have to obtain frequencies for which impedance of this circuit loose 3dB to its maximal value

So i've tried pretty much everything but still i can't find out how to use this RL given while there is no L or C given!

Remember , that in fact the crystals are mechanical elements , ( not electrical ) . if crystals are electrical , those specifications are impossible . for example , do you know about mechanical SSB filters ( with pretty high Q and narrow band )
Best Lucks
Goldsmith

The Anthropic Principle

We use Quartz because it has the property of a high Q which is an ideal property for an oscillator. Otherwise we would look for a better material with a higher Q.
Effectively, the high Q is a result of the narrow bandwidth of the resonant frequency at which Quartz vibrates when subjected to electric current or vise versa.
The resonant frequency is a result of the properties of the elasticity in the molecular bonds of the material.
So, the high Q can be attributed to the properties of the molecular bonds and most likely to the high purity of the molecules and the repetitive and constant patterns in crystal materials.

Thanks guys! Guess i know everything about quartz . But what about that:

There is a LC circuit with no L,C given. We have only f0 = 1MHz and Q0=20. That would be easy, but there is a load connected - RL = 50kOHM. We have to obtain frequencies for which impedance of this circuit loose 3dB to its maximal value? Anyone has an idea?

Series or parallel circuit? :grin:

parallel sorry

There is insufficient information to solve this problem: you need to know either L or C so that you can solve the winding resistance of the coil.
From there, you can Thevenize the equivalent parallel resistance of the tank circuit. Then calculate the new Q and BW with the load Thevenized.
As is, it cannot be solved because the equivalent parallel resistance will vary depending on the choice of L and C to form the 1Mhz tank circuit.
For example with L=1uH I get BW=50.1kHz but with L=100mH => BW=12.6Mhz
I am of course assuming that by Q0=20 you are referring to the Q of the coil without the load connected and that the Q of the tank will be lowered when the load is connected.
Otherwise, quite simply BW=Fr/Q = 1Mhz/20 = 50 and thus Fl = 950kHz and Fh = 1050kHz