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Three Phase Rectifier
Some basic observation
1) KVL shows only 1 diode in top half bridgeconduct at a time (D1, D3, D5) with anodconnected to phase
2) KVL shows only 1 diode in bottom half briwill conduct at a time (D2, D4, D6) withcathode connected to phase
3) As a consequence D1 and D4 can not conat a time nor D3 and D6 or D5 and D2
4) Output voltage across the load is the one the line-to-line voltage of the source, e.g.D1 and D2 on, the output voltage is Vac,which is highest L- L voltage.
5) Six combination of line to line voltage (3-p
taken 2 at a time). Highest line to line voltake place at 360/6=60 degree apart.6) The fundamental frequency of the output
voltage is 6ω.
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Three Phase Waveforms
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Diode Rectifier
vpn: The Voltage at point P with respect to ac voltage neutral point n.
vnn: The voltage at the negative dc terminal n.
Since Id flows continuously, at any time, vpn and vnn can be obtained in terms ofone of the ac input voltage van, vbn and vcn.
Applying Kirchhoff voltage law, the dc side voltage is,
The waveform vd consist six segment per cycle of line frequency. Each segmentbelongs to one of the six line to line voltage combination.
Each diode conduct for 1200 i.e. 2 Π /3.
Considering phase a current is,
nn pnd vvv −=
conductingis4or1diodeneitherwhen0
conductingis4diodewhen-
conductionis1diodewhen
=
=
=
d
d a
Ι
I i
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Diode Rectifier
The current in a conducting diode is the same as the load current. To determine the current
each phase of the source, the Kirchhoff’s current law is applied at node a, b, and c,
Each diode conducts one-third
of the time, resulting in
Apparent power from the
three-phase source is
The periodic output voltage is define as v0(ωt) = Vm, L-L sin(ωt) for Π /3 ≤ ωt ≤2Π /3 with perΠ /3. The Fourier series for the out put voltage is expressed as,
25
63
41
-
D Dc
D Db
D Da
iii
iii
iii
−=
−=
=
rms s I I ,01 78.061
==
π
13,11,7,5,h
h
II s
sh
=
=1
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Three Phase Rectifier Average Voltage
Where Vm, L-L is the peak line to line vo
of three phase source, which is √ 2 VL-L
The amplitude of ac voltage terms are
The Fourier series of the currents in pha of the ac line is,
13,...11,7,5,
1
=
=
h
h
I
I s
sh
dds IIπ
I 78.061
1 ==
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Three Phase Waveforms
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Three Phase Waveforms
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Three Phase Controlled Rectifier
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Three Phase Controlled Rectifier Average Voltage
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12 Pulse Rectifier
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12 Pulse Rectifier
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12 Pulse Rectifier
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12 Pulse Rectifier
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Problems three phase controlled rectifier
C i Si l Ph B id R ifi
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Commutation: Single Phase Bridge Rectifier
C i Si l Ph B id R ifi
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Commutation: Single Phase Bridge Rectifier
C i Si l Ph B id R ifi
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Commutation: Single Phase Bridge Rectifier
C t ti Th Ph B id R tifi
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Commutation: Three Phase Bridge Rectifier
C t ti Th Ph B id R tifi
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Commutation: Three Phase Bridge Rectifier
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Commutation: Three Phase Bridge Rectifier
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Commutation: Single Phase Controlled Bridge Rectifier
Generally ac side inductance can not ignore in the practical converter. So for control switcdelay angle alpha, the current commutation takes a finite commutation interval u. During
commutation interval, all switch are in conduct. So vd = 0 and vLs = vs.
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Commutation: Single Phase Controlled Bridge Rectifier
Generally ac side inductance can not ignore in the practical converter. So for control switcdelay angle alpha, the current commutation takes a finite commutation interval u. During t
Commutation interval, all switch are in conduct. So vd = 0 and vLs = vs.
Commutation: Three Phase Bridge Rectifier
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Commutation: Three Phase Bridge Rectifier
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Commutation: Controlled Bridge Rectifier
Generally ac side inductance can not ignore in the practical converter. So for control switchdelay angle alpha, the current commutation takes a finite commutation interval u. At ωt =alpha, current begins to switch 5 to 1. During the Commutation interval, switch 5 and 1conducting Simultaneously, the phase voltage van and vcn are shorted together through Ls.During commutation interval,
Where,
The reduction volt-radian area A u due to commutation interval,
=
The average output voltage is reduced by A u,/(Π /3)
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Commutation: Controlled Bridge Rectifier
During the current commutation, phase a and c are short together i.e. the phase voltage vand vcn are shorted together through Ls. Therefore, during commutation,
also
Therefore,
Since Id (ia+ic) isassumed constanduring commutatinterval
Even through, commutation interval u is not required for to calculate Vd, it is required to ensure reliable operation in the inverter mode,
=
Where
Integrating betweenrecognizing that during this interval ia changes from zero to Id,results in,