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chap013

# chap013 - Exercise Solutions-Ell3 Wuhan = v V5300 = 15 —...

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Unformatted text preview: Exercise Solutions -Ell3 Wuhan) = v* - V5300.) = 15 — 0.5 = 11.4 v 14:0(min); 4Vag(on) + V+ = 4(06) —- 15 = -12.6 v -12.6 5 thcm) 514.4 v E114 .. Vﬁmax) ; V+ — 21/336111) = 1s .- 2(0.6} mum) =13.a v mm) = 3V3g(on) + V‘ a 3(D.6) — 15 mm.) ; -13.: v --13.2 < V; < 13.8 V II. Vanna) = 5 -1.2 = 3.0 v Vo(min) E was + V' = 3(0.6) - s = -32 v --3.2 < V! ‘2 3.5 V 313.5 Icy =3 Ia: g 9.5 #151 _ _ 9.5 RA _ Ia. _ 152-. m _ 0.0475 11A :1! [m = IE3 =47.5 11A 513.5 cu 5—205 — - Inn—(79.01-..“ _ fur _ 0.72 x 10") Va: — V'rln( [s ) _(0.026]ln( m_u = 0.550 V 30 I”! = 30 - 210.65! 40 #IaizaoJlﬂmA VEE“ = 0.650 V 11,-1.3. .—. V-rln (“u”) Ian 1cm By rm and error: 1cm - 16.9 11A. 1:106} = (0.025)1n(°'713) Value = V3211 - IGIﬂR-t = 0.380 - (0.01%)“) =- Vm, ; 0.550 v [as 5 L2“ = \$29- = 9.45 11A I 9.45 x 10" V535 = V:- In = (0.0264111 =9- VEE‘ = 0.537 V 513.7 lrc 13,4: OJ 531.114 V33 0.13 x 10'J = 10‘“ exp V? 0.10 10‘3 VD = V:- In 0.13 x 10" = (0.035) Ill. VD = 0.6140 VEE = 2V2: 3—: 1.228 V V 2 Ian = Iczo = ISBIP( I?!" __ _“ 0.6140 -axlﬂ exp(u_oza) IQ” = IQ“ = 0.541 11121 E118 I _10—o.s—0.s— -10 as: — ‘0 =3- IQEE = 0.4.7 1111’. Ictoﬁd = Vrln in?) cm OAT 1cm“) = (0.026)ln Syn-inlan = a Ten: Ia. = T =? Ia. = 8.5 EA Jena = (0-75)Iur =3» 12”: a 0.353 111A [1:134 = (0.25Hnsr =5 IQ”, = 0.118 InA Ru = R4: + Rm: R __ 1‘21; + Rndllﬂoua 3“ ' 1+0. rm + 3.011 lﬂoua 1 + .3? 30133 = r0108 8 92.6 k9 Row =‘~ r011[1 +§m1r(Reﬂfxn)] = 283 kn Rm- From previous calculations Rm: = 1.51 kn Rad = 31'; + 3522 = 301.144) + 1.51 =1.30 1:9 Run = 1'01“ = 273 m HHV-r =£200t10£25t = 1.“ kn Ian 5 RE“ = #1419 201 R. = R. + 33,. =27+14.1 = \$41!] fir“ = E1340 For Q5 we. have V555 = V56. = 1.05 V So Vsndsat) = 1.00 — 0.5 = 0.56 v For M1 and J”: I ID = “:- = Kp(Vam +V1r)z 0'039' = 0125(ng -0.5}' =? V551 = 0.898 ‘0' 50 maximum input wing: = V+ - Vsmhﬂ} - V551 = 5 — 0.56 — 0.893 =- Vin max! = 3.54 V For M3. K, = (525)(20) = 125 M IV’ I 39.7 100 = 3'1 = — 00 39.1 T = 121%” ‘ 0'10)1 V,w = 0.5 V :> V65 = 0.890 V Vsnﬂm) = 0.593 — 0.5 = 0.395 V Venr(miu) = V" + V65: 4- Vsm(sn) - V5.5. = —5 + 0.800 + 0.395 -- 0.590 Vengmia} = 4.50 V 4.00 5 Vmﬁm) 5 3.54 v E1111 Wham) = V” - Kendal) V500 = Vsas = 1.06 V Vsm(su} = 1.05 - 0.5 = 0.56 v mama} = 5 - 0.50 4.44 V %(min) = V" + Vos-r‘m) V4357 = 1.00 => V9513“) = 1.06 — 0.5 = 0.56 Vohuin) = -5 + 0.56 = —4.44 -4.44 ( V9 < i.“ V E1112 (111%: 111,, K” =125mw= K1111”, w”? = m 5 + 5 - V565 100 12.5(V5265 — V505 + = 10 — V555 12.519313 — 11.51313 — 5.575 = o 11.5 t 11111.5)2 + 11(12.5)(6.575} 0.125111%, —o.5)‘ = Vsas = 2112.5) V555 =1.33 V 111m 0 — .33 Iﬁsp=fq=lm=fm=1 1 #:8631115; “In Ian =11): = ID: = 112+ = I—Q- =43-35 11A ______..._____2_____.._ MIX" = K1,, =125pAW‘ 1 1 = = — = ——————- = 1 153 1:1 ’3’ '3‘ 11,, [o.oz)(o.o4335) Input stage gain AI = {zxrliﬁ '(rvllrﬂ) = 1j2(1).1:15)(0.01;57) -(1 1531|1153) 2 A‘ = 84.9 Transconductance of M, g_, = 74m = 2,f(ozso)(o.o1167) = 0294 and 1V 1 1 r" z r“ = :1; = [o.oz}(o.0367) = 577 In Second stage gain A»: = 1.111..) = [0294)(57711577) :1 4,, = 84.8 0W1 gain = 14.1-14.2 =(84.9)(s4.s)= A = 7,200 E1113 ID, = In = 25 1.11 k' W r 40 g_, =g_,=2 Etta)!” =2. [3— 25x25] 2 \$.1=S.s=22‘ﬂAW k; W so g... =2 LEI-ﬂ!” = 2 [7 25)(25) :0 g," =316yA1‘V r111 =r96 =5: =rolo = 1 11—5 (0.02x25} ' Cha 1:: : rci 1 tie S R4 = gl‘(rulrol°) 2 (2209)”) = 396 w R116 = 3-s(ro5X’-4 ﬂrﬂ) = 3 16(1)(]l2)= 421 m Then A, = g_,(Rm'LRM)= 224(4211is95) => A, = 64,158 513.14 (3-) Ad = Bgnl(roélral) g_, =1 131m =2 [ﬂyzoxm 2 2 g_,=4oo,mw 1 1 r =..——=——-——=0.333Mﬂ "‘ 1,106 (o.ez)(1so) 1 1 =0331Mﬂ r“ = LID. = (o.oz)(1so) A, = 3(4oo)(o.333[10333) => .1, = 200 1 a”) f” ‘—m.(C.-.+c,) where R, = rip” = 033350.333 MD. 1 f” ’ 2::(033?.|]03:13)x10‘15:le1 3 f m = 477 kHz f“, .11, 4477111010110): 95.4 MHz E1115 (3) From Exercise 1.3.14, g”. = 400.1114U’ "a: =ralo = roll = 0-333 m r96 = ' r k' W 40 3qu =2 Tynan = 2 203050) 3' 3.11=490MW f k' W 30 3.11 = 2 [YIXE]IDII = 2 20m”) 5 3.1: = 693 FA! V R... = smotwdl =(490R0333X0333) = 543 Mn Rm .—. gldﬂrmr") = (693X0333X0333) = 76.8 Mn ‘41: = Bgnl(Ral0“Ro'l!) = 3(400)(54.3“76.8) :5 A4 = 38,172 0’) R0 = RomnRolz = 54.3“763 = 31.3 W 1' =__——‘—— "’ 2n(31.3x10‘)(2x10"‘) j”, . A, =[25x10’X38_172) = 95.4 ME: = 250 kHz E1116 (9 A: =gnl('R-IIRII) From Example 13.10, g. =316MW.R.. =316Mﬂ Now R06 3 g-6(r06)(rﬂlkl|) rel =1Mn,r,. = 0.5Mn I“ so =_.__=—--—=:-i.923mAIV 3" V, 0.026 r“ = 13—“ = E9- = 1.619!!! ICG 50 Then R,6 =(1.923)(1600)(0.ﬂ|1) = 1026 Mn ,4‘ =(316){1026||316):> Ad = 75,343 0:) f..= ‘ f,” = 329 Hz {,0 -A‘ = (329)(TG,343) :5 25.1 M2 E13.” V+ — V'- = V331. + V385 + V35? + IIRI = 0.6 + 0.0 + 0.6 + {0.24)(a) = 3.72 V 50 Law 313.13 PU Q:- and RI V39 2 V337 +Ilﬂ‘ = 0.5 + 1:5) PM MI: 1: =Kp(Vﬂ ‘H’n-F I; = 0.30%.: - 1.0)2 By will and cant: VIE = 2.54 v M 21r(3IG||1016)x10‘x2x10'“ 3 E1119 1-3;: = 163).“? 9.: REF]: v .. r“: = 131/:- = \$200)! 0.026! [cm 0.20 = 26 kn .ﬂlz = Tau-[3 + (1 + 5)}?5'13 = 25 + 201“) = 127 kﬂ r — ; - ; — 500 1.0 “1" " Mm. ‘ (0.02)(0.1) ‘ VA 50 ''01.: — fen — E -- 500 kn _ Icu _ 0.1 9"” ' V1- ' 0.026 ' 3'85 MN V . r“: =f4=w=53kn c1: 0-1 Run. = I‘llll [1 + UMII(rf12||R-'ﬁ)} = 500[1 + (3.35)(52|ID.5)] = 1453 ha AI = {my '(rllﬂF-II‘RII) = \f2(0.6)(0.2) - {5001453"227} =(0.490)(1+1) = A1: 59.1 513.20 For J}; biased in the main]: mginn => Ica =15“ = 300 .03. Q1. Q2. Q4: 113mm =5 {Cl =Ic; =IGJ =30!) EA lcc n'c ' u' a] sis and De ' 2“d '15 n olutions Man 111 Chapter 13 - 1 Problem Solutlons (c) f", = and zydeﬂc" 13.3 CM =[10)(1+2313) = 23,140 pF (3.} A, = g_,(r,, ||r,. “116) R4 = \$45.11;. = 4|}4|1.16 = 0.734 MI: _£c_1_ 2° =——-—-1-———--=7.71Hz 5"“ ' V, _ 0.026 ﬂungmw f” 240.730.:10‘)(2s,140x10"=) 1 = h: E 2 4 Mg Gain-Bandwidth Product = (7.71)(159x10‘) =5 ' 1c, 20 12.3 MHz V“ so —-— = —— = — = 4 Mn r“ [a 20 13.4 R6 = r“ +(1+ﬁ_)[R1ﬁrﬂ} a. Q: act: as the pmwction dcvica. m . r _ (120)(0.025) _mm b' Sm “ p (a) ‘7 _ 02 - ' 13.5 If If: assume Vas(0n) = 0.7 V. (11:11 I“ EM=E=QD3OMA R1 20 mn=0.7+0.7+50+5 r“ =(120)(0.026) :1“ m 0.030 80 breakdown volmge z 56.3 v. Then 2104+ 121 2 15.6 =>1.16MQ 13.6 2: ( )[ 01 ] 15-0.6—0.5-[-15) en (a) I“, = ———_— = 050 A, = 769(4[|4H1.16) :5 A, = 565 Rs Now = R, = 57.6 m .. _ _ _ R1 = V 1 £821 VI - [J07 — {ﬂuff} 01 "' _ﬁuroT[Rl +r‘7 n‘ IC'III‘R‘. 1" n {cm 0 0215 050 V, = ' 1 R = 2.44 m = -ﬁ.(l+ﬁ.)r.v[ K )1“ and I» = -—‘ R‘ 0.030 "( 0.030) :3 «'——— R1 ".1 Rm 5 06 The“ (b) 1m=_“._-L‘('5)=, 1W=Q153m A‘ s V_._~ﬁ.(1+ﬁ.)r.1[ 5 ) 57-6 —— ' V.. Ru R. + 11.1 12,4144) =(0.026) 1.1[0'153 ] v; 00 1cm '-r=g=53=4°°m Bytrial and error, IMEZIJM 5“ 4 x x 1 m 120 121 400 20 = I s 050 m A” 1160 [20+15.6)= (‘0 I 050 104 A» = 4313 V3, = V,1n[ 15"” ) = (0.025)1n[—16’5_;—) => Overall gain = A, - A”1 =(565X—2813): " = 0.541V = V A = -1.59x10' M — Then 15—0.64!-0.641—(—15) 30 0.026 = _....____ = 57.4 m (b)&=2rg| 311d rn=%=lo4m 050 a ""“—“—RS R, = 200 m 1; z “026 [:1[ 05° J :5 R. = 2.44 ha —--—-- 0.030 0.030 0.03 0:10" VIE-m = 0.025 :1 V's-m = 0567 V (11) From Problem 13.6, I E 0.15 nu! ‘ 0.15:10" VI!“ = Vat: = 0'026l"[ -|4 ) s — 0509 — 0.609 —(—s) __..___._.____ a 57.4 = 0.609V Then I”, = I“, = 0.153 nu! Then {m 2211 M ﬁ'um Problem 13.6 13.3 5—0.5—D.6-j—5! a. In! = \$0 => I555 = 0.22 mA I [62034 = VTln( u?) 1cm 1mm = (0.02311. 1cm By till and error: 1c“ ; 14.2 EA Icu = 0.75Iup => IQ” 30.165 mA 1m“ = 0.251115? =- Im“ = 0.055 mA b. Using Example 13.4 f.“ = 31.5 In“ R'; = 50||[31.5 + (201)[o.1)] = 5D||5L6 = 25.4 to 1'1“: 5 any? a 1:1: 0.155 10.165210.1!+ 0.6 “ Icu — no + 5 — 0.0132 mA 1"“. I 3“ H1 The: 12.; a 394 + (2011mm) => 5.5 M11 r" s 732 kn 5.... = =- 0273 MN r“ = '0‘:ng = 7.04 M11 Thu! 3..., :- miu + (o.273}(11|132}] = 3.95 M0 50 '°‘ ' 0.071 " 7"" Mn 7.1 _ Ad = - (T.D‘lll3-96"5.a) or A1 = —627 Gain of differential amp 5mg: Using Example [3.5, and mglecting the input resistance to the output stage: _ v. _ 5n Rm: - [ma — 0.165 _ 303 kn A = —[209)(201)(503(303) "' {5500)[50 + 31.5 + (201)(o.1)] or 44...; = —-1090 Gain of second stage 13.9 1cm = 19 WA From Equation (13.6) ﬁ§=+2ap+2J [[10|2+2(19)+2] I =2! _ =2 ..____._..__ C" [ﬂ}+3ﬁp‘+2 I [10}‘-.-3[10)+2 122 = 2 — I[132] So 132 21' = *- = :- (19(122) 20 a6 pA W fm=....._.2£.§._..= —-?-o-.-a-g---=>Icg=1?.l3 11A 1 _ — ( + ﬁr) (” 10} [5-, 17.13 I3: =-' -""' = p? m =- Im = 1313 EA 1 10.23 ____=..._ z =o.934s A I" (1 + 39) 11 9 ——-——L" Ic. = I(1 :2») = (10.23}(%) =§ [cg = 9.345 #3. 1.3.10 V35 - V”. = Vagﬁan} + Ich) = 0.6 + (011095)“) = 0.5095 0.5095 50 I3? = => Ic-r =12.2 p15. In =Icu == 1911A Hus;- = 0.72 31A 1513 = 11212; = 0.72 mA Icn = [38 16A Power = (W' — V“)[Ic.- + Ic. + Icg +fnsp + I51; 4- 11:11] = 30[0.0122 + 0.019 + 0.012 +0.T2 + 0.72 + 0.138] = PURE! = 43.8 mW Current suppliad by V" and V’ = [or + Ice + Ics + IRS.” + IE1: + [611 =lﬂmA 13.” (a) v_(min) = -15+0.6+0.6 +0.5+ 0.6 = —12.6 V v_(rnax) = +15—.6 = 14.4 V 50 42.65 v, \$14.4 V (b) ernin) = -S + 4(06) = -2.6 V vw(max) = 5-0.6 = 4.4 V So -2.6\$ Va S 4.41" 13.12 vao=V‘=-15 v.01GblSBV01ﬂﬂeonuilpuned low. Ind Q10 and Q19 meﬁecﬁvely moﬂ’. A: a ﬁrst Ippmximﬁun - 0.6 1:11 = = 22.2 m 22.2 Ian = W - 0.111 M Thu Icu = 1613.4. - Ian = 0-13 — 0.1“ = 0.069 mA Now V3511 = Vrln (Ian) 15' 0.069 x 10'3 =0.589 V A: a second nppmximation 0.589 Icu - =1- Ig“ = 21.8 m 21.8 [an — m — 0.109 mA And Ln, = 0.13 — 0.109 a IQ“ = 0.071 11111 _—_—_—.JLEQIELm—S—QMQEE 13.13 I. Neglecting hue emu: ID = 1121.15 Tum V35 = 2V9 = 2Vr ln Is 0.25 x 10':I = 2(0.026) ln OI’ V35 = 1.2089 V V 1511: = Icy = 15:11:]: ( 2' _ -1. 1.2009 -5x 10 exp(2(o_o26)) So [2.3 = IQ: = 0.625 mA 0. For II: = 5 V, M n. E E = 1.25 mA _i—_ A5 a ﬁrst appmximalion [car 2 i; = 1.25 m 1.25 x 10‘3 = _ = . V V35" (0.026)1n( 5 K mm ) 0 E225 Neglecﬂn; but: meats. V“ = 1.2089 V Than Vgap = 1.2089 - 0.6225 = 0.5864. V 0.5364 0.026 I”. 2 5 1-: 10““ exp ( ) :1 (c,- = 0.312 12:21 As a second approximation. IcN =1}. 4- Ian: = 1.25 + 0.31 =1» I“; a 1.56 111A .5 '3 V35," = [0.026) Ln 5 x mm ) = 0.52020 v Vggp = 1.2039 - 0.62325 = 0.5806 V -11 0.5306 = — = ' Ia? 5x10 112:1)(9026) In: ‘1st [3.14 V53 ' 1.15? “ --—-"= —=54.23kn R1 + R: [0 DIEM: 0.018 0.9 1 Va: = VT In = (0.026) [11 (ﬂ—( EMS) 0.162 x 10" = (U.026)ln w V5; :3 0.5112 V Val: = ( R" )Vaa R: + Ra R: 0.5112 = (64.25) (1.15?) 50 R! = 33.95 kn Then R] = 30.32 m 13.”! (a) A: = ‘3.("u“’o¢l&z) From example 13.4 95 = =365 IV =5.26 MI! 3" 0.026 ’"A ' r“ Now I" = rd = 526 m Assuming R. = O, we ﬁnd Rn = rlla +(l ins-JR; = 329 +(2o1)(5c-]|9.63) => 1.95 M51 Then A, = 43651524361195) z: A, = —409 (1)) From Equmion (13.20}, A. _ -ﬁ.(1+ﬁ.)R.(R...HR..ILR...) R..{R. +[r... +(1+p.)R.]} For R, =0, R, =195Mﬂ Using the results of Emple 13.5 _ -2oo(201)(50)(92.cﬁ4oso1|92_6) = fl" ’ (19so){so+9.53} 21,, = -792 [3.16 LetIcm = 40 ML then In = In = 20 3A. Using Enamel: 13.5. Ra: = 4.07 M9 r11 =Q3361§322fﬂ=260 kn 0.020 = -—-— = '1 9...; 0.025 O 69 mA/V 50 me - m# 2.5 MG “that 3”“ - 2.5[1 + (0.769)(11|260)] = 4.4.2 Mzn in fan 3 ﬁ- 2.5 A! = - (TuiIRmIHR-‘zl 20 = -(0.026)(2'5"4-42II4.07] So _ A4 = -353 13.17 NOW 7w}! + Ra: = _.—-— R 2 Reid 1 + lip and n R: + Ra“. Assume series resistance of Qua and Q19 is small. Then ROI = T01341ch + 1' when Ran ___ 7:23 Run! 0153 l+ﬁP and Rm? = TonU +§mIT(RIIlTIITn Using results ﬁ'om Example I16. rfll = fry; 3" gm” = 20.3 mAIV run = 92.6 kn Then Rm = 92.6[1 + (20.8)(o.1||9.53)] = 233 m 50 r0“; = -—n.54 = 93.5 kn Then _ Rm = 7.22 + 233 92.6 = 1-51 m 51 R411 = r0134.“ch = 27B||1.51 = 1.50 119 ﬂaw-..” P’L‘ = Thu: IR.“ = 0-55 + L50 = 0.0422 1:9 51 Ol’ H.“ = 42.2 n Then Ra =42.2+ZT=> §£=69J29 13.18 R“ - 2[fu +(1+ﬁ“)(1f+':59)] Assume ﬂ“ = 200 and B1» = 10 Then m = {2001(00261 = w m 0.0095 (10)(0.020] 1'.r = ——-—-=21’.4 kn Then Rid =2[547+ 201 27.4 ] 11 of Bid = 2.095 M“ (a A: this frequency. IA”): = 1. so 1: 356,790 f 1 1+(m) -1f1+(o.3s4}’ _ 335.275 I)“ 1+(iﬂ ar-L-aaszTS-éf—IB‘E'VIE 5.43‘ ' "' ‘ ” Mampuieu f ft=m\$£1=5MHz [3.20 I. CW 9-»: Ind 9m: 19.. = K” = = (12.5)(10) =125MIV‘ So [(2 gml = gm; =2 K” F -"-' 2 = 0.09975 mA/V If is increased to 50. then I9. = 16,, =(50X10) = \$0er 50 gm. = 9...; = 2x/(0.5)(0.0199} = 0.1995 mAfV b. Gain of ﬁrst stage Ad = 9m1[fozﬂfm] = {0.1995)(5025l]5025) 01' A! = 501 Voltage 505.0 of second stage remains the same. or A»: = 251 Then A. = A‘ 41,, = (501)(251) 0! Ag: 125. 751'. 13.22 a. x,=(xo)(20)=2oomw==ozm.uv‘ _ _lU—Vsa-!-10! In.” —Issr - 200 = 0pm.; - 1.0)2 20 — Vsa = (0.2)(200)(1v§G — avg-a + 2.25) 400%; 419:!” + 70 = o 119 0: JUN): — 4(40)(70) V5“: 2(40} 3' Via = 2.17 V Then -m-1n IMF - T => ME: M3,. Mg. Mu meshed mammal: I2 =Ipt=fngr=89£ 11A. rug = = = l 12 M9 XpIp {0-02) 392.2) I 1 1.“ = = __.....__ = 124 Mg ‘I'hul .14 = ./2(200){39.2) - (1.1211234) Dr A! = 141 Slum-dyad voltage gain. of aacond stage: A»: = 9mT(TOT“1'OI) K_, =(2oxzo) = 400 pm” 5a 9...: = 2.;K_.1m =2./(o.4)(o.0392) = 0.373 1 1 '°' ' Jul-I97 ' (n.02)(n.0392) " 551 m f0: =3 1 l = 1121 1...}... = (o.o1)(o.0392) 30 A.“ =(u.31's)(1121||5s1) =1. .1“ = 141 Mammgemn A. = A. - A... = (141M141) =- A... = 19, 3111 13.23 Small-sign]. volume gain of input m A: =‘J2Kpllﬂ '(rozlru) , K” =(10)(10)=1oomW' 1 1 m=—= =1oonkn' A; (0.111) 1 1 fol == = = 2000 m 1.. (0.005) 111m A. = ./2[o.1)(u.'2) - nonauzano) OT A“ = [33 Swill-aim voltage pin of and mm: An = ym7(fnrllf«) g", = (zoxzo) = 400 [.14 W‘ 50 9.... :2 ml... :2 (0.4)(02) = 0.566 MIN? 1 -—1—--=500kn 1.1.19. "' (o.o1](o.2) 1 1 = — = _..—..—.-— = lﬂﬂu "’7 1.1m. (c.nos)(o.2) To: = 50 A»: =(o.5sa)(1ono||5n01=.~. 21.; = 139 Then Overall voltage gain is A. = A1144": = {l33}(189] = A... = 25.1.37 13.24 J. _ ; ’5 ‘ 2:3..0. what: Reg = foallfa: and C; = 01(1 + |A.;[) We can ﬁnd that Av: = 251 and fun = re; = 5.025 Now 3... = 5.025|[5.925 = 2.51 MG and C. = 12(1 + 251] = 3024 13F 50 f _ __._._1____ ’D ' 2:12.51 1-: 106mm x 10-“) or [pa = 21.0 Hz 13.25 1 I” ‘ 2:3,,0. We Ree = Tadlﬂn From Problem 13.22. +02 = 1.12 M51. r... = 2.21 MD MA... = 141 ———_——QhﬂDJ§Llli—mm§ﬂmm§ Sn 13.25 B _ 1 (a) A: =3m(RulRol) “(i-mill“) " 1°“ " G g_, = 2JK,,Im = 2‘ ((0510025) :3 224 pAIV 2-266 10"-C 1 A C 2 gm=gﬂ " ' x _ ‘( H "‘D‘ ‘(H ] gﬁ=2‘((05){0_025):>224mw 01' 1 1 0 =133 F r, =r =r =r‘, =—=—-—-—=2.67Mﬂ _g_1=_ ' “6 " '° Hm (0.015)(25] 1 1 = = _._— 133 ME) 1126 r“ um (0.0151(50) 2’ Ru '-"- foTHl’nl Now We can ﬁnd that R”. = g,,(r,,r,,o) = (224)(2.67}(2.57) =1597 MQ m = m = 2.52 Mﬂ RmS = 3*(rﬁ)(r“|]r,,) =(224)(2.67)(2.67|1133)=:. m R“ =531MD. Then R0 = 2.52:1252 A, = (224)(53Iu|597) z: A, = 89,264 0! R =3 R =5311597 R =393Mn Ro=Lzamn 0’) a «no: H =5 __._ (c) 1' ~— ‘ ——-——-—' =. 13.27 "’ 2mg 2n(398x10‘)(5x10"1) a. f”:I = 80 Hz GBW=(89,264)(80) = GBW = 7.14 MHz 13 .29 1 1 (a) r. =r =r' =—=———= m ‘ “' '° 1,13, (o.oz)(2s) - 1 1 = — = —— = 2,67 m r‘" AJD (0.015125) 1 l r“ ' 1,1,” ' (0.0153(50) ' 133M!) 3-: =21K? 331(3) = 41.3 m = 3.“ V9 = (9m1V,,L)(ro:||rog) g“ =2W=6um v: = vs... + w. “1'41 V0 = 9m1(fmllfnz )(V: - Va] R0 = RHIIR-I =[gush-6X54l!’u}][gul:ruroion l W f W o: { I ) Deﬁne X. = and X6 = [vi-)5 = 9m: Tea! 1'02 _ Av 1 4. mg,” "m, Then 3, = [632X5(2.67)(l.33||2)K41.8)ﬁ(2)(2)] 22 539X X =134.8X 167.2X =—-'—'———‘—— b. IX +3?!“ Vial. = Eff:- + 3:3? and V91] = *Vx 5n 1 + _ l _ _ 2:'Z,SZ’:9X..’1'Ii - r01 ru: I‘ll-3.13.."(4I-3X1 —‘—1343X5+1672X1 =10,ooo Now Xs=m = m=0574xl 13.30 Let V‘ =SV,V' =—-5V P=1,(10)=3::I,=0.3m11 21m =0.lmA=lOOyA 1 z r“ = r’” = (0.02)(50) r = 1 " (0.015)(50) r = 1 " (0.019000) r =1MD ll =133MQ = 0.667 MI 35 W 3-11 =2 I150) = 592X1 :3}: when: XI = I Assume all width-to-lcngth ratios are the same. gmG =2 [8%) %)(50) = 39.43] Now R. = = [Entruxrulro1)][gu(raro1o)] -_— [119.4x1(133}(0.65111)1592x,{1)(1)] _ (47.6X1X592XI) = [471511. ][592X,] _ W So R, = 2114):, Now AI, = g_,R_ = (592X1)(26.4X,) = 25,000 Scumx:=%=1a forallmlstors 13.31 (a) A. = 33.40401) 1 1 r =,. =___.=—=0:141 Mn °‘ " AIDE (0.015)(90) gm =2 %)1m :2 (500x30) = 245 Ml? A, = (3)(245){0.741|0.741) z» A, = 272 anual (b) R. = r..||r_. = 0.741||o.141=> R_ = 371m 1 1 (c) f” " mac — 2::{371x10‘)(sx10"‘) 3 f”, = 353 kHz GBW = (272)(35_3x103] => GBW = 23.3 MHz 13.32 1 (0.02)(25]{40) = 0'5 Mn 1 r“ = (0.015)(2.5)(40) A4 = Bg-1(rodlrd) 400 = [2.5)gm(05“0.667) :: g_, = 560 11.4 W gm = 560 = 2J[3?01%)(40) => = 49 Assume all (WI!) ratios are the same except for M, and 114,. {g =1225 5 5 (a) r06 = = 0.657 MD (b) Assume the bias voltages are P" =\$V ,V' =—SV. 15¢ HE Féuﬂ *anrsﬁa-mu: MB 4- M14 ‘éw ~5'1/ W W ._ =—- :4 mmelLl 9 30 1 19 = [T 49x14," 415) = so: V“. = 0.702 V Then 80 W 1 =30: _ .— V _05 In! [2 L)! cm ) For four transistors Var = —1°'°'7°2 = 2325V so = [%I%)c(2325— 0:3)2 :1, [i151 = 0.50 1 =_ R =osuo§67=0286m ('3) I” 2m; ‘ l = ____._.. = [85 H f” 212(286xIO'X1x10'“) z GBW = [400)(135x10’) = 14 m: 13.33 (a) From previous results, we can write R." ’1 g—I0[ra|lr96] Rot: = gml2(rollrm) A: = Bgnl(RurollRal 1) Now 1 1 l 1 1',” =5, =W=W=Oﬁﬁ7m Assume all transistors have the same width-to- length ratios except for M5 and M‘. W Let --- =Jl'1 (L) Then g_,., =2 \$1.0m”) =2 ’(25)(4o) = 83.67X g." =2 5311139,: =2 ([EZEJX’CESXN) = 1265X g‘ = 2 [%]X’[4o) .—. 80X Then R... = (83.67X}[05)( 0.5) = 20.9.? Inn Rm =(126.5X)(0.657)(o.667) = 553X MD We want 20,000 = (25)(sox)[2o.9x||sas x] (ZDSXXSGJX) — z 1 "200 20.9X+563X] max (1:) Assume bias voltagesare I" =SV,V‘ =_5V + s v Mb F r UE leﬂnlsfsams g i :fKEF ‘ : =5’Bu Yea-«r We 4- MA ‘éﬂ ’EV Assume = = 6.56 L A L a 30 2 IO = so = [T 6.56)(Vm — 05) 2 Va“ = L052 V Need 5 transismrs in series Var = 10-1952 =13“, Then ‘ 80 W 1 I =80: — —— 139—05 m [2 L]; ) =~ W — =120 [L15 1 (C) )3... = m Where R. = RmIRm Now 1a,", = 20.9J656 = 535 ME} Rm = 563J656 =144 Mn Then R, = 535nm = 39 Ma fM=—————l—=136kHz 2::(3sﬂo‘)(3x10'”) GBW = (20_ooo)(136x103) :> GBW = 272 m2 Des' [3.34 A: = 3-(Ma)‘[r.11(Ma )l'h‘Qzﬂ g,(M,) = 2 “[522 25}(100) = 447 m f V 1 l M = ——-—- = -——-—-—---— = 500 kﬂ "'4 3) um (0.02)(0.1) V 120 fﬂ(Q2) = Ti- : 3-1"- = kﬂ co - Then A, =447(05|12)=> A, =1ss 13.35 An! = g-(Mz) '[5:(Mt)ﬂraz(gz g_(M,) = 2 [%}25)(100) = 632 AA 1 V 1 1 = —— = —— = 57 m raw” ,um (0.015)(0.1) V 30 r, (Qz)=—4-=—=300kﬂ ‘ 15, 0.1 A, =(632)(0.66‘7l|0.80) = A, = 230 13.36 Iw=2oopA K_=x,=o.smAW= A_=A, =0.015V“ where. Ru = 3.10 draw) Ru = Sith'hﬂ’n) Now g_ = 2,fK,I,,, = 2,ﬂos}(0.1} = 0.447 M W 1 1 = _ = —— = 667 "' 1,1,, (0.015)(0.1) m =667kﬂ 0.1 —-— = 3.346 MA I V 0.026 E“ It :1 |§~ ll r - “.1...— = -1— — “ ,um (0.015)(02) ' 1 1 r“ = 1,1,, = (0.015)(0.1) = 667 m g_. = 2,fx,1,,. = 2 (05x01) = 0.447 m I V l""lvscliti n ' ns u 50 Rd =(0.447)[667)(66?) :9 193.9 M!) 11,, = (3.346X800)(333ﬂ667) = 633.4 Mn Then A, = 447(1989ﬂ6814) z: A, = 63,865 13.37 Assumebiased at V‘ =10V,V' =-10V P = 3!,,,,(20) = 10 :9 [m =16? ,0; Ad =g.1(RusuRal}=251000 k;=so,uAW‘ ,k; =3SyA/V‘ 2., =0.015V" ,1, =0.02V" Assume =22[—F—V—) L J, L , Roi = gﬂ[ralrnlo) Rog = gnud{rad)(r94lP-") 1 1 1 = 0.50 MQ °' 2.1,!” f k' W 35 = _-' _ 1 =2 — 2.2 X2 33.3 3.: 2 2 L1 m [2 ) ( ) =113.3X h X“: —" W are r0: rﬁ=£=£=0360Mn 1,, 33.3 r —__1__—__'.1_=040Mn “'AJD, (QOISXIS?) ' r,-;=—l——-—=o.50m ‘ 11,10. (0.021333) "u—r g_,=2 [*4 E] In. =2 2211x1013: 1 N w R: = (3204)(0.960)[O.40|0.60]= 7311 M!) R, =[113.3X)(0.60)(0.60)= 4031' M: Then A, = 25,000 =(113.3X)[7331403X] 30,110): _(113'3X)[738 +403X] which yields X = 2.48 Of =(2,2)(6.l6)=12.3 P I133 For Wuhan). mum: Vc3(Q5) = D. Then V5=15-0.5-0.6= 1-3.8" 0.236 159 =11)“; = T 3 0.113 IRA Using parameters given in Example 13.11 f1 um V = A—V =J_—+l.4=2.l7V " K, ”’ 0.20 Than vmtmu}=13.8 —— 2.17 =» 9:...(mu) = 11.5 V For Hm(min.). assume VD(M9)=VD(mr)=V-ﬂ “’17 = 2.17 — 1.4 = 0.?7 V Now Vmo = Imo(n.5) + 0.6 + 19140.5) - 15 = 0.113 + 0.6 — 15 : Vm = —14..2s V Then = -14.23 + Vsn(m...
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