Open loop gain and open loop phase for an oscillator

 I would recommend that you use a negative resistance analysis in lieu of the methods you are proposing. It is far more intutive and allows one to examine the impact of using different crystal units. The method can be used for both large and small signal simulations and substitues a current source for the quartz crystal unit to examine the impedance the quartz unit experiences in the Pierce oscillator.

Shawn

  Thank you for your help, I'm trying that now.

 I would certainly not recommend using the first method that was proposed of opening the loop. The stb analysis however useful (it's not quite the same as Middlebrook, but shares some similarities). It is not measuring "open loop" gain and phase, but actually the loop gain and phase (the loop is not opened with this analysis, so it can't give you the open loop gain).

For an oscillator which has significant large signal behaviour, stb may not be appropriate - pstb may be better. 

You might want to take  a look at <MMSIMinstDir>/tools/spectre/examples/SpectreRF_workshop and the file PstbAN.pdf (there's also a database to try out, Pstb.tar.Z). This covers using Periodic Stability on an oscillator, and comparing it with stb (for an oscillator which is not that linear).

The negative resistance idea is also a good approach in many cases.

Regards,

Andrew.

 Thank you very much for your help.

I'm trying the negative resistance right now.

However one of my colleague asked me if I could tell him why the first method doesn't work, and I must admit that I really don't know. It looks good on paper.

So if it's not too much (as it's not really related to my topic), why is the first method not good for measuring open loop gain?

Thank you again.

Breaking the loop in a feedback system (particularly if there is high gain) means that you have to work quite hard to ensure that the operating point is correct - small offsets in the system could mean that your amplifier hits the rails and so the operating point is not correct, and hence the AC response is wrong (well, it's the small signal AC response about the wrong bias point). People traditionally have either used large inductors/capacitors to keep the loop closed during the DC analysis/low frequency to get the operating point correct, and then open during the small-signal AC analysis. The problem with that is that the L anc C can then interfere with your circuit.

The other approach is to use spectre's switch component (spt1switch in analog) to ideally close the loop during DC analysis and ideally open it during AC. The downside here is that you still have to get the loading of the loop correct when open, so you end up replicating the circuit - and sometimes this is really tough, particularly when the loading is dependent upon the precise condition of the loop.

So the stb analysis should give you a more accurate answer, and is massively easier to use, than either of the above two approaches used to open the loop.

Regards,

Andrew.

 Thank you very much for your help!

I'm trying both technics now to compare the results.

 Hello again,

I'm experiencing some troubles with the PSTB analysis. In the example it says that I need to run a PSS first, however my PSS does not converge (even with a 10mS extra time to stabilize). I was wondering if there was another way.

Thank you

You have to use PSS before PSTB - you'll need to give more information (such as the spectre output log) to be able to determine why your PSS is not converging. Maybe going via customer support would be a good idea so that we can take a proper look?

Regards,

Andrew.

I must admit that it must be hard to help me while I give you vague answers.

I don't know how to use the customer support, as an intern I'm not even sure I can use it, I will ask my manager as soon as possible.

Meanwhile I'm focusing more on the negative resistance approach, do you know if I can find a tutorial for this approach using CADENCE?

 From what I understood, I need to replace the quartz model by a current source. The voltage drop across the current source will give me an impedance, then I should be able to extract the negative resistance from the impedance I'm reading. Am I correct? (I still need to study this method to fully understand but I think it's the broad idea).

Thank you for your help.

 This technique is independent of Cadence and can be used with any circuit simulator or circuit analysis.

Your concept is correct. I would recommend placing the C0 of the quartz model and the parasitics on each pad in your oscillator amplifier. In other words, place the current source across the two the input terminals of the oscillating amplifier where the two terminals of the quartz crystal unit are normally placed. Remove the quartz crystal unit mode, but include the C0 of the resonator and any parasitics on either side of the quartz crystal unit (due to board, pads, etc). Assuming the DC operating point forces the oscillating amplifier into its high gain region, the real part of the resulting voltage response using an AC value of 1 A in the current source will illustrate the "negative resistance" charactersitic of the oscillating amplifier as a function of frequency. The imaginary part provides an estimate of the reactance of the oscillating amplifier. From the real part, and knowing the range of quartz crystal unit series resistance, by inspection you can tell if sufficient gain exists to support oscillation. The gain is the absolute value of the negative resistance over the quartz crystal unit's series resistance. If there were no non-linear effects, which is not the case as the oscillator relies on those to limit the amplitude of oscillation, the exact frequency of oscillation could be determined by solving the equation Xamp + Xresonator = 0 where Xamp is the reactance of the oscillating amplifier and Xresonator is the sum of the Ls and Cs reactances of the quartz crystal unit.

I hope this helps.

Shawn