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. 


Expert's answer

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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