Issue related to PZ Analysis

Discussion in 'Cadence' started by Samiran, Nov 10, 2010.

  1. Samiran

    Samiran Guest

    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.
     
    Samiran, Nov 10, 2010
    #1
  2. Samiran

    langisdexiM Guest

    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.


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


     
    langisdexiM, Dec 28, 2010
    #2
Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.