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.
POWER POINT PRESENTATION on Three Phase Diode Rectifier By Nitish kumar singh EEE 36 GURU GOBIND SINGH EDUCATIONAL SOCIETYS TECHNICAL CAMPUS
Three phase half wave rectifier Three phase mid point 6-pulse rectifier Three phase bridge rectifier Three phase 12-pulse rectifier
VD1 D2 D3 R Va Vc VbnA B C A B C io Vo Fig.1 Three Phase half-wave diode rectifier with common cathode arrangement
Rectifier element is Diode. Diode can only conduct at the highest +ve instantaneous voltage. D1 will conduct for t =30 to t =150 D2 will conduct for t =150 to t =270 D3 will conduct for t =270 to t =390 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 0 30 150 270 0 Vc Va Vb Vc Fig.(a) Fig.(b) Fig.(c) Vo Voltage of neutral n Voltage of terminal P Vmp 0.5 Vmp Fig.
0 30 150 270 ic ia ib ic io 90 210 330 is= ia i D1 0 Fig.(d) Load current or output current. Fig.(e) Source current Fig.
0 D3 D1 D2 D3 150 390270 -1.5 Vmp -3 Vmp VD1 30 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 Vo When Diode D1 conduct: Vo = Va Therefore , VD1 = Va Va = 0 When diode D2 conduct : Vo = Vb Therefore , VD1 = Va Vb At t = 180, Vb=0.866 Vmp , Va = 0 VD1 = - 0.866 Vmp At t = 210, Vb= Vmp , Va = -0.5 Vmp VD1 = -1.5 Vmp At t = 240, Vb=0.866 Vmp , Va = 0.866 Vmp VD1 = - 3 Vmp At t = 270, Vb=0.5 Vmp , Va = - Vmp VD1 = -1.5 Vmp HOW
When Diode D3 conducts : VD1 = Va - Vc At t = 300, Va=-0.866 Vmp , Vc = 0.866Vmp VD1 = - 3 Vmp At t = 330, Va=-0.5 Vmp , Vc = Vmp VD1 = - 1.5 Vmp At t = 360, Va=0 , Vc = 0.866Vmp VD1 = -0.866 Vmp At t = 390, Va= 0.5 Vmp , Vc = 0.5Vmp VD1 = 0 EQUATIONS OF PHASE VOLTAGES FROM WHERE ABOVE VALUES ARE OBTAINED--- Va = Vmp sin (t) Vb = Vmp sin(t 120) Vc = Vmp sin(t 240)
Average output voltage V0 =(1/periodicity) Vmp sint d(t) =(3/2) Vmp sint d(t) = (33/2)Vmp R.M.S value of output voltage(Vor) =[3/2 Vmp sint d(t)] = 0.84068 Vmp Ripple Voltage =(Vrms Vavg.) = 0.151 Vmp Form Factor = Vor/Vo = 1.0165 R.M.S value of O/P current (Ir) = 0.84068Vmp/R = 0.84068 Imp Pdc = Vo Io = (33/2) Vmp Imp 2 1 5/6 /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 sint d(t)] = 0.4854 Imp VA rating of transformer = 3Vs Is = 0.707Vmp 0.4854 Imp =1.0295 Vmp Imp Transformer 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 120. 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 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 b c D1 D5D3 D4 D2D6 R Va Vc Vb Vo ia ic ib n Fig. Three phase Bridge rectifier using Diodes CIRCUIT DIAGRAM
Vo t0 Vcb Vab Vac Vbc Vba Vca Vcb 90 360270180 Fig.2(c) output voltage waveform ia or is 0 30 270210 15090 330 390 iab iac 0 iD1 Fig.2(d) Input current waveform Fig.2(e) Diode curent waveform through D1 Fig.
0 150 390270 -1.5 Vmp -3 Vmp or Vml VD1 30 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(t+30) d(t) =(3/) VmL sin(t+30) d(t) = (3/)VmL = (32/ )VL = (36/)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 sint d(t)] = 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 2 1 /2 /6 /3 2/3 22 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 sint d(t)] = 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 2 1/2 /3 2/3 2
Disassembled automobile alternator , showing the six diodes that comprise a full-wave three-phase bridge rectifier.
Specification : 1. Off state voltage max. =1.6kV 2. Rated current =160A 3. Max. Forward impulse current =1.2kA 4. Thermal resistance =0.16K/W 5. 4000 Vrms isolating voltage Additional information : 1. Totally lead (Pb) Free 2. Gross weight=0.23kg 3 Phase Bridge rectifier