{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chap006

# chap006 - Electronic Circuit Analysis and DesignI 2nd...

This preview shows pages 1–22. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Electronic Circuit Analysis and DesignI 2nd edition Solutions Manual Chapter 6 ExerCisa SOlutionS C. If. = 0.4 sin a)! = #4, = -[3.9|(0.4)sinut 56.1 lid, = -3.56 sing! g_ =2K—(VG£_V1N) and 'AI V951 = 6 «- 3.56 = 3.4-1 1 _ K {V )2 VGS1 = 2.39 + 0.4 = 3.39 D - n 63‘ TN 0-75 = 05W“ 433): =3 V53 = 2.025 V VG“ “ VTN = 3-29 “ 2 = 1-2? g =2(05)(2.025-0£) a g = [22 mA IV 50 V05! >V631‘Vm =5 13mm EGA- "- : [Mn (Va: " 1"1'.'tl)1].1 = [(0.01)(05)(2.015—o.s)’]4 = 5:133“) 7:12-IDQ{E}=’!iDQ=0-533 [ILA 109 = [90,56 ‘1‘”er a. Van = Van *15030 E61 0.33: = 2(1/55 - 1}1 =. 1/5; = 1.55 v gm=1'.it'u(‘.fﬁJ 4’“) and ID=KI(VG\$‘V7-N): _.__...._ ,1 . =2K v —v =22 1.65-1 =VcS-Vm = 7?— and b g" r[ 56 l TPD ( X ) :3 gm 2 '16 mA/V. rg = 3c r, =[RK.(VG, vmﬂ‘ 4152]" A, = ”—° = -9mﬁo = -(216)(G) r,=[(oals)(2)]‘= n=333m ”54.11555. 56.3- 136 5 _ = V — 2 I I' [be K11 G: TM) 109 = Kn(VLT-S- TN): =VGI-Vm = ?’Q~ 0.4 = 0.5(Vas — 2}’ => V2: = 2.39 v . VDSQ = VDD - IDQRD = 10 — (0.4)(10) g“ =2K.(Vm-Vm)=2K. 10\$ :3» Vpsq = S V K b. g_ = Mania, — v”) = 2(05)(239 - 2) a g... = 0.39 mA/V 55.6 r°=[A!DQ]-l, A=0#ro=ou __ Y — ﬂ— , = ."—°_. = —g...Rn = —(o.59)(1o) 2:124”! ”is 1’- 5 A, = —ﬂ.9 Electronic Circuit Analysi§ and Design, 2'“ edition Solutions Manual 040 = ' :9 = 0.104 a”) '7 231035)“ "——— g_ = z (Ejm' = 2,}(05)(0.75) = 1.22 mA IV For (a). g... = 33 = (1.22)(0.153) = g_,, = 0.137 M IV For (1)). gm, = (1.22)(o.104)= g“, = 0.127 m IV 56.? 3- fog = Kn (V65 " m)! = (0‘25)(2 ‘03): => I52 = “.35 [11.4. Vpsq = Van - Ipqﬂp = 5 - (0.35H5} => 1352 = 3.2 V b. g" = 1K,(VG, — m) = 2(0.25)(2—03) => 9... =0.6 mA/V. re =ac c. A. = 1 = -§mRD = —[o.s)(5; 56.8 v. = 9,, = 0.1 sin wt i4 = y..u,, = (D.6}{O.]Jsinwt ii = 0.053inwt mA n, = (—3}(D.1}sin.ut = -0.Ssin .ul “min = IDQ +1} = 036+ 0.065inwt =ip EA. up: = VDSQ + 94. = 3.3 - 0.3sin‘r! = Va: 56.9 1"ng = 3 V and for; = 0.5 m.-"L aac=5013=39=4m I” = K,(vm -|v,,|)1 0.5 =10}; -1y2 :: v5.5 = 1.7.1 \’ => V5.3 = 5 - 1.71 = V95”- : 3.29 \- -4-u = “gm RD g.=2 Igloo =2 W05} 9... = 1.41 mA/\" A. = —(1.413(4J:.4. = -5.54 A, = a = —w.4 = _D.465in-..~:I = _5_“ Vi I". II'I => v. = 0.08168inu! 56.10 1 a. Vsa = 9 - Ians. [are = KP(V5'G 'lvﬂl) Vsc = 9 —(21(1.2)(vsa — 2)” = 9 - 2.4(V525 - 4Vsc + i) 2.4vs’a — 3.51/55} 0.6 = a 3.5 :L- \/[3.5)’ - 4(2.4}(o.5) 2(2.4) v55 = 3.51V. Ipq = 2(3.51 — 2)1 a log = 4.56 mA Vsoq = 9 + 9 — IDQU-z + 1) = 13 — [4.56)[2.2) : V559 = 7.97 V V55 = V], = ngsaRo A. = —ngo = -(6.04}(1) => A,: = -6.D4. H" , Lu ? =23 K =(2o)(25)=soomrv= '55: g+15=35V=>V5=-25V Au="§m-RD y... = 2(o.5)(2.5 — 1.5} = 1 mA/V ForA.=—4.0=~Ro=4kﬂ. VD = 5 - (0.5m) = a v = Vpsq = 3 — (—2.5) = 5.5 V E612 3.. mm Ra = Vcs = Vns => transistor biased in SEE. region In = K.(Va: ‘me = K-(Vns ‘Vm )1 Va: =Vw ‘Ian =vm— K.R.(vo. - m)‘ Test Your understanding Chapter 6: Egcrcise Solutiong V95- =15 — (0.151(1unvn; — 1.11)1 K = 1;, + g... V..Rs = V“ = v: _ 1+9 R- =15 — 1.50/35 - 16%;; + 3.24) W: __ —9' V RD In a — m 5H 1.51/3.S - 4-4V“ — 10.11 = o g_ = 215.0%; —v,,,) = 2(03](-1.35+2) V 1.4:h1/(43ﬂ1 +(4)(1.5}{1u.11) D _ s _ = 1.01 mA/V 2(1-51 A 3 ya = —ngo _ —(1.043(T.331 =5 VDSq =4.45 v ” V. 1 +ng5 ' 1 +(1.o-1)(4) hm = (0.1539145 .15}2 a. 190 =1.os mA =:- A. = —1.38 6.14 b. Neglecdng client of RC,- : E a. 5: [MRS +Vm. and A» = -!m(RD"R1) Ina = K,(Vsa +1117): 3" = 219M: 4!“) = 2(0.15)(4.45-1.3) =, g z u 795 mAfV 0.3 = 0.50%; + 0.5)’ => V5,; = 11.455 v A. = -(o.795)(10[15) =:~ A. 2 4.55 5 = (0.5185 + 0.455 = R5 = 5.57 m Van =10 -— [chLe + RD} 1:. Kg =» establishes Va; = V05 =5- ésscntiaﬂy no cﬁ'cct 3 =10 _ [O‘BHS‘ﬁj t‘RD} on small-signal voltage gain. RD = W2: :5 R2 = 3.08 kn V V IDQ = 0.3[2 —- V55}: = -—55— = i 3.2(1 — {V513 + V5113) = VSG 3.21%; -13.avsa + 12.5 .—. n 13.3 t \/{13.s)' — 4(3.2)(12.8) ' = v = mv, = - "v.- R Vsa 26m) 0 9V 9{Rni1rol 9' ( ollfo) vs; = 1.35 v =9 19¢, = 0.3(2 — 1.35:1 A» = 7° = -9m[Rollro} = M g_ =2K,(v,a +11") =2(05)(o.455+os) = 1.21 mAJV b. Vase = Von - IDQKRD + RS} _ 1 _ 1 ro-m—mﬁ—SZJM} 6 =10 - (Hamlin + 4} A. = —(1.27}(3.oa|152.5) = A, = 4.12. _ 10 - (11333)“) - 6 _ RD — 0.338 => RQ —- 7.53 1:51 56.15 C. V0 = Univ-giro v; =v,,+v;,==.v,,=v.—La So 1'3 =9mruﬂ’. -V21! 1 _ 39._ gmrn _ (411591 ‘ " " 1". 1+ gmro 1+ [HIE-0} E cctrouic lrcuit Anal sis and Desi n 2“‘1 edition Snlutibns Manual v ' . 1" +§mv'; = l'_: Ind Lil = —“, h=gmu+—=Ro=f0”—=501|i :9 £9 E 0.25 kn WW! 35 =4 kﬂza- AN 3 9”("0| RS} 1+9mfrnllﬂsi (43(3‘TI = ' = .7 ﬂ . I, = --—--— TDIIR; 30"“ J k s ‘i 1+[4H3-1‘] = A. = 0.937 15.6.16 V1259 = V00 - IaaRs 5=1 10—(1. 5)Rs=>RE=3.33kﬂ 1m =K_V{G, —v,,,) =5 15: (”(1113—05) Voszzmv =va_v5_vc..5. R: R: V = _ . 314-32) DD 400 ID So R; = 280.8 [(0. R1 = 1192 1:9 =§VG=T.02V=( Neglecting R5.- . A” = gmtﬂsﬂrn) 1+§m(R-5'|1To} 1.. = [2190)" = [(0.015)(1.5}]" = 44-4 kn Rsllro = 3.33II44.4 = 3.1 R10 3,, = Lila!” =2 (1X15) =25er {2.451(32) _ v- =Wﬁfu—DJB‘I 9—,." Rsllr "2—15 1333115“ =0. I1108||3.1 =2 R4: = 0.35 m 55.17 a . 9.3 __ V“ = (R. fay“ = (mu-9.3)“ =0.531\-‘ Im=K.-V(sa -|an) = ‘90”: VG‘IVWI} _ 5 -— 1':- .. R;- Thm 10.015102 — 0.521- 0.5)” = 5 _ 2;. 20’s - 1.151)1 = 5 - 1/: 2W3: ~ 2-752)”: +1.90?) = 5 _ 0;. 2V5“ — 4521',- — 1.19 z 0 1.52 2 Jun)” + 4(2)(:.19) 2(2) Vs=2.50V=rqu= gu=zm_ W: 0394120111! v. = 5—5'25=0.5mA A»: 3-K: . R1ER2 l+g_R, !«'.',HR,+R£r 4 . . = (039 15) ‘ 707u93 :3 5:11.770 1+(0594)(5) 70.7ﬂ93+05 — Neglecting R .. A, =08” mzﬁsl 0—m=° 51" 0-594 = R5 = 0.915 m B618 (:1) gﬂ=2 Knlm=>2=2 K_(DB]=> K =l.25mAlV2 Kn =#"—C"vE=L=25 (0.020)?) 9, L L 50 E£=¢525 Jr...— 1 DQ=KIM Wm): I’D-8:1-25‘V05'2}: =:» 1‘9: = 2 B \' 1:. r0 2 ”mar: [(.)001).(0 s )1“ = 25 m ___ 9m[mi|RL‘1 ' 1+QM(FD1|RL) rDHRL = 1'35i|4= 3.55 _ (2113.30) __ A... — ‘—'"_1 + (21(3 00 =. .1. — 0.556 Rﬂ=gl— ro=-1125=>Ra~0.5kn 136.19 2 92 = p(VSG ”WI?” 3 =2(vs._-, - 2)2 =- V55 = 3.22 v 5 — V 5 —- 3.22 Ipq = £556 = 3 = R5 a» R: = 0.593 kn = wool“ .-. [(0.02)(a)}" = 10.7 m g_=2 Kg” =2 (2X3)=4.9mAIV Hmﬁ'nllﬂs} 1 +9...(rol1.Rs) To‘IRS' =15.T||0.593 = 0.573 kg (2.9)(0573) ‘ A = —— .1. = 073-. " 1+ (4.9)(0513) :——-‘—— ForRL=o¢. Av: If A. is reduced by 1.0% =- A. = 0.73? — 0.073? = 0.653 Stu-(rullRSIIRLI ,1, = -——-———-—-——-— l+!m(fa1lREIIRL) Test Your undergtanding Chapter 6: Exercise Solutions LG! TOIIRSIIRL = .1: (4.9): l + (1.9): t = 9.402 = 0.573"RJ_ 0.573RL EL 4- 0.573 :1 (0.57:1 - 0.002151 = {0.402)(0.573) 0.563 = = 0.663 = 4.9::[1— 0.663) = 0.102 = R: =1.35 kn 136.20 .,.-.- 3L=005m=gm=2asmAn ll’0 “In — = R R z 2. 4 = R .. r. all 1 DIM—4+1“) [4 — 2.133,; = {2.4)(4) =- R9 = 5 kn gﬂ=2 Kl‘rm 235=2JK_(05)= K. =4.09 m4.le 1 log: K.(Va5 ‘Vm) 0.5 = 4.09{Vas —1)"’ =1. V135 =1.35 v :Vsz—IJSVI i-‘b=5—(0.5){6)=2V We have V“ =3.35 >VG§ —V,,, = 1.35— 1 = 035V :0 Biased in the saturation region E621 Vi! = gravidﬂﬂllﬁl.) and V”. = V. A» = PM‘RDIIRL] 5— V =_;:£= K rvmt: "Iva-D s - 1:25 ={11{4)(V.=a - 0.01” a - V15 = 411/515 -1003“; + 0.04} 411-c-5.41'_-a—2.=14=0 s 4::- 1/15 41 + 1411-1112 44) [be V =——————-——__. 3G 2(4) 11.33 = 1.7] V I” = ° 11'”: 0.022 111A gﬂ=2JK Im=21“1)(0.822)=l.81m14lv A. =11.01](2t|4] =(10111(1.311= m __ = "" I91". ‘ "—131=4I|0-°°- 9W Rm = Rs 56.22 C W K". = __’”- .. [__] =(0.020)(30)= 1.15:1:21/1'2 2 L , K_ = '” C" £[%lm =(0—020}(1) 0.020 mmvz —- - =- —— => = -3.94 A“ Kn 0. 020 53—— The transition point is dctcrmined from Kn "can — vma = van _ V1741." I "F”(vas: vmn} n2 v55: —- 0.3 = [5 — 0.8-) — {8.94HVGSI — 0.5) (s - 0.51 + (3.941(00) +0.3 ”55‘ = 1 +3.04 pas, = 1.22 V For Q-point in middle of saturation rcgion VG: _ 1.22 — 0.5 _— 2 +0.33VG\$=1.UIV =E—"C" K:_[W1K“ =.(0015)(2)= 0.030mmv2 -36 K.)=(36( 0.)03o —103 ”0111!1 1.08 =(0'015)(E]1 =1 (%]. = 72 The transition point is found from Vcsg—l={19-1]‘{5NPGSI-n 10—1+5+l =2.29V 1+5 ”GS: = For (2-point in nﬁddl: of saturation region 2.29 - 1 Vac: 2 -.-1=>Vr-[c=1.615 \- E624 (a) Transition points: For 14,: v0, = v“, -|vm|=s— 1.2 = 313v For M1: K.1[{Vm )1“ + hm” = Kakvm )20 + ‘12P” - v“ D] zsqvi. +(0-011V3u] .. ”[02” 1+ (0 .(01) 5-) (0- 03%)] Icctronic Circuit Anal 5i and Desi 2nd edition Solutions Manual Hing“ +(o.01)v;‘]= l512 — 03:11am“M {0.00% mg,“ +0.00l4-4vm —0512 = 0 which yields um i 0.388 V Then middle. of saturation region 3.3 - 0.388 [log = + 0.358 :3 VDEQ! = 2_CI9‘I V Knlkvcsl ' VTND}2(1+ Alva )] = K..[(Vm.)‘(1 + 12%» - ”011] ﬂout/cg; — 0.512(1+[u.01][2.094])‘_ = 25[(1.2]’(1+[o.01}[5 — 2.0941): m[(va,-, — 0.31%].02093‘; =1.m (v55, — 0.3}: = 0.145 :3- VEE] = 1.15 v b_ 100 = Kuhn”. —o.s)’(l+(o.01)(2.094))] JDQ =.-(0.25][(0.145)2(1.02094]] => IDQ = 37.0 HA 'Fgml = m = wmtrmnrm 3.1: 2K41(Vcsa "Va-no) = 2(o.25)(1.13 — 0.5) = 0.19 mA/V —n.19 c. A. A”: (G.037)(ﬂ.01 +0.01) =’ ﬂ 15.6.25 1 Ho = Rs: “ —' 9”: 9m: = 0.532 mA/V, R5; = 3 kn 1 - Boas“ m—a|11.a3=§2_1.32m 56.26 I.. fog: = 211124.. Vqug = ID V J'aq:-ﬂ.—-:=10=ER\$: =>R§3=5lcﬂ a lag: 3 Ka2(v551 ‘de 2 =1(V::52 — if =- V501 = 141V 9 Va; = 3.41 V [0 - 31%! “than an. = ——;-—— = am =13 m For VpJ-qi =10 V =- Vs; = 3.1] -10 = -6.39 V Then R51= w =, 11., = 1.71 m IDI = .1013: - I'm)1 2 = [(1,651 - 2): =9 V651 = 3.11 \' R: 65‘ (R1+R,)£ 3 100131 R: _ 1 121+ R, “' R1 Rm 3-“ = income) — (mm) 1 => RL = 556 kn 586R; _ _ _ _ - EST :2, * 2"” =° We - 200m: —(2001(a85} => R7 = 304 kﬁ' b‘ Sm=21aniIDQI =2'JU‘X2) =:' gm = 9...; = 2.63 mA/V From Exampl: 6-16 A _ -9Ml§m2RDl(-RFJHRL) k |+§maiﬂsallRLi REINRL = Si|4 = 2.22 m —{2.83)(2.83)(3.3)(2.22) A“ 2 1+ (2.53)(2.22) => 14., = —8.06 R... = —1-|| £5: = —1-||5 = 0.353115 9M2 2.53 =- R0 = 0.330 kn E627 "' IDQI= nI(Vo:I'Vm:)1 1= Lewes, — 2)“ => V051 = V652 = 2.91 \-' E5 = 10 kn =3 V51 = Iqus - 10 = {1)(10) -- IO =0 _ _ Ra Va; -23: _ (—R: +3“ EJUD) = (3%) [10} a R; = 145.5 kn Vase: = 3.5 =- V5: =15 V =- 3.5+ 2.91 :9- Vm = 6.41 Pa: = (m)(10) =s.41 R2 + R3 = 320.3 = R: +1i5.3 =9 R3=175 kn T113513:+Ra+ﬂa=500=ﬂ1+175+l‘£5.5 =5 R] = 179.5 Hi How Vs: = 3.5 =- VD: = Vs: +Vanz =3.5+3.5=TV Test u nde tandin 10—? 503.9: 1 #RQ=3|¢Q 11. From Examplc 6-18: A» = ‘91!“ RD g_l = afﬁrm = 2 (1.2)(1) = 2.19 mA {V A. = —(2.19}(3) = A. = —6.57 156.28 From Example 6-13: 9... = 2.98 mAJV. I'o = 42.1119 MIR; = 420||1so =123 kn .“ 12.1132. . "" ‘ (£11111: +3.)“ / 126 " 123 + 23 )W = 0.36314 -ym"}.1rollﬂoilﬂa1 1.; = -12331111.3331(42.11|1.1||4} At: 42.37 :3 A -—3 -....— [42.1;|1.31) =.- 415711135) 3 [L E629 V3 = [1303.3 = [1.2}(2.7} = 1.24 V V1: = V: + Vosq = 3.24 +12 =1524 — :2 RD =20—121—H3 RE =3.97kn Va: 2 In - 1055(1 — W) 2 1.2: 4(1 - V—E’i) =3 515-1: 0.352 V.» Va,- :1: (3.4521141 = 4.333 1},- = 1.;— + 12;; = 3.24 —1.333 = 1.33 v . R: _ R: _ vs a (R1+Rz){201_ (5)123] — 1.33 =9 R; =47 kn. R; = 453 kn l 1 ° = 11119.; (3.0115102) =16? m _ 21055(1_ Vas‘ _ 2(_4)(1_ 1.356) 901 - (—VF} . — l' . VP 3 3 = L46 mA/Y A. = *9mEToIIRDEIRL} = 41451037133711.” =3 1-1... = —2.37 E630 _ R: ‘ V51 (121+ R1)(VDD} Ch ter 6: Exercise Solutions v ‘ v —v 1 1nm=1955(1-_g) =5(1__G‘_.i) VP 2 172 V511 51 2 = -— --— =6 —— 3(1 2 +2) 2 73 20—1! andlpq1= 51 4. 2 “1111-1120;“TI = (\$4.15) V2 20 - 1/5,: 24({4 - 7.61151 + 57.73) = EVE-1 —132.4V51 +1335.24 6V§1 — 181.4%. +1366.24 = 3 131.4 i 1/(13131’ — 4(3)(1333.24} v = 51 2(6) V5; =14.2 V = V551 =11: —14.2 = 3 V > V; SO V5] = 15.0 9 V331 = 17.2 — 15 =1.2<Vp—Q 20-16 0111qu: :Ioo1=1mA V5901 = 30 - 19011351 + Rm} =20-(1113) ==v V5199: =12 V v3. = 133.113. =111<11= 4 v = Va: Ina: = 1033(1 - -—.i = 61/}. —— 731's, + 215 = o 1/ 73:1: 1/1711” —4(5}(2161 ’2 2(6) => V5: = 7.59 V or = 5.05 V For V5; = 5.03 V => V65; = 4 -- 5.08 = -1.03 transistor biascd 0N lcctronic Circuit nal is and Desi 2nd edition Solutions Manual [Baa-J Elf—E: [DQI =1.11mA b Von=2u-Vsa =20-5.05:V253 =1‘L9V I} h. V32 = PMIVIMHDI = “Emivu‘RDl V0 = gm2V¢.:(RnIIRL) V92 = VII: + H) =5 Vail = A— 1 + 9m2(R\$21|RL) V0 ‘PMIRDI A..=—-' V. — 1+ 9m2(R32ll-RL) ll 2I953( V65) gnu 1- —."' IVPI 1'9- - Hi]. _ 1:3 _ v _ 2 {I 2 ) — 2 4 mad gm = 2(25 ( _, _) = 2.15 mA/V Than A, -[2 “(4) = —2 as = A” Va: 2 Vas_ . 2—3[l-v—r) =- VP —0.a V55 = [man—3.5} =- v5; = -1.75 -V 3 - (-10) A150 Ina: Re 2:131:22: 35:5.aam Rs 2fnss( V65) 2(3)( 1.75) 5!". = 1— -— = .— ]_ .— lvpi VP 3. 3.5 .—. 2.29 mA/V m = m = 50 m v. = V9. +ngsVy = V}. = WET A“ = 1/3 = ngS-"ru _ (2.29)[5.as|]50] V- 1+ymRero " 1+{2.29)[s.33||su] a .m— gmtasnm _ _ = 0.911— 0.186 = 0.745 1+9m(Rs\iRu (2.29){R5!RL) _ 7 1+(2.29)(RsI!RL) _ 0‘ 45 {2.29)(RslIRL)(1 - 0.745) = 0.745 =§ RSIIRL = 1.28 kn 5.38121: 9-1-23 5.55“!- R; => (5.33 — 1.28)R;_ = {1.38)(5.85} => R; = 1.54 kn Eectronic icuit Anal sis and Des’ n 2“d edition Soluti ns Manual Chapter 6 Problem Solutions #ﬂcﬂi W ‘ (a) s..=2 Kai’s =2 [HZ—17:)!” 0.5: 2 (o.%020)(—)(0.5):E=12.5 W (b) 1., =[&§L‘XI)(VGS ’Vm}1 0.5 = (0.02)(12.5)(sz _ 0.0)2 :- Vi: = 2.21 V 6.2 J“ C... W ‘a’ “=2 “”2 [ ’2 II)“ (“129): = (10)(¥)(100} => 3:): = 0.025 (b) 1 =[#.2 C“I—W-)(vm +v,, 100- - (10)(0.525)(Vsc - 1.2): :- VEE = 4.2 V 6.E 6.3 ID = Kn(Vcs ‘ m):(1+lvm) 1m. 1+1VDs1 3.4 1+MID} ___,=——-—..——_ —— 1m 1+Ame =’ 3.0 ‘ 1+,\(.=.) 3.4[1+ 5A] = 3011+ 10).] 3.4 — 3.0 = Aux-10 - (3.4) - 5) =0. A = QﬂﬁQS 1 1 '“ " 1'1; ' (0.0300)(3) =’ 515—1—949—3- 6.4 :D=1 1 6.5 A” = -gm(rnllﬂg} = -[I}(50“10) E? (L. = -3.33 E=W=~ID=Qﬁ3EmA 6‘6 v v 10 5 _ DD - Dsq ___ — ‘- R” ‘ Inc 0.5 For Vase = ‘1 V for ample. #(IC ('1 10:2- =““2—( 1% )(Wm‘ Hm 0.5:(0.03:](%)( 2-0.B} W 115 :—= . L b, g_=2 K‘Im =2K.[Vasq 42,.) 9,, = 2(0.030){11.5}(2 — 0.0} =5 5..., = 0.035 mMV [W = (0.030)[11.0)(2 - 0.0)2 = 0.501 mA 1 _ ' 1 =:» Mag ‘ {0.015)(0.501} 1'0: r0 = 133 RQ c. A, = —gm[ruliﬂn) = “(0-335)(133HS} =-.~ A, = —6.30 6.7 Kati; = K_[v,, 5mm]1 _ K 1/1 sin m: l ‘1 sin1 (or = 1[1 —cos?.wt] 1 So Knv:,= K—‘f” [—1— cosZcut] Ratio of signal at 200 to that at 01: K v1 up ~coslax ZKHVm—wm), sinax The coefﬁcient of this expression is than: _"e__._ 4(vGm - v," ) 5.8 0.01 = —Y£---- 4(VGSQ ' Vm) 50 V9: = (0.013(4}(3 — 1) => VI' = 0.03 V so .- - .-= -V.-= - R,||R,+R,, (son) (0962)“ I; = —g_(0.962)(r;[Rn) = ~(1)(0.952)(50l10) =9 :1, = -B.02 :RQ=BkQ Electronic ircuit Anal is and Desi n 2"cl edition 6.10 A» = -9m(rollﬁnl — 1a a -gm(100||5) =3! 9... = 2.1 mAfV 6.1] R 2 = V a. VG (121+ R2) DD 200 “G = (mill?) * 4-3 V V Va: 2 In: -£-—-—= K..(Vas" 1'») RS 4.8 - Vas ={1)(2)(V§3 — ”’65 + 4) ZVés — 7V6; +3.2 = o 7 i ‘Im' — ¢{2)(3.2] _ 2 95 V 2(2) In = (mus - 2)’ =3» I; = 0.920 mA VD: = Von - IDERD + 35') —.— 12 .- {o.92)(3 + 2) =2- V2: = 7.4 V (b) V. _ -Igul 51(RDIRL) 1+gnR5 whm V113 |R . zoolzon R,|]R,+R5, =3oo|l200+2 ' 120 — V . 4 120+2 (093 ) Then 1+ g_R, g_ = 2(1)(2.96-2) = 1.92 M IV 30 A. ___ -(1.92}(o.934)(3||3) [1.033(2) = ﬂ Va: {-ng 1.5.1 \ 51.4525: 11—. = ‘ —1 &' = - 'D 3.5 m Ay“ Aip = 0.92 mA => {Aral = (o.92)(3.5) = 3.22 a 6.44 V E-ﬂm 1143+? Ex. Solutions Manual 6.I2 a" [DO a 3 mA :- V5 =11;qu = (3)035] = 1.5 V [m = 1mg"55 41m): 3 = (ZNVGS - 2)“ =- Vas = 3.22 V Va = Va: +-Vs = 3.22 +1.5 = 4.72 v R: l = v = .._. m VG. (RI + R2) DD R: R. 'VDD 4.72 = Riemann) =:. 3.: 535 m 1 636R: 636 + R: 'leRDIiRL) 1+gm-RS 9". = 2(2H332 - 2) = (.55 nah/V - "(4'88)” '5) =-.» A. = —2.03 ” ‘ 1+ (4.353015) b. Av: 6.13 From Problem 6-1.1: ID = 0.920 mA '55- = 7.4 V gm =13? mAIV Av = —SM(RD||RL{R_IHJ;I—1|T—Rs:] =4L1:2)(3|3)[2:0"———°(:§l030032 ]= 422310.934] .4, =-2.83 c232. -l 1.5 k9 :- IAmsl = (D.92)(1.5) = 1.38 =:> 2.76 V pubic-peak output voltage swing big = - (3.91:5. Alp = 0.92 mA 6.1-4 Chapter 6: Pmblcm Solutions Vnsq = W - IDGRD — (-Vasql Output Value: Swing- — Vnsq - V5593] =[v* ~1 m11,,+1»'m]—( 0"ch —v m) =V’-[DQR,,+Vm 5° Ian=laa =‘_“' AVosl l kk‘-l=—r[v*—1m2 R +vm] 5kg. for: = 5 THIS ‘- J'IJ<.'1(1‘:")~1- 1] = 1.2-2IDQ 2190 =9 [no =04 mA=IQ 3,. = 2.111.159 = 2.!(05}(0.4) = 0.894 mAI‘ v 1 1 '° 11m”{n.o1)(o.1)"25°m Av = —Qm(-RD "35“,“) = _{n.891}{1ol|1o|12511} = A. = -1.38 6.15 3- 1r1) = 51'an: ""21111)2 2 = “v.” - (—1))3 V55 .1 -0.293 V :- Vs = 0.293 V = InRs = (2)35 =5 R; = 0.146 H2 VD = VD: + V:- = 6 + 0.293 = 6.293 R1: =W#RE =1.85kﬂ _ A” = -9m(ﬁnllﬂL] b l+ymRs 1-321? (V6.1' 111:) g," = 2(4)(— —o.293 + 1) = 5.66 mA/V A. = —(5.ss)(1.35| 2} =1» A. = —2.93 1+ (5.561(0146) 2.2-1.4 A... = (151911111552) = (211125112) =1.92 v 1.92 .'=—-=D.645=>V.=U.645V A” 2.95 _ 6.1.6 1- V1251: = V121) - [130(R1: + Rs) 2.5 = 5 -— 190‘? + Rs} I _ 2.5 no _ 2+3; —V Im=K.[Vcs‘Vm)2=_E§i I _ 4.535 = Va: — --IoaRs — 2 + Rs 2 K“ 4.512, “Vm = 25 2+R5 2+R, 2 . 4 1 _ 2.5R5 = 2.3 2+Rs 2+3: 4 2+R3—2.SR_.7 ’_ 2.5 2+3; _ 2+ Rs (2—1.51133 4— =15 2+Rs 4(4 —- 685 + 2.25113.) = 2.5(2 + R3} 9121. — 26.53.; +11 = 0 25.5 :1: 1/065)" - 1{9y(11) R5 --——-——-——— = 2(9) R5 = 0.5 R“ or 2.44 1:9 But 12; = 2.44 119 => VG; = -1.37 Cu: 05'. =9- Rg = 0.51:9, [pg =1.D wit. 1.. A. = -9«-{RDIIRLJ l +9m-RS 3,. =2 I11!m =1.I(4)(1) =4mAiV .. = ~(H21I2) a A. = —1.33 1+ (1)(11.5) 5.17 I I. 5 = Ians +Vsna +IDQRD “5 5 = Iqus-I-E +Ipq(10} -5 1 1. Inq=R +10 ——._._._E.._.... Vs = Vsoq + Iano - 5 -"- Vsaq 2. 1+ Inc-(10) = Vsco 3. _ I Inc — K,(Vm ’2) Electronic ggircuit Analysig and Dcsigg, 2"I1 egjtion Sgllmggs Mgual 4 Choose. 35:10 knafna = 20 =0.‘2Dmt\ Vscq = 1 +{o.2)[1'(1) = a v 0.20: K,(3—2)‘ = K, =o.20nw152 b. log = (0.20m — 2)2 = 0.20 mA g_ = 2 KP!” = 21/(o.2)(0.2) = 0.4 mA I v A, = —gm(.ﬂp"RL) = -(D.4)(IDHIO) 3 A. = -2.0 c. Choose. R5 = 20 kn =3 Ipq = :7] = 0.133 mA "5.5:; =1+{ﬁ.133}[10)= 2.33 V 0.133: 1:42.33-2)‘ = K, =122mmv1 g... = 2 (1.22){o.133) = W A, = -(o.aoe)(m||1o) = A“ = —4.03 A Large: gain can be achieved. 6.18 L :m =1: K,(vm 1114,.)2 1 = 50/59.; — 1.5)3 => V564 = 1.95 V R; = 5 —11-95 => RE '2 3.05 kn Vgnq = 10 —- IpQ(Rs + Ru) 5 =10 — (1)1335 + Rn} 9 RE 21.95 kn b. g_=2 K,1m=L/(5)(1)=4.47m1v A, = —y..[Rg[[R;,} == —(4.4T)(1.95}|2) =3» A" = -—{.41 c. 1 0.937 m => Impsl _-. (1)0351} = 0.937 v Swing in output voltage = 1.91 V pubic-peak Ain = - -Al¢gs 5.19 Ina = Kn(VGIa ”VTN) g_ = 2 If"!m 2.2 = 2% = K. = 0.202 "1.11le 6 = o.202(z.a Wm)” =1 vm = £2.65 v 2 Vpsq =13 -IDQ[Rg+ RD] RS+RD=13;'° =13: 20:35:1113—39 A = _9m(RDIIRL] ” Hymns Rp-l '2‘2(Rn +1) = 1+ (2.2)(1.33 - RD] 2.2RD 1+Ro (3.93 -- 2.235)“ + R5) = 2.2511) 3.93 + 1.7311; — 2.21::2 = 2.2129 2.21:}, + 0.4735 — 3.9: = o —o.47 +1/(0.4T)’ + 4(2.2)(a.93) RD _ 212.2} =a~ R9 =1.23 kit a; = 0.10 m vc=vﬁ+vs= 2.3+{s)(o.1)=3.4v 1 1 V = __ . "I . — __ — _ 5 R1 R. Van RI. (100)08) 3 4 => R1 = 529 kn 5293: 529 + R: 1 + 3.93 - 2.25:; = 2200: R2 =123 kﬂ ’" / /;1 i 4 1435:") lg,“ V. V1 = 9+Vsa~ VIA-7m} = Vsa *‘Vrp V -V Vina =J—:EIM+VSD(5‘") =(9+Vm)”(vsc+vrr} 2 = 9 +21.5 + Vsc _ 1.5 Vqu = 3.75 + V55 = 9 + V35 - Inqﬂn 2 I” = KAVSG +Vn.) + (V50 + V11») Cha ter 6: Problem Solution Set RD : 0.1.3.1, = 2 H? 3.75 = 9 — Ipq[2) => 1m; = 2.625 11141 b; g_ = 2 m1” = 2.!(2)(2.625) = 453 MIA IV 1 1 ° = um; = (0.0l)(2.625) Open circuit. A» = *9m(Rnll"01 A. = —4.55(2:1as.1)==» A, = 4.70 l' = 38.1kﬂ a. With R L A. = -4.58(2|120||38.I)=» A. z -195 = Cums: = 5.62% 6.21 ‘- ’09 = Kr-{Vsc +Vrr)2 2: (0-5NV5'G + 2‘]: =2: V53 = G V 10 —0 R = 5 2 Vn=u—vm=o—s=>ﬁp=:5_—(ﬂ q .- =>R5=5kﬂ #RE=21(Q Av = -§mtRD u. g. =2J.r(,,.'De =2,a‘(05}(2) =2mAW A. = 42m; =- A. = -4.0 6.22 A“: —Fm(RDHRJ—} '95.: == Van - Inq{Rs+ RD] 10 =20-[1}(R5+ RD] =5 Rs +Rp =10 kn LGIRQ=8HL Rc=2kﬂ A... = -10 = —g.,.[8|[20) g, = 1.75m1v =2 Kg” =2.{Kl(1} => K_ =o.755mrv= V: = IDQRS = (HR) = 3 V :m = Riva. - V". )' =5 l: 0.766(VG, — 2)2 : V55 = 3.31 V VG = vas+v5 =3.31+2 =s.31 1 VG=ﬁT'RIn‘VDD 1 - _ =3! R—1(200)(2{Jj = 3.31 =5 RI = :53 kn T53R1 m- —200=ﬁ R3 =27? kn 6.13 gmru (“(50) A” = _--- = —-—-—- v = _ 1+ gmrn 1+ [4)(50) ” LE5 Ra=illro=i 50:30:0249m 9m 4 “—_“— A = m 2 «5012.5; ” 1+ammnRs} 1+4(50||2.5) 4(233) l+4(2.35} R0 = L 9m =~ Ra = 0.225 m z A, = 0.905 I. fa R5 = I 50 2.5 6 .24- R2 _ 396 "' VG ‘ (R. +R2)VD° _ (396+124o)(10] => VG = 2.42 V 10~(vm+2.42) “=0 “—3— S 7.58 — V55 = (2)(4...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern