0132130084_ism14

0132130084_ism14 - CHAPTER 14 Exercises E14.1(a iA = vA RA iB = vB RB iF = iA iB = vA v o = RF iF = RF RA(b For the vA source RinA(c Similarly RinB

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520 CHAPTER 14 Exercises E14.1 (a) A R v i = B = F + = + = + = = o (b) For the source, = = in . (c) Similarly . in = (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 = = in mA 1 1 2 = = V 10 2 2 = =
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521 mA 10 = = L o R v i mA 11 2 = = x (b) mA 5 1 1 = = in mA 5 1 2 = = V 5 2 2 3 = = mA 5 3 3 3 = = mA 10 3 2 4 = + = V 15 2 2 4 4 = = E14.3 Direct application of circuit laws gives 1 1 1 = , 1 2 = , and 2 2 3 = . From the previous three equations, we obtain 1 1 1 2 3 2 = = . Then applying circuit laws gives 3 3 3 = , 4 2 4 = , 4 3 5 + = , and . 5 5 =
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522 These equations yield 2 4 5 3 3 5 v R o = . Then substituting values and using the fact that , 2 1 3 = we find . 2 4 2 1 = E14.4 (a) in 2 2 in i s = + = (Because of the summing-point restraint, . 0 2 = ) 0 1 in 1 = = (Because . in = ) 0 2 1 in = = 0 1 3 = = in 3 3 = + = Thus, 1 in + = = A and . in in in = = (b) (Note: We assume that . 3 2 1 = = ) in 1 in 1 = = in 2 in 2 = = in 2 1 in 2 = + = 2 in = 1 in 1 3 = = in in 1 3 3 3 = = = 1 in = =
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523 E14.5 From the circuit, we can write in v F = , R i = , and o = . From these equations, we find that in = . Then because is independent of L , we conclude that the output impedance of the amplifier is infinite. Also in is infinite because in is zero. E14.6 (a) in 1 = 1 1 1 = 1 1 1 2 2 + = 1 2 2 = 2 1 3 + = 2 3 2 + = Using the above equations we eventually find that 2 1 2 1 2 in 3 1 + + = = A (b) Substituting the values given, we find = 131. (c) Because in = 0, the input resistance is infinite. (d) Because in = is independent of , the output resistance is zero.
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524 E14.7 We have 1 2 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 = × + × = + = 706 . 9 01 . 1 9 . 49 500 . 0 99 . 0 499 max 1 max min 2 min = × + × = + = E14.8 Applying basic circuit principles, we obtain: 1 1 1 1 v i + = 1 2 = = 2 2 B + = f + = o = From these equations, we eventually find 2 2 1 1 1 2 + + = E14.9 Many correct answers exist. A good solution is the circuit of Figure 14.11 in the book with . 19 1 2 We could use standard 1%-tolerance resistors with nominal values of 1 1 = k and 1 . 19 2 = k . E14.10 Many correct answers exist. A good solution is the circuit of Figure 14.18 in the book with 20 1 and ). ( 25 1 2 + We could use
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525 standard 1%-tolerance resistors with nominal values of 20 1 = R k and 515 2 = k . E14.11 Many correct selections of component values can be found that meet the desired specifications. One possibility is the circuit of Figure 14.19 with: = 1 a 453-k fixed resistor in series with a 100-k trimmer (nominal design value is 500 k ) B is the same as 1 499 2 = k 5 . 1 = A M 5 . 1 = f M After constructing the circuit we could adjust the trimmers to achieve the desired gains. E14.12 kHz 40 100 40 10 5 0 0 0 = × = = = CL BOL OL t BCL The corresponding Bode plot is shown in Figure 14.22 in the book. E14.13 (a) kHz 9 . 198 ) 4 ( 2 10 5 2 SR 6 = × = = π π om FP V (b) The input frequency is less than and the current limit of the op amp is not exceeded, so the maximum output amplitude is 4 V.
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This note was uploaded on 11/20/2011 for the course ELE 220 taught by Professor Anas during the Spring '11 term at American Dubai.

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0132130084_ism14 - CHAPTER 14 Exercises E14.1(a iA = vA RA iB = vB RB iF = iA iB = vA v o = RF iF = RF RA(b For the vA source RinA(c Similarly RinB

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