HW#5 Solution

As the capacitor charges up exponentially the gain

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Unformatted text preview: T = (R1+R2) × vIN/R1 = vIN(1+R2/R1) [non- inverting amplifier gain]. The left hand side of R3 is held at vIN, hence I3 = (vOUT – vIN)/R3 = vIN (1+R2/R1 – 1)/R3 = vINR2/R1R3. This current cannot flow into the v- terminal of the op- amp, hence it must flow into the vIN battery. 2.43 a. Assume the op- amp operates in the linear region and maintains a virtual short between v+ and v- . It then follows that v- is held at ground potential. If VIN is a step function of magnitude Vs, then i1=(VIN/R1)exp(- t/R1C1), where i1 flows into R1. This same current must also flow thr...
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This note was uploaded on 02/13/2014 for the course EE 2230 taught by Professor Staff during the Spring '08 term at LSU.

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