Explanation Given: The given circuit is Calculation: The circuit can be drawn as The current through inverting and non-inverting terminal of an ideal op-amp is zero. The voltage at the non-inverting terminal is zero because it is connected to ground. v 2 = 0 V Concept of virtual ground states that the voltage at inverting and non-inverting terminals are equal. That is, the voltage at the inverting terminal will be v 1 = v 2 = 0 V According to Ohm’s law, the value of input current will be i 1 = 1 − v 1 1 × 10 3 i 1 = 1 − 0 1 × 10 3 i 1 = 1 × 10 − 3 i 1 = 1 mA Apply Kirchhoff’s current law at the inverting node, i 1 = 0 + i 2 i 2 = i 1 i 2 = 1 × 10 − 3 i 2 = 1 mA Applying Kirchhoff’s voltage law (KVL) in the figure above, − v 1 + i 2 ( 10 × 10 3 ) + v o = 0 − 0 + ( 1 × 10 − 3 ) ( 10 × 10 3 ) + v o = 0 10 + v o = 0 v o = − 10 V Applying ohm’s law, the value of load current i L will be i L = v o 1 × 10 3 i L = − 10 1 × 10 3 i L = − 10 × 10 3 i L = − 10 mA Apply Kirchhoff’s current law, the value of output current i o will be i o = i L − i 2 i o = − 10 × 10 − 3 − 1 × 10 − 3 i o = − 11 × 10 − 3 i

Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition

7th Edition

ISBN: 9780199339136

Author: Adel S. Sedra, Kenneth C. Smith

Publisher: Oxford University Press

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Chapter 2.2, Problem 2.6E

To determine

The values of vo,i1,i2,iL and io .

The values of voltage gain, power gain, current gain.

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Chapter 2.2, Problem 2.5E

Chapter 2.2, Problem D2.7E

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Chapter 2 Solutions

Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition

Ch. 2.1 - Prob. 2.1E Ch. 2.1 - Prob. 2.2E Ch. 2.1 - Prob. 2.3E Ch. 2.2 - Prob. D2.4E Ch. 2.2 - Prob. 2.5E Ch. 2.2 - Prob. 2.6E Ch. 2.2 - Prob. D2.7E Ch. 2.2 - Prob. D2.8E Ch. 2.3 - Prob. 2.9E Ch. 2.3 - Prob. 2.10E

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