Figure 3 shows a fullwave bridge rectifier with a load resis

Figure 3 shows a full-wave bridge rectifier with a load resistance R = 1 k Ohm. The 10:1 transformer steps down a 120 V_rms 60 Hz voltage to 12 V_rms. The four diodes in the bridge rectifier can be modeled with 0.7 V drop for any current. What is the peak value of the rectified voltage across the load? For what fraction of a 60 Hz cycle does each diode conduct? What is the average voltage across the load? What is the average current through the load R?

Solution

Ans) The peak value of output voltage is

Vo(peak)=Vs(peak)-2V_D=12*sqrt(2)-(2*0.7)=15.57 V

Vo(peak)=15.57 V

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The conduction angle of diode is

theta=180-2*sin^-1(0.7/12*sqrt(2))=180-2(2.36)=175.27 deg

The fraction of 60 Hz cycle the diode conducts is =(175.27/360) *100=48.68% cycle each diode conducts

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The average voltage across load is

Vo=2Vo(peak)/pi=2*15.57/pi=9.91 V

Vo=9.91 V

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The average current through the load is

I_o=Vo/R=9.91/1k=9.91 mA

I_o=9.91 mA

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The equation editor has a problem so I typed here itself

sqrt means square root