Oscillating the ring oscillator in feedback system

[surianova]

feedback system

hi all,

Just a question for all of you regarding feedback system. In the ring osillator by using odd number of inverter, the system won't be able to osillate unless i put initialize voltage (or .IC out 0 in Hpice). Is it in the feedback system, it needs a rising voltage to the node so that the feedback system can work either in positive or negative feedback?

 

[Analog_IC]

Re: feedback system

I donot think that .IC is must to set. I had done simulation for ring osc. without using .ic in spice .sp file.
If u see fail output in u r .mt file u can try setting node to gnd using .IC

 

[terryssw]

Re: feedback system

An oscillator have an unstable operating point. You must have some trigerring source to cause the oscillator to move away the unstable operating point then it can oscillate. Usually in real systems, such source can be thermal noise or some power turn-on events. In simulations, numberical error serves as noise source, which can also trigger the oscillations. However, if your time step is not small enough, your round-off error will not sufficient to cause it oscillate and then you must use initial pulse or initial condition to activate it.

Try the following methods:
1. make simulation time longer. there may be some very small oscillation that is growing initially and you need longer time for the oscillation to reach steady state.

2. Control smaller time step:

3. Use initial condition.

 

[surianova]

Re: feedback system

An oscillator have an unstable operating point. You must have some trigerring source to cause the oscillator to move away the unstable operating point then it can oscillate. Usually in real systems, such source can be thermal noise or some power turn-on events. In simulations, numberical error serves as noise source, which can also trigger the oscillations. However, if your time step is not small enough, your round-off error will not sufficient to cause it oscillate and then you must use initial pulse or initial condition to activate it.

Try the following methods:
1. make simulation time longer. there may be some very small oscillation that is growing initially and you need longer time for the oscillation to reach steady state.

2. Control smaller time step:

3. Use initial condition.

How about negative feedback? need initialize condition or not?

 

[sutapanaki]

Re: feedback system

No, you don't need initial conditions to simulate you negative feedback system. You may need .IC, however, if your schematic is large and complex, just to help it find the operating point or start running, but this is a different matter. Do run a step responce of your system to see if it settles well, meaning not oscillating.

 

[JNekas]

feedback system

Add a noise source with smallest value of voltage or current at any point of your circuit.

 

[lastdance]

Re: feedback system

Terry, I think what you meant is an oscillator has a stable operating point. One need to move away from this op to start the oscillation. Therefore, to ensure that the smallest amount of noise can kick start the oscillator, ensure sufficient gain. U can also try ramping the supply up quickly to get it to start.

 

[jiangwp]

Re: feedback system

The phenomenon is like the stat-up circuit of Bandgap.
There are two solution for the differential equation of ring oscilator. we should give the matix a initial solution.But the middle solution(vdd/2) is not stable for system, so it is not exist in the practice circuit(chip), we should not add a addition start-up circuit.

 

[terryssw]

Re: feedback system

Terry, I think what you meant is an oscillator has a stable operating point. One need to move away from this op to start the oscillation. Therefore, to ensure that the smallest amount of noise can kick start the oscillator, ensure sufficient gain. U can also try ramping the supply up quickly to get it to start.

Hi lastdance,

All oscillator should have unstable operating point. If the operating point is stable , then if you have some noise source to force the oscillator to deviate the operating point, then the stable negative feedback will push it back to its operating points and the oscillator will never oscillates. You can have a look on Kundert's book for more information.

Also dear surianova:
Oscillator should work with some kind of positive feedback. Negative stand alone can only produce stable output, and the oscillator cannot oscillate with negative feedback.