Solution manual for microelectronic circuits 3rd edition by rashid

Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ Chapter 1.1 From Eq (1.1) VLSB = 1.2 From Eq (1.1) VLSB = 1.3 VFS = 16 = 76 mV n 2 From Eq (1.1) VLSB = 1.4 VFS 3.5 = = 136.7 mV 2n VFS 3.5 = = 53.4 µ V 2n 216 From Eq (1.2) vo = (1x 2−1 + x −2 + 1x −3 ) x = (0.5 + 0.125) x = 0.625 x V vo = 3.125V 1.5 From Eq (1.2) vo = (1 x 2−1 + x 2−2 + x 2−3 + x 2−4 ) x 5V vo = (0.5 + 0.25 + 0.125) x 5V = 4.375 V 1.6 vAB = 6.5 + x 10−3 Sin 200 π t VDC = 6.5 V I DC = VDC 6.5 = = 6.5 mA RL 1K Ω vab = x 10−3 Sin 200 π t V ia = vab x 10 −3 Sin 200 π t = = x 10−3 Sin 200 π t mA RL 1KΩ vAB = 6.5 + x 10 −3 Sin 200 π t V Vab = 6.52 + ( x 10 −3 ) ≃ 6.5 V x 10−3 I a = 6.5 + ( ) mA ≃ 6.5 mA 2 © 2017 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ 1.7 Given vDC = 7.5 + 10 x 10 −3 Sin 2000 π t V VDC = 7.5 V I DC = VDC 7.5 = = 7.5 mA RL K Ω vab = 10 x 10−3 Sin 2000 π t ia = 10 x 10−3 Sin 2000 π t = 10 x 10 −3 Sin 2000 π t mA 1K −3 10 x 10 Vab = (7.5) + ( ) ≃ 7.5V 2 iA = 7.5 mA + 10 x 10−3 Sin 2000 π t mA 10 x 10−3 ) mA I a = (7.5) + ( I a ≃ 7.5 mA 1.8 vs = 1.5 + 12 x 10−3 Sin ω t V vo = 7.5 + 2.5Sin ωt V From Eq (1.9) AV = 7.5 =5 1.5 From Eq (1.10) Av = 2.5 = 208.3 12 x 10 −3 1.9 vs = 2.5 + 20 x 10 −3 Sin ω t vo = 7.5 + 4.5 x 10 −3 Sin ω t From Eq (1.9) AV = 7.5 =3 2.5 From Eq (1.10) Av = 1.10 (a) 4.5 = 225 20 x 10−3 From Eq (1.12) vo PL RL 100 K Ω Power gain AP = = = = 312.5 (40 x 10−3 ) Pi vs 50 K Ω Ri â 2017 Cengage Learningđ May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ (b) For RL = Ri = 50 K Ω AP = PL vo = = = 625 Pi vs (40 x 10−3 ) From Eq (1.11) vo io RL AI = = = = 25 is vs 40 mV Ri 1.11 (a) From Eq (1.12) vo AP = vs 1.22 RL = Ri 147 K Ω = 153 (80 x 10−3 ) 100 K Ω (b) RL = Ri = 100 K Ω vo Ai = vs RL Ri = 1.2 = 15 80 x 10−3 1.12 (a) Av = io = Ai = vo(peak) = vi(peak) 6.5 50 × 10 –3 = 1.30 × 103 = 130 or 42.28 dB vo 6.5 sin1000 π t = = 1.3 sin (1000π t) mA RL 5000 io(peak) iL(peak) = 1.3 × 10 –3 × 10 –6 = 1300 or 62.28 dB Ap = Av ⋅ Ai = 130 × 1300 = 169 × 103 or 52.28 dB (10 log Po/Pi) Ri = (b) vi(peak) ii(peak) = 50 × 10 –3 × 10 –6 Pdc = VCC ICC + VEE IEE = 50 kΩ = 15 (15 + 15) mW = 450 mW vo(peak) io(peak) 6.5 × 1.3 × 10 –3 ⋅ = 2 = 4.225 mW PL = Pi = vi(peak) ii(peak) 50 × 10 –3 × × 10 –6 ⋅ = = 25 nW 2 â 2017 Cengage Learningđ May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ η= (c) PL 4.225 × 10 –3 = ≃ 0.938% Pdc + Pi 450 × 10 –3 + 25 × 10−9 Av vi(max) = Vo(max) = VCC vi(max) = 15 = 115.4 mV 130 1.13 Vo = 5.3 V at VI = 21 mV, Vo = 5.8 V at VI = 27 mV ∆Vo = 5.8 – 5.3 = 0.5 V, ∆V I = 27 – 21 = mV (a) Av = ∆Vo 0.5 = 83.3 or 38.4 dB = ∆VI × 10 –3 (b) Adc = Vo 5.5 = = 229 or 47.2 dB VI 24 × 10 –3 (c) Vo –V(min) Av ≤ VF – 24 mV ≤ Vo(max) – Vo Av – 5.5 + 11 – 5.5 ≤ VI – 24 mV ≤ 83.3 83.3 – 42 mV ≤ VI – 24 mV ≤ 66 mV – 18 mV ≤ VI ≤ 90 mV 1.14 (a) (b) io = vo 2V = = 0.2 mA RL 10 kΩ ii = vi mV = = 10 nA Ri 100 kΩ Ai = io 0.2 × 10 –3 = = × 104 or 86 dB ii 10 × 10 –9 Av = vo = = × 103 vi × 10 –3 Ap = Av Ai = × 103 × × 104 = × 107 or 152 dB vi = ii Ri = × 10–3 × 100 = 10–1 V vo = io RL = 100 × 10–3 × 103 = 100 V v 100 Av = o = = 1000 or 60 dB vi 0.1 Ai = io 100 × 10 –3 = = 100 or 40 dB ii 10 –3 Ap = Av Ai = 1000 × 100 = 105 or 50 dB © 2017 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ 1.15 (a) From Eq (1.23) Av = = Avo (1 + Rs Ri ) (1 + Ro RL ) 150 (1 + 200 1800) (1 + 50 4700) = 133.6 From Eq (1.24) Ai = Avo Ri 150 × 1800 = = 56.84 RL + Ro 4700 + 50 From Eq (1.25) Ap = Av Ai = 133.6 × 56.84 = 7593.82 (b) Problem 1.15 Amplifier VS AC 100MV RS 200 RI 1.8K E1 150 R0 50 RL 4.7K TF V(4) VS End 1.16 For maximum power transfer Ro = RL = 50 Ω PL = vo io = (Av vi) (Ai ii) = Avo ⋅ vi A R ⋅ vo i ii (1 + RS Ri ) (1 + Ro RL ) RL + Ro = Avo Ri RL vi Avo ⋅ vi ⋅ ( Ri + Rs )( RL + Ro ) ( RL + Ro ) = = = ∴ PL(max) = Avo Ri RL ⋅ vi2 ( RL + Ro ) ( Ri + Rs ) Avo Ri RL ⋅ ( RL + Ro ) ( Ri + Rs ) Ri vs ( Ri + Ro ) 2 Avo RL Ri3 vS2 ( RL + Ro ) ( Ri + Rs )3 1502 × 50 × 18003 (100 × 10 –3 )2 1002 (1800 + 200)3 = 820 mW © 2017 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ 1.17 Ro ∆ vo = vo RL + R o Ro ∆ vo = 0.15 = vo 1.5 k + R o Ro = 1.5 × 103 × 0.15 = 264.7 Ω 0.85 1.18 vo = Avo vi (a) = R i ⋅ Vs RL RL = Avo ⋅ RL + R o RL + R o Ri + Rs 200 × 22 × 105 × 50 × 10 –3 ( 22 + 20) (105 + 1500) = 5.16 V (b) From Problem 1.16 PL = = Avo RL Ri3 Vs ( RL + R o ) ( R i + R S )3 2002 × 22 × 1015 × (50 × 10 –3 )2 (22 + 20) ⋅ (105 + 1500)3 = 1.19 W (c) Av = vo 5.16 = = 103.2 vs 50 × 10 –3 (d) Ai = io 5.16 , io = = 122.86 mA 22 + 20 is is = Ai = vs 50 × 10 –3 = = 4.93 × 10 –7 A Rs + Ri 1500 + 105 122.86 × 10 –3 4.93 × 10 –7 = 249 × 103 (e) Ap = Av Ai = 103.2 × 249 × 103 = 25.7 ì 106 1.19 is àA = vs 10 ×10 –3 = Rs + Ri 2.5k + Ri Ri > 7500 Ω From Eq (1.27) Ro ∆ vo = , RL ranging from kΩ to 10 kΩ vo RL + R o © 2017 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ For ∆ vo ≤ 0.5% vo Ro = , Ro ≤ 10 Ω 1000 RL + R o Av = 5V = 500 or 53.98 dB 10 mV From Eq (1.23) 500 = ∴ (1 + Rs A vo R i ) (1 + Ro RL ) = A vo (1 + 2.5 k 7.5 k ) (1 + 10 2000) Avo = 669.8 1.20 From Eq (1.23) Av = Avo (1 + Rs Ri ) (1 + R o RL ) Variation in Av will be contributed by Avo, RS, and RL Assume equal contribution by each Hence the value of Ro that will keep the variation in gain within 0.5% for variation in RL from kΩ to 20 kΩ can be found from 5k 20 k = × 0.995 5k + Ro 20 k + Ro Ro = 20 k (1 – 0.995) > 33.5 Ω × 0.995 – 1.21 (a) By Kirchoff’s current law at node A Rs R A is if + Vi ii Ro Ri – Avo Vi Rx is = ii + if = vi vi – Avo vi + Ri R + Ro  1 – Avo  = vi  +   Ri R + Ro  Rx = vi = is Ri + (1 − Avo ) ( R + Ro ) © 2017 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ (b) For Ri = 50 k, Ro = 75 Ω, Avo = 2, R = 10 k Rx = 50 k + (1 − 2) (10 k + 75) = – 12.62 kΩ Vs 20 mV = – 1.8 µA is = = Rs + Rx 1.5 kΩ – 12.62 kΩ vs is 20 mV = = – kΩ – 2.5 µA Rx = – k – 1.5 k = – 9.5 kΩ (c) For is = 2.5 µA, Rs + Rx = ∴ 1.22 Ais = 200, Ri = 150 Ω, Ro = 2.5 kΩ RL = 100 Ω, is = mA, Rs = 47 kΩ (a) From Eq (1.30) Ai = Ais 200 = (1 + Ri Rs ) (1 + RL Ro ) (1 + 150 47 k ) (1 + 100 2500) = 191.7 Av = Ai RL 100 = 191.7 × = 0.4078 Rs 47 k Ap = Av Ai = 191.7 × 0.407 = 78.17 (b) Problem 1.22a Amplifier VS AC 1MV RS 47K RI 150 F1 VX 200 VX 0V VY DC 0V R0 2.5K RL 100 TF V(s) VS END © 2017 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ Problem 1.22b Amplifier IS AC RS 47K RI 150 F1 VX VX DC VY DC R0 2.5K RL 100 TF I(VY) IS END 1MA 200 0V 0V 1.23 From Eqs (1.28) and (1.29) io = = A is R o R o + RL ⋅ Rs ⋅ is Rs + Ri 10 × 22 × 103 22 × 103 + 150 × 100 × 103 105 + 50 × 50 × 10–3 = 4.96 A 1.24 ii Rs is + Vi AIisi Ri Ro – (a) Output resistance = = open-circuit voltage short-circuit current 12 100 × 10 –3 = 120 Ω Ais ii = short circuit current = 100 × 10–3 A ii = is Ais = Av = Rs × 10 –6 × 100 k = ≃ × 10–6 A Ri + Rs 50 + 100 k 100 × 10 –3 × 10 –6 = 20 × 103 vo Ro R = Ais iL ⋅ L vs Ro + RL is Rs â 2017 Cengage Learningđ May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ = 100 × 10–3 × 120 2700 × 120 + 2700 × 10 –6 × 105 = 22.98 (b) Ai = 20 × 103 = 850.6 (1 + 50 100 k ) (1 + 2.7 k 120) (c) Ap = Av Ai = 22.98 × 850.6 = 19,547 1.25 Following Example 1.4 we have 0.99 Ro Ro = Ro + 20 Ro + 500 Ro (1 – 0.99) = 500 × 0.99 – 20 Ro = 47.5 kΩ For Ri 0.99 × 100 k 10 k = 100 k × Ri 10 k + Ri Ri = 111.23 Ω Ais = 20 × 10 –3 100 × 10 –6 = 200 A/A 1.26 Assume Ai varies equally due to contribution from Ais Rs, and Ro Ro = 0.995, Ro + RL Ro = Ro = 0.995 Ro + 100 100 × 0.995 > 19.9 kΩ – 0.995 Similarly Rs 100 k = 0.995, = 0.995 Rs + Ri 100 k + Ri Ri ≤ 503 Ω ∴ Ai = Ais (1 + Ri Rs ) ( RL Ro ) 50 = Ais ≃ Ais (1 + 503 100 k ) (1 + 100 19.9 k ) Ais ≃ 50 10 © 2017 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ 1.31 Rs Assume 1% for Ri and 1% for Ro io + + vi – – Vs Gm vi Ri Ro RL Ri > 99, Ri (1 – 0.99) > 0.99 Rs Ri + Rs For Rs = kΩ, Ri > 0.99 × k = 99 kΩ 0.01 Similarly Ro > 0.99, Ro + RL Ro > 0.99 × 200 0.01 Ro > 19.8 kΩ 1.32 For Rm = 20 Ω to 100 Ω 0.99 = Ro + 20 Ro + 100 Ro = 7.9 k Ω Io Gm Vi Ro Rm For Rs varying from kΩ to kΩ 0.99 = , Ri (1 – 0.99) = k × 0.99 – k Ri + 2k Ri + 5k For Ri = 295 kΩ vs = 10 V, Io = 100 mA Gm = Io 100 mA = = 10 mA/V vs 10 V Transconductance amplifier 14 â 2017 Cengage Learningđ May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronic Circuits 3rd Edition by Rashi Full file at https://TestbankDirect.eu/ 1.33 Using Eq (1.41) Z mo Zm = = (1 + Ro RL ) (1 + Ri Rs ) 0.5 k (1 + 4.7 k 4.7 k ) (1 + 1.5 k 10 k ) = 217.39 V/A From Eq (1.42) Z mo RL 0.5 × 103 × 4.7 × 103 = ( Rs + Ri ) ( RL + Ro ) 10 × 103 + 1.5 × 103 4.7 × 103 + 4.7 × 103 Av = ( = 21.739 × )( ) 10–3 Rs 500 104 × 50 × 10−3 ⋅ is = = mA Rs + Ri 10 + 1.5 × 10 11.5 × 103 ii = From Eq (1.39) vo = Zmo ii RL 500 4.7 × 103 = 0.5 k × × RL + R0 11.5 × 103 4.7 × 103 + 4.7 × 103 ( ) = 10.87 V 10.87 = 2.31 × 10–3 A io = 4.7 × 103 io 2.3 × 10 –3 = = 0.046 is 50 × 10 –3 Ai = 1.34 Since the output variation should be kept within ±2%, variation of effective transimpedance Zm should be kept to ±2% According to Eq (1.41) the variation in Zm will be contributed by Zmo and Ro Assume equal contribution to the variation Let Ri = 10 Ω Then Zmo = 10 V 20 cm × = 1000 V/A ± 1% cm 100 mA The value of Ro that will keep gain variations within 1% for variation of RL from kΩ to 10 kΩ is 0.99 × 10 k 2k = 10 k + Ro k + Ro Ro ≅ 25 Ω Design specifications are Zmo = 1000 V/A, Ro ≤ 25 Ω, and Ri ≤ 10 Ω 1.35 Let Ri = 10 k