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The value of σ in cadence monte carlo simulation

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Full Member level 2
Dec 19, 2006
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I use two simple voltage sources for input offset in my design.
How can I set their σ value in cadence monte carlo simulation? I want to use gaussian function.


Depends on how much of the Gauss distribution you want to cover: with ±1σ you include ≈68% of all possible cases under it (you could also say: 68% confidence level, or trustability), with ±3σ about 99.7% of all cases are covered, see e.g. this Wiki explanation.

I don't know how I can apply it on monte carlo simulation of cadence. I use "tools/monte carlo" icon.


See the C@dence Monte Carlo tutorial below, which shows the setup for several examples. In the statistics { process, mismatch } section, the parameter distribution type (gauss, lnorm, ...) and the standard deviation (std) are stated. std is identical to σ, see the a.m. Wiki explanation.

View attachment Cadence_Monte_Carlo_simulation_tutorial.pdf

Don't be bothered by the low std values shown there e.g. on p. 13; these are example values only; the correct σ values should be given by the foundry. For the variation of your own parameters - like the variation value of a voltage source - you can state an arbitrary σ=std value, which should comply to the expected variation.

The mean value of the parameters is to be given in the parameters section.
Also, don't forget that if one voltage source has standard deviation sigma1, and another voltage source - sigma2, then the difference or the sum of these voltages (assuming they are statistically independent) is sigma=sqrt(sigma1^2+sigma2^2). (it's not clear form the description of the initial post how do these two voltages define the input offset).
Actually - for input offset effect analysis - it would be sufficient to put one single voltage source in series with one of the two inputs of a differential input amplifier. Mean value = 0 , and a (process-dependent) ±1σ value of -say- 10mV .

If you then run an MC analysis with -e.g.- 400 runs over a ±2σ distribution, you cover about 95% of all possible cases - with a max. error of 5% - normal (=Gauss) distribution assumed.
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