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Download Q. 1 Two inverse parallel connected SCRs are used as a contactor
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POWER ELECTRONICS TUTORIAL ER PO W POWER CONVERTERS ICS TRON ELEC AC-AC PHASE ANGLE CONTROL : R +L VT1ak iT1ak iinin(t) iout iT2ak vin (t ) VT2ak VR R Vout VL Q. 1 L Two inverse parallel connected SCRs are used as a contactor to connect an inductive load to a 240 V rms, 50 Hz ac supply. The load is an 8 resistor in series with a 0.12 H inductor. Determine the phase angle at which the SCRs should be first triggered in order that there is no turn-on current transient. ( = 78o) Q. 2 An a.c. regulator is used to control the power delivered to a load consisting of a 100 resistor in series with a 550 mH inductor. The regulator SCRs receive gate drive from the instant of triggering until the following input-voltage zero-crossing. Determine the minimum range of firing delay necessary to allow control of output power over the maximum possible range. ( = 60o to 180o ) Q. 3 A single-phase ac controller supplies a series R-L load with R = 5 and L = 5 mH. The input voltage is 240 V rms, 50 Hz. The SCR firing delay angles ) are symmetrical and equal to 60o. Determine EET307 a. the scr conduction angle b. the rms output voltage c. the rms output current d. the scr rms current e. the scr average current f. the input power factor POWER ELECTRONICS TUTORIAL PETac-ac2: R+L 1 Prof R T Kennedy VT1ak iT1ak Q. 4 iinin(t) iout iT2ak VT2ak vin (t ) VR R Vout VL L Vin,rms f Z cos 240 V 50 Hz 2 0.5 (i) If the circuit shown is used as a contactor to connect the load to the supply determine the trigger delay to minimize turn-on transient current. (60o) (ii) If the SCR conduction angle 120o in the AC regulator determine using performance curves . (a) 60o , the SCR trigger delay angle (b) rms output voltage 162.92 V (c) ac switch rms current 55.545 A (d) SCR rms current 38.94 A (e) input power factor 0.26 (iii) sketch the circuit waveforms using PEW: PAC:AC:R VT1ak Q. 5 iT1ak iinin(t) vin (t ) 90o iout iT2ak 270o VR R VT2ak VL Vin,rms f L 240 V 50 Hz 18 mH R 10 Vout L (i) The circuit shown is used as an AC regulator to control the load power. Using performance curves determine 230 V rms! (a) , the SCR conduction angle 120o (b) output power 1555 W (c) individual SCR average current 4.53 A (d) input current 12.47 A (e) input power factor 0.542 (ii) Show with the aid of circuit waveforms (PEW: PAC: AC : R+L) that there is a need to consider snubber circuits in parallel with the SCRs. (iii) Determine the permissible control range for EET307 POWER ELECTRONICS TUTORIAL PETac-ac2: R+L ( 30o 180o ) 2 Prof R T Kennedy Q. 6 VT1ak iT1ak 90o iout iinin(t) 270 VT2ak o vin (t ) iT2ak VR R Vin,rms f L 240 V 50 Hz 18 mH R 10 Vout VL L (a) The circuit shown is used as an AC regulator to control the load power. Using performance curves determine (i) the SCR conduction angle 140o (ii) the output power 290.95 W (b) If the inductor is removed determine the new firing delay angle for the same load power ( 1300 ) EET307 POWER ELECTRONICS TUTORIAL PETac-ac2: R+L 3 Prof R T Kennedy