Question about ring VCO start up

[wany]

ring oscillator start up circuit

I am design a differential ring oscillator. I want to know if i need a start up circuit in my Delay cell? Thank you for your help

 

[ravirajdv]

vco startup

Hi Wany,
As I undersand, the ring oscillator is controlled by the Voltage (VCO) and hence there should be no problem with the startup (I don't see any problems with the VCO that I have).
There can be issues with the startup if it is self-biased or does not have any inputs.

Regards,
RDV

 

[FvM]

vco start

Oscillator startup is only a simulation problem.

 

[wany]

ring oscillator start up

Thank you for ravirajdv and FvM. I have read a paper there is a start-up circuit in the VCO. I have some quesiton about it. The fig is the schematic. When Vc(control voltage) is 0 . M5 is closed . And M11 and M4 is opened because start up circuit. Then there is current through M1 and M2. The VCO will has a frequency when Vc is 0. But i think it is not right output.

So i want to know if the current source in the ring VCO need a start up circuit?

 

[oermens]

in simulation (with cadence spectre)...

1. add cap to ground on one of the output nodes of value 0Fand give it initial condition of Vdd
2. in ADE goto simulation->convergence aids->initial condition and set the initial voltage on one of the output nodes.
3. set VDD as a ramp input
4. exponentially decaying current source at one node

etc...

in actual measurement, turning on vdd and any electircal noise should be enough to start oscillation. refer to niknejads lecutres on oscillator design: https://webcast.berkeley.edu/course_details.php?seriesid=1906978242

 

[wany]

Thank you oermens. My question is if the current source need a start up circuit in VCO.