Basic pole-zero analysis

Hello! I am having trouble while trying to perform a basic pole-zero
analysis. I am simulating a simple 1-transistor, resistor-loaded
common source amplifier. I want to compare the results of the poles
and zeros with the ones I get from my hand calculations.

The problem is that the results I am getting seem rather awkward. I
suspect this has to do with the following warning that I am getting
when performing the simulations:

*************
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.
*************

Does this mean that spectre is unable to perform pole-zero analysis
for circuits including BSIM instances? If so, how could I get around
this?

Otherwise, Is there any other approach to finding the poles and zeros
of practical circuits (including BSIM instances?) using spectre?

Thanks in advance for any help, as usual!

Regards,
Jorge.

 

Oh, I forgot to add that the AC analysis seems to agree with the pole
& zero locations predicted by the pz analysis... would this
categorically confirm that are my hand calculations the ones that are
plain wrong? :S

 

Dear Jorge,

I Honestly never ever seen this error before. This does sound very odd
for me though. I recall having used PZ analysis on BSIM3V3 devices
without any worries. My best Advice is to get in touch with your
Cadence AE. Don't forget to mention you Spectre version/subversion.
To be honest with you, I don't really like the Spectre PZ analysis,
just because it does give you the poles/zeros information only. It's
sometimes a bit hard to interpret as well. This is a pure personal of
view. When I do stability analysis, I'm always after the Gain/Phase
margins, that's why I was always doing an AC analysis by breaking the
DC loop. You can also determine the Poles and Zeros with the AC
Analysis. Spectre does provide a stability analysis 'stb' as well.
Please look at the Spectre doc for more information. I would recommend
this K. Kundert's Paper for a better understanding :
http://www.kenkundert.com/docs/cd2001-01.pdf.

BTW, give it a look at the designer's guide forum. There are loads of
threads discussing the stability analysis over there. Really
interesting.

I recall having asked one of my great friends at my former company 7
years ago about the Stability analysis of my LDO regulator (I was
beginning by that time). He advised me to print out the Transient
waveform of my output and see if it does oscillate or not. It was his
friendly way to get rid of me when busy ;-)

Hope this help !
Kind Regards,
Riad.

 

Ohh Dear, there is a little misspelling mistake in my previous post.
It's definitely not the DC loop to break ! but rather the AC loop. In
fact, one need a DC operating point to run small signal analysis. I do
apologize for this :-[

Riad.

 

Riad KACED wrote, on 10/01/08 12:11:

And if you are running "stb" analysis, there's no need for any loop to get
broken during the AC analysis; the method is a direct one which means that the
loop remains closed all the time, which has _lots_ of advantages.

Back to the original message. This comes about if the components contain any
(say) frequency-dependent capacitance. Often you see this with distributed
components like transmission lines - whilst the impedance will naturally vary
with frequency, if the capacitance varies with frequency, then it has to
approximate it at a certain frequency. You can tell pz which frequency to use
(it's on the ADE analysis form). See this snipped from "spectre -h pz" :


In case there are frequency dependent components, poles and zeros are computed
by approximating those components as equivalent conductances and capacitances
evaluated at 1Hz.

(Note: A frequency dependent component means the capacitance or conductance
equivalent representation of the component is frequency varying. Examples are
transmission lines or bjts with excess phases. A linear capacitor is not a
frequency dependent component.)


17 freq (Hz) Frequency at which components will be evaluated in
setting up the linearized network.

Regards,

Andrew.