By Helmut E. Graeb
What you’ll locate here's a interesting compendium of primary challenge formulations of analog layout centering and sizing. This crucial paintings presents a differentiated wisdom concerning the projects of analog layout centering and sizing. particularly, worst-case situations are formulated and analyzed. This paintings is correct on the crossing aspect among procedure and layout expertise, and is either reference paintings and textbook for figuring out CAD equipment in analog sizing.
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Additional info for Analog Design Centering and Sizing
This leads to the formulation of a deterministic optimization method for the statistically analyzed yield, which could be denoted as deterministic statisticalyield optimization approach. 1, which leads to a deterministic geometric-yield optimization approach. 9%, which are described by terms like six-sigma design. 7. 9. Analog sizing is a systematic, iterative process that applies mathematical methods of numerical optimization. The solution methods can be characterized in different ways, which are summarized in Table 3.
Xr,k . . xr,nxr ]T ∈ Rnxr are parameters that are subject to a range of values they can acquire. Typical examples of range parameters are so-called operating parameters that model operating conditions, as for instance supply voltage or temperature. Range parameters differ from statistical parameters in that they do not have probability data in form of statistical distributions on top of their given ranges of values. For the temperature for instance, a range of -40◦ C to 125◦ C can be specified.
Example 3. This example is of the same type as Example 2. Given are a random variable y originating from a standard normal distribution (47) and a quadratic function (48) that maps y onto z: 1 2 1 √ e− 2 y 2π pdfy (y) = (47) z = y2 (48) We are looking for the probability density function pdfz (z). t. R → R+ . In this case we can add the probabilities of all infinitesimal intervals dy that contribute to the probability √ √ of an infinitesimal interval dz. Two values, y (1) = + z and y (2) = − z, lead to the same value z and hence two infinitesimal intervals in y contribute to the probability of dz: pdfz (z) · dz = pdfy y (1) · dy (1) + pdfy y (2) · dy (2) (49) This leads us to: pdfz (z) = pdfy y (1) (z) ∂y (1) + pdfy y (2) (z) ∂z = 1 1 √ e− 2 z · 2π = 1 1 1 √ · z − 2 · e− 2 z 2π ∂y (2) ∂z 1 1 1 −1 z 2 + − z− 2 2 2 (50) (50) denotes the probability density function of a χ2 (chi-square)-distribution with one degree of freedom.