HW1-Solutions_Chapter 1

# HW1-Solutions_Chapter 1 - Homework Solution(Chapter 1...

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Unformatted text preview: Homework Solution (Chapter 1) Electronics Devices (332:361) Fall 2010 EXERCISE 1.10 Voltage: gain = 20 log 100 = 40 :13 1.14 Avg =40 :13 = 100 ‘WV Currant gain = 20 Log 1000 = 60 (15 U: RL 2 Power gain = 10 log AP= 10105 (AVAE) PL = E = (A”HU'RL+R9) RL =10: 105:50d3 1 0g =v§x£100x;) 10003-251}? 1 + 1 10 -5 ? v?“ 1.12 va=l>< 6 #10 v=1ouv P£=ﬂ= * 10 + 10 R; 10,000 ~15 2 P 2.5:)? 1951:3an = ﬁlo—xi.) =10'”w v Apsf= A 2=2.5><10‘wa 10 i 10 1:; IO log AP: 44 (13 With the buffer ampliﬁer: 1-19 R: ._ 1 X R‘ x A x R!“ Ila "" Rl+ I w RL+Rﬂ 1+1 10+10 2 2 pL=Eazﬂ=6zsmw EL 10 . v 0.25 V = —° = —— = 0.25 VN Voltage gam v: 1 v =—12dB ﬂ Power gain (AP) E P_ where PL = 6.25 111W and P; = viz}, 1V _ 2: =05 A “"0511 and “ 1MQ+1MQ ’" Thugr Pi = 0.5 X 0+5 = and, _3 6.25 x10 = 25 X“): 0.25 x 10'“ 10103 AP = 44 dB AP: vi, = ibr#+(ﬂ+ l)i&Rf = ib[rn.+ (ﬁ—I- 1)R,] But 11;, = v; and t}, = ix. thus RinE\$=ﬂ=rx+(B+l)Rp I: ll, PROBLEM . ' ode ‘elds . 1.16 (a) NOde equatm“ at the 39mm n 3" Now, I... II. and 13 can be easﬂy found as __iS-V_lS—2.375= 2. Using the fact that the sum oi: the voitage drops across ’1 " 10 " 10 13625 m L06 ma RI and R3 equals 15 V, we wnte I _ ID— V = = L125 m =l 13 mA l5=IIR1+13R3 1" 5 S . =1011+Ut+19><2 13=1=ZEE=L1375111A==L19M = 121.1-212 R31 2 Method (b) is much preferred; faSIer. more insightful and less prone to errors. In general. one attempts to identify the least possible number of variables and write the corresponding minimum number of equations. That is. Similarly. the voltage drops across R2 and R3 add up to 10 V, thus 10 = Isz + [3R3 =5I2+UI +12)1<2 which yields Dividins {1) by (1) Rim 2!. + 7n= 10 (2) _' “Rs” 10} .-.. 3 _ . 1+ 1111;111:011 Equattons (l) and (2) can be solved together by multt- =3 R “136m 1 ‘ ' ‘ p ymg (2) by 6‘ Substituting in {2) gives I 1211 + 4211: 60 (3) 11,:- 3315 my Now. subtracting (1) from (3) yields 40!; = 45 :21 I; = 1.125 m Substituting in (i) giVes 31:10-7 x1.125 mA = I, = 1.0625 mm I3=II+I2 11250 v"- R‘ XA x R" 311:1. v: “a V=I333 (a) 3.? _ xA x 10R... = 1.1375 x 2 = 13,750 V v, 10R, + R, "“ 103., + R” T0 Summarize: = 39 x 10 x 19 = 3.26 wv 1.=1.06mA Iz=l-13mﬁ 11 11 I =1.19mA V=2-33V . = (b) A node :quation at the common node can he wnttﬁﬁ or! 20 mg 8.26 18.3 dB R in [611115 0f V35 b E =' R: a 15_v+10_v=1 U5 R5+RIXAv”xRo+Ra R1 RI R3 = 0.5 11101:: 0.5 = 2.5 wv Thus ‘ 15_v+1o_v=g or.2010g2.5=3tlB 19 5 2 (C) E, __ R,/10 x A x 110/10 =‘ 0-3“: 35 v, (R,/10)+ R, "a (110/111) + R... 11 =—><le——=0.033va 1.44 ZOIth Av”: 40 (113 a: Av“: 100 VA! + 1MB 1:, A =93 I! 100 " llick)":me+10 =90.9wv or. 10103 90.9 = 39.1 dB 2 AP: m2 = Aﬁx 10“ = 3.3 x107 WIW iii/1 MQ or 1010g(8.3 x 10"): 79.1 dB. For a peak output sine-wave current of 100 Q, the peak output voltage will be 100 mA at 100 f1 = 10 V. Correspondineg 1:. will be a sine wave with a peak value of 10 V/Au = 10/ 90.9 or an rms value of 10/“(903 x Ji) = 0.03 v. 2 Corresponding output power = { 10/215) /100 £1 = 0.5 W 1.56 Current ampliﬁer. To limit the change in i9 resulting from R, varying over the range 1 to 10 1:9, to 10% we select RE sufficiently.r low so that. R,- ERIN-“HO .rhUS. R; = 9 To limit the change in t"ﬂ as RL changes from 1 to 10 REP... 10 10% we select Ra sufficiently' large; Ru 2 IORLM 111115. R... = 100 1(1) NOW for f, = 10 _. R ~ R _ 5 3mm AF :1 tam“ Rsmin +Ri JR0+RLm Thus. —3 —5 1000 100 1.61 Apply kirehhoff’s current law in the mesh on the left 111 471+ n.1,. + RE(Bi,, + i.) = 0 5:51qu +113- + DRE] Uh : New vi. = —|3 >< ihXRL Ur? : _ BRL vb rat + + 1.63 Using the voltage divider rule =10 A gm 0 gmvi R1L = Skﬂ . 7 R . L iio 1 .65 + + R R V:- C V0 Vi C V0 {3) (b) 1.67 Using tHe voltage-divider rule. l HSC J f V = V — m (a) ” ’(1 are + R E 2 1 VI, 1 + SCR where k=1 Cu?! + R1) 1 which is from Table 1.2ie of the hi gh-pase type with mg 3 from table 1.2 it is low pass. R1+ R2 “ one1 + RE) ﬁx ([3) v“ = v! R As afurther veriﬁcation thetthie is ahigh—pase R + i network and HS) is a hi gh-peee transfer function, SC weessumeasL-rzbﬂ,T(s)=}D;andaeS——}OG+ ﬂ} = ﬂ T(5) = REEL»?l + R2).Also, from the circuit V! i + SCH observe as s—em, (.1 MC) —-}U and E3 = S VOIVI. = ream?L +R1).Now,for V, S + 1 " E R1 = 10 kﬂﬁg = 40 kﬂ,and C = 0.] nF. where k=l f 2 E i 1 r; —E 3 ten = éfrornteble 1.2itie high pass. 27]- 211' X 0-1 X 10 (10 + 40} X [0 = 31.8 Hz K 40 l = — = m— = 05’?va IV DESIGN EXERCISE 1.24 i xn K; = Rt ___l_._ R: 5 Overall power gain a v 5 x If m I R3+Ri\$+; _ ﬁrm) i }|' SC C (R, + RE) 2 which is a HP STC function. / 1000 1 2 = —— S 100 Hz = -————---——— f3” 2eC(R,+ R.) [10x10'3)x(ﬂ.1x10'6) C a ————L-3——— = 0.16 as Ji Ji 2::(l+9)10 x100 =9xlotwm PROBLEM 1'49 . Required overall voltage gain. = 2 V/ 10 mV = 200 (T1115 takes into acct. the power dissipated in the internal V/V. Each stage is capable of providing a maxi- resistance of the spurce.) mum voltage gain of 10 (the open—circuit gain value). For :1 stages in cascade the maximum (unattainable) voltage gain in 10". We thus see that (c) If (Aware) has itspeak value limited to 5 v. the larg- est value of R0 is found from - . . h l R ' we need at. least stag-es For 3 stages t e overal . 5 X L = 3 = R0 = gRL = 667 Q voltage gam obtamed IS RL + .Rﬂ U_o = 10 X 10 X '1 Q x 10 X _ 10 (If Ra were greater than this value, the output voltage I’3 10 + 10 1 + 10 I + 10 across RL would be less than 3 V.) X 10 X ﬂ {(1) For Rf = 90 kn. and R, = 667 E1. the required value of A c h f 2 206-6 WV 1,” an e ound frog) 1 Thus. three stages sufﬁce and provide a gain 300 VJV = 90 10 x A". 3-: my slightly larger than required. The output voltage + + ' actually obtained is 10 mV X 206.6 = 2.07 V. = A1,“ = 555.7 VN (e) ai=100 m (13:105 a} aﬂ=1oom1x1ozm 3on=_.1_99_xgu xﬂ 100+10 ' 10004-100 =1: Aug = 363 VN 1.68 Using the voltage divider rule, (:1) Required voltage gain 5 i’ b : —-—RL—— 5 = = V 0.01 v 300 N RL 3 {b} {The smallest R. allowed is obtained from _ RL + Res J. _1.___ C(RL + RS) 0,1 M = ‘0 "1V __., Riv. Rf = 100 m which is of the high-pass STC type (see Table 1.2) R, + Rl' Thus R,- = 90 m. ' RL 1 For Ri=90 ko, 5,.=0.1 ,uApealt. and K 2 RL + R3 03’” = C(RL + as) /R - ‘ Overall current gain = v" * L For f E 10 H2; III a If) "I 3 mA 4 ————— i: 10 = = 3 x 0.1 M 10 MA 217C(RL + RS) => C E 1 2e- ><10(20 + 5)><103 Thus. the smallest value of C that will do the job is C = 0.64 uF. ...
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## This note was uploaded on 10/23/2011 for the course EE 101 taught by Professor Hoa during the Spring '11 term at American InterContinental University.

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HW1-Solutions_Chapter 1 - Homework Solution(Chapter 1...

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