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FACULTY OF ELECTRICAL ENGINEERING UNIVERSITI TEKNOLOGI MARA
EPO642 POWER ELECTRONIC II
ASSIGNMENT 1
MUHAMMAD KHUDRI BIN HALIM BASHAH 2007270412 EEB7AP
DR. MOHAMMAD NAWAWI SEROJI
ASSIGNMENT 1: A full-bridge controlled rectifier produces a continuous and discontinuous (intermittent) output voltage waveform when feeding a purely resistive load. 1) Identify the range of firing angle that produces both waveforms
For continuous waveform condition; 0° ≤ α ≤ 60° At α = 30°,
Figure 1: For continuous waveform condition at α = 30°
For discontinuous waveform condition; 60°
α
120°
At α = 90°,
Figure 2: For discontinuous waveform condition at α = 30°
2) Derive the average output voltage equations for both operations (continuous and discontinuous, Vo)
For continuous waveform:
VDC
=
6
1
3
V LL cos(t )d (t )
6
1 6 = V LL sin(t ) 3 6
3V LL = sin( ) sin( ) 6 6
3V LL = (sin cos cos sin ) (sin cos cos sin ) 6 6 6 6
=
3V LL
=
3V LL
(0.5cos 0.866 sin ) (0.5 cos 0.866 sin )
cos
For discontinuous waveform,
VDC
=
2
1
3
V
LL
cos(t )d (t )
6
1 2 = V LL sin(t ) 3 6
3V LL = sin sin( ) 2 6
3V LL = sin [sin( ) cos cos( ) sin ) 2 6 6
=
3V LL
1 0.5 cos 0.866 sin )
3V LL 1 cos( ) = 3
=
3V LL
1 cos( 60)
3) Verify your answers using PSIM simulation. State any assumptions you have made i.e. values of V s, f, R etc.) Answer: Assume;
VLL(rms) =415V
R load=100Ω
Frequency = 50Hz
Pulse width = 120
Figure 3: PSIM simulation circuit diagram of three phase full bridge controlled rectifier purely resistive load with 60 degree firing angle.
Let α = 30°
Figure 4: Firing angle at α = 30°. From simulation; VDC = 483.914V From formula for continuous current condition;
VDC =
3V LL
cos
= 415 x √2 =
3 2 (415)
cos 30
= 485.3613 V
Let α = 90°
Figure 5:Firing angle at α = 90°. From simulation; VDC = 73.7533V From formula for discontinuous current condition;
VDC=
3V LL
1 cos( 60)
= 415 x √2 =
3 2 (415)
1 cos(90 60)
= 75.0856 V
EPO642 POWER ELECTRONIC II
ASSIGNMENT 1
MUHAMMAD KHUDRI BIN HALIM BASHAH 2007270412 EEB7AP
DR. MOHAMMAD NAWAWI SEROJI
ASSIGNMENT 1: A full-bridge controlled rectifier produces a continuous and discontinuous (intermittent) output voltage waveform when feeding a purely resistive load. 1) Identify the range of firing angle that produces both waveforms
For continuous waveform condition; 0° ≤ α ≤ 60° At α = 30°,
Figure 1: For continuous waveform condition at α = 30°
For discontinuous waveform condition; 60°
α
120°
At α = 90°,
Figure 2: For discontinuous waveform condition at α = 30°
2) Derive the average output voltage equations for both operations (continuous and discontinuous, Vo)
For continuous waveform:
VDC
=
6
1
3
V LL cos(t )d (t )
6
1 6 = V LL sin(t ) 3 6
3V LL = sin( ) sin( ) 6 6
3V LL = (sin cos cos sin ) (sin cos cos sin ) 6 6 6 6
=
3V LL
=
3V LL
(0.5cos 0.866 sin ) (0.5 cos 0.866 sin )
cos
For discontinuous waveform,
VDC
=
2
1
3
V
LL
cos(t )d (t )
6
1 2 = V LL sin(t ) 3 6
3V LL = sin sin( ) 2 6
3V LL = sin [sin( ) cos cos( ) sin ) 2 6 6
=
3V LL
1 0.5 cos 0.866 sin )
3V LL 1 cos( ) = 3
=
3V LL
1 cos( 60)
3) Verify your answers using PSIM simulation. State any assumptions you have made i.e. values of V s, f, R etc.) Answer: Assume;
VLL(rms) =415V
R load=100Ω
Frequency = 50Hz
Pulse width = 120
Figure 3: PSIM simulation circuit diagram of three phase full bridge controlled rectifier purely resistive load with 60 degree firing angle.
Let α = 30°
Figure 4: Firing angle at α = 30°. From simulation; VDC = 483.914V From formula for continuous current condition;
VDC =
3V LL
cos
= 415 x √2 =
3 2 (415)
cos 30
= 485.3613 V
Let α = 90°
Figure 5:Firing angle at α = 90°. From simulation; VDC = 73.7533V From formula for discontinuous current condition;
VDC=
3V LL
1 cos( 60)
= 415 x √2 =
3 2 (415)
1 cos(90 60)
= 75.0856 V
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