Question #184245

A single-phase asymmetrical semi-converter feeds an R-L load with R=50 Ω and a large L= 10 mH so that load current is constant. The source voltage is 230 V, 50 Hz. For a firing angle of 600, determine: (i) Average value of output voltage and output current. (ii) Average and rms values of diode, thyristor and source currents.(iii) Input power factor. 


1
Expert's answer
2021-04-26T02:03:54-0400

i)Average value of output voltage and output current.

VO=vmπ=2302π(1+cos60)=155.31VV_O=\frac{v_m}{\pi}=\frac{230 \sqrt{2}}{\pi}(1+cos 60)=155.31V

Io=VoR=155.3150=3.11AI_o=\frac{V_o}{R}=\frac{155.31}{50}=3.11A


(ii) Average and rms values of diode, thyristor and source currents.

Ipavg=3.11×ππ32π=1.04AI_pavg=3.11\times\frac{\pi-\frac{\pi}{3}}{2\pi}=1.04A

IDAvg=3.11×π+π32π=2.07AI_DAvg=3.11\times\frac{\pi+\frac{\pi}{3}}{2\pi}=2.07A

ITrms=3.11×ππ/32π=1.79AI_Trms=3.11\times\sqrt{\frac{\pi-\pi /3}{2\pi}}=1.79A

IDrms=3.11×π+π/32π=2.54AI_Drms=3.11\times\sqrt{\frac{\pi+\pi /3}{2\pi}}=2.54A

Source current IDAvg=3.11×π+π32π=2.07AI_DAvg=3.11\times\frac{\pi+\frac{\pi}{3}}{2\pi}=2.07A

IDrms=3.11×π+π/32π=2.54AI_Drms=3.11\times\sqrt{\frac{\pi+\pi /3}{2\pi}}=2.54A

(iii) Input power factor. 

ISIISrcosα2=2.422.54cos602=0.825\frac{I_{SI}}{I_{Sr}}cos\frac{\alpha}{2}=\frac{2.42}{2.54}cos\frac{60}{2}=0.825


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