Three phase full wave rectifier
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Transcript of Three phase full wave rectifier
THREE PHASE FULL WAVE rectifier
Submittd to submitted by Mr. dharmendra
vinay singh Upaddahya
1404620904
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.
Three-Phase, Full-Bridge Rectifier
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
Summary
• Line-frequency diode rectifiers converts line-frequency ac into dc in an uncontrolled manner
• Various diodes rectifier circuits have been discussed
• Three-phase rectifiers are preferable in most respects over the single-phase ones
• Rectifiers inject large amounts of harmonic currents into the utility systems – remedies would have to be implemented