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cadence pz analysis

jasonxian
jasonxian over 11 years ago

for a two stage opa as  Allen  described,with a compensation cap,with the cadence pz analysis,found that  there are 5 poles,5 zeros ,is that right? somebody said that should cancel some pole-zero pair,how should I do that?

Is the pz analysis correct? How others usually do?

 that I have another question,I found that nmos's Kp is 6 times  of pmos' in tsmc 0.18um technology when I design a rail to rail constant-gm opa ,but in books ,it is usually 2-3 times.Is it because of in sub-micron process ,some complicated effect?

Thanks!  

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  • Andrew Beckett
    Andrew Beckett over 11 years ago

    Are you (jasonxian), pyroblast and ianx all on the same course? It's a bit suspicious there have been three questions about "Kp" in the last few weeks (and it's never come up before, so I smell a rat). Kp was a model parameter for old (very basic) MOS models representing the transconductance. There was some mention in other posts of "Kn" - but as far as I know the parameter was always Kp. In modern transistors, it is far more complex and so if you want the transconductance, I'd have thought you'd use "gm" from the operating point output. I don't know how you're determining the "Kp" of these devices - it's not a model parameter (unless you have a very strange 0.18 model deck).

    As for your pz analysis, I have no idea whether there should be 5 poles and zeros. Note that the documentation for pz tells you about pole-zero cancellation. From the output of "spectre -h pz", I see:

    Pole-zero cancellation is performed when a neighboring pole-zero pair is located within `absdiff' distance. The distance
    is also determined relatively as `reldiff' times the magnitude of the pole or zero. Spectre uses the larger value of the
    two distances for cancellation. A subtle note on resistance: by default, a lower bound of resistance is enforced; you
    may remove this limitation by defining the resistor parameter `rac'. This may affect pz results.

    pz analysis should be correct. But I don't know how you're using it...

    Regards,

    Andrew.

     

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