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I'm reading "Op Amps for everyone" (http://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf) and I'm currently studying the chapter about the compensation of op.amps. but I'm missing a passage. At page 111 after explaining why a loading capacitance can add another pole which can shift the phase and decrease stability, it is written that if Zo << Zf the closed loop transfer function is easy to calculate because the loading capacitance IS enclosed in the feedbackloop.
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I mean, if the output impedance could be considered a "shortcut" then the output capacitance is excluded from the feedback loop. In fact the next page it is written that the closed loop transfer function is the same as the non inverting op amp one. What I'm missing? Thank you.

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    \$\begingroup\$ if the output impedance could be considered a "shortcut" then the output capacitance is excluded from the feedback loop You mean: If \$Z_o = 0\$ then any (capacitive) loading on the output can be ignored. It is still in the loop but the influence of any output load becomes zero when \$Z_o = 0\$. In the real world \$Z_o\$ is never 0 so there will be a pole. \$\endgroup\$ – Bimpelrekkie Oct 15 '17 at 14:05
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I agree with you that the wording sometimes is a bit misleading.

(1) "...it is written that if Zo << Zf the closed loop transfer function is easy to calculate because the loading capacitance IS enclosed in the feedbackloop."

That means: Because CL is considered as part of the feedback loop - and is, therefore, part of the time constant that contains Zo - the CALCULATION will be easy and CL will have no influence while letting Zo<< Zf.

(2) As far as your last sentence is concerned: The author just states that for Zo << Zf the resulting transfer function is identical to the the function already derived earlier for ideal conditions. That´s all!

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