HW#5 Solution

This same current must be pulled down through r2 and

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Unformatted text preview: 9 KΩ) = 21.4 KΩ. Lowest possible gain = (198 KΩ +10.1 KΩ)/ (10.1 KΩ) = 20.6 KΩ 2.20 a. Assume the op- amp operates in its linear region, so that v+=v- . By KVL, the voltage drop across R1 equals vIN and the current, i1 through R1 equals vIN/R1. This same current must be pulled down through R2 and the VR source because i- =0. The output, equal to VR plus the drops across R1 and R2, becomes vOUT = i1R1 + VR + i1R2 = (vIN/R1)(R1+R2) +VR = vIN(1+R2/R1) +VR = 6vIN +VR, where R2/R1 = 5. b. In this second case, where again v+ = v- , vIN appears across both VR and R1, so that i1 through R1 becomes, via KVL and Ohm’s law, i1 = (vIN – V...
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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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