双语建模实验一答案.docx
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双语建模实验一答案
1-phaseDiodeBridgeRectifiers(DBRECT1)
1,ExecuteDBRECT1toobtainvs,isandvdwaveforms
Iswaveform
Vs
vs
waveform
Vdwaveform
2,FromtheresultsoftheFourieranalysiscontainedintheoutputfileDBRECT1.OUT,
calculatetheinputpowerfactorandthedisplacementpowerfactor.
DPF=COS(-9.959)=0.985,THD=0.8906
PARAMLS_VALUE=1.0000E-03
PHASE(DEG)=-9.959E+00PF=
=0.7356
PARAMLS_VALUE=2.0000E-03
PHASE(DEG)=-1.760E+01
DPF=COS(-17.60)=0.953,THD=7.219879E+01PERCENT
PF=
=0.7728
PARAMLS_VALUE=3.0000E-03
PHASE(DEG)=-2.198E+01
DPF=COS(-21.98)=0.9273,THD=6.303744E+01PERCENT
PF=
=0.7845
3,MakeuseoftheFourieranalysisinDBRECT1.OUTtoplotis,is1,is3andis5.Superimposethedistortioncurrentcomponentidistortion=is-is1ontheaboveplot.
Answer:
is,is1,is3andis5waveform
Is:
Is1
Is3
Is5
Is-Is1waveform
Allwaveforms:
4.CalculateIcap(thermscurrentthoughthefiltercapacitor)asaratiooftheaverageloadcurrentIload.
5,Plotthecurrentandvoltageassociatedwithoneofthediodes,forexample,d1,and
obtaintheaverageandthermsvaluesofthecurrentasaratiooftheaverageload
current.
D1voltage
D1current
AveraAverageD1currentasaratiooftheaverageloadcurrent
D1,thermsvaluesofthecurrentasaratiooftheaverageloadcurrent.
6,VaryLsasaparametertoinvestigateitsinfluenceontheinputdisplacementpowerfactor,theinputpowerfactor,%THD,andthepeak-peakrippleinthedcvoltagevd.
Ls=1mh
DPF=COS(-9.959)=0.985;THD=8.906256E+01PERCENT(THD=0.8906256);
PF
==0.7356;
Vd=
Ls=1mh
DPF=COS(-17.60)=0.953;THD=7.219879E+01PERCENT(THD=0.7219879);
PF=
=0.773
Vd=
Ls=3mh
DPF=COS(-21.98)=0.9273;THD=6.303744E+01PERCENT(THD=0.6303744)
PF=
=0.784
Vd=
7,VarythefiltercapacitorCdtoinvestigateitsinfluenceonthepercentagerippleinvd,inputdisplacementpowerfactorand%THD.Plotthepercentage∆Vd(peak-to-peak)/Vd(average)asafunctionofCd
whenweincreasethevalueoffiltercapacitorCd,wewillseethethepercentagerippleinvddecrease.
whenweincreasethevalueoffiltercapacitorCd,wewillseethethepercentagerippleininputdisplacementpowerfactordecrease.
PHASE(DEG)DPF=COSPHASE(DEG)
-9.959E+000.985
-1.627E+010.960
-1.811E+010.950
-1.896E+010.946
-1.946E+010.942
whenweincreasethevalueoffiltercapacitorCd,wewillseethethepercentageripplein%THDdecrease.
%THD
8.906256E+01PERCENT
8.491219E+01PERCENT
8.351652E+01PERCENT
8.284880E+01PERCENT
8.240134E+01PERCENT
Plotthepercentage∆Vd(peak-to-peak)/Vd(average)asafunctionofCd
8,Varytheloadpowertoinvestigateitsinfluenceontheaveragedcvoltage.
averageVdvoltage.
9,Inthenominalcircuitinputfile,removethelimitonthemaximumtimestepduringthesimulationandobserveitsinfluenceonthecircuitwaveforms.
Answer:
Whenweremovethelimitonthemaximumtimestep,weseethewaveforms
Vs:
Isbeforeremove
Isafterremove
Vd:
10,Obtainthevs,isandvdwaveformsduringthestartuptransientwhenthefiltercapacitorisinitiallynotcharged.Obtainthepeakinrushcurrentasaratioofthepeakcurrentinsteadystate.Varytheswitchinginstantbysimplyvaryingthephaseangleθofthesourcevs.
(1),whenthefiltercapacitorisinitiallynotcharged.
Vs
Is
Vd
(2)thepeakinrushcurrentasaratioofthepeakcurrentinsteadystate
Thepeakinrushcurrentis110A,whilethepeakcurrentinsteadystateis35A,sotheresultis3.14
(3)Varytheswitchinginstantbysimplyvaryingthephaseangleθofthesourcevs,forexample,thephaseangleischangedto30degree
Vs
Icd
Vd
11,ReplacethedcsideofthediodebridgebyacurrentsourceId=10A,corresponding
toaverylargeLd.MakeLsalmostequaltozero.ObtainVd(average).
Answer:
wereplacethedcsideofthediodebridgebyacurrentsourceId=10A,andchangethevalueofLd
Is1000mH,morelargerthantheLs,soVd(average)is
12,MakeLs=3mHinProblem10andobtainVd(average),displacementpowerfactor,
powerfactor,%THD,andthecurrentcommutationinterval.
Answer:
(1)whenLs=3mH,andthefiltercapacitorisinitiallynotcharged.Weobtainthevs,is
andvdwaveformsduringthestartuptransient
Vs
Is
thepeakinrushcurrentasaratioofthepeakcurrentinsteadystate
Thepeakinrushcurrentis83A,whilethepeakcurrentinsteadystateis16A,sotheresultis5.19
Vd(average)
(2)DPF=COS(-2.199E+01)=0.927
(3)PF=
=0.780
(4)%THD=6.426583E+01PERCENT=0.6426583
(5)thecurrentcommutationinterval.
Id1waveformis,sothecurrentcommutationinterval.is12ms.