V o = Gm v i (Ro RL ) Ri = Gm v s (Ro RL ) Ri + Rs Thus, vo Ri = Gm (Ro RL ) vs Ri + Rs Thus, io Rs Ro = Ais is Rs + Ri Ro + RL Ex: 1.19 Ro Rs Ro = Ais is Ro + RL Rs + Ri Ro + RL RL vo vo ii Rs = × = Rm × is ii is RL + Ro Ri + Rs ∴ Range of voltage gain is from 409 V/V to 810 V/V. = 81.8 V/V This figure belongs to Exercise 1.15. Ri = 1 M Ro = 10 Av o = Av 1 ×Av 2 ×Av 3 = 9.9×90.9×1 = 900 V/V The overall voltage gain Ri RL vo = × Av o × vs Ri + Rs RL + Ro = 0.909 × 90.9 × 9.9 × 0.909 744 V/V v L = 744 × 1 mV = 744 mV Ex: 1.17 Using voltage amplifier model, the three-stage amplifier can be represented as Output voltage with load connected = Av o v i 0.8 = V i2 v i1 v i2 = × = 9.9 × 0.909 = 9 V/V vs v i1 vsĮx: 1.13 Open-circuit (no load) output voltage = Av o v i Where PL = 6.25 mW and Pi = v i i1, v i = 0.5 V and ii =Įx: 1.16 Refer the solution to Example 1.3 in the text. Fraction of energy in fundamental = 8/π 2 = 0.81 Fraction of energy in first five harmonics 1 1 8 = 0.93 = 2 1+ + π 9 25 Fraction of energy in first seven harmonics 8 1 1 1 = 2 1+ + + = 0.95 π 9 25 49 Fraction of energy in first nine harmonics 8 1 1 1 1 = 2 1+ + + + = 0.96 π 9 25 49 81 Note that 90% of the energy of the square wave is in the first three harmonics, that is, in the fundamental and the third harmonic.Įx: 1.11 Pdc = 15 × 8 = 120 mW √ (6/ 2)2 = 18 mW PL = 1 Pdissipated = 120 − 18 = 102 mW 18 PL × 100 = 15% × 100 = Pdc 120 10 Ex: 1.12 v o = 1 × 6 10−5 V = 10 µV 10 + 10 η=
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SEDRA AND SMITH MICROELECTRONIC CIRCUITS SOLUTIONS SERIES
R π 9 25 49 It can be shown by direct calculation that the infinite series in the parentheses has a sum that approaches π 2 /8 thus P becomes V 2/R as found from direct calculation.Thus, the values are spaced 0.25 V apart. V A = 0 V ⇒ D = 0000 v A = 1 V ⇒ D = 0000Įx: 1.9 (a) D can represent 15 equally-spaced values between 0 and 3.75 V. V A = 0.25 V ⇒ D = 0000 v A = 3.75 V ⇒ D = 0000 (b) (i) 1 level spacing: 20 × +0.25 = +0.25 VĮx: 1.7 If 6 MHz is allocated for each channel, then 470 MHz to 608 MHz will accommodate 806 − 470 = 23 channels 6 Since the broadcast band starts with channel 14, it will go from channel 14 to channel 36. Microelectronic Circuits, 8th International Edition isc =Įx: 1.3 Using voltage divider: RL Rs + RL v oc = is (t) × Rs When output terminals are short-circuited, as in Fig.
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Solutions to Exercises within the ChapterĮx: 1.1 When output terminals are open-circuited, as in Fig. Vincent Gaudet University of Waterloo New York Oxford OXFORD UNIVERSITY PRESS Tonny Chan Carusone University of Toronto