Fully differential OTA's PSRR simulation

[didid]

I wonder my PSRR simulation is accurate in Cadence virtuoso.

When we want to get CMRR, we calculate CM_gain and Diff_Gain. Then CMRR is the value of dB scale of (Diff_Gain)/(CM_Gain).
In PSRR simulation, I also put GND into input stage of fully differential OTA. Then, I put VDD source just like the picture below.
I used ideal voltage amplifier with the gain of 1/2 to see the single output.

I get the AC simulation results with the calculator of 'dB20((VF("/VOUT3") / VF("/VDD1")))'
Then I get the graph as you can see below.

Does this graph mean PSRR+? Or should I calculate (Diff_Gain)-(the graph above)?
How can I get PSRR of my OTA?

 

[D.A.(Tony)Stewart]

Try again with mismatched tolerance error

 

[KlausST]

Hi,

nice. -320dB!!!
you don´t have a feel what this means?

it´s 0.000000000000001%

it´s maybe a factor of 10,000,000,000 better of what I have seen.

Klaus