Half-Wave Rectifier with Ripple Capacitor We can see from the output of the rect
ID: 2081292 • Letter: H
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Half-Wave Rectifier with Ripple Capacitor We can see from the output of the rectifier in the previous section that the output voltage waveform is far from a constant value. We can smooth this rectified AC voltage waveform by using the charging and discharging properties of a capacitor. A capacitor is a device that stores electrical energy. By connecting a capacitor across the output of the bridge rectifier, the capacitor will charge on the rising part of every positive half-cycle and discharge across the load circuit during the remainder of the cycle. In other words, when the AC voltage starts to reduce, the diode turns off and the capacitor supplies the load current until the next positive half-cycle. Unless you have a very large capacitor and a very low power load (i.e. a high load resistance) there will be some noticeable ripple voltage. Some ripple voltage is quite acceptable. It is possible to show that a rough approximation for the relationship between the capacitor value and the load current l_L, period T and ripple voltage V_r is given by: C_r = I_L T/V_r The output voltage from a full-wave bridge rectifier together with a ripple capacitor would be approximately DC. It is termed an unregulated DC voltage source not only because there is still some ripple, but also because there might be an overall voltage drop due to internal resistance in the transformer at high current loads. In this section we will examine the effect of adding a capacitor to a half-wave rectifier and measure the output voltage ripple (that is, the peak to peak voltage) Do not switch on the mains power supply yet! Caution 1: The large capacitors are of the electrolytic type as shown in Figure 2.19. TheseExplanation / Answer
the calculation of vr depents on the equation
vr=Iload/(fC)
Iload is load current
f is ripple frequency
and C is capacitor value
substituting given values
f=50hz
Iload=300 mA
C=100,330,1000,2200 micro F
vr for ,c=100microF=>60 v , substitute in eqn , ie 300*10-3/(50*100*10-6)=60v
vr for ,c= 330 micro F=>18v
vr for c=1000micro F=>6v
vr for c=2200microF=>2.7v
measured value is done in laboratory and result should be compared
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