One of the significant advantage of three phase transformer is that the rectified voltage do not falls to zero even when no smoothing arrangement is used. Therefore, if in place of single phase transformer we use three phase transformer the ripple factor can be reduced up to a large extent. Thus, the Smoothing circuit is used, filter works as a smoothing circuit for rectifier system.īut after this smoothing process the rectifier voltage falls to zero at some point. If this pulsating DC voltage is used in several applications it lead to poor performance of the device. These ripples are the AC components in the DC voltage. Thus, in such types of arrangement we need smoothing circuit in order to remove these ripples. While in case of half wave rectifier the value is quite large but in full wave rectifier too the value of rectifier is significantly large. In both the cases the value of ripple factor cannot be neglected. In case of half wave rectifier the ripple factor is 1.21 and in case of full wave rectifier the ripple factor is 0.482. This means the book was giving the formula for the average voltage output not the rms voltage output, and from above it was also for a delta input not a wye input.The drawback of this arrangement is high ripple factor. So from this we can see that if we calculate the average voltage we get a different result than if we calculate the rms voltage. The conversion factor for a three phase system peak voltage to rms voltage is:Īssuming a Delta input with line to line voltages of 170v peak we get: There is another ambiguity here however, in whether we are looking for the average value of the output or the rms value of the output. What this means is that the 'equation' is for a delta input, not a wye, so the drawing is incorrect and should be shown with all three voltage sources in series, not with their neutrals connected together. The factor for peak to average for the three phase system is the same, 3/pi, so multiplying this factor times 170 we get 162.3 volts average. The line to line voltage vLL is only 120vrms then, and that means 170v peak. Now if the drawing was drawn as a Delta and not a Wye, then the interpretation is entirely different. Not including diode drops (as many of these problems do) this is the average voltage that would be measured out of the circuit shown in the drawing. The factor for peak to average for a three phase system is 3/pi, so when we multiply this factor times 294 we get 280.7 volts average.
The line to line voltage vLL of the Wye configuration is sqrt(3) times the line voltage vL, so the line to line voltage is not 120vrms it is 207.8vrms which means the peak is 294v peak. Had it been drawn as a delta, then the line to line voltage vL would have been equal to 120vrms or 170v peak. The drawing is of a Wye input not a delta, so Vs must be interpreted as the line voltage vL not the line to line voltage vLL. The line voltage vL=Vs of the two are the same, but the line to line voltages (vLL) are not the same. The thing is, the drawing shows a Wye input, not a Delta input.
#BRIDGE RECTIFIER CALCULATOR FULL WAVE HOW TO#
There is a technical inaccuracy in the drawing which causes some ambiguity on how to interpret this when combined with the 'given equation' (really the factor).