Chapter 1-Microelectronic Circuits 5th Ed sedra-Electronics-Solution Manual, Exercises of Electronics

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CHAPTER 1—PROBLEM SOLUTIONS 1.1 fa) f 10.mA ¥ b Rave =10 ka ao t 1ma icy V=IR= 1) mA x 10 kit = 100 ¥ Me TOM oy, eta ett A ® Rr i008 Nowe: Volts, milliamps, and kilo-ohms constitute consistent set of units, 1.2 (a) P= PR = (30x 107)? x Px 10'=0.9 W Thus, ® should have a 1-W rating (b) Pe PR=(d0x 10% 1x 1 = 16 Ww Thus, the resistor should have a 2-W rating, «10% 10'= 0.09 W 1 AW rating (c) P= PR=B XA , the resistor should hay (4107) 10% 10? = 0.16 W jor should have a |-W rating R= 20741 * 10") = 0.4 W the resistor should he W ra ng HPAL « 10") = 0.421 W  suffice ni Thus, a rating of |W should ¢ though | W would be py » allow for tolerances and measurement errors 10mA x1 kQ=10¥ 1.3 (a) VER ; k= 10 Po?R=(10mAy ymW (b) R= W1= 10VA mA = 10k P=Vis10V x1 mA=10mW (c) J P/VETWADV S014 R= W1=10VA0LA= 100 2 (d) V= PT = 0.1 W410 mA = 100 mW/10 mA = 10 V R= Wi=10WI0mA=1kQ () P= PRT z De .f000 mW/T kK VaiR=31.6mAx 1 kQ Note? ¥ of units 31.6 mA 31.6 mA, KQ, and mW constitute a consiste 14 a oS We ot go to i ot yy b Liaw le resistance values. 1.5 Shunting the 10 k9 by a resistor of value R result inw mbination having 4 resistance OR R ee mR +O action, Fe 2 R= 990KR R+ % reduction, 0.95 = R= 190k 10 0% reduction, R R+10 reduction, R R+ lo ting the 10 KQ by O90 => R= 90 ke For a 509 050 = R= 10 ko Stu (a) 1 MEQ result in 10x 1000 10° 1000410 LOT 7 9.9 KO, a 1% reduction; (b) 100 KS results in 10x 10) _ 10 1o+10 1 = 9,09 ki, 09,196 reduction (e) 10kQ resultsin =—!0_ = 5 69, 0.50% rediiction 10+ 10 R R+R, and look back inte To find Ry, we short ire node X, 68k  LI Connect esistor R in parallel with R,. To make ), = 0.21 (and thus the current through &, 0.8/), R should be such 027% 1kOQ = O.8/R > R=2502 = 024] O8ry be =D kee R (b) Connect a resistor in series with the load resist R lance of the load branch by 105%, the current division ratio to its added series resistance must be 10% To make the current throu; by a resistance Ry of vi it will be 249; thus gh R equal to 14 we shunt & such that the current through specified maximum of | V, we have to shunt R, with a resistor R whose v 1 ation of Ry and & the problem can be salved in two ways (a) Connect w resistor Ry across R, of value such that i¢ is such that the paral- Ry Ry = R/2, thus S102, Thus, RR, RR, RSLUIKQ AKO uit, value 1.1 kQ utilizing only one additional acurrent divider across  } loki 7 AAA ° 1OKen 10k LO KER i i 1 ' ) okN 4 4—o AW © gion 10 kf ok inequivatent: (lO 16)" 615KR 4  G15 kft Whe — L "T Now, when a resistance of 1.5 kQ is com between 4 and ground ~ Ls kn 0.97 G15 +15 OlmA 1.16 (a) Node equation at the common node yields hah +l Using the fact that the sum of the volta f, and Ry equals 15 V, we write ISS AR, +R 1h 4+ YD 121, +2 H10v 10.ke yy} Sk 2kt 12h, +21,=15 l) the voltage drops across Ry and Ry add up to 10-¥, thus: LO= ER + 1B ty (4 +4) x2 which yields ! 2h + T= 10 (2) Equations (1) and (2) can be solved t plying (2) by 6, 12h, + 425 = 60 a) Now, subtracting (1) from (3) yields = L125mA gives 2h =10- 7% 1.125 mA => f= 1.0625 mA h=hth 525 + 1.1250 \ 11875 x2= 2.3750 ¥ h=L06mA b= 113mA heLigmA ¥=238¥ (b) A node equation at the common node can be written 4 Vas 15. ‘ Vv R RORY Thus, in be easily found as 375 1 = 1.0625 mA = 1.06 nA = 1.125 mA = 1.13 mA = 1.1875 mA = 1.19 mA preferred; faster, more insightful Ne altempts to st possible number of variables and write ing minimum number of equations, and less prone to ¢ identify the h vfs. In general 4.9800 me a Yy=1 bo SHA x IKR= 5.64 mV iF fi 2-2-2722  orton current J, can be found as 3 ja hm SES 135 ua 28.6 ke s-| m¥/PC which is the sensor, presum voltage ved output voltage specified b 1.24 Th one half Thus, R in y-cireuit conditions that is without a a= 10 nected, 1t follows that that sensor intemal mice 100A ¢ must be equal 1o Ry, it.» 10 KO, i 1.25 ov * 4 0.1 MM = 100 ka — 100 fA 4 ~201¥ \ 4 12,2 10 WA R OAV 20.01 MQ= 10 kQ ' Rs s 10 HA 1.23 p-circuit 4 (ip = 0) voltage k — i, 8 Short-cireult (2, = O) eusrent Thus, 4.36 R, i ‘ ee and Ry, sepresents the input resist =09%, Substituting in (2) 7 1.27 Case a? (rads) F (Hz) T (s) a 6.28 x 10° 110" 1x10" b 1x1? 1.59% 10" 6.28 10°" © 6.23.x10!° tio’ 1x10 dar xie 6 1.67 x 107 © 6.28 «104 tx1o 1x10" f 6.a8-x10° bx 10° Ixio* 1.28 (a) View = 117 x V2 = 165°V (b) =33.9/,/2= 24 CC) Vyen. = 220% 2 311-V (0) Vpeay = 2204/2 = 311 KW 1.29 (a) v= 10sin (2x 10"), V (b) 0 = 120,72 sin (Linx 60), ¥ (c) v= 0,1 sin (10008), ¥ (d) #0. sin Qrx 10H, V 1.40 Compuring the given waveform to that described by Eg, 1.2.we observe that the given waveform has an amplitude of 0.5 VW (LV peak-to-peak) and its lev shifted up by 0.5 V (the first term in the equ ‘Thus the eform look as follows. a a - Average valuc = 0.5 V Peak-to-peak value = 1 V Lowest value= V Highest value = 1 V 2 3 Period T= 1 = 27 = 19% Jo #4 1.31 The two harmonies have the ratio 126/98 = 9/7. ‘Thus, these are the 7th and 9th harmonies. From Eq, 1.2. we note that the-amplitudes of these two tonics will have the ratio 7 10 9, which is confirmed by the measurement reported. Thus the fundamental will have a frequency of 98/7 or 14 kHz and peak Amplitude of 63 x7 =441 mV. The rms valu: fundamemal will be 441/./2 = 312 mV, To fi peak-to-peak amplitude of the square-wave we note that 4V/ m= 441 Peak-to-peak am six = 693 mv — = 74 is 10 ly audible by harmonic must be limited 10 20 kHz; will be 4 kHz. At the low end tthe fifth and the fifth must be ao lower than 20 Hz. Correspondingly will be at 4 Hz. a relatively young 1.33 If the a tude ¢ ¢ Re wave to c R will be V7 that delivered by a sine wave of peak amplitude Y then R. If this power is to equal Via 2 (RZaf Thus, is pendent of fre: quency 134 Decimal 0 5 8 g TELOOL 1.38 b, 6, 6 by Value Represented 6000 +0 ool 41 o1d #2 ool +3 0100 +4 9101 +35 -0 loot 1010 o1t 100 2  Nove that there are two possible 0000 and 1000. Fora Vv the range 43,5-V can be represe Inpur Steps Code “5 ‘ o10 1 vidi 6 To 1.36 (a) For N bits there will be from 0 10 Veg. Thus there will be (2 from 0 to Vyp with the step size given by Step size 1 This is the sponding hang the LSB, It is the v resolution of the ADC ¢b) The maximum error when the analog signal value is at the middle of Step w step, Thus the maxi- T num error is This is known as the quantizatio 10.¥ tc) <5 mV at 2" 122000 2" 2 2001 ForN=11, Résolutic Quantization error 1.37 When-d;=1, the ith switch i 2R) flows to the nys Cornesp current (¥, the sum of all the curn (b) by is the LSB 6 is the MSB = 1.96875 mA to the LSB changing from 0 to 1 the Correspondin es by 10/5 172° = 0.03125 mA, output char 2 will be 44,100 samples per secon! with : sresentod by 16 bits, Thas the through: speed will be 44,100 % 16 = 7.056 % 10° bits per my 10¥_ v, 1oOm¥ 40 dB 100 WAV JR, OV ALO ON A ae 4 100 ju 100 jc LOO AYA 60d 20 log 1000 i A, = 100 x 1000 at 10 Log. 10° M_2V > 2% 10° Viv vy 10pV or, 20 log 2x 10" = 106 dB 2¥/10 KO 100 nA Z _ 0.2.x 10 10 nA 100 10 2000 ATA | i u x 10% x 2000 x10! WAY of IO log A, = 86:dB % WV Linvv V x, 1 or, 20 og 10 i,_ t/R, _ 10 V/10 2 Lima 1000 AA  "I L46 20 log A, = 40.dB = A, = 100 VV = R,/10). Nota good design! Neverthe- is less, ifthe source were connected directly 4o the load, On ° WA —o: ° * R+R, IMO 100, 100.8 —— Oe _ 100 92+ 100 ke ~ 0.001 V/V 100 ke = 100% 10 100+ 10 s09VN log 90.9 = 49.1 dB ; which is clearly a much worse situ: w,/100 0 2 MQ br 10 log (8.3 x 107) = 79.1 dB. ion, Indeed inse: ng the amplifier increases the gain by a fi 8.3/0,001 = $300, A He * 10 = 8.3% 10" Wr 1.48 For 4 peak output sine-waye cument of 100 2, the 100 ke on output voltage will be 100 mA x 100 2-= 10 V po M0 — 90 gly v; will be a sine wave with a peak ‘ | value of 10 WA, = 10/909 or an ms value of Ow Ima 3 y won » 10(90.9 x 2) = 0.08 V. xy Corresponding output power = (107./2)°/ 100 © & Shine walVkew i MO 4, 100 La? oT Ms 100 kD 100 2+100 100k teh aa tiy SAN AAA <> = li ‘ KO Sy, ) x /oltage gait O83. V/V or — tomy (kh Sy 1000, 100 f Voltage gain O83 WV or -L6dB ~ 100 2 . = = = 0 083% 11 10 . Cunront gain = ag 7 O83 ¥ 1.1 x 10 10 kQ 100-9 I OT 1000 x 10 v, 1OkD+ 100kR 100 + 10 100 = = x 1000 <= = $.26 Viv 110 jig °° ‘The signal loses about 9096. of its strength when co nected to the amplifier input (because 2, = R,/10). Also; the output signal of the amplifier loses approxim tely = 90% of its. strength when the load is connee |  1.50 Case (a) S-A-B-L 100 k2 wo. AN — $< 7 i ' | y , 4 0 joo.t u% 10 m¥ uit ' ' ' = = b 5 8 , a 10 100% x1 lO 10+ 100 0+ 10 100+ 100 0A or 20 log 4.1 = 12.3.dB Case (b) S-B-A-lL ‘ \ ' 1000 WkKa | ww ww Te) ' + ' ' oe lok via WO, 100.8 tly 10 m¥ is ' oe ' t 1 = i = ; = : A 1 5 ‘ ' ; _100 WK gg — 1 100+ kQ+ 100.2 100 + 10k =05x =O5 VV or -6dB Thus, obviously case (a) ie.. SABL is preferred LS  10 mV = 200 VV mun Required overall voltage gain = 2V Each stage is capable of providing gain of 10 (th cascade the maximum (uy 10", We thus sce that y ue), For a stages in open-circuit gain able} volta need at least 3 stages. For gain in 3 stages, the overall voltage pain obtained is Foe 10 gy 10 og 10 gy uy 10410 1+10 T+10 T+ 206.6 V/V Thus, three stages suffice and provi pain stig per than required. The output voliage actually obtain is 10 m¥ x 206.6 = 2.07 ¥ 1,53 R= Wk a oo ~ ‘ [ AL % Ky a 21K TT pe (a) Required voltage gain = © = = OVI (b) The smallest &, ullowed is obtained from 10 mV O1 pas R,+ 8, = 100 kQ 7% - (=)/rom Mx 107) | (0.1.x 107 BIR = 910° way he: power dissipated in the imernal uc.) ted to $V, the lurp= Re 53 y= 2, = 667 eee R, = 3h, = 667.0 (If R, wore greater th he output volt across &, would be less than 3 V.) (@) For R)= 90 kA and &, = 667 92; the required value fA found from 300 W/V = — Ox 4 + 10 1+ 0.667 > A, = 555.7 VV (e) Ry= 100k (Lx 10° 9) R,= 100 (1 10" ) 300 Oe 10% | DO + 100 = 363.V/V 200.10 AW —0 Loa 00 200  100 ket pA 1OkN Fy x (100 Jf10) x 10 an RFR, 1 +s (RHR) s transfer function is: of the STC low-pass type ha de gain R= R,/(R,+R,) and a 3-0B fre- fy = 1/ CCR, HR,). R, = 20k, Ry = 80 kG, and 1x 10" =B8B V/V or 58.908 10 Spe, 2200 AIA or = 1.25% 10" madls a = f A ae ae : ae x aR 10" x = 10x 10° meres +10) 100+ 10, Pac 1asx 10” ns haeis = 19.36% 10° WAVY or 62.9 dB o 2s . i 1.66. Using the voltage-divider rule, Overall eurrent gain = : H i c v/ Ry _ as(4x Ana —o- Fe we DS BER) ; ThA 10° =2000A/A or 66dB Vv, Ry Vv, 1.65 Using the voltage divider rule se x, Mf L ——— COR| +R) om Table 1.2 is of the high-pass type with io | ‘As.a further verification that this is.a hig and 7(s) is a high-pass transfer function we observe-a #=0, Tls)= 0. and thatas s TUS) Ri “Also, from the circuit observe as at s+ =, (L440) + 0 and V,/V; = R,AR, + R;). Now, for Ry = 10 kQ, Ry = 10 KO, 0.1 WP, and 1 28 29x 0.1 « 1010-440) x 10" = 318 He K 4001 |? oxy] = = 10+40,5 = O57 W/V 1.67 Using the voltage divider rule, R, ‘ A i © re) 4 : which is of the high-pass STC type (see Table 1.2) with Ri a= —— _ R+R, CRFR) For fy $10 He | __<19 aCiR aR) el Ix 10(20+ 5) x 10 Thus, the smallest vah 0.64 uF. 1.68 The given measured data indicate that 1 fier has a low-pass STC frequency respon frequency gain of 40 dB, and a 3-dB frequency c 10" Hz. From our knowledge of the Bede plots low-pass STC networks (Figure . . .) sve can complete the Table entries and sketch the amplifier freq response. ¢ of C that will do the jobis C= 20 dB/devade oO TD 10°10" y (Atay F(z} Im (d3} 27 degrees) 1000 40 $.7° lot 7 45" 10° 20 88,39 108 0 ~5¢y our knowledge af the Bode plots of STC orks we see that this of 40 dB, a low-frequency 107 Hz, requency response of the low-pass STC ° Hz. We thus can sketch the ampli- and higi amplifi response and o high type with fia % & 8 3 a = = a _ lof (Hz) to? y 10 10? 10° 10 10° 10° 107 108 0 20 37 40 40 40 37 20 0 =-_oe ee teieirterirrneeg™e™: fF & @»  nsfer function is that of three identical STC LP circuits in cascade (but with no kt ing effects since the bu zero output resistances) the overall gain wil 3 dB below the value at de at the frequency for which the gain of each STC cireuit is 1 dB down. This fre quency is found as follows: The transfer function of each STC circuit is Tis} = where ‘Thus, anaes = 0.51 ay =0S1/CR 1.73 RK, = 100 kQ, s by o very high f must be much larger than C). T find C, from AB frequency is reduced 6 MHz to 120 kHz} C, . neglecting C, we 120 kHz. = 2n€ Thévenin alent at trode A equ k ANA Shunt capacitor If inal 3-dB frequency (6 MHz) is attributable (oC, then 6 MHz = “are 2m 690108 x 10° = 0.26 pF 1.74 SB frequency is ¢ value of C must be much cr parnsilic capacitance originally between A and ground), Further- miore, iLinust be that Cis now the dorninant determinant of the amplifier 3-dB frequency (ic,, it is dominating wer whatever may be happening at nade B or any: where else in the amplifier). Thus, we can write 150 kHz =! TAC(Ray ia) l anx 150% 10° 1x 10" = 1.06 ko 100 ka, Thus &,, = 1.07 kQ Similarly, for node B, => (R,, ER, Now 1 16 kHz = -——-—__— ** TAC (Ra Ry) = R08, = ———____, 2ex 15x10 x1 «10 =10.6 k& =119kQ  greater er equal to (1 — 2/100), Thus Ay (3) Sea (1-5 Jai) Substituting &, = 10 k&2 and x = 20% in (1) results in = 40kQ Substituting fim = 3 MHz, C, = 10 pP and A, = 10 kd in Bq, (2) results in ; H3ka tmx 3x 10° x 10x10? - — 10 Ry = 10-kQ, and Substituting Ay = 80, x 11.9 kG, eq. (3) results in zo — = 18.85 mA/V Ge = " 2) 1 1 Sm | (1 11,3) x 10 ( il i : If the more practical value of &, = 10 kf is used then 20 mAsV where Rice and Z; hile It is obviously more convenient to work in terms of admittances. Therefore we express V/V in the alter nate form i ¥, Wt ¥s and substitute ¥,=(1/R))+5C, and Yy = (1/Ry) + sC; to obtain +8C, +s5(C,+C,) x Ry transfer function will be independent of frequency (5) if the second factor reduces to unity. This in turn will bappen if men Lp LY CR, C, + CAR, Ry) which can be simplified as follows thy CyRy = CR, When this co compensated, plies, the attenuator is said to be fer function is given by ition V Cy, y C+, which, using Eq. (1) above can be expressed in the alte form Md I Vi ke Rey Ry Thus when the al CR two resistors the transmissi wuator is compensated (C)R, = an be determined either by its its two capacitors. C,, Cz and on of Frequency, its transmission is nota fu 1.80 The HP STC circuit whose response determines: the frequency response of the amplifier in the low- frequency range has a-phase angle of 11.4° at f= 100 Hz. ion for 271 je) from Table 1.2 we = fo = 20.16 He. re The LP STC circuit w nse determines the fi nse at the high-frequency end has a kHz. Using the relation- in Table 1.2 we obtain for the HA? > fo = 4959-4 He  0 — Vir = 0.8 Vow NM), = Vip = Vig, =(0:4 —0.1)¥ pg 0.6 Vp = 0.2Vinn 3¥ pp th of transition region = Vy_— Vi, = 0.2¥ pp for a minimum NM of 1¥ = 0.2¥pp= | Similarly. atf= | kHz the drop in gain is caused by the LP STC ork. The drop in gain is =o 184 20 log — = -U.i h 7 1000 (a) Worse case NMyy= Vey, win = Vy = 2A 4 \G9594 Worse case NM, = Vii. may — Vz = 0.8 — 0 The gain drops by 3 dB at sie Fesicack Fwy A 15345 1]-= 10 mW the two STC networks, ¢ f= 4959.4 Hy 20 (ec) Dynamic power dissipation = fe Vio 18100 NMy= Von—Y lox 45x10 25 = La mW 1 (op epCtypicat) = AC tigi + foun) = A(T 11) = 9 fp( maximum) Las #2 1.85 5¥v (8) NM = Vou ~ Vai = Vii Vo (b) In the transi on regio: =3.5(¥;- 1.5) Vou 8 ¥ 5 NM = Vaie— if NM, = Vy — Vo, =01455-V Vo = Vj) 4.5¥5=9.2! if ° i (b) Voy =5--N (0.2 LOR =5-—0.4N NM_p=5-O4N-2=3-O04N=04 (c) Slope =-3.5  10) G) Pp, og = (S-0.1)°2.2 EO = 10.9 mW (il) P, = 5X(0.2%6) = 6 mv = 0.10 ns. 1.86 (a) Vo, =O d=25V we NM p= Vy = Von = 2.3 ns NMy= Vou Vin= 8-2 Yj. LOW V (hy VoG)=0-(0- se 2 5¢ For toi Volt) = Se tre = -(10"7010") In = = 0.69 ns For thy, Volt) = Se =4.5¥ 1 Rye 0.0lns Volt) = Se “205 V =2.3 ns fous = hy ty = 22 ns [he ty av | 1 1 i + — /5=CR QED. - © ‘ont Voy) at t= ty, Uy = 55e = 35 ns 5(Vow+ Ver) = Vox- (Vay —Varve 0.69CR QED