240_homework_5_soln

240_homework_5_soln - ECE 240 Electrical Engineering...

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ECE 240 – Electrical Engineering Fundamentals Solutions to Homework #5 1. Writing KCL at the inverting input, which has the node voltage, V 1 : V 1 1 K + V 1 V 0 19 K = 0 V 0 = 20 V 1 = 0.2 V 2. Label the nodes at the two op amp inputs as V 3 . Writing KCL at the non-inverting and inverting inputs yield the following: V 3 V 1 R 1 + V 3 R 2 = 0 V 3 V 0 R 4 + V 3 R 3 = 0 A v = V 0 V 1 = 1 + R 4 R 3 1 + R 1 R 2 3. a. Label the nodes at the two op amp inputs as V 1 . Writing KCL at the non-inverting and inverting inputs yield the following: V 1 12 1 K + V 1 3 K = 0 V 1 = 9 I 0 = 9 6 K = 1.5 mA b. Removing the op amp and writing KCL at the node above the 6K: V 1 12 1 K + V 1 3 K + V 1 6 K = 0 V 1 = 8 I 0 = 8 6 K = 1.33 mA 4. The inverting inputs of the two op amps are labeled as V 1 and V 2 . The output of the left op amp is labeled as V 3 . Writing KCL at both inverting inputs: V 1 V 3 R 2 + V 1 R 1 = 0 V 2 V 3 R 3 + V 2 V 0 R 4 = 0 V 0 = 1 + R 4 R 3 V 2 R 4 R 3 1 + R 2 R 1 V 1 5. The inverting input of the op amp is at 0 volts.
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This note was uploaded on 02/27/2010 for the course ECE 240 taught by Professor Drsharlenekatz during the Spring '10 term at CSU Northridge.

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240_homework_5_soln - ECE 240 Electrical Engineering...

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