Chapter 14

Chapter 14 - CHAPTER 14 Solutions for Exercises E14.1 (a)...

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CHAPTER 14 Solutions for Exercises E14.1 (a) A A A R v i = B B B R v i = B B A A B A F R v R v i i i + = + = + - = - = B B A A F F F o R v R v R i R v (b) For the v A source, A A A A R i v R = = in . (c) Similarly . in B B R R = (d) In part (a) we found that the output voltage is independent of the load resistance. Therefore, the output resistance is zero. E14.2 (a) mA 1 1 1 = = R v i in mA 1 1 2 = = i i V 10 2 2 - = - = i R v o 1
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mA 10 - = = L o o R v i mA 11 2 - = - = i i i o x (b) mA 5 1 1 = = R v i in mA 5 1 2 = = i i V 5 2 2 3 = = i R v mA 5 3 3 3 = = R v i mA 10 3 2 4 = + = i i i V 15 2 2 4 4 - = - - = i R i R v o E14.3 Direct application of circuit laws gives 1 1 1 R v i = , 1 2 i i = , and 2 2 3 i R v - = . From the previous three equations, we obtain 1 1 1 2 3 2v v R R v - = - = . Then applying circuit laws gives 3 3 3 R v i = , 4 2 4 R v i = , 4 3 5 i i i + = , and . 5 5 i R v o - = These equations yield 2 4 5 3 3 5 v R R v R R v o - - = . Then substituting values and using the fact that , 2 1 3 v v - = we find . 2 4 2 1 v v v o - = 2
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E14.4 (a) in 2 2 in v i R v v s = + = (Because of the summing-point restraint, . 0 2 = i ) 0 1 in 1 = - = R v v i s (Because . in v v s = ) 0 2 1 in = - = i i i 0 1 3 = = i i in 3 3 v v i R v s o = + = Thus, 1 in + = = v v A o v and . in in in = = i v R (b) (Note: We assume that . 3 2 1 R R R = = ) R v R v i in 1 in 1 = = R v R v i in 2 in 2 = = R v i i i in 2 1 in 2 = + = 2 in R R = 1 in 1 3 R v i i = = in in 1 3 3 3 v v R R i R v o - = - = - = 1 in - = = v v A o v 3
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E14.5 From the circuit, we can write in v v F = , F F F R v i = , and F o i i = . From these equations, we find that F o R v i in = . Then because i o is independent of R L , we conclude that the output impedance of the amplifier is infinite. Also R in is infinite because i in is zero. E14.6 (a) in 1 v v = 1 1 1 R v i = 1 1 1 2 2 i R i R v + = 1 2 2 R v i = 2 1 3 i i i + = 2 3 2 v i R v o + = Using the above equations we eventually find that 2 1 2 1 2 in 3 1 + + = = R R R R v v A o v (b) Substituting the values given, we find A v = 131. (c) Because i in = 0, the input resistance is infinite. (d) Because in v o v A v = is independent of R L , the output resistance is zero. 4
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We have 1 2 R R R A s vs + - = from which we conclude that 20 . 10 99 . 0 9 . 49 0 01 . 1 499 min 1 min max 2 max - = × + × - = + - = R R R A s vs 706 . 9 01 . 1 9 . 49 500 . 0 99 . 0 499 max 1 max min 2 min - = × + × - = + - = R R R A s vs E14.8 Applying basic circuit principles, we obtain: 1 1 1 1 s R R v i + = 1 2 i R v A - = A A A R v i = 2 2 s B B R R v i + = B A f i i i + = f f o i R v - = From these equations, we eventually find 2 2 1 1 1 2 v R R R v R R R R R v B s f A f s o + - + = E14.9 Many correct answers exist. A good solution is the circuit of Figure 14.11 in the book with . 19 1 2 R R 2245 We could use standard 1%-tolerance resistors with nominal values of 1 1 = R k and 1 . 19 2 = R k . 5
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Chapter 14 - CHAPTER 14 Solutions for Exercises E14.1 (a)...

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