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6.002x

Week4ExerciseSolutions

S7E1Usingthevalue:vI=5;

Part1

TofindthevoltagevA,youhavetohavethecurrenti Asatisfythetwoequations:

iA=(5-vA)/2

iA=10*(1-e-vA/5

)

Tosolve,setthemequal,andsolve.

Sincetheexponentialtermchangesfaster,then,youarrangetheequationfortheexponentialtermwith

vAtobethevariableandpluginyour"guess"ofv Ainthenon-exponentialtermandsolveforv A:exp[-vA/5]=1-(vI-(vA_guess))/20,orexp[-vA/5]=1-(5-(vA_guess))/20Letssaythattheinitialguessisv A_guess=1(letscalltheinitialguessv A0),then,solvethefollowing

equationtogetthefirstiterationofv A,whichwillbecometheimprovedvA_guessforthenextiteration

(letscalltheresultandtheimprovedv A_guessvA1):exp[-vA1/5]=1-(5-vA0)/20,orexp[-vA1/5]=1-(5-1)/20Tosolveforit(ortogetabetterconvergence),takenaturallogonbothsidestogetmyv A1:vA1=-5*ln(1-(5-vA0)/20)ThisgivesvA1=1.116.Continueoninthismanneruntilv Aconverges.UsingExceltohelpwiththecalculation,hereisthesequenceforv A0,vA1,vA2,...

1.115718 1.079686 1.090878 1.087399 1.08848 1.088144So,youcanseethatitconvergesto1.088.

Howtocheckyouranswer:Plotthetwoequations,andfindtheintersection.

Part2

Takethederivativeoftheequationdictatingthebehaviortofindtheratio:

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(-vA+vi)=10*(1-e^(-vA/5))

-dvA+dvi=20*(1/5*e^(-vA/5))dvA

dvA/dvi=1/(1+4*e^(-vA/5))

PluginthevAfoundabove,andyouhavetherationtobe0.237.

Part3

Welookfortheincrementalresistancedv A/diA:

iA=10*(1-e^(-vA/5))

diA=10*(1/5)*e^(-vA/5)*dvA

dvA/dvi=1/(2*e^(-vA/5))

PluginginthevAfoundinpart1,wehavetheincrementalresistanceas0.622.

S7E2Usingthevalues:i=v

3,I=4.0A,R=8.2ohms.

Part1

Findv:I=v/R+i=v/R+v3

Thus,wehave4.0=v/8.2+v3,andsolvingforvgivesusv=1.56.

Part2

Theoperatingcurrentisfoundviai=v3.Wehave1.56

3as3.81.

Part3

Theincrementalresistancedv/dicanbefoundbytakingthederivativeofi=v3.Youhavedi=3v

2dv,so

dv/di=1/3v2=0.137.

S7E3

Part1

Takethederivativetofindincrementalresistance:

diD=I0/VT*e^(VD/VT)dVD,

dVD/diD=VT/I0*e^(-VD/VT)

Therefore,whenbiasedatV D,wehavetheincrementalresistanceV T/I0*e^(-VD/VT).

Part2

VT=26mV,I0=81014

A,R=3.9k,VD/iD=100.

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WesettheincrementalresistanceV D/iD=VT/I0*e^(-VD/VT)=100,andsolveforVD.

WefindVDtobe0.57.

FromVDwecanfindV I.SinceweknowthatVD/iDis100,wefindiDtobe0.00026.

Wethensolve(VIVD)/R=iDtofindVItobe1.58.

Part3

VD/iD=VT/I0*e^(-VD/VT)=1000.

Solving,wefindVDtobe0.51.

Thus,weknowthatiD=0.00026.

WefindVIbysolving(VIVD)/R=iD,andwehaveVI=0.61.

S8E0

Part1

vo=Ro*(-iD)=-RoK1vB=-RoK1VI

Part2

vo=Ro*(-iD)=-RoK2iB=-RoK2VI/RI

S8E1

UsingR=850.0,VS=5.0,K=0.088.

v=i*R=K/v2*R.

Solveforvwehavev=4.213.

vo=VSv=0.787.

S8E2

Part1andPart2

TofindRTH,setindependentsourcestozero,akasetI O=0(opencircuit).Wethushavetworesistorsin

parallel.

RTH=R1*R2/(R1+R2)=398.

TofindVTHwethensolvei*R 1+alpha*i-(IOi)*R2=0,andfindItobe0.0019,andthusV TH=(IOi)*

R2=1.6.