hw9_sol - ECE 342 Summer 2011 - HW9 Solutions k Vo = − R2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 342 Summer 2011 - HW9 Solutions k Vo = − R2 VI = − 10k (−0.5) = 5V R1 1 −0.5−0 i1 = 1k = −0.5mA 5 i3 = 0−k = −0.5mA = i1 10 5 i2 = 2k = 2.5mA i4 = i2 − i3 = 2.5 − (−0.5) = 3mA The additional current comes from the opamp's power supply (not shown in the symbol). 1 P2.111 (Eq. 2.53 from the Textbook) (Eq. 2.52 from the Textbook) P2.114 from (2.52) from (2.53) 96 V/V 96 x 8 = 768kHz 24kHz 768/24 = 32 V/V ...
View Full Document

This note was uploaded on 11/06/2011 for the course ECE 342 taught by Professor Nareshshanbhag during the Spring '11 term at University of Illinois, Urbana Champaign.

Ask a homework question - tutors are online