电力电子建模与仿真实验3文档格式.docx
- 文档编号:21016013
- 上传时间:2023-01-26
- 格式:DOCX
- 页数:16
- 大小:594.98KB
电力电子建模与仿真实验3文档格式.docx
《电力电子建模与仿真实验3文档格式.docx》由会员分享,可在线阅读,更多相关《电力电子建模与仿真实验3文档格式.docx(16页珍藏版)》请在冰豆网上搜索。
DCCOMPONENT=-5.821027E+00
HARMONICFREQUENCYFOURIERNORMALIZEDPHASENORMALIZED
NO(HZ)COMPONENTCOMPONENT(DEG)PHASE(DEG)
11.000E+053.924E+001.000E+001.357E+020.000E+00
22.000E+052.829E+007.210E-01-1.787E+02-4.501E+02
33.000E+051.412E+003.598E-01-1.331E+02-5.402E+02
44.000E+051.131E-012.883E-02-8.969E+01-6.325E+02
55.000E+057.158E-011.824E-011.387E+02-5.398E+02
66.000E+059.350E-012.383E-01-1.760E+02-9.902E+02
77.000E+056.445E-011.642E-01-1.309E+02-1.081E+03
88.000E+051.131E-012.882E-02-8.985E+01-1.175E+03
99.000E+053.560E-019.073E-021.424E+02-1.079E+03
101.000E+065.514E-011.405E-01-1.732E+02-1.530E+03
111.100E+064.319E-011.101E-01-1.289E+02-1.622E+03
121.200E+061.131E-012.882E-02-8.990E+01-1.718E+03
131.300E+062.157E-015.496E-021.472E+02-1.617E+03
141.400E+063.836E-019.777E-02-1.703E+02-2.070E+03
151.500E+063.301E-018.412E-02-1.270E+02-2.162E+03
161.600E+061.131E-012.882E-02-8.993E+01-2.261E+03
171.700E+061.404E-013.579E-021.540E+02-2.153E+03
181.800E+062.878E-017.333E-02-1.670E+02-2.609E+03
191.900E+062.690E-016.856E-02-1.252E+02-2.703E+03
202.000E+061.131E-012.882E-02-8.995E+01-2.804E+03
212.100E+069.410E-022.398E-021.641E+02-2.686E+03
222.200E+062.246E-015.724E-02-1.632E+02-3.149E+03
232.300E+062.273E-015.794E-02-1.234E+02-3.244E+03
Vo(avg)=5.821V
Voi1/Vo=0.476
Voi2/Vo=0.0.344
Voi3/Vo=0.172
Voi4/Vo=0.013
Voi5/Vo=0.087
……
3.Increasetheloadresistanceto10Ω.Obtain
and
waveformsinthe
discontinuousconductionmode[Hint:
useV(0)=5.8Vand
(0)=0].Checkifthe
resultsagreewiththefollowingequation:
Where
VLwaveform
iLwaveform
FOURIERCOMPONENTSOFTRANSIENTRESPONSEV(R_Rload)
DCCOMPONENT=6.894733E+00
Vo=6.8947VVd=8VVo/Vd=0.8618
FOURIERCOMPONENTSOFTRANSIENTRESPONSEI(R_Rload)
DCCOMPONENT=6.894733E-01
Io=0.6895A
=2AD=0.75
=0.8671
4.Obtainthepeak-to-peakrippleintheoutputvoltageandchecktoseeiftheresults
agreewiththeanalyticalcalculations.
DCCOMPONENT=5.683625E+00
Vo=5.6836V
=5.7061V-5.6649V=41.186mV
=
Vo=35.522mV
5.Calculatethermsvalueofthecurrentthroughtheoutputcapacitorasaratioofthe
averageloadcurrent
OURIERCOMPONENTSOFTRANSIENTRESPONSEI(R_Rload)
DCCOMPONENT=1.136725E+01
Io(avg)=11.367A
FOURIERCOMPONENTSOFTRANSIENTRESPONSEI(C_C1)
DCCOMPONENT=2.564849E-03
HARMONICFREQUENCYFOURIERNORMALIZEDPHASENORMALIZED
11.000E+051.265E+001.000E+00-1.333E+020.000E+00
TOTALHARMONICDISTORTION=3.792972E+01PERCENT
Icdis=0.3393AIc1(rms)=0.9567A
Ic1/Io=0.0842
6.Calculatethepeak-to-peakrippleintheoutputvoltageinthepresenceoftheoutput
capacitorEquivalentSeriesResistance(ESR)[SuggestedESR=100mΩ].Plotthe
rippleacrossC,ESRandthetotalripplein
=5.8263V-5.5523V=274.006mV
ThewaveformofVC
ThewaveformofVESR
ThewaveformofVo
Full-bridge,bipolar-voltage-switchingdc-dcConverter
1.ObtainthefollowingwaveformsusingFBBSDCDC:
(a)
(b)
(a)Vowaveform
Iowaveform
Powaveform
(b)Vowaveform
Idwaveform
2.Calculatepeak-to-peakripplein
Io=3.5311A-764.312mA=2.7668A
3.BymeansofFourieranalysis,calculatetheaveragevalueandtheharmoniccomponentsin
.Obtainthermsvalueoftheripplein
andcheckitwiththeanalyticalcalculations.
FOURIERCOMPONENTSOFTRANSIENTRESPONSEV(N00396,N00826)
DCCOMPONENT=8.228991E+01
12.000E+042.012E+021.000E+009.001E+010.000E+00
24.000E+041.234E+026.132E-01-9.000E+01-2.700E+02
36.000E+043.376E+011.678E-019.000E+01-1.800E+02
48.000E+043.075E+011.528E-019.000E+01-2.700E+02
51.000E+055.051E+012.510E-01-9.000E+01-5.400E+02
61.200E+053.112E+011.547E-019.000E+01-4.501E+02
71.400E+053.451E+001.715E-029.001E+01-5.401E+02
81.600E+052.716E+011.350E-01-9.000E+01-8.101E+02
91.800E+052.698E+011.341E-019.000E+01-7.201E+02
102.000E+058.000E+003.976E-02-9.000E+01-9.901E+02
112.200E+051.327E+016.596E-02-9.000E+01-1.080E+03
122.400E+052.169E+011.078E-019.000E+01-9.901E+02
132.600E+051.332E+016.620E-02-9.000E+01-1.260E+03
142.800E+053.521E+001.749E-02-9.000E+01-1.350E+03
153.000E+051.570E+017.802E-029.000E+01-1.260E+03
163.200E+051.503E+017.466E-02-9.000E+01-1.530E+03
173.400E+053.428E+001.704E-029.000E+01-1.440E+03
183.600E+059.517E+004.729E-029.000E+01-1.530E+03
193.800E+051.423E+017.069E-02-9.000E+01-1.800E+03
204.000E+058.000E+003.975E-029.000E+01-1.710E+03
214.200E+053.641E+001.810E-029.000E+01-1.800E+03
224.400E+051.167E+015.799E-02-9.000E+01-2.070E+03
234.600E+051.041E+015.174E-029.000E+01-1.980E+03
TOTALHARMONICDISTORTION=7.748038E+01PERCENT
Vo(avg)=82.229V
V0=vd*√(1-(2*D-1)^2)=200*√(1-(2*0.708-1)^2)=181.873v
4.BymeansofFourieranalysis,calculatetheaveragevalueof
andthermsvalueoftheripple.
FOURIERCOMPONENTSOFTRANSIENTRESPONSEI(V_V1)
DCCOMPONENT=-9.035826E-01
12.000E+042.248E+001.000E+00-1.065E+020.000E+00
24.000E+041.367E+006.083E-011.050E+023.180E+02
36.000E+046.684E-012.973E-01-3.265E+012.869E+02
48.000E+044.938E-012.197E-01-1.383E+022.877E+02
51.000E+055.409E-012.406E-018.654E+016.191E+02
61.200E+053.940E-011.753E-01-5.714E+015.819E+02
71.400E+052.541E-011.130E-01-1.713E+025.743E+02
81.600E+053.131E-011.393E-016.905E+019.211E+02
91.800E+052.983E-011.327E-01-7.570E+018.829E+02
102.000E+051.915E-018.521E-021.533E+021.218E+03
112.200E+051.939E-018.627E-024.719E+011.219E+03
122.400E+052.312E-011.029E-01-9.104E+011.187E+03
132.600E+051.777E-017.903E-021.272E+021.512E+03
142.800E+051.341E-015.964E-021.713E+011.508E+03
153.000E+051.772E-017.882E-02-1.075E+021.490E+03
163.200E+051.691E-017.522E-021.073E+021.811E+03
173.400E+051.123E-014.997E-02-1.889E+011.792E+03
183.600E+051.278E-015.686E-02-1.271E+021.790E+03
193.800E+051.523E-016.774E-029.208E+012.116E+03
204.000E+051.156E-015.142E-02-4.709E+012.083E+03
214.200E+059.581E-024.262E-02-1.553E+022.081E+03
224.400E+051.298E-015.775E-027.573E+012.419E+03
234.600E+051.197E-015.326E-02-6.873E+012.381E+03
TOTALHARMONICDISTORTION=8.443615E+01PERCENT
Id(avg)=-0.9036A
THD=0.8443Id1=1.590AIdis=1.342A
Id(rms)=2.081A
5.With
=0and
=0,
=0V.Therefore,
=0.Calculatethefollowing[Hint:
use
(0)=-1.67A]:
and
waveforms.
(b)peak-to-peakripplein
.Compareitwithitsanalyticalvalue,andthatin
Problem2.
(c)Inpart(a),labeltheintervalsduringwhichvariousdevicesareconducting.
(a)Vowaveform
Io=750.579mA-(-2.5622A)=3.3127A
When
=0,D=0.5,
=3.3333A
(c)
6.Intheregenerativemode,thepowerflowsfromtheloadtothedc-busat
.Let
=79.5V,
=10Ainthereversedirection,and
=79.5-0.37x10=75.8V.Therefore,
Calculateparts(a)through(c)ofProblem5[Hint:
(0)=-11.67A].
Io=823.762mA-(-1.9895)=2.8133A
D=0.6985,
=2.8080A
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- 电力 电子 建模 仿真 实验