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Miller's Theorem for a capacitor can be written as follows: enter image description here

In the book 'Design of Analog CMOS Integrated Circuits' by Behzad Razavi, I read the following: enter image description here

How does the author say that the capacitance at node E will be (1+Av2)Cc, shouldn't it be (1-Av2)Cc?

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  • \$\begingroup\$ Yes it is. I voted to close. \$\endgroup\$ Oct 27, 2022 at 21:54

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It depends on whether \$A_{V_{2}}\$ is considered negative or positive. I have seen the assumption either way. The problem here is that the treatment is inconsistent. The Miller theorem expects a negative gain which is often expressed as -A thus "assuming" A is positive. To make that clear it should be written as -|A|.

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How does the author say that the capacitance at node E will be (1+Av2)Cc, shouldn't it be (1-Av2)Cc?

You are correct. I guess what happens here is that Av1 and Av2 are negative numbers (because the figure represents a two-stage op-amp), so 1-Av2 would be the same as 1+|Av2|. Perhaps the author is used to thinking of the gain magnitude here.

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