Hi,

I'd like to know how long it takes for oscillator to reach steady-state when it's suddenly powered up. Is there any reference I can find?

Thanks

Sapphire

I thought Q determines the turn-off time more, but not sure how it's related to the turn-on time.

For turn-on time, I tried to use equivalent negative resistance at start-up condition, but this number will approach zero when the oscillator reaches steady-state. But I don't know how to represent its change along this process.

Perhaps the following additional information can help to estimate the start-up time because of the relationship between filters and oscillators:

As a rough estimate the settling time is app.
bandpass:    tau= 1/bandwidth
lowpass:      tau=1/2*bandwidth  

Of course, bandwidth is correlated with Q (and Q with damping resistors and capacitors).

I like to repeat that the start-up time is correlated with the Q of the frequency determining circuitry. This has been proven by experiment as well as simulation. But, in addition there are some other influencing parameters as mentioned before.
(comment to the reply above: negative resistance oscillators are a specific version of a two-pole network. The question was a general one including four-pole oscillators).

With simulation, that's very easy to determine the start-up time.

I am working on a paper, so it's helpful to put some analysis with equations. It's also interesting to quantify the problem.
Right now, I just find out the time constant at the start-up condition, and use this number as a bound for estimating the turn-on time. If I can move a step further to find the average time constant along the transient, that's so great.

Hi buddypoor,

Thanks for your information. Could you please tell me which page of the applicatio note is related with the question?

Regards,

Sapphire

Thanks buddypoor. I read that part of the AN, and seems it is mainly introducing some simulation skills for crystal oscillator.

I got some new thoughts about the turn-on transient. The Q of resonator is related with turn-on time through the parallel tank resistance. High resistance means large time constant, which therefore makes the turn-on time longer. First calculate the average oscillator negative resistance along the transient, and parallel this number with the tank resistance, and then multiply with the tank capacitance to have the equivalent time constant.

The first step is a little tricky.

|Rneg| = (Rp+(-Rneg1))/n,     Rp is the tank resistance, Rneg1 is the small-signal diff pair negative resistance, Rneg is the average negative resistance
                                            n is an empirical constant and should be greater than 2.  
                                            I think n is equal to 2.5

More discussion is welcome

I wrote a relatively long response to this, but unfortunately I lost it due to an unfortunate copy/paste incident.  :-/  So, I'll here's a short version - please ask if you need details.

In any LC-based oscillator, the losses in the tank are caused by nonidealities such as parasitic resistance.  These are modeled with tank Q - the higher the better.  This tank Q also represents the "tank resistance" Rp sapphire mentioned; a quick approximation would be Rp=Qtank*Ztank, where Ztank=2*pi*fres*L=1/(2*pi*fres*C) and fres=1/(2*pi*L*C).

To achieve and sustain oscillation a method of cancelling or compensating for the tank losses is needed - that's what the negative resistance is for.  Assuming it is sufficient to overcome the losses, the oscillation amplitude will increase from cycle to cycle.  This would be similar to an RC discharge, where the voltage slowly discharges, and the voltage could be calculated at any time point if we knew the RC time constant and the initial value, only that this time the value of R would be negative.

And this is where it gets tricky; to know how the amplitude changes over time, we need the time constant, the oscillation frequency etc.  But to know the exact value of the amplitude, we would also need to know the initial value.  In practice, the "initial value" depends on the amount of noise in the circuit, and that is somewhat hard to analyze - because of this, there is no simple answer to your question.  However, the trends are definitely there - the higher your tank Q is, the faster the amplitude will ramp up, and the "more effective" negative resistance (e.g., the mode current you burn in your active transistor), the faster the amplitude will ramp up.

Overall, the only way to make the amplitude predictable is if you could have a well-defined initial value for the circuit at start-up (e.g., having an uneven initial bias voltages for a differential negative-R stage in a differential oscillator) - you could use that as the starting point for the calculation instead of a much less reliable estimation of noise in the circuit.