tools My courses My books My folder Career Life -K (a) M -K V -W= RE K+1 777 177 (b) In Figure (b), the inverting voltag

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tools My courses My books My folder Career Life -K (a) M -K V -W= RE K+1 777 177 (b) In Figure (b), the inverting voltag

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Tools My Courses My Books My Folder Career Life K A M K V W Re K 1 777 177 B In Figure B The Inverting Voltag 1
Tools My Courses My Books My Folder Career Life K A M K V W Re K 1 777 177 B In Figure B The Inverting Voltag 1 (79.05 KiB) Viewed 19 times
explain the calculations
tools My courses My books My folder Career Life -K (a) M -K V -W= RE K+1 777 177 (b) In Figure (b), the inverting voltage amplifier has a gain of K with input quantities of vand i The output voltage is Vo The equivalent input resistance can be found as follows. First, by definition of op-amp open- loop voltage gain, -K, V. =-K. At the input node, apply KCL (Kirchhoff's current law). V-V-1=0 R; Substituting for voi vi-(-K-v)_(1+K)- 1 - 1 R R Rearrange the above equation for input resistance, vli, and we have Miller's Theorem: Rg 1-K For an amplifier with output resistance, the equivalent shunt contribution from RAS K Four V. (-K)-v; - -V:-(1+K)/R () R=KTA Rock From the output, Rappears to be slightly less than its actual value for large K. From the input, Rappears to be 1/(1+K) times its actual value, causing input resistance to be much reduced and providing a low- resistance path for i. For infinite K. the input node is a virtual ground, as it is for the ideal inverting op-amp. The equivalent circuit resulting from Miller's theorem is shown above in (b). The circuit has been converted by the Miller transform to an equivalent circuit that does not have an input-output bridging resistance.
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