HW3-SOLUTIONS

# HW3-SOLUTIONS - Exercise Ex 4.10 a Using iteration Diode...

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Unformatted text preview: Exercise: Ex: 4.10 a. Using iteration Diode has 0.7 V drop at 1 mA current. Assume VD : 0.7 V I = 5 — 0.? D 10 k Use equation 4.5 and note that v1: 0.7V. 11: lrnA : 0.43 mil. R=10kﬂ VCC = 5 V VD I V2 — V] = 2.3 X Vrlog(I—2) l :2 1 First iteration V2 2 0.7 + 2.3 >< 25 x 10—31%??? = 0.679 v Second iteration I _ 5 - 0.679 2 ._ .._.____ = 0.432 mA 10 k V2 = 0.7 + 2.3 x 25.3 x 10-3]Og(0.z:32) = 0.63"} V: 0.68 V we get almost the same voltage '. The iteration yields ID : 0.43 ma. VD = 0.68 V I). Use constant voltage drop model VD = 0.7 V constant voltage drop I = 5—0-7 9 10k = 0.43 rnA 4.12 a + .. .1! 1: V ram + I = «ma-1+5 1-5 -EN’ '1 - h liZm Vor- Q-S-ﬁ-o-rDDS x50: 9-?61/ CO UT. 4_.13 _ Ga: Vsslnbot l V at 0.31: = 5' "> U33- dagtﬂﬁ :Vpo 0.5L 2” W <32- (2%) 1. 1.. + NI: NO Load Vz=§_-l_\_}_ a V1: =Z-‘_-1:° = Cardwﬁén ski-{'5 at 6: 24+. and Jtrmtﬁﬁhsr out 1%0.‘ 6' ML For Line EWTIDl-J 4th.! cor-ulch mush. '13 : 20051:: I 0- a -= I 5. ° 01 m i z 7+ 2, - —' ——-——— _—_ 0") V0 M = J... I j 2-"- Vaslngp—VDO Lin-Q Eﬂﬂuhkﬂﬂ: ﬂ = ,_-L_ :33.9 M“ I“ i —' ._ '6' 01. 24001-1 V __ gﬁvscﬁqb _ Vmip] ¢=e For Lao-mi €03th =‘L[U5Cos°-V Eu we 3%... : mt53m- r'a ﬁr S 5(' 3 250.0. lﬂh “VM(T”2.Q) ] 45 .. __ a: van 5:“ - bud: Lose: embr ang 1T"‘Ze'f_:1r 5% "'=." :5) U9; ﬁVﬂ = 3,35- ___ THE 2'“ 211‘ * = is; -. 32 4.19 u- : QE-D, . nu. Funk (libel-Q. mrrxnt News. at 'Hw. plank 4-inch UBH‘Rﬂ-L. é . 2. p "'3 V5 " VIM: n. l-cd'wﬂ I'D 13+ R (a = 1231—04 [DD :. ‘63 tape 4.20 Rib-1" ‘19 “ﬂu. mane. mailﬁtér D'F F‘La- 326(1) V0333: “SSH-d} - VD A? 2.!" w-e- :(ﬁﬁvsmsgs — WDLw )1; ; 2%”VD b (C) T’kam-rmt :- VsSJné-Vp \ K m; 3 hi 3 lZJ'i—O? v R loo 7-"— @3119: TIME. av‘l‘fmt is an”: Ink-Jun. @w 9) 365-6) 1'- ZB' wkuL 6' in. “HM. 0.9314. n“: which "HA-L "II-Fit a;ij maths \ID :5) Va. Sine 1" VI) '5' :. Sm" I a =- 2.3m“ 81.513. a ¥°r - 5 I'S 0-“: '4"; +5 (b) Peak diode current = 13—36%:th e bo'Hm dnohs are. cut 2 V5“ 2V5- : lZfi --1.4 95,5. ERA 0'0: 03; R —*“1m = 156 mA 010-:- Lifts-BU «ND-L MOI-5 Pw=v3—Vﬂ=12ﬁ—m=m.3v ' 95: + 2 5\/ 10k!!— 4— D7.- .Dl 0‘: 5" Ev _ .114 g .0442. Ex:4.21 t t "' . “Eli,” ‘\ Input I I 2 1r—EI = lgzvg — 2vﬂm J 29)} 1T But c056 at cushr - B) m *1 ’11" F H} “~=’ 11' 2V g Vﬂ‘avg = f _ =2><12J§ 11' = —1.4 = 9.4V PROBLENI = 2V Conductin g 2m lr=2—(—5) 2 kﬂ = 3.5 mA —5V ('3) 5V 2knl1=5 _(+” Conductin 2 g =2mA +1 V +2 V 4.9 (a) I: 0.3 mA l9l v=ov E 10er 0—](0—3)=o.3 mA Conducting (b) 4.10 (a) (10l|20)kﬂ v —."1"' 20 I =4VI zokn - 4 r = ——--——-—— = on A (10” zo)+2o 5'" 20 V = —x4=3v (lﬂll20)+20 (b) 5 V: — *- 3= v-osv 2 — v + lonlo 5 so. Cutﬂff 10||10 5m to x s 10 + 10 = 3 v 4.23 The voltage across three diodes in series is 2.4 V; thus the voltage across each diode must be 0.8 V. VD! if]. Using ID : lge . the required current is found to be 7.9 mA. If 1 mA is drawn away from the circuit. ID will be 6.9 mA. which would give VD: 0.?94 V, giving an output voltage of 2.39 V. Thus the change in output voltage is — 10.15 mV. 4.29 For a diode conducting a constant current. the diode voltage decreases by approximately 2 m‘v’ per increase of 1° C. A decrease in VB by 100 mV corresponds to a junction temperature increase of 50" C . The power dissipation is given by: PD = (15 A)(U.6 V) = 9 W. The thermal resistance is given by: 5—7" = 50°C = 5.56°CKW PD 9w —E"* I}; l” R . _ 1.1L . _‘ 0.? i: ‘D.pcak LR "50 R a —120~/§ ‘ 0'7 = 3.33 m 50 Reverseveltage = 120\$: = 159.? v. The design is essentially the same since the sup- ply 1lreltlslge :22:- U.7 V 4.38 Refer to example 3.2 ~ CONSTANT VOLTAGE DROP MODEL {a} I’D. 10--(—IU)—0.7 15 — 10 +1.29(IO)+ {17 = 3.6 V = 1.29 mA 4.39 (3) 5V lUkﬂ v 4,: -5V v = 5+0.7 = —4W I __ 5 +4.3 10 = 0.93 mA ('3) 5v 101d! v Cutoff +; —S\’ V = S—IUO) = 5V 1' = 0A (0) 5V 4.: 1/ 10m w5v I = “+5 = 9.3m 10 (d) Cuteff —5V V = 2-0.? =1-3V I : l.3—(-5! 2 = 3.15mA (13) 5V 21d} IV D] 2V D2 Cutoff I: 5—1'7 =1.6SmA 2 V =1+07 =1.7V 4.41 (a) 0.46 — 0 3 : 0.16 mA (b) 3 v M : 0353 "m 10 kﬂ @ —|.235 + 0.7 = —a.53s [email protected] 04.} V=—3+ﬂ%%ﬁ v = — 1.235 v (3) —3v 4.44 use experiential diode medel te find the percent- age change in the current. In" VT ID = {Se 5 w: — VI 1W... amvr L3 = g = 9 im Fer +5 m'v' change i 1 5:25 _—”- = e = 1.221 iat ‘i‘achange : I'm—I'an 100 = 1.221—1MOO in. l I 22.1% Fer —5 rnV change i 5H5 E = e = 0.818 in] %change = E—‘Hf In" >4 100 I [L]? _ I X 100 [D1 = -— 18.] % Maximum allowable voltage signal change when the current change is limited It) i 10%. Se the current varies frem 0.9 tn 1.] ﬁVI‘VT ‘DI For 0.9 AV = 25 ln(0.9) = "2.63 mV FerLl AV = 25 ln(l.l) = +2.38 m‘v’ Fer i 10% current change the vellagc signal change is frem "2.63 mV te +2.38 mV 4.51 Representing diode by the small signal resiss tanees. the circuit is ﬁr + U; C v” v, = rd :17" Small-signal model D b. If m diodes are in series i % = £ Z Mr E? — 5C = 1 517+ mrd+R mVT+1R 1!: rd + i“ I + SCI-ﬂ: mvr sf; 2 — mVr + (W — 0.7m) 5C1!"1r Phase shift = —tan '[ 1 i] e. Form = 1 “f = V—f = 2.68 mwv = —[3n "‘(mcrd) av VT + V - 0.7 VT Form = 3 = -tan " (toe—J ’ ME = ———————————mvi = 94mVN For phase shift of —45° 3V "1V1" + m — m X 0'7 _ ﬁV —45 =~tan'!(21r><100><103><10>(10QXM) For—Jiﬁm—V 1 IL mA =i=15?p.A ie “EX Vim—o? 5va Now [varies from 11%? pA to 15? x 10 pA Io VT + V+ — 0.? "1A Range is 15.? pa in 1570 as {01325 x ___19_“_9:3_] 5 5 III—V _ _ ID 0.025 + l0 — 0.? ma Range of phase shift IS 84.3“ to —5.71 '3 ID a 4.98? mA 4.52 {0:5 mA v+ R2v+~0.7=10—0.7 IL. 5 mA R = 1.86161 R Diode should be 5 mA diode e. For m diodes connected in series I I v" — 0.7 m + V” D R and rd = m X E In __ — So now av” = -— I ' Ir, .1. + J... R mrd AV+ R + rd R + E“ In + ID i 11+ + 0.? n1 "3V? 2 VT = -mVT 1 In + VT ID mVT + l + + — V -0.7 For no load I = V V” m R = _mVT V+—0.7m .. Mo = Vi In W — 0.7 m + mVT av” VT + (VL — 0.7) 4.60 V' 2 VIII] + WET L 9.1: V30 + 5 x 23 X 10—3 v... = 3.96 V "v2 = vm + 5Q = 8.96 >< 5:: V. = 9.01 v n=nov For I: = 10 unit For I: = 100 mA 4.65 using the constant voltage drop model: ideal (a) on = Us + 0.7 V. For Us __ —[}.7 V on, I 0. for v5 _ -0.7 V —12V l (c) The diode conducts at an angle 6 = sin {33) = 3.34” &5tops at 1T M 8 = 176.165" Thus the conduction angle is 11' — 26 = 1733]” or 3.025 rad. rr—ﬂ yam = — J (lZsindJ —0.7) dd:- 3 :—1[—12eos¢ — 0.7m?“ 2n = 3.412 X Zeosﬂ — 0.7(1': — 23)] 211' = “4.572 V (d) Penk current. in diode is: l2 — 0.7 15x io‘ : 253 “m (e) F'l‘lar occurs when "US is at its the peak and vi; = 0. PIV =12V 4.88 kﬂ Each diode has 0.? V drop when conducting The Zener has 8.2 V drop when conducting. So the limiter thresholds are :(2 x 0.7 + 8.2} r :96 V V: (V) DE SIGN PROBLENIS Ex: 4.15 +15 V org = ZDmV : 200 AIL lmA a. In this problem Total small signal resistance of the four diodes = 20 ﬂ 2!} FUTeHChdiOdE rd = E = VT 25 rnV Butrd = -——=>5 =-—— —— In In diode = i = 0.75 v 4 WV in. : {Se I 15 = I” = ———-—5 = 4.7 x10““’A vaT eorﬁrzs x in _3 diff” : 5—iL = Sﬂl = 4mA Across each diode the voltage drop is V” : VTInGI—j') 15 = 25 x 10‘3 x 1n( 4 x 10—3 ) 4.7 >< 10"“ = 0.7443 v Voltage drop across 4 diodes = 4 x 0.7443 = 2.977 V so change in V0 = 3 - 2.977 = 23 mV Th. MIHIMUM R ‘1']: 242m mrnnt SkOU‘A bl... Currant can b1- aLs \arﬂL A5 IE)ka «)4: should 5419.52.“ R sa‘l'Lut m'u‘l'k TL“; “EMMA, a. new current 0.8.. Bank .15 avidablt . TIM); 1"le current" sknoli b0... 20 m R: = 4g,0_9_ zomk .___._--——--_.._.. MMJIQUM pawl!- driiipahd I‘M. 'I'LL (£19691. occurs whit It..=-0 £5 "PMX: ZIIJXHD“3 x B'é :: Hva Problems: 4.11 m Ari—1.5!— amt-.51 VoH‘a J1. appucar;n5 across, Hag dége :5 83%” 4-0 The Peak mpol',‘ “Heat IzoJZ = m .7v 4.35 UD=%I_;D.?_5\j R 0 3V to:- I‘ge. D/ndr CD _ J7 kg: 6% u \ d “‘J .‘a Ln. =LDL= LDI ﬂ “UT m; 2. Ir ell-root? = \$3361.“: 9 - “viz-=1 ==- "’3 =094?UL D 1.33? 4.43 'L a p U; a TIM: Auriga 1'5 N am Sinai. TM.- supﬁka, ¢>>qu 4.53 V1- : -m:£l) a V0 K ‘3’ 3'4“ a I; I'“ IL Jr; + 117" Mo 4 2 ~03 HQ 62 a D. I", - -‘ (5)61Ma519c1:1t_03_ _T_n_._..,__tl_:g__ D R 1:03”! InnV-r “’80 Q :: HUT Ur ' I : ﬂmn » In mnV'r +, I 104. Jam: Vr'o‘7’" V __ _ l =- —mnVr Err-04m EEO ﬂ 4—5."— ID V+—o'?mfman L R Q . 2 " I 4.54 In 4.332... Diode has 0.7 V drnp at 10 mA current For §\_}o_ < 5 ML 30:].5thenRL= 15011 L. “ am In 2 10 >< 10"”L = 139"”“1‘Rﬁ —;ro-st‘ x _w“_o_.1__ﬁ < giro-*3 :43 = 6.9} x 10"‘A ID '0 - 0'? + 0:05 \ [0‘2 Voltage {Imp acruss each dinde : = 0.75 v In : 11¢va = 6.9} x 10‘” x eﬂ'ﬁm'ﬂlﬁ ID 7/ ? ' 947mﬁ=>Ip =- [Omn- = 73.9 mm :L =1.5x150 = 10mA R: n. - -. #l' '°.g?=qso.<z m” m “4‘ ?‘°”“ 1" «— zom- 1m . (C) 31th! 101 drodﬂs til):— V+—K0.;m g G: CONT. A I: Iﬂ+1L = 73.9mA410mA =33.9 mA I 5*” = 41.7 n 83.9 mA Use small Signal model to ﬁnd veltage we when lead resistor. R,_. has lewer values When lead is disconnected all the current! flows threugh the diede. ID = I = 33.9 mA 199 = V—ran—D] = 0.025 X ln[ ’3 ya = 0.753 V Se Ne lead. v0 = 2 VI, = 2 X 0.753 = 1.506 V. Increase in voltage = 1.506 — 1.5 = 0.006 V New lead is changed 33.9 x10 3‘) 6.913410 '5 32.210011: {L = 1:5. =15mA 100 The diode current reduced by =15—10 = 5mA sat/'0 = —5 maler = —1.7 mv 1.5 R,=75.0:: I =-——=20mA ' i 75 Diede current reduced by 20— 10 = 10 mA Avg 2 —10 rd : —lﬂ><ﬂ.34 = —3.4 mV E = 301m 50 Diede Current reduced by 30— 10 = 20 mA RL=SUﬂ: IL ‘1an .ujuiailzéu :. —- 120 Muir-1A 4.62 Fits-r 'DEMQN -— ‘N supply can Many Suffijcurﬁmt “1%:20Mﬂ' amt.“ .. = "0.19.. w Lime Eauldhon : b o = E but: G'PK ; .L. 5+‘HO .. V #4 . ______. 3 5 If S’EG‘DND DESIGN v mrrmi' 'Frﬂﬂl 7V “1er Ia= O'Zsmﬁ VI '=- VII: 3 V30 '- daisy/4k Va-o Got-4 V}: Vw+' car I31" 6-3":- V'ao1-5 "°'°1- V3.9: mu :* R5 =- q.2k-Q__ D. ____..—-—_-__-—__.."'——"-_-—-- Lunar ﬁrs-awn»): gfa ; 31-50 [50}, ?50r92ﬂo .: . MU 75 ‘4 T {63 15V: 10% Va: V1.0? GI? ‘H = Vw'rwhﬂg R 35:2 V20 = ‘3' 8‘3V V1- 12» Liz. IDM’f L lua- CONT. _ wag, 34-34.— 3o(o.o|): Q-ISV In: ‘3"3/IkJL '-'-= q'lémﬂ' In. = to + 9-13 -.-. 17-13,..A .l. R: =- 1943 Va=CI-I4v V2: 3'?3 +30 «EH - VIE—w ’“ﬁ’B—[rﬁa +- 15 t 045]: 7"”t°"3" :0 :OWSY varl'allojn ;n aofpobwlbc Hd-f-xka 'Hu. cat'th a at, dadb’lnj =-- we, + 3 15d: .. V? °( \$2.0 350 m9 I!» -BV _ ‘2 “3'3 R“ IS-S-r __ moons a, 00-0—- Boc: 383V = 538-0.. 4.71 Refer In Fig 4.23. For Zvﬂo «1:: :4: VS, 2 “ting _ EV5_ ZVDU _ (21) Far yam? = [U V v3 = 3x114 =11le 2 ([3) For vﬂ'avg = 100 V V5 = gx 101.4 = 159.3v Tums Ratio 159.3 4.72 m...— baa 3'2}, Far who <4 Vs. I 6E): “a :- = —]20‘& = 1.065 to 1 21? a. “ZVDD =%}"V5"l"f‘ Vs=Iz* "’"4 =15‘b3v [HI-“5.3.11.6: BEE. :1 4.78 120 Wm 60 Hz PW = VP ~ v00 — 1432 + VP = Vﬂ,avg+VF = 15 +163!r = 31.7V Fora 50% safer margin PWr = 1.5 X 31.7 = 47.6V (d) jam = “(1 + 1, /W) F ' Us} ° ' using IL:UL;E=%wehave “9 ’3“. W PM: I-s-(32-'+) == 2:931 . __ 1 ( 2mg) ______ {DJ E — — 1+1'r 50 2 ' ﬂ 2 1.36A 2(VP_ pm) 8:6 (33:4) E(1 +27r mar] incurs 5-”- IL- ( 1+ 77— VP /2Vr‘ 150 =2_61 _-_- (é A ISO (xv-11’ ) an, 1: 6*?35: _ .9.— (a) [OI-UL 0,) I W + 4— + :‘T‘D J1: d; U}. .— F 9‘! MP1 J.an I: .5 6:) mid. + + o; i g :13 ———_ ...
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HW3-SOLUTIONS - Exercise Ex 4.10 a Using iteration Diode...

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