How to simulater Oscillator

[Bin_Wang]

I have been design a Oscillator for 32.768K, and when I simulate the ciruit ,the oscillator can work ,but when the simulate time is long ,such as 500ms and more
the amplitude will decrease from 3.3V to zero(the power supply is 3.3V), could some one give me advice ,thanks

 

[yangbala]

How much simulate time about your Oscillator for 32.768K? I think that you must check your ciruit again or post conditions and results of simulation.

 

[BigBoss]

To simulate an oscillator there two ways..
PSS ( Periodic Steady State Analysis ) simulation in Cadence or transient simulation still in Cadence..

Both will give the same result if you increase the accuracy especially in transient simulation.

To decrease the amplitude from 3.3V to 0 in time domain ( if you have meant that ) you may use negative feedback in oscillator biasing so that when oscillator reaches steady state response ( you can use time delay ) then biasing starts to decrease the biasing or current of the oscillator core the oscillator stops.But when it slows down, frequency changing may occur...

There may be many ways to stop it...

 

[Hughes]

Do you mean oscillation disapears after a long time?
You may change the timestep to a smaller value (e.g. 10ps).

 

[huojinsi]

Your oscillator's frequency is 32.768K,how does the frequency has been simulated? Did you simulate the result with transient analysis or not? If the 32.768K was not gained under the conditon of transient analysis,maybe your oscillator circuit cann't oscillate. You should analysis your oscillator circuit well.

 

[power-twq]

I think the gain of amplifier may be too small.

best regards

I have been design a Oscillator for 32.768K, and when I simulate the ciruit ,the oscillator can work ,but when the simulate time is long ,such as 500ms and more
the amplitude will decrease from 3.3V to zero(the power supply is 3.3V), could some one give me advice ,thanks

 

[windman]

You can use the command .ic to initialize the current flow through the effective inductance of the quartz model,thus the circuit may works quickly to reach the stable amplitude.
Why should you initialize the current? As you might know,the Qulity value of a quartz is very high,normally it's above 0.1million,so the energy storing procedure of the inductance would be very slow when power on.

If it still stops from oscillating finally, the feedback resistor cross the input and output of the amplifier might not be reasonable,you can try to adjust it and make sure the bias point of the input and the output of the amplifier are around about vdd/2, the amplifier should work in the linear region.

e.g.
.IC I(L1.X1)=10uA
I assume L1 is the effective inductance in the quartz's model, X1 is the quartz.

Any further problems ,you are welcome to contact me via email skyman88@gmail.com

 

[sergeyr77]

It means that gain of corresponding amplifier at working frequency is low.
I usually make AC analyses of open loop to check Nyquist criterion- it saves a
plenty of time.

 

[yli]

I think that the close loop gain of the Oscillator is smaller than 1. Then the oscillation will stop after a period of time.

 

[leo_o2]

I suggest you having a try like this:
Seperate your simulation to two periods: 300ms for each period.

 

[tecsiun]

I think you can decrease the timestep of simulation,and initial condition is added to the crytal oscillator 'ic=xmA' If you want speed up the simulation.