why spectre.out doesn't include the corner offset-frequency of the vco?

Dear all,

I am a freshman in CDN simulation. When I follow a workshop of Hartley VCO simulation, I didn't get the same result as what the material PPT shows to me. I only convert the model to tsmc.18 as I don't have the example's original model.

I just do as the workshop asks me to. I don't know why. Who can help me!

Here is part of the spectre.out file:

************************************************** ***********************
Periodic Noise Analysis `jitter1': freq = 1.12416 GHz + (1 kHz -> 10 MHz)
************************************************** ***********************
Using the operating-point information generated by PSS analysis `pss'.
The estimated line width of the oscillator is 2.09534 mHz.
jitter1: freq = 2.512 kHz (10 %), step = 1.512 kHz (10 %)
jitter1: freq = 6.31 kHz (20 %), step = 3.798 kHz (10 %)
jitter1: freq = 15.85 kHz (30 %), step = 9.539 kHz (10 %)
jitter1: freq = 39.81 kHz (40 %), step = 23.96 kHz (10 %)
jitter1: freq = 100 kHz (50 %), step = 60.19 kHz (10 %)
jitter1: freq = 251.2 kHz (60 %), step = 151.2 kHz (10 %)
jitter1: freq = 631 kHz (70 %), step = 379.8 kHz (10 %)
jitter1: freq = 1.585 MHz (80 %), step = 953.9 kHz (10 %)
jitter1: freq = 3.981 MHz (90 %), step = 2.396 MHz (10 %)
jitter1: freq = 10 MHz (100 %), step = 6.019 MHz (10 %)
Total time required for pnoise analysis `jitter1' was 500 ms.


************************************************** *************************
Periodic Noise Analysis `jitter2': freq = 1.12416 GHz + (-1 kHz -> -10 MHz)
************************************************** *************************
Using the operating-point information generated by PSS analysis `pss'.
jitter2: freq = -2.512 kHz (10 %), step = -1.512 kHz (10 %)
jitter2: freq = -6.31 kHz (20 %), step = -3.798 kHz (10 %)
jitter2: freq = -15.85 kHz (30 %), step = -9.539 kHz (10 %)
jitter2: freq = -39.81 kHz (40 %), step = -23.96 kHz (10 %)
jitter2: freq = -100 kHz (50 %), step = -60.19 kHz (10 %)
jitter2: freq = -251.2 kHz (60 %), step = -151.2 kHz (10 %)
jitter2: freq = -631 kHz (70 %), step = -379.8 kHz (10 %)
jitter2: freq = -1.585 MHz (80 %), step = -953.9 kHz (10 %)
jitter2: freq = -3.981 MHz (90 %), step = -2.396 MHz (10 %)
jitter2: freq = -10 MHz (100 %), step = -6.019 MHz (10 %)
Total time required for pnoise analysis `jitter2' was 700 ms.

There should be one line tells the saturation freq, like " the estimated phase noise saturation freq is blabla..., doesn't it?

Regards,

Justin

Hi,
That's normal. As the technology that you use is not the same, results will not be the same.