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Message started by sapphire on May 30^th, 2008, 1:06pm

Title: how to predict oscillator's start-up time?
Post by sapphire on May 30^th, 2008, 1:06pm

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

Title: Re: how to predict oscillator's start-up time?
Post by buddypoor on May 31^st, 2008, 4:13am

Good question, but hard to answer.
The only answer I can give is as follows:
As a general rule one can expect that an oscillator exhibiting a frequency determining circuitry with a high pole Q will need a longer time period to settle if compared with a low Q circuit.
But, in addition, the time depends on other parameters like bias conditions, amount of "over design" (loop gain >1), kind of amplitude stabilization,...

Title: Re: how to predict oscillator's start-up time?
Post by sapphire on Jun 1^st, 2008, 9:52pm

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.

Title: Re: how to predict oscillator's start-up time?
Post by chenyan on Jun 2^nd, 2008, 1:07am

start up time depends on gm the active part provides, the load resistance and the tank capacitance, of course the steady state amplitude also. It is difficult to predict accurately as gm is nonlinear of both voltage and time, but first order approx. is easy.

Anyway, can't you simulate that with transient?

Title: Re: how to predict oscillator's start-up time?
Post by buddypoor on Jun 2^nd, 2008, 3:01am

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).

Title: Re: how to predict oscillator's start-up time?
Post by chenyan on Jun 2^nd, 2008, 5:36am

I think it is not directly related to the passive Q. The time constant should be like -(Rneg+Rl)C. However, Rneg is quite difficult to evaluate rigorously since it is nonlinear function of voltage， people normally relies on simulation

Title: Re: how to predict oscillator's start-up time?
Post by buddypoor on Jun 2^nd, 2008, 10:19am

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).

Title: Re: how to predict oscillator's start-up time?
Post by loose-electron on Jun 2^nd, 2008, 10:26am

IMHO you cant really answer the question. Too many unknowns, so to speak. Generally bias and control circuits are much lower in BW than the oscillator itself and frequently dominate the power cyclng process.

Why not do a cold start simulation and see what the time constants are in the system?

Jerry

Title: Re: how to predict oscillator's start-up time?
Post by sapphire on Jun 2^nd, 2008, 12:01pm

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.

Title: Re: how to predict oscillator's start-up time?
Post by buddypoor on Jun 3^rd, 2008, 12:31am

Hi sapphire,

in preparing your paper perhaps the following information is useful for you:

There is an application note from MICROSIM (former vendor of PSpice) dealing with high-Q oscillators.
This AN is contained in a booklet with several other ANs and can be downloaded via the following link

http://www.it.uom.gr/project/digital/appnts.pdf

In this AN my presumption is confirmed that the start-up time is proportional to the Q factor.
Regards

Title: Re: how to predict oscillator's start-up time?
Post by sapphire on Jun 3^rd, 2008, 11:32am

Hi buddypoor,

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

Regards,

Sapphire

Title: Re: how to predict oscillator's start-up time?
Post by buddypoor on Jun 3^rd, 2008, 11:39am

In my edition it is on page 225.
The name of the AN is "Simulating High-Q circuits using open loop response"

Title: Re: how to predict oscillator's start-up time?
Post by sapphire on Jun 3^rd, 2008, 2:54pm

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

Title: Re: how to predict oscillator's start-up time?
Post by buddypoor on Jun 4^th, 2008, 12:13am

Unfortunately, at the start of this topic you have forgotten to mention that you were speaking of an oscillator based on a LC tank circuit with negative resistance in parallel. So I´ve tried to give a general answer because I believe there must be a general one - independent on the specific case.
As far as your approach and your formulas are concerned - I have to think about it.

Title: Re: how to predict oscillator's start-up time?
Post by Hyvonen on Jun 4^th, 2008, 3:16am

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.

Title: Re: how to predict oscillator's start-up time?
Post by sapphire on Jun 6^th, 2008, 10:32am

Hi Hyvonen

Thanks for your detailed explanation. I have one comments about your reply. You said that higher Q means faster transient. Is that true? Higher Q is good for better phase noise at steady state. But higher Q also means larger time constant, which translates into longer transient period into steady-state. That's also the reason whey crystal oscillator is difficult to simulate as mentioned in the AN buddypoor referred.

Title: Re: how to predict oscillator's start-up time?
Post by Hyvonen on Jun 6^th, 2008, 10:00pm

Hi Sapphire,

Yes; it's true that having a higher Q means faster oscillation ramp-up (if everything else is equal). This can be analyzed in many different ways, but here's how I do it:

Oscillator tank actually consists of a parallel network of L, C, Rtank and Rneg. Rtank represents the tank Q - the higher the tank Q, the higher Rtank is (approximately, Rtank=Q*wL=Q/wC for high-Q tanks). Rneg is the "negative resistance" - the mechanism through which the active device provides energy into the tank, keeping the oscillation going. The parallel combination of the two resistances is Rtot=Rtank||Rneg - if the parallel Rneg is smaller than Rtank, Rtot is negative (meaning the oscillation amplitude grows over time).

If you write down the differential equation for the tank and solve the transient response, you get something like:

v(t)=e^-(1/(2RtotC))*(c1cos(w0t)+c2sin(w0t))

where w0 is the oscillation frequency and c1 and c2 are constants (dependent on initial conditions). Note that in our case, Rtot is negative (for growing oscillation amplitude), so e^-(1/(2RtotC))=e^(1/(2|Rtot|C)), indicating exponential growth in amplitude.

To make the amplitude ramp up faster, we need to increase the value of 1/(2|Rtot|C)=0.5*(1/C)*(|1/Rneg|-1/Rtank). If the only knob we have to turn is the tank Q, increasing it is the best way to improve the ramp-up speed - in the extreme case of infinite Q, 1/Rtank=1/(QwL) approaches zero, so 0.5*(1/C)*(|1/Rneg|-1/Rtank) approaches 0.5*(1/C)*|1/Rneg|. If Rtank is anything less than infinite, it would only reduce the value of 0.5*(1/C)*(|1/Rneg|-1/Rtank), reducing the oscillation ramp-up speed.

(Another way to improve the ramp-up is to make the value of Rneg really small while keeping it negative... that would really increase the value of 0.5*(1/C)*(|1/Rneg|-1/Rtank). This is another unintuitive result, but it's all coming from the way negative resistances behave.)

One way to look at this is to consider the ideal case of an infinite tank Q of the RLC tank. Here, your Rtank would be infinite (since it's connected in parallel it would be like an open circuit), and all you would have left is the L and the C. Then, you connect a "negative R" component in parallel with it to provide some energy into the tank that gets the oscillation going. If your tank wasn't ideal (i.e., finite Q), you would lose some of that energy in the non-idealities (parasitic resistance...). If it is ideal (infinite Q), the negative resistance can do its job to increase the oscillation amplitude without any losses.

To me, these things are always a bit confusing - parallel resistances, series resistances... tank Q, inductor Q, capacitor Q, negative/positive resistance... resistance/admittance... All of it can be analyzed through math and network theory, but for quick roundabout answers, I rather go through energy: negative resistance supplies energy into the tank and positive resistance sucks it away (and converts it into heat). The more effective the negative resistance is (the smaller the parallel negative resistance is), the faster energy is supplied into the tank, and the more "effective" the positive resistance is (the smaller the parallel positive resistance is), the faster energy is sucked away.

The explanation above is an incoherent mess. :o But if you have any questions, let me know, and I'll try to elaborate.

Title: Re: how to predict oscillator's start-up time?
Post by Hyvonen on Jun 6^th, 2008, 10:09pm

Forgot to mention two things:

1) You can rather easily verify that increasing tank Q increases the ramp-up speed in simulations. Just make your "inductor Rser" (=inductor series resistance, used to model the inductor Q) really small, and you'll see the amplitude ramp up like mad.

2) The constants c1 and c2 in the differential equation solution are dependent on initial conditions. This is what I was trying to point to when I said it's difficult to exactly predict the ramp-up time, since it depends on the initial condition. If you set v(t=0)=0, you can't satisfy the initial condition unless c1&c2 are both zero (no oscillation, ever). So, you have to have some initial energy in your system (noise...) to have non-zero c1&c2, and to get any oscillation going. (By the way, this is partly why it's difficult to get oscillators start up in simulators... if the simulator is perfectly accurate and no noise is applied, the oscillation doesn't start up because of this very reason... you need to have an inaccurate simulator and/or initial condition for the oscillator nodes to get the oscillation going)

Title: Re: how to predict oscillator's start-up time?
Post by buddypoor on Jun 7^th, 2008, 3:21am

Hyvonen wrote on Jun 6^th, 2008, 10:09pm:

Forgot to mention two things:

1) You can rather easily verify that increasing tank Q increases the ramp-up time.........

Hyvonen: There seems to be a typing error as your position is just the opposite: Q prop. to speed rather than time.

But it is to be mentioned that - as suspected by me formerly (my replies 1, 4 and 6) - it can be proven that the start-up time of oscillators is proportional to the network Q. However, it seems that this holds only for four-pole oscillators with an active feedback element.

If Hyvonen is correct in his argumentation (and I cannot argue against it) there is obviously a difference between two-pole negative-resistance and four-pole oscillator circuits - as far as the start-up time ic concerned. That seems to be a fact worth mentioning. I think it was a very interesting question not answered up to now in any textbook.

Title: Re: how to predict oscillator's start-up time?
Post by Hyvonen on Jun 7^th, 2008, 11:22am

Buddypoor: yes - I had a typo there; I edited the post to say "speed" instead of time. Thanks for finding the error! :)

I also slightly updated the equations in the main post - there were sign typos in some of the equations. The discussion was correct, though.

I'm not familiar with four-pole oscillators. Are crystal oscillators four-pole...?

Title: Re: how to predict oscillator's start-up time?
Post by buddypoor on Jun 8^th, 2008, 1:50am

Hyvonen wrote on Jun 7^th, 2008, 11:22am:

I'm not familiar with four-pole oscillators. Are crystal oscillators four-pole...?

Four-pole oscillators consist of a frequency selective network (bandpass, lowpass, highpass, notch, crystal) and an amplifier, all connected in a closed loop to produce a loop gain Aloop=1 (Barkhausen condition).
The well known WIEN oscillator belongs to this group.

Title: Re: how to predict oscillator's start-up time?
Post by sapphire on Jun 10^th, 2008, 1:01pm

Hi Hyvonen,

Thank you very much for the elaboration. I need some time to understand. Would you please tell a little more about how the damping voltage equation is derived? I think we are more close to the final goal - derive an analytical model to predict the turn-on time.

Regards,

Sapphire

Hyvonen wrote on Jun 6^th, 2008, 10:00pm:

Title: Re: how to predict oscillator's start-up time?
Post by Hyvonen on Jun 10^th, 2008, 3:33pm

You can set up a differential equation for the parallel LRC circuit (where R=Rneg||Rtank) by using Kirchoff's current law:

ic(t)+ir(t)+il(t)=0
<=> C∂v(t)/∂t+1/R*v(t)+1/L∫v(t)∂t=0
<=> C*D^2+1/R*D+1/L=0 (where D=∂v(t)/∂t)

You can solve this second-order equation (see
http://www.analyzemath.com/calculus/Differential_Equations/solve_second_order_3.html for help), with the assumption that the voltage response is oscillatory (i.e., second order solution is a complex pair). This yields the result

v(t)=e^-(t/(2RtotC))*(c1cos(w0t)+c2sin(w0t)) (note that I forgot the "t" from the original equation - that was a typo)

Now, we should be able to determine one of the two constants by defining the initial voltage at t=0 to be something (some amount of noise); that would give us c1 (sin0=0). I'm not sure how to obtain the second constant, though... if anyone has a suggestion, that would be great! :)

Title: Re: how to predict oscillator's start-up time?
Post by sapphire on Jun 10^th, 2008, 9:39pm

Thanks for the reply. I have derived the transient voltage using inverse fourier transform by abstracting the differential pair as a negative resistance and absobing it into the RLC tank. It has exactly the same format as the one you gives! What I also find is that the constant C1and C2 is related to each other. So if we know the initial damping voltage, we can determine C1 and C2 together.

Based on the equation, I think your comment about Q is correct.

There are still two questions needed to be addressed in order to give a closed-form analytical solution.

1. how to determine the initial damping voltage. If the oscillator is just kept running in steady-state, then this analysis is not meaningful. If the osicllator is turned on/off and used as an OOK modulator, then the analysis is helpful. Supposed the oscillator is suddenly powered on, what's the initial damping voltage and how to determine it?

2. As the osicllator goes from small-signal to steady-state, the negative-resistance is also changing non-linearly. That means the function has a time-varying parameters. How to represent this change into the equation and/or give a approximately accurate result?

Hi buddypoor, what's the differenence between 2-port oscillator and 4-port oscillator? Is the common differential LC VCO 2-port or 4-port oscillator?

Thanks for all the discussion!

Hyvonen wrote on Jun 10^th, 2008, 3:33pm:

Title: Re: how to predict oscillator's start-up time?
Post by Hyvonen on Jun 11^th, 2008, 2:30pm

Sapphire,

1. By 'initial damping voltage', do you mean the tank voltage at t=0 (initial condition)? The word 'damping' confused me. :) Initial voltage is undefined, but the oscillator will start up due to noise in the circuit - the exact ramp time is hard to predict, though. A couple of posts ago I suggested providing a well-defined initial voltage (e.g., charging up the tank capacitor to some voltage before turning on the diff. pair); this would give you an initial voltage to start from and the ramp would be possible to calculate.

2. This is true; the negative-R value will get larger until the oscillation amplitude doesn't increase anymore; this is much harder to analyze. You'd need to analyze this using nonlinear differential equations (as opposed to the simple linear differential equation I used), and I cannot help you there. What is it that you're aiming to solve? The time it takes for the oscillator to reach a certain oscillation amplitude?

Title: Re: how to predict oscillator's start-up time?
Post by sapphire on Jun 12^th, 2008, 12:08pm

"initial damping voltage" is referred to the voltage amplitude of first cycle when oscillator is powered on. So it may also depends on how fast the power supply is turned on. The oscillator is not starting up from noise voltage, but from the initial damping voltage. This is kind of initial condition. Your suggestion is valuable, and I can try to relate the bias current and tank characteristic with the initial damping voltage.

In this scenarioes, I guess the transient voltage function only contains sin(wt) term, which means C1 is zero. Because the process can be modelled as a step input or 1/s in laplace domain. How do you think?

For the nonlinear negative resistance, yes, it's very hard to represent it in the linear equation. My goal is to find an approximate expression to model the turn-on time: the time from power on to steady-state oscillation. Hopefully it can help understand the transient process and guide the design a little bit. I am trying to talk some math people to see if there is some simple analytical solution to nonlinear differential equation.

Hyvonen wrote on Jun 11^th, 2008, 2:30pm:

Title: Re: how to predict oscillator's start-up time?
Post by buddypoor on Jun 13^th, 2008, 12:33am

[quote author=sapphire link=1212177977/15#23 date=1213159181]T
Hi buddypoor, what's the differenence between 2-port oscillator and 4-port oscillator? Is the common differential LC VCO 2-port or 4-port oscillator?
[quote author=Hyvonen link=1212177977/15#22 date=1213137185]

Watch your typo: Not 2-port and 4-port, but instead 2-pole resp. 4-pole topology.

Four-pole oscillators consist of a frequency selective network (bandpass, lowpass, highpass, notch, crystal) and an amplifier, all connected in a closed loop to produce a loop gain Aloop=1 (Barkhausen condition).
The well known WIEN oscillator belongs to this group.