In Razavi's Design of Analog CMOS Integrated Circuits, how does he manipulate equation 8.5 into 8.6 below?
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2 Answers
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I think it's high school math. Take \$\frac{1}{AB}=x\$. Then multiply the num. and denom. of the relation for \$\frac{Y}{X}\$ by x: You get \$\frac{Y}{X}=\frac{1}{B}.(1+x)^{-1}\$. Using binomial approximation and noting that x is much less than the unity, the approximation becomes \$ \frac{Y}{X} \approx\ \frac{1}{B}[1-x]\$ or \$ \frac{Y}{X}\ \approx\ \frac{1}{B}[1-\frac{1}{AB}]\$. Note that this approximation neglects second and higher order terms in the Taylor series of the function \$\frac{Y}{X}\$. \end
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\$\begingroup\$ I think it's high school math My thoughts exactly but I was too lazy to prove it. Indeed terms are neglected and also in these approximations certain assumptions are made, for example that \$A\$ is large. Then we end up with \$\frac1 \beta\$ which is easy, insightful and also "close enough" for practical situations. \$\endgroup\$ Dec 19, 2017 at 8:16
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This is one way to get the approximation given in (8.6).
However, for me I don't like that approach much. Usually assign T as a loop gain and then make approximation depending on magnitude of T compared to 1.
