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Solution Manual For Electronics, 2nd Edition

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    Contents Chapter l....i.ieieeeeeeeenenenanceeanenoaanonassasasssaaanal Chapter 2....ceceeeneeenneneneasoseseseassansasaaannaess38 Chapter 3.....i.ieieieececieteeieenereacessccsnacaccassessssl00 Chapter 4........icececiieiieeneececcccsoncnnncnaseesss.154 Chapter 5...cuiiiienenreeeeanenecassssssanaasseanasassasa203 CHAPLET Bue eeeeeeeeeeeeecaseesneeeannecenanennnennsasa244 CHAPLET 7 reneen vee eennnnnnnnnenaeeeeeeeennnnnnnnnnssa284 Chapter B8......cieieeeeieteeneencecersecsssonenancesssa328 Chapter 9.....ieriiieeeeereeenenneasccecesnenennaassssa382 Chapter 10. .tuuceenueeaneeaneeaneenanrenneeenneeenns. ddl Chapter 1l......ceeeereecneeeaccccnascconncnneonenesess.d467 Chapter 12.........iiiteeneenenecnencssseascanscnnsacassabll

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    Chapter 1 (a) For a noninverting amplifier a, = +50 and we have: Vo (t) = Avi (t) = 50 x 0.1sin(2000mt) = 5sin(20007nt) (b) For an inverting amplifier A, = -50 and we have: vo (t) = A vy (E) = -50 x 0.1sin(2000mt) = -5sin(2000mt) Re Sn +1, Soon i Vs O v. R: R Q0xY 2 “ bn <> 750 R; R. R; 1 L i 4 A = o————— xX A X =——=—» = 300 A. =A =-=10 vs ~ Rg + Ry vo © RF Rj i v Rp, v BL _ Oo _ _ _ _ 6 A, Ty TA xg sg - 375 G =A; = 3.75 x 10 i L o For maximum power transfer to the load, we must have Rp, = R, = 25 Q. Then as in Exercise 1.2 we find A, = 250, A; = 2 x 10%, and G = 5 x 10°. Rs, Ron Ros + + + + YU, R, @ {TAuvee Kis AvosVis Uy 1 + StudyXxy

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    R, R, i2 i3 A =A = A == A = 5357 vo vol Rp + Rio vo2 Roa + Ris 'vo3 R; = R;; = 1000 @ R, = R_, = 300 Q R, R; i2 i1 A = A os A 5 A = 4348 vo vo3 Ros + Ris vo2 Roy + Riq vol Ry = Ry; = 3000 Q R, = R,, = 100 0 P = (1.54) x (15V) = 22.5 W Py =P +P; =P =20.5W Ps m= 5° x 100% = 11.1% Ss Ro= loon o Re {7 R.=3005 4 kn A, = v_/v. = 2000 J 1500 v of’ "i Ry + Rg Aggp = 20loglAl = 63.5 dB R, c=(a)%t=1.5x10" Vv R L Gup = 10logG = 71.8 AB 2 + StudyXY

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    PB, = 10log|-B| = 1010g|5107>| = 23.0 aBw dBW Tw I= : P 5x10” P4Bm = 10log Tow = 1oleg ("257 = 10log(5) = 6.99 dBm v 20log|_%| = 23 » v_=1023/20_ 14.03 v Tv x Ai Ji + - R; P + Open 7; Ra <8 Py Vs circait - or load. v 200i.R [=] 1 0 a = 0C2=-__"10°0 _ R, = 1000 @ R_=20Q vo vi 1;Ry 1 o Az Aose —> —> + R; 4) vi Soo 7, — Son I SR msc Vv. 5001, 1 1 R; = 500 Q R= 500 3 + StudyXxy

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    A; 4 + + 2 & Ho Ro Us 4 id Dec iM 0.05; loa _ v G__ V.R _ _ooc _ "msc'io _ — Rnoc = 71; v./R; CnscRoRy = 500 kA i iT (a) We have R_ << R, and R, >> R_. This is a nearly ideal Ss i L o voltage amplifier. (b) We have R_ >> R. and R, << R_. This is a nearly ideal Ss i L o Y current amplifier. (c) We have R_ << R; and R, << R_. This is a nearly ideal Ss i L 0 24 transconductance amplifier. (d) We have Ry >> R; and Rp >> Rg. This is a nearly ideal transresistance amplifier. (e) We have R_ = R, and R. << R_. This is not close to any Ss 1 L o ideal amplifier. Exercise 1.14 Aun = Voen/Vien = 0-1/1 = 0.1 A.naB = 20logla | = -20 dB 1A] 3 = d _ 50 x 10° _ CMRR 35 = 20logz—— = 20log—pG—5— = 114 dB cm 4 + StudyXY

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    Exercise 1.15 (@) Vig = Vip "Vip, TLV -1 = Viem = 2(Via * Vip) SOV A. + A _ _ _ tM 2 Vo = AVip T AVi; = —3 A /2 + A_/2 _ i! 2/% 1 Aq = Vo/Vig = TTF TA) (6) Vig = Vip "Vip SOV = 1 - Viem = 7(Vip * Vip) 1 VY Vo = AyVip T AVip TAC Ay Aen = Vo/Viem T Bp 7 2; (¢) 1A] A. + A _ aq 1 2 _ 201 _ CMRR = 2010g— = 20log 3G. A 20log (100-101) | = 40 dB cm 1 2 Problem 1.1 Some examples of electronic systems are electronic brakes, printers, cash registers, microwave ovens, CD players, airport landing systems, electronic door locks, and so forth. Problem 1.2 Electronic system blocks include amplifiers, filters, signal sources, wave-shaping circuits, digital logic functions, digital . memories, power supplies, and converters. Problem 1.3 Some electronic systems process information in electronic form and some power (hopefully as little as possible) is consumed. In power electronics, the power delivered to a load is the main concern. 5 + StudyXY

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    Conversion of analog signals to digital form is a two-step process. First, the signal is sampled at periodic points in time. Second, each sample is approximately represented by a codeword. (11 Samples 110 \ {lol Andlog I 100 Jou Signed /1 ojo loo | Ooo _ Of A> ofilo Glog Goo ool ©o jo cool 0©0e00 Jol 0 (io! Digital ins) we Pale Problem 1.5 Provided that it is not too large in amplitude, noise can be completely removed from a digital signal. Noise tends to accumulate in analog signals. Digital circuits tend to be easier than analog circuits to implement with integrated techniques. Thus extremely complex digital systems are feasible while equally complex analog systems are not. Digital systems are more adaptable than analog systems to a variety of uses. 6 + StudyXY

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    Problem 1.6 Number of bits per second = 16 x 44.1 x 10° = 705.6 kbit/s (for monaural) (1.411 Mbits/s are used for stereo.) Number of amplitude zones = 216 = 65,536 -5- (5) _ A = ~€58536 152.6 uv Minimum sampling rate = 2fy = 200 sample/s _ 10 mv _ . . _ 10 _ N = 0.01 mV — 1000 which requires k 10 at least (2 1024) Number of bits per second = 200 x 10 = 2 kbit/s Problem 1.8 See Figure 1.6 in the book. Because we are limited to resistors, the only option is a resistive voltage divider as shown below. ©-0.34 =RIIR — Ry, ! I 2 + R + = Load = “y — R, — Vi We denote the nominal values of the resistors as Ry and R,. The highest load voltage (at most 6 V) occurs when I = 0, when the resistor in parallel with the load has its highest value (which is 1.05R,), and when the resistor in series with the source has its lowest value (which is 0.95R,;). To achieve the desired no-load voltage we need 7 + StudyXY

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    1.05R, | PRE 0.95R, + 1.05R, Solving for Ry/ we have Ry = 0.6786 Ry (1) The smallest load voltage (at least 4 V) occurs with I, = 0.3 and resistance values of 0.95R, and 1.05R,. For these values, the Thévenin voltage is 0.95 R. Vv. = 14 m2 th 1.05R, + 0.95R 1 2 and the load voltage is Vy = 4 = Ven T Renn 0.95 R. 0.95(1.05) R.R 4=14 — 2 _ 0.3 — 12 (2) 1.05R; + 0.95R, 1.05Ry + 0.95R, Using Equation (1) to substitute for R, in Equation (2) and solving we obtain: R, = 11.06 Q 1 Then from Equation (1) we obtain: R, = 7.507 Q Thus we could use the standard nominal values of Ry = 11 Q and Ry = 7.5 Q. System engineers design the block diagrams of systems including specifications for each block. Circuit designers design the circuits for each block. Process engineers design the fabrication processes. Semiconductor physicists research fundamental processes used in electronic devices. 8 + StudyXY

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    Problem 1.11 The components of integrated circuits and their interconnections are manufactured concurrently on a semiconductor wafer by a sequence of photolithographic processing steps. The components of a discrete circuit are manufactured separately and then interconnected, usually on a circuit board. Often overall cost can be reduced by integrating the system onto as few chips as possible because chip cost is nearly independent of complexity (within certain bounds). Problem 1.12 The area consumed by each transistor is (10 pm) x (10 um) = 107% n?. The chip area is (2 cm) x (2 cm) = 4 x 107% n?. Thus the number of transistors that can be placed on the chip is (4 x 107%) 71070) = 4 x 10° transistors/chip Problem 1.13 Gain is a negative number for an inverting amplifier, and the output signal is an inverted version of the input signal. Gain is a positive number for a noninverting amplifier, and the output signal has the same polarity as the input signal at each instant of time. Problem 1.14 "Loading effects" refer to the fact that the input voltage of an amplifier is less than the internal source voltage because of the voltage drop across the internal source impedance. Also the amplifier output voltage is less than the open-circuit voltage gain times the input voltage because of the voltage drop across the output impedance of the amplifier. 9 + StudyXY

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    Problem 1.15 Ry=50RA R.=1a + Vv + R, Ro s= (OO) wv > 4 H Ju L100 jo) m Vamos - ha v R. o L A = =A ——=_ = 50 Vv A vo Rp + R, vs A vo Ry + Ro Ry + Ry R. = 1 _ 6 A; =A, g = 1.25 x 10 L G = AA. = 6.25 x 107 voi Using the unity-gain amplifier we have: R5=/00kn Ro=/00n + “0 Ov wl 50 SV - rome 5 R. Rp —- 1 — = v2 - AS = Vo R, TR. R +& _ 152 Vrms and Py = Vo/Ry, = 45.9 mW i s 'L <) With the source connected directly to the load, we have: 10 + StudyXY

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    Ry =/00kn A ° + ORLY ° sv ’ rms Rp Vo = Vs mw" 2-5mVrms L Ss P= vV?/R = 125 nW o o’"L Thus the output power is much larger if the unity-gain amplifier is used. P. = V2 JR. = 0.333 pW in in’ Tin P =vV%/R =3.135 W [3] o/Ry, h = = 12 G = Po/Pin = 9.376 x 10 Ra + + Vi R o> Av; V. Ro ~ i J] _ ~Y=loov; _ 10ka Ry 4 10 A = 90 =A =F = 100 ——M—— v vo Ry + Ry rR, + 104 Solving we find that R, = 1.11 kQ 11 + StudyXY

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    Problem 1.19 With the switch open we have: R. R. i L Vo=50nv=v, ——1 aA __ > (1) o SR, + 10° vo Rp "+ Rj With the switch closed we have: RL Vo = 100 mV = Vo A pp (2) L o Dividing the respective sides of Equation (1) by those of Equation (2), we have: 50 mv _ Ry 100 mV R. + 10° i Solving we obtain R; = 1 MQ. Problem 1.20 If we cascade two amplifiers A and B the equivalent circuit is: A B —_———— > R + Roa + Rog Alun {7 Asia Rig Avon Vin The open-circuit voltage gain of the cascaded amplifier is: R. iB A = A A Ts vo VOA'voB Roa + Rig Problem 1.21 See the figure shown in the solution for Problem 1.20. When the amplifiers are cascaded in the order A-B, we have: 12 + StudyXY

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    Ry = Ria = 3 kQ Ry = Rop =20Q R. iB 4 A =A ES TF 5 = 4.998 x 10 vo VOA"voB Roa + Rip On the other hand for the B-A cascade we have: Ry = Rip = 1 MQ Ry = Roa = 400 Q R. iA 4 A = A A Tp — = 4.967 x 10 vo VOA"VOB R + Ria Kea [M4 n Ro . + S + + {5 2 2; Vea = AvoaVis °° Res n? 2 Vv R../n = °B _ iB A |e] Via [Aven “RE. +R /n? mx | vos OA iB R..n = a iB [sel [vn ) | vos | AC OA iB 13 + StudyXxy

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    2 2 ala, —0= Ja « |a % Rip = " RoaRip dn 'VoA VvoB m’rR__ +R )2 OA iB Solving for n we have: Rip n= |= OA Problem 1.23 The internal source impedance is: R= open-circuit voltage _ 20 x 1073 = 20 kQ Ss short-circuit current 1076 The desired voltage gain is required to be at least: v Ape =g2 = —X0 _~s00 Ss 20 x 10 If we cascade n stages, connect the source to the input, and connect the load to the output, the voltage gain is given by: n-1 PRU. SE N.S x SL an vs Ry + Rg Ry + Ry Rp + Ry vo Substituting values and reducing we obtain: A, = 0.02381 x (0.9091)™71 x (10) By trial and err we determine that we must have n = 5 to achieve A in excess of 500. vs Problem 1.24 To avoid excessive loading effects at the input, we should choose the first stage such that its input resistance is larger than the source resistance. Therefore we choose type A as the input stage. To avoid excessive loading effects at the output, 14 + StudyXY

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