3 phase diode rectifiers/power electronics

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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 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
  • D5 D1 D3 D5 D6 D2 D4 D6 Vo t0 Vcb Vab Vac Vbc Vba Vca Vcb 90 360270180 Fig.2(a) Fig.2(c) Fig.2(b) Fig.
  • 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