why crystal oscillator(high Q system) is hard to begin its oscillation

Hi Guys
It seems like crystal oscillate is hard to begin its oscillation. DOes anyone know the reason for it?
I would image a high quality factor will be easy to begin oscillation. For example, in LC oscillate. The cretiera is gm*Rseris*Q^2>1. Which means the higher the quality factor, the easer to meet its cretieria. Any suggestions?
Thanks,

The start-up time of a high crystal oscillator will be considerably higher, which isn't exactly the same as "hard to begin it's oscillation". The start-up time is caused by the fact, that the meachanical oscillator is only loosely coupled to the electrical circuit and needs some time to build up it's energy.

so what determine the startup time?
Well, compare a ring oscillator and LC oscillator. Normally LC oscillator have a high quality factor. Which one will be ease to begin oscillation?
ALso, how about time needed to begin oscillation?
Thanks a lot FVM:)


---------- Post added at 22:42 ---------- Previous post was at 22:04 ----------

also, can you comment the start-up time and ease to oscillation between ring oscillator and LC oscillator?

My feeling is that ring oscillator might be easier to begin oscillation than LC oscillator. but ring oscillator is a low q system.
Very confused.
thanks

Besides resonator Q, the amplifier loop gain is strongly affecting the startup time. To achieve good oscillator performance (e.g frequency stability, low phase jitter), the loop gain will be selected only a small amount above unity. A ring oscillator will usually have excess loop gain and show fast attacking oscillations. In steady state, the loop gain is reduced by driving the amplifier stages into nonlinear compression.

Loop gain is the only factor to consider. The crystal resonator is usually loosely coupled in the loop (in order to get a sharper resonance), thus the loop gain is only slightly above unity though it has very high Q.

thanks guys. It make sense.
since ring oscillator has a much bigger loop gain than LC oscillator.
So how about the relationship between tank Q and loop gain and phase noise.
Generally high Q means better phase noise. Will loop gain play a role in determine the phase noise?
thanks,
Bo


---------- Post added at 00:16 ---------- Previous post was at 00:14 ----------

ALso, maybe not very improtant, why crystal oscillator need a small loop gain and need to be loosely coupled? Does it mean in order to get a high Q, small loop gain is necessary?
thanks

In a feedback oscillator the noise will be amplified repeatedly until a single oscillation frequency is produced.
Because the crystal (with high Q factor) is part of the feedback loop, the bandwidth of the loop is narrow. This mean, lower initial noise power (than wide BW), and needs more time to arrive to certain power (to oscillate).
This is the reason than not only using crystals, but higher the Q factor of the resonator, greater the time to start the oscillation.

Loose coupling of the resonator is a built-in property of usual crystals. It's represented by a very high characteristic impedance in the equivalent circuit. Even a bad designed oscillator circuit can't easily reduce it under a certain amount. As a consequence, the start-up time can't be arbitrarily shortened, although loop gain has some influence.

High Q and low excess loop gain is however strongly wanted to achieve minimum phase noise.

Another factor influencing the start-up time is an initial circuit state different from equilibrium, e.g. provided by a supply voltage transient. With a perfectly balanced initial state, the oscillation amplitude has to rise from nV scale circuit noise.

Thanks, yes, it makes sense the noise power will be lower due to the high Q bandwidth.



---------- Post added at 21:59 ---------- Previous post was at 21:57 ----------

Thank you, it is very helpfil. I understand high Q is necessary for low phase noise. But why the loop gain also affect the phase noise? I mean we definitely need the loop gain to be larger than 1. But why it has to be close to 1 to get a good phase noise? Any comments?
Thanks

With a high loop gain, the amplifier is driven more into the nonlinear part of it's characteristic, increasing waveform distortion and also intermodulation between amplifier noise and oscillator signal. This intermodulation is a major source of oscillator phase noise.

Intermodulation is not the major source of the oscillator phase noise, and also is not the only source.
In the Leeson phase noise prediction equation there is nothing mentioned about intermodulation or IM products.

Saying it's "a major" source implies that it's not the only one, isn't it?

Apparently, there are different opinions in literature if it's a major source. I found it considered a major source in a classical crystal oscillator literature, the "Quarzkochbuch", unfortunately only available in german. I'm not familiar with "Leesons equation", but it's generally accepted, that upconversion of amplifier 1/f noise is an important source of phase noise. If so, I think it's obvious that amplifier non-linearity will cause additional modulation products and thus increase phase noise.

The discussion was however about the role of excess loop gain. So what can be said in this regard according to "Leesons equation"?

Let's play with words. Could be major 99.9%, and minor 0.1%. So in this situation "major" is almost like the only one.

1/f noise doesn't have anything with intermods...not even with the gain of the amplifier. And as I said, intermods generally doesn't have big impact in phase noise of an oscillator.

On the other hand, if you didn't hear about Leeson equation when is about oscillator phase noise, sorry but is no reason to continue the discussion.

Although I'm not a native English speaker, I know, that major can have other meanings, I'm sure you know, too. Think e.g. of major comanies...
So you mean, that the role of 1/f noise for phase noise is a linear effect? I wonder how?
That's up to you. You'll notice, that I had asked a question in my previous post, or more exactly repeated a question of the original poster. Besides fighting my erroneous believe, it won't be bad to contribute to the original discussion.