3 phase bridge uncontrolled rectifier, resultant waveform

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SUMMARY

The discussion focuses on understanding the resultant waveform of a three-phase uncontrolled rectifier, specifically addressing the discrepancies between the expected and actual voltage peaks during the rectification process. Participants reference C.W. Lander's power electronics text to clarify the voltage/time relationship and the behavior of the rectified waveform. Key points include the importance of recognizing phase shifts and the correct interpretation of diagrams illustrating the waveforms. The ripple frequency resulting from the rectification process is established as 360 Hz, derived from the 60 Hz line frequency and the characteristics of a six-pulse diode rectifier.

PREREQUISITES
  • Understanding of three-phase power systems
  • Familiarity with uncontrolled rectifiers and their operation
  • Knowledge of voltage/time relationships in electrical circuits
  • Basic principles of waveform analysis and interpretation
NEXT STEPS
  • Study the operation of six-pulse diode rectifiers in detail
  • Learn about the effects of phase shifts in three-phase systems
  • Research ripple calculation methods for rectified waveforms
  • Examine C.W. Lander's power electronics textbook for deeper insights
USEFUL FOR

Electrical engineers, power electronics students, and professionals involved in the design and analysis of rectifier circuits will benefit from this discussion.

Bringitondown
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Hi all,

Having difficulty understanding the above resultant wave form, been looking at power electronics (C.W.Lander) but don't quite get the addition of the wave forms (positive and negative cycles)can anyone point me to some reading of theory I am missing for this, thanks.
 

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I don't see how you cannot understand it if you understand the voltage/time relationship between phases and you understand how rectifiers work. Maybe you need to take a step back?
 
What I don't understand is why the rectified waveform does not have the positive voltage peaks (a b c) occurring at the same phase angle as the original 3 phase positive voltages
This is also happening with the flipped negative voltages. Pointed out with the dotted lines in the image above.
 
I think this might be the answer I'm looking for. Need to do the sums to prove I think
 

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Well upon closer inspection I don't think that diagram is completely correct. The upper diagram shows one complete cycle. The lower shows slightly more than one cycle so the vertical dotted lines do not line up correctly. However, the basic shape of the rectified waveform will look pretty much as shown. Sloppy diagram in my opinion.
 
Thanks for your opinion, these were my initial thoughts but I thought the book couldn't be wrong.
 
Diagram is right.
 
Please explain how.
 
I will have to double check the book but it's back in the library just now but from memory the resultant voltage at the load is the difference between the line voltage and the phase voltage that would appear at the top of the load. I have annotated a chopped page from that book, please see attachment.
 
  • #10
Attachment
 

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  • #11
That attachment is pretty much the same. Wrong. Notice the waveform in the top that starts at zero, goes positive, goes back through zero, goes negative and completes on cycle where the drawing ends at zero. Now do the same on the bottom except of course it does not swing negative, it goes back positive but it DOES NOT end up at zero at the same place the top one does. It is wrong.
 
  • #12
I disagree - the diagrams are correct, could be better notated however.

Ref Post #10

The top is Van, Vbn, Vcn, the second is |Vca|, |Vab|, |Vbc|

Since we are only concerned when the Load ( a 6 pule diode rectifier) is positive biased - the second waveform is folded to show the 6 points in the cycle where different pairs of diodes are conducting. The Horizontal axis is 0V, and we "see" phase shift, but we are now talking about Vphase to phase vs Vp-n..

Example -Red point 1
Van is at its peak. However |Vba| is decreasing, and |Vca| is increasing. At this point where Vbn = Vcn (not zero) ... Vbc=0 the bottom diodes in the B and C legs of the rectifier commutate, and the Top Rectifier in A stays forward boards here ( but this is the ideal case, - important to not make real world assumptions based on ideal conditions at points like this )

So the ripple (@ 60Hz line) is 60 hz * pulses (3 Phase * 2 "polarities" ) = 360 Hz
 
  • #13
Thanks for the ripple calc, useful and explained well.
 

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