Ch14s - Chapter 14 Problem Solutions 14.1 Ad = vo = -80 vi...

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Chapter 14 Problem Solutions 14.1 80 (max) 4.5 (max) 56.25 mV o d i oi v A v vv == = = So 56.25 (max) 39.77 mV 2 ir m s v 14.2 (a) 2 4.5 0.028125 mA 160 4.5 4.5 mA 1 L i i Output Circuit 4.528 mA = 4.5 0.05625 V 80 o ii v A =− = (b) 4.5 15 mA (min) 300 o o LL L v i RR R ≈= = 14.3 (1) 2 V o v = (2) 2 12.5 mV v = (3) 4 21 0 OL A (4) 1 8 V v μ = (5) 1000 OL A = 14.4 From Eq. (14.4)
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21 2 1 2 3 1 2 1 2 33 1 2 1 2 1 / 1 11 / 15 1 0 15 1 15 1 0 0 15.0075 (0.9925) 15.12 CL oL RR A R AR R R R R R R R R R R = ⎛⎞ ++ ⎜⎟ ⎝⎠ −= × = ×× = = 14.5 10 00 1 12 0 1 0 0 1 and so that 111 I L i L I i vv v vA v R v v A v v v RRR =+ = =− += So 0 0 1 2 1 1 1 I Li v v A R R R ⎡⎤ + + + ⎢⎥ ⎣⎦ Then 0 1 2012 (1 / ) 1 1 1 CL I v R A v RARRR == + From Equation (14.20) for L R =∞ and 0 0 R = 0 2 (1 ) 1 1 L if i A R + =+⋅ a. For 1 k i R 3 3 (1/ 20) 1 100 20 100 1 10 0.05 [0.01 1.06 10 ] CL A = + = or
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3 4.52 11 1 1 0 90.8 0 0 CL if if A R R =− + =+ b. For 10 k i R 3 4 (1/ 20) 1 100 20 100 10 10 0.05 [0.01 1.6 10 ] CL A = ⎡⎤ ⎛⎞ ++ + ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ = or 4.92 CL A 3 1 1 0 98.9 10 100 if if R R + c. For 100 k i R 3 5 111 100 20 100 100 10 0.05 [0.01 7 10 ] CL A = + = or 3 4.965 1 1 0 99.8 100 100 CL if if A R R + 14.6 2 1 2 1 1 1 o CL i OL R R v A v R AR + == For the ideal: 2 1 0.10 15 0 0.002 R R += = ( ) (0.10)(1 0.001) 0.0999 o v actual = So
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0.0999 50 49.95 1 0.002 1( 5 0 ) OL A == + which yields 1000 OL A = 14.7 From Equation (14.18) 2 1 1 1 2 1 111 OL o o vf Lo A RR v A v RRR ⎛⎞ −− ⎜⎟ ⎝⎠ ++ Or 3 3 3 11 51 0 1 0 0 (4.99999 10 ) 11 1 1.11 10 1 100 4.504495 10 o o vv v × −× =⋅ =− × Now 1 1 1 i i K vR v =≡ Then 1 i vv K R v −= which yields 1 1 1 i v v KR = + Now, from Equation (14.20) 3 3 1 151 0 10 10 100 1 10 100 5.0011 10 (0.1) (0.01) 45.15495 1.11 K ⎡⎤ + ⎢⎥ =+ ⎣⎦ × = Then () ( ) 1 45.15495 10 1 452.5495 ii v + We find 3 1 4.504495 10 452.5495 i o v v × Or 1 1 9.9536 o vf i v A v For the second stage, L R =∞
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3 3 21 1 3 11 1 1 1 51 0 1 0 0 4.950485 10 0 0 1 5 1 0 49.61485 1 10 100 1 100 1 (49.61485)(10) 1 497.1485 o oo o vv v K v v KR ⎛⎞ × −− ⎜⎟ ⎝⎠ ′′ =⋅ = × + ⎡⎤ ⎢⎥ ≡+ = + ⎣⎦ == = ++ Then 3 2 1 4.950485 10 9.95776 497.1485 o o v v −× So 2 ( 9.9536)( 9.95776) 99.12 o vf vf i v AA v = 14.8 a. 10 31 2 0 I i vv v RRR R = + (1) 0 1 2 2 3 1 I ii v v v RRRR R RR = + 000 0 1 02 0 Ld L A v RR R = (2) or 0 1 0 2 0 111 L d L A v v v RRR R R =+ 1 3 I di i vR + (3) So substituting numbers: 0 1 10 20 10 40 40 10 20 I v v v 1 (1) or [0.15833] [0.025] [0.03333] I v 4 1 0 (10 ) 11 1 1 0.5 40 40 0.5 d v v v (2)
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or [ ] [ ] () 4 01 3.025 0.025 2 10 d vv v =+ × 1 1 20 0.6667 10 20 I dI v ⎛⎞ =⋅ = ⎜⎟ + ⎝⎠ (3) So [ ] [ ] ( ) 4 1 3.025 0.025 2 10 0.6667 I v v × (2) or [ ] 44 3.025 1.333 10 1.333 10 I v =×−× From (1): 10 0.1579 0.2105 I v Then [ ] 00 34 0 3.025 1.333 10 1.333 10 0.1579 0.2105 2.1078 10 1.0524 10 II I v v + ⎡⎤ ⎣⎦ ×= × or 0 4.993 CL I v A v == To find : if R Use Equation (14.27) 3 1 2 1 0.5 0.5 1 14 0 10 1 1 0.5 0.5 0.5 1 10 40 1 40 40 (40) (1.5125) {(0.125)(1.5125) 0.0003125} 25 I d I d i v v iv v ++ ⎧⎫ + + −− ⎨⎬ ⎩⎭ =− or (1.5125) {0.18875} 25 I Id v Now (20) iI vi Ri and 1 (20) vvi So (1.5125) [ (20)] [0.18875] 25 (20) [505.3] (0.18875) I I i i = or 2677 k I I v i Now 10 2677 2.687 M if if RR To determine 0 : f R Using Equation (14.36) 3 0 2 1 11 1 1 0 40 0.5 1 1 10 20 L f i A R ⎢⎥ = + + or 0 3.5 f R Then 0 1 k 3.5 f R Ω 0 3.49 f R b. Using Equation (14.16) 3 5 (1 0 ) ( 0 . 0 5 )% 10 CL CL CL CL dA dA AA 14.9
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00 0 0 0 Ld I i vA vvv RR −− += and 0 dI vv v =− So 0 0 0 0 44 0 0 () 0 11 1 1( 1 0 )1 ( 1 0 ) 0.2 0.2 100 100 0.2 [5.000501 10 ] [5.000001 10 ] L I I ii LL I I I v v R R AA RRR −⋅ + = ⎡⎤ ++= + ⎢⎥ ⎣⎦ 1 + ×= × So 0 0.9999 CL I v A v == b. Set 0 I v = 0 0 0 and d i L i v v iv v A =+ = + Then 0 0 1
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Ch14s - Chapter 14 Problem Solutions 14.1 Ad = vo = -80 vi...

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