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Figure 3-7.Peak and average values for a half-wave rectifier

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3-6 Since a half-wave rectifier conducts once for each full cycle of input voltage, the frequency of the pulses is the same as the frequency of the input sine wave. The output pulse frequency is called RIPPLE FREQUENCY. If the rectifier circuit is supplied power from a 60-hertz ac line voltage, 60 pulses of load current will occur each second. Therefore, THE RIPPLE FREQUENCY OF A HALF-WAVE RECTIFIER IS THE SAME AS THE LINE FREQUENCY. If a series of current pulses like those obtained from a half-wave rectifier is applied to a load resistance, an average amount of power will be dissipated over a given period of time. This average dc power is determined by the amplitude of the pulses and the time delay between pulses. The higher the peak amplitude of the pulses or the less the time between pulses, the greater the average dc power supplied to the load. To determine average dc voltage (Eavg), it is necessary to know the average value of the pulses and the peak value of load voltage. This is illustrated in figure 3-7. Figure 3-7.—Peak and average values for a half-wave rectifier. Since current and voltage waveforms in a half-wave rectifier circuit are essentially half sine waves, we can develop a conversion factor. The formula for average value was discussed earlier in NEETS, module 2. By now you should know that the average value for a full sine wave is .637 times its peak or maximum value. Therefore, if you want the average value of a half-wave rectifier output, you should multiply half the value of .637 (.318) times the peak or maximum voltage, as expressed in the following equation: Eavg (the average load voltage) = .318 × Emax Where: Emax = The peak value of the load voltage pulse In most applications the drop across the rectifier tube is small compared to the load voltage, so we can assume Emax in our equation to be the same as the peak value of the input sine wave. Since the load current has the same wave shape as the load voltage, we can modify the equation so that it applies to the load current. Thus, Iavg (the average load current) = .318 × Imax Where: Imax = The peak load current If a line is drawn through the rectified waveform at a point that is 0.318 of the distance from zero to maximum, the waveform will be divided so that area A is equal to area B (fig. 3-7). Therefore, current or


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