Unformatted text preview: R)/R1. The output becomes vOUT = i1R1 +VR + i1R2 = (vIN – vR)(R1+R2)/R1 + VR = vIN(1+R2/R1)+VR(R2/R1) = 6vIN + 5VR. 2.24 The input current source is connected directly to the v+ terminal. Assume linear op amp operation, so that v+ = v . The left hand side of RF thus rests at ground potential. Meanwhile, iIN cannot flow into the v+ terminal, but flows instead through RF, yielding a drop iINRF that is positive from left to right. Applying KVL to the v+ terminal yields vOUT =  RFiIN. The value of Rin is defined by the ratio vIN/iIN, where in this case vIN = v+ v ≈ 0; hence Rin ≈ 0. 2.26 R2 and R1 provide negative feedback, holding v to the same value as v+ = vIN. Hence vOU...
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 Spring '08
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 Amplifier, 1%, Vin

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