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ENGR40_Homework_4_Solution_2011_Autumn

# ENGR40_Homework_4_Solution_2011_Autumn - Homework#4...

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Homework #4 Solution October 27, 2011 Point Distribution of Homework #4 (55 points total) Basic rule: 1. All calculation mistakes get 1 point off. 2. Mistakes in setting up necessary equations such as KCL get 1 point off if KCL is applied at the right nodes. 3. Applying KCL at wrong nodes, such as the output of opamp without a condition of the output current, gets no point. 4. If the answer for later parts of successive problems is affected by a wrong answer from the previous part but the approach and the calculation are right, it gets the full point. Point Distribution 1. U&M 4.20 (5 points total) 2. U&M 4.38 (10 points total) (a) 4 points (b) 4 points (c) 2 points 3. Inverting Amplifier (15 points total) (a) 5 points (b) 10 points 4. U&M 4.43 (5 points total) 5. U&M 5.10 (10 points total) (a) 3 points (b) 2 points (c) 3 points (d) 2 points 6. U&M 5.17 (5 points total) 7. U&M 5.22 (5 points total) (a) 2 points (b) 2 points (c) 1 points 1

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Problem 4.20 Determine the linear range of the source v s in the circuit of Fig. P4.20. v o v s 200 Ω 400 Ω 1.2 k Ω V cc = 12 V 2 V + _ + _ Figure P4.20: Circuit for Problem 4.20. Solution: The circuit is a standard summing amplifier. v o = R f R 1 v 1 R f R 2 v 2 = 1 . 2 × 10 3 200 v s 1 . 2 × 10 3 400 × 2 = 6 v s 6 (V) . For v o = V cc = 12 V, 12 = 6 v s 6 , or v s = 3 V . For v o = V cc = 12 V, 12 = 6 v s 6 , or v s = 1 V . Hence, the linear range of v s is 3 V v s 1 V .
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ENGR40_Homework_4_Solution_2011_Autumn - Homework#4...

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