Issue related to PZ Analysis

Wed Nov 10, 2010 12:41 pm   

Dear All,

I am trying to perform an ac simulation of an opamp written in
veriloga model. The opamp is modeled to have two open-loop poles. I
have given the pole at 10 KHz and 100 MHz. Simulation result for PZ
analysis is in compliance with the theoretical values.

But when I added another pole in the model for example at 500 MHz
(along with the above-said location of other two dominant poles), PZ
analysis is returning location of only one pole at 600 MHz. Where
might this pole @600 MHz be coming from? I am confused. Please help.

In the veriloga file the transfer function adding second and third
poles are written as:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
real den[0:3];
den[0]=-2*3.1416*pole2;
den[1]=0;
den[2]=-2*3.1416*pole3;
den[3]=0;
vout = laplace_zp(V(n1), { }, den);
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

** first pole is added by a resistance-capacitance combination written
explicitly.

Regards,
Samiran.

Mon Nov 22, 2010 10:35 am   

Samiran wrote, on 11/10/10 10:41:

Hi Samiran,

I'm not sure that the pz analysis is able to do the analysis on a behavioural
model using laplace_zp. Anyway, I'd suggest you log a service request at
http://support.cadence.com to go into this in more detail.

Best Regards,

Andrew.


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Tue Dec 28, 2010 8:19 pm   

Hi Andrew,

I have browsed a few posts by you and Riad on PZ analysis. For a four
stage complex amplifier I needed accurate pole zero locations. However
I got the typical frequency dependent error. Then i tried a simpler
two stage, miller compensated amplifier to check my pole zero plot.
However i have band of unexpected poles and zero, rather and two-three-
or-four poles, and fewer zeros.

Also, from "spectre -h "pz" ", i couldnt get appropriate information
if there is some normalization factor here ? What would constant
factor be in that case?


****************
PZ Analysis `pz'
****************

Warning from spectre during PZ analysis `pz'.
BSIM3v3 MOS Transistor - frequency dependent components are
present in the circuit, approximated as AC equivalents at 1.000000e
+00Hz for pz analysis.

Poles (Hz)

Real Imaginary
Qfactor

1 -4.23400e+05 0.00000e+00
5.00000e-01
2 -3.53485e+07 0.00000e+00
5.00000e-01
3 -3.93851e+08 0.00000e+00
5.00000e-01
4 -4.18103e+08 0.00000e+00
5.00000e-01
5 -4.63590e+08 +/- 1.33845e+08
5.20422e-01
6 -1.53371e+10 0.00000e+00
5.00000e-01
7 -4.80746e+10 0.00000e+00
5.00000e-01
8 -6.28470e+10 0.00000e+00
5.00000e-01
9 -1.14155e+11 0.00000e+00
5.00000e-01
10 -1.57689e+11 0.00000e+00
5.00000e-01
11 -1.80225e+11 0.00000e+00
5.00000e-01
12 -1.92236e+11 0.00000e+00
5.00000e-01
13 -6.45862e+11 0.00000e+00
5.00000e-01

Zeros (Hz)
at V(vOut,0)/Vdiff

Real Imaginary
Qfactor

1 -3.74331e+07 0.00000e+00
5.00000e-01
2 -1.56942e+08 0.00000e+00
5.00000e-01
3 -5.13000e+08 0.00000e+00
5.00000e-01
4 -8.54096e+08 0.00000e+00
5.00000e-01
5 2.87459e+09 0.00000e+00
-5.00000e-01
6 6.41163e+09 0.00000e+00
-5.00000e-01
7 -1.53453e+10 0.00000e+00
5.00000e-01
8 -5.48343e+10 0.00000e+00
5.00000e-01
9 -5.49539e+10 0.00000e+00
5.00000e-01
10 -1.15266e+11 0.00000e+00
5.00000e-01
11 -1.54998e+11 0.00000e+00
5.00000e-01
12 -1.87114e+11 0.00000e+00
5.00000e-01

Constant factor = 4.52065e+21

DC gain = 7.84084e+01

Accumulated DC solution time = 0 s.
Intrinsic pz analysis time = 10 ms.
Total time required for pz analysis `pz' was 10 ms.


***************************************************************************************************************


On Nov 22, 10:35 am, Andrew Beckett <andr...@DcEaLdEeTnEcTe.HcIoSm>
wrote: