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# CQ5_pody - ECE 201 Class Quiz 5 Names R2 i2 v2 i1 R1 iN iP...

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ECE 201 Class Quiz 5 Names _______________________ _______________________ _______________________ Ideal operational amplifier model (0 V, 0A): i N = 0; i P = 0; v P – v N = 0. Take v S = 1 V, R 1 = 1 k ! , R 2 = 10 k ! , and an ideal operational amplfier. Take the load to be a 1 k ! resistor and determine the following. v P = v N = v 1 = i 1 = i 2 = v 2 = v L = i L = i O = v S + R 1 i 1 - + + v L R 2 i 2 i O Load i L i N i P ! ! + v 1 - + v 2 -

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ECE 201 Spring 2008 1 An ideal op amp requires v P - v N = 0, or v P = v N . So v N = 0. Determine v N v N By inspection, v P = 0. Remember that v P is the node voltage at the positive input terminal of the op amp expressed relative to ground. In this circuit, the positive input terminal is connected directly to ground. So v P = 0. v P Determine v P Given: R S =1 k ! R F = 10 k ! V S = 1 V ideal op amp Load = 1 k ! resistor v S + R 1 i 1 - + + v L R 2 i 2 i O Load i L i N i P ! ! + v 1 - + v 2 -
ECE 201 Spring 2008 2 Can use fundamental property of node voltages: v 1 = v S - v N = v S - 0 = 1 V. Have already determined that v P = 0 and v N = 0. Next, v 1 = ? Given: R S =1 k ! R F = 10 k ! V S = 1 V ideal op amp Load = 1 k ! resistor v S + R 1 i 1 - + + v L R 2 i 2 i O Load i L i N i P ! ! + v 1 - + v 2 - Or, one could write KVL around the loop, taking into account that the voltage drop across the op amp input terminals is zero. -v S + v 1 - 0 = 0, so v 1 = v S = 1V. Next, determine the value of i

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