建模与仿真实验

课程名称电力电子建模与仿真

学生姓名

学号

专业班级电气专业

指导教师

2012年5月31日

PWM, bipolar-voltage-switching, 1-phase

(1PHBSINV)

The experiment circuit diagram as above.

Problems

1. Obtain the following waveforms using 1Phbsinv: (a) o v and o i .

Figure 1-1Output voltage （o v ） and current （o i ） waveform

(b) o v and d i .

Electronics Modeling Simplified using PSpice (Release 9)," . Time

0s

5ms 10ms 15ms 20ms 25ms 30ms 35ms 40ms 45ms 50ms

I(V5)

-20A

0A

20A

V1(L1) - V2(V5)

-400V

0V

400V

SEL>>

Time

0s

5ms 10ms 15ms 20ms 25ms 30ms 35ms 40ms 45ms 50ms

I(V1)

-20A

0A

20A

V1(L1) - V2(V5)

-400V

0V

400V

SEL>>

Figure 1-2Output voltage （o v ） and the input current （i d ） waveform

(c) o v , o i and o p .

Figure 1-3 Output voltage （o v ）,current （o i ）,and power （o p ） of the waveform

2. Obtain v01 by means of Fourier analysis of the o v waveform. Compare 1o v with its precalculated nominal value.

Output voltage to Fourier analysis.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(N00396,N00826)

DC COMPONENT = -1.364900E+00

HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED

NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)

1 4.000E+01 2.152E+0

2 1.000E+00 7.240E+01 0.000E+00 2 8.000E+01 8.215E-01 3.817E-0

3 -1.373E+02 -2.821E+02 3 1.200E+02 1.087E+00 5.052E-03 -8.038E+01 -2.976E+02

4 1.600E+02 6.943E-01 3.226E-03 -8.213E+01 -3.717E+02

5 2.000E+02 5.693E-01 2.645E-03 -9.786E+01 -4.599E+02

6 2.400E+02 6.443E-01 2.993E-03 8.586E+01 -3.485E+02

7 2.800E+02 9.711E-01 4.512E-03 3.420E+01 -4.726E+02

8 3.200E+02 1.027E+00 4.772E-03 9.653E+01 -4.827E+02

9 3.600E+02 1.231E+00 5.722E-03 1.556E+02 -4.960E+02 10 4.000E+02 1.534E+00 7.125E-03 3.867E+01 -6.853E+02 11 4.400E+02 6.245E-01 2.902E-03 1.273E+02 -6.690E+02 12 4.800E+02 3.197E-01 1.485E-03 -3.498E+01 -9.037E+02

Time

0s

5ms

10ms 15ms 20ms 25ms 30ms 35ms 40ms 45ms 50ms

I(V5) * (V1(L1)-V2(V5))

-5.0KW

0W 5.0KW I(V5)

-20A

0A 20A V1(L1) - V2(V5)-400V

0V 400V SEL>>

13 5.200E+02 1.613E+00 7.495E-03 -1.635E+02 -1.105E+03 14 5.600E+02 1.071E+00 4.974E-03 -1.276E+01 -1.026E+03 15 6.000E+02 2.202E-01 1.023E-03 7.034E+01 -1.016E+03 16 6.400E+02 1.064E+00 4.943E-03 -4.897E+01 -1.207E+03 17 6.800E+02 6.628E-01 3.079E-03 -1.572E+02 -1.388E+03 18 7.200E+02 1.169E+00 5.433E-03 -8.197E+01 -1.385E+03 19 7.600E+02 8.240E-01 3.828E-03 -8.104E+01 -1.457E+03 20 8.000E+02 1.059E+00 4.919E-03 1.142E+02 -1.334E+03

TOTAL HARMONIC DISTORTION = 1.969687E+00 PERCENT

V 01.peak =215.2V

V o1=

01.peak

2

=

2

=152.17V

V o 1（rms ）=153.33V

Through the comparison we found Fourier analysis get V o1 and its precalculated nominal value V o 1（rms ） approximately equal .

3. Using the results of Problem 2, obtain the ripple component ripple v waveform in the output voltage.

Figure 1-4the ripple component V ripple waveform in the output voltage.

4. Obtain 1o i by means of Fourier analysis of the o i waveform. Compare 1o i with itsprecalculated nominal value.

For the output current of Fourier analysis as follows.

FOURIER COMPONENTS OF TRANSIENT RESPONSE I(V_V5)

DC COMPONENT = -4.719698E-01

HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED

Time

0s

5ms 10ms

15ms 20ms 25ms 30ms

V1(L1) - V2(V5) -215.2 * SIN(2*pi*40*Time+72.4)

-500V

0V

500V