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Stability Bode plot vs. pole/zero analysis in Spectre

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tshiu

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Dear all:

When I simulate a LDO circuit by Spectre, I find a dominat pole at low frequency on Bode plot.
But in pz analysis of Spectre, there's no pole at low frequency.
Why the two results between Bode plot and pz analysis are quite different?
Does the miller cap compensation will effect the pz analysis results?

In my pz analysis:
Output Probe Instance: iprobe (between Vout and Vf)
Input Voltage Source: vdc (with ac=1V and generate the DC reference for err Amp)
Component Eval. Freq (Hz): 1
 

dick_freebird

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I find the stability analysis (stb) tool convenient yet untrustworthy. If there
is anything "funny" it will give you results that don't make sense without
saying why. Always look at the simple gain / phase frequency analysis plot
if you have any doubt at all about reasonableness.
 
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jimito13

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iprobe is used for stb analysis.I would suggest that you don't use it and at the setup form of pz analysis just declare the output(s) of your circuit.
Also check the options tab of the analysis,there is a choise there that let's you define the maximum frequency that the simulator will take into account during the
analysis run.Maybe you have specified a very low freq. there and that's why you can't see the pole you are expecting.

Does the miller cap compensation will effect the pz analysis results?
The pz will analyze your circuit and will illustrate the result of the applied method of compensation.
 
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tshiu

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iprobe is used for stb analysis.I would suggest that you don't use it and at the setup form of pz analysis just declare the output(s) of your circuit.
Also check the options tab of the analysis,there is a choise there that let's you define the maximum frequency that the simulator will take into account during the
analysis run.Maybe you have specified a very low freq. there and that's why you can't see the pole you are expecting.


The pz will analyze your circuit and will illustrate the result of the applied method of compensation.
Spectre can't be just declared the outputs, and it enforce me to declare input voltage or current source.
Then I replace the iprobe with a large inductor in order to break the feedback loop at ac analysis.
But the pz results still can't match Bode plot.
When output voltage is declared, Positive Output Node: Vout, Negative Output Node: ground,
there is no pole and zero in the result.
Or when output probe is set as the large inductor, also the pole and zero doesn't match Bode plot.
In the Bode plot, the dominant pole is at -2e-1 Hz
But at pz analysis, there're 2 conjugate poles at -3e-11 +/- j3e-10 Hz, and 2 zeros at -1e-27 Hz, -2e7 Hz
Is any thing wrong in my pz analysis setting?
 

jimito13

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When you perform the pz analysis you don't have to break any loop.Just declare inputs,outputs run and plot.

Please upload your test schematic and analysis' setup forms to see how can we help you.
 

tshiu

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When you perform the pz analysis you don't have to break any loop.Just declare inputs,outputs run and plot.

Please upload your test schematic and analysis' setup forms to see how can we help you.
The attachments are the different results between Bode's plot and .pz analysis.
Is there any thing wrong in the pz analysis setting?
 

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jimito13

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Where is the input of your circuit?
The input source of pz analysis should be the input source that you make use to calculate the bode diagram of your circuit,thus a sine source or port and not a dc voltage source.
 

tshiu

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Where is the input of your circuit?
The input source of pz analysis should be the input source that you make use to calculate the bode diagram of your circuit,thus a sine source or port and not a dc voltage source.
Input voltage source is set as DC=0.6V, ac=1V.
I use hspiceD to simulate the same ckt, and
I find that the hspcieD pz analysis results can match with Bode's plot create by Spectre.
The situstion is the same as the problem below:
https://www.edaboard.com/thread48269.html
 

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