HW9 Solution - Ω 9 V − 8 k Ω a b v s − At node a 4 v...

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Chapter 5, Problem 8 Chapter 5, Solution 8. (a) If v a and v b are the voltages at the inverting and noninverting terminals of the op amp, v a = v b = 0 1mA = k v 5 0 0 v 0 = -5.00V (b) - + - + 7V - + 3V 10 k Ω 2 k Ω i a v b v a + v o - (a) + v o - i a (b) 10 k Ω + v a - + - 3V Since v a = v b = 7V and i a = 0, no current flows through the 10 k Ω resistor. From Fig. (b), -v a + 3 + v 0 = 0 v 0 = v a - 3 = 7 - 3 = 4.00V
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Chapter 5, Problem 10 Chapter 5, Problem 10 Solution: Since no current enters the op amp, the voltage at the non-inverting input of the op amp is v s . Also, v + = v - = v s . Hence v s = v o 3 20 10 10 o v = + s o v v = 3.00
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Chapter 5, Problem 13. Find v o and i o in the circuit of Fig. 5.52. Chapter 5, Solution 13. Figure 5.52 for Prob. 5.13 i o + 50 k Ω + v o 10 k Ω 100 k Ω 1 V + a b 10 k Ω 90 k Ω i 2 i 1 By voltage division, v a = V 9 . 0 ) 1 ( 100 90 = v b = 3 v v 150 50 o o = But v a = v b 9 . 0 3 v 0 = v o = 2.70V i o = i 1 + i 2 = = + k 150 v k 10 v o o 0.27mA + 0.018mA = 288 μ A
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Chapter 5, Problem 20. In the circuit in Fig. 5.59, calculate v o if v s = 0. Figure 5.59 Chapter 5, Solution 20. + 2 k Ω + v o 4 k Ω 4 k
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Unformatted text preview: Ω 9 V + − 8 k Ω a b v s + − At node a, 4 v v 8 v v 4 v 9 b a o a a − + − = − 18 = 5v a – v o- 2v b (1) At node b, 2 v v 4 v v o b b a − = − v a = 3v b- 2v o (2) But v b = v s = 0; (2) becomes v a = –2v o and (1) becomes -18 = -10v o – v o v o = -18/(11) = -1.64V Chapter 5, Problem 28 Chapter 5, Problem 28 solution: +-10V 50k Ω 10k Ω 20k Ω i + V o-v+ = v- = 10V And V v v k k k v v . 60 ) 10 ( 6 6 50 10 10 = = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = + i = v o /20k = 60/20k = 3.00mA Chapter 5, Problem 37 Chapter 5, Problem 37 solution: Use nodal analysis at the inverting input of the op-amp. V v k v k V k V k V v v But k v v k V v k V v k V v o o 00 . 3 30 30 3 20 2 10 1 , 30 30 ) 3 ( 20 2 10 1 − = = − + + − + − = = = − + − − + − + − − + − − − −...
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HW9 Solution - Ω 9 V − 8 k Ω a b v s − At node a 4 v...

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