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3 phase diode rectifier ppt.It clears your concepts how it works.Hope you will gain knowledge about the nature of diode in different combinations.
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POWER POINT PRESENTATION
on Three Phase Diode Rectifier
By – Nitish kumar singh EEE
36
GURU GOBIND SINGH EDUCATIONAL SOCIETY’S TECHNICAL CAMPUS
• Three phase half wave rectifier
• Three phase mid point 6-pulse rectifier
• Three phase bridge rectifier
• Three phase 12-pulse rectifier
CLASSIFICATION OF THREE PHASE DIODE RECTIFIER
Three Phase Half wave Diode Rectifier
VD1
D2
D3
R
Va
VcVbnA
B
C
A
B
C
Common cathode terminal
io
Vo
Fig.1 Three Phase half-wave diode rectifier with common cathode arrangement
Three Phase Half wave Diode Rectifier
Rectifier element is Diode.
Diode can only conduct at the highest +ve instantaneous voltage.
D1 will conduct for ᾠt =30⁰ to ᾠt =150⁰
D2 will conduct for ᾠt =150⁰ to ᾠt =270⁰
D3 will conduct for ᾠt =270⁰ to ᾠt =390⁰
A Diode with the highest positive voltage will begin to conduct at the cross over point of the 3-PHASE supply.
Working
D3 D1 D2 D3
030⁰ 150⁰ 270⁰
0
Vc Va Vb Vc
Fig.(a)
Fig.(b)
Fig.(c)
Vo
Voltage of neutral ‘n’
Voltage of terminal ‘P’
Conduction of Diodes in proper sequence
Vmp
0.5 Vmp
Fig.
030⁰ 150⁰ 270⁰
icia ib icio
Imp= Vmp /R
90⁰ 210⁰ 330⁰
is= ia i D1
0
is= ia= iD1
Fig.(d) Load current or output current.
Fig.(e) Source current
Fig.
0
D3 D1 D2 D3
150⁰ 390⁰270⁰
-1.5 Vmp
-√3 Vmp
VD1
30⁰
Fig.(f) Voltage across Diode D1
Fig.
Variation of voltage across Diode D1
Voltage variation across diode D1 can be obtained by applying KVL to the loop consisiting of diode D1, Phase ‘a’ winding and load R.
So, -VD1 -Vo + Va = 0 or VD1 = Va – VoWhen Diode D1 conduct:Vo = Va Therefore , VD1 = Va – Va = 0When diode D2 conduct :Vo = Vb
Therefore , VD1 = Va – Vb
At ᾠt = 180⁰, Vb=0.866 Vmp , Va = 0 VD1 = - 0.866 Vmp
At ᾠt = 210⁰, Vb= Vmp , Va = -0.5 Vmp VD1 = -1.5 Vmp
At ᾠt = 240⁰, Vb=0.866 Vmp , Va = 0.866 Vmp VD1 = - √3 Vmp
At ᾠt = 270⁰, Vb=0.5 Vmp , Va = - Vmp VD1 = -1.5 Vmp
HOW
When Diode D3 conducts : VD1 = Va - Vc
At ᾠt = 300⁰, Va=-0.866 Vmp , Vc = 0.866Vmp VD1 = - √3 Vmp
At ᾠt = 330⁰, Va=-0.5 Vmp , Vc = Vmp VD1 = - 1.5 Vmp
At ᾠt = 360⁰, Va=0 , Vc = 0.866Vmp VD1 = -0.866 Vmp
At ᾠt = 390⁰, Va= 0.5 Vmp , Vc = 0.5Vmp VD1 = 0 EQUATIONS OF PHASE VOLTAGES FROM WHERE ABOVE
VALUES ARE OBTAINED--- Va = Vmp sin (ᾠt)
Vb = Vmp sin(ᾠt – 120⁰)
Vc = Vmp sin(ᾠt – 240⁰)
Average output voltage V0 =(1/periodicity) ∫Vmp sinᾠt d(ᾠt)
=(3/2∏) ∫Vmp sinᾠt d(ᾠt)
= (3√3/2∏)Vmp
R.M.S value of output voltage(Vor) =[3/2∏ ∫Vmp sinᾠt d(ᾠt)] = 0.84068 Vmp
Ripple Voltage =√(Vrms – Vavg.) = 0.151 VmpForm Factor = Vor/Vo = 1.0165R.M.S value of O/P current (Ir) = 0.84068Vmp/R = 0.84068 Imp
Pdc = Vo Io = (3√3/2∏) Vmp Imp
ᾳ2
ᾳ15∏/6
∏/6
5∏/6
∏/6
22 1/2
2 2
2
Pac = Vor Ir = 0.84068 Vmp Imp = 0.706743Vmp Imp
Rectifier efficiency = Pdc/Pac = 0.9677
% Rectifier efficiency = 0.9677 ×100 = 96.77
Rms value of source voltage(Vs) = Vmp/√2 = 0.707 Vmp
Rms value of source current(Is) = [1/2∏ ∫Imp sinᾠt d(ᾠt)] = 0.4854 Imp
VA rating of transformer = 3Vs Is = 0.707Vmp × 0.4854 Imp =1.0295 Vmp ImpTransformer Utilization Factor(TUF)= (Pdc / Transformer VA Rating) = 0.684VmpImp/1.0295 VmpImp
= 0.6644
2
22 1/2∏/6
5∏/6
NOTE
• PIV for each diode = √3 Vmp.• Each Diode conduct for 120⁰.• There are three pulses of output voltage, or output
current, during one cycle of input Voltage.Therefore it is called 3-phase three-pulse diode rectifier.
• Current in the transformer secondary is unidirectional,therefore,DC exists in the transformer secondary current. As a result transformer core gets saturated leading to more iron losses and reduced efficiency.
THREE PHASE BRIDGE RECTIFIER
• USING 6 DIODES.• UPPER DIODES D1,D3,D5 CONSTITUTES +ve GROUP.• LOWER DIODES D4,D6,D2 CONSTITUTES –ve GROUP.• THREE PHASE T/F FEEDING THE BRIDGE IS CONNECTED
IN DELTA-STAR .
CONSTRUCTION
Positive group of Diodes conduct When these have the most positive anode.
Negative group of diodes conduct if these have the most negative anode.
WORKING
+Ve group -Ve group
This group will conduct during +ve half cycle of I/P source.
This group will conduct during -ve half cycle of I/P source.
A
B
C
A
B
C
a
bc
D1 D5D3
D4 D2D6
R
Va
Vc Vb
Vo
ia
ic
ibn
Fig. Three phase Bridge rectifier using Diodes
CIRCUIT DIAGRAM
D5 D1 D3 D5
D6 D2 D4 D6
Vo
ᾠt0
Vcb Vab Vac Vbc Vba Vca Vcb
90⁰ 360⁰270⁰180⁰
Fig.2(a)
Fig.2(c)
Fig.2(b)
Fig.
Vo
ᾠt0
Vcb Vab Vac Vbc Vba Vca Vcb
90⁰ 360⁰270⁰180⁰
Fig.2(c) output voltage waveform
ia or is
030⁰
270⁰210⁰
150⁰90⁰330⁰
390⁰
iab iac
0
iD1
Vml/R = √3Vmp/R = Iml
Fig.2(d) Input current waveform
Fig.2(e) Diode curent waveform through D1
Fig.
0150⁰ 390⁰270⁰
-1.5 Vmp
-√3 Vmp or Vml
VD1
30⁰
D5 D1 D3 D5
D6 D2 D4 D6
Fig .2(f) Voltage variation across Diode D1.
Voltage variation across D1 can be obtained in a similar manner as in the case of 3-phase half wave diode rectifier.
Fig.
Average output voltage V0 =(1/periodicity) ∫VmL sin(ᾠt+30⁰) d(ᾠt)
=(3/∏) ∫VmL sin(ᾠt+30⁰) d(ᾠt)
= (3/∏)VmL = (3√2/ ∏)VL = (3√6/∏)Vp
Where, VmL = maximum value of line voltage
VL = rms value of line voltage Vp = rms value of phase voltage
R.M.S value of output voltage(Vor) =[3/∏ ∫VmL sinᾠt d(ᾠt)] = 0.9558 VmL
Ripple Voltage (Vr) = √(Vrms – Vavg.) = 0.0408 VmL
Voltage ripple factor (VRF) = Vr/Vo = 0.0408 VmL/(3/∏)VmL = 0.0427 or 4.27%
Form Factor = Vor/Vo = 1.0009
R.M.S value of O/P current (Ior) = 0.9558Vml/R = 0.9558 ImL
ᾳ2
ᾳ1
∏/2
∏/6
∏/3
2∏/322 1/2
2 2
Pdc = Vo Io = (3/∏) VmL ImL
Pac = Vr Ir = 0.9558 VmL ImL
Rectifier efficiency = Pdc/Pac = 0.9982
% Rectifier efficiency = 0.9982 ×100 = 99.82%
Rms value of source voltage(Vs) = Vmp/√2 = VmL/√6 (Since, VmL= √3Vmp)
Rms value of line current(Is) = rms value of T/F secondary current
= [2/∏ ∫ ImL sinᾠt d(ᾠt)] = 0.7804 ImL
VA rating of transformer = 3Vs Is = 3 (VmL/√6) × 0.7804 ImL
= 0.955791 VmL ImL
Transformer Utilization Factor(TUF)= (Pdc / Transformer VA Rating) = (3/∏)^2 /0.955791 = 0.9541
2
2 21/2
∏/3
2∏/3
2
Working of 3 phase bridge rectifier
Disassembled automobile alternator , showing the six diodes that comprise a full-wave three-phase bridge rectifier.
Specification :1. Off state voltage max. =1.6kV2. Rated current =160A3. Max. Forward impulse current =1.2kA 4. Thermal resistance =0.16K/W5. 4000 Vrms isolating voltage
Additional information :1. Totally lead (Pb) Free2. Gross weight=0.23kg
3 Phase Bridge rectifier