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I put the following circuit in a simulator to see how it works. I understood it but I have a question:

enter image description here

At the time of the vertical green line, The simulator shows me that the current flows from the supply of phase 1 because it is the most positive. And then all the current goes to the supply of Phase 3 because it is the most negative. But Phase 2 is also negative and I expect that some of the current should return back to it. I know that Phase 2 is not the most negative but I think a small amount of current should return back to it.

I watched a video on youtube, I learned also that the current flows form the most positive to the most negative and they did not mention phase 2 although it is not zero voltage !! it has some negative voltage that the time of the green vertical line.

enter image description here

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  • \$\begingroup\$ "but I think a small amount of current should return back to it." so the question is why you think that \$\endgroup\$
    – PlasmaHH
    Sep 18, 2015 at 10:37
  • \$\begingroup\$ but where is the rectified output?? \$\endgroup\$
    – Lokanath
    Sep 18, 2015 at 11:08
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    \$\begingroup\$ @PlasmaHH I think that because the red phase is not zero volt it has a negative volt so it should draw current as well as any other negative volt phases. Andy aka says that's because of the diodes and I'm convened. \$\endgroup\$ Sep 18, 2015 at 11:37
  • \$\begingroup\$ @Lokanath The rectified output is the voltage difference between the most positive phase and the most negative phase. It is known that the wave form is six output pluses and it is called "six pulse rectifier". \$\endgroup\$ Sep 18, 2015 at 11:40

4 Answers 4

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Think about two diodes feeding a load resistor. Don't think about 3-phase until after you understand this: -

Circuit diagram where A and B supply a load resistor through separate diodes

If "A" is 10 Vots and "B" is 9 volts, "A" supplies all the current to the load. "B" can't supply any because its diode is reverse biased.

This is what happens in a 3 phase rectifier. Think about the scenario when the phase is about 130 degrees: both phase 1 and phase 2 will be at a positive voltage but only one of the phases will supply current to the load because its diode is the only forward biased diode. Same with negative sides of the voltage waveform - the most negative phase's diode will be the only one forward biased.

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  • \$\begingroup\$ Thank you very much for your great answer. You provided an Excellent illustration. Now, it is clear to me :) \$\endgroup\$ Sep 18, 2015 at 11:47
  • \$\begingroup\$ This is a clear example with a schematic. My power electronics professor always told us that if you don't understand 3-phases, see if you understand 1-phase \$\endgroup\$
    – Doombot
    Sep 18, 2015 at 19:23
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Which abstraction level do you want to use?

If your diodes are ideal, only the most positive and most negative legs feeding the 3-phase rectifier will carry any current, because the other diodes will block perfectly.

With non-ideal diodes, the third leg will indeed carry some very small current. But it will (at the moment in time you indicate by the green line) flow out of the red phase towards the blue phase.

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This is the way a full wave bridge works. Obviously nobody can answer why you think it should be different, but it isn't and it shouldn't.

Diodes (at least ideal ones, close enough for purposes here) are like IF statements. If the voltage is positive, they will conduct. If the voltage is negative, they won't. The current will flow from max(phases) to min(phases). In you example, phase 2 is has neither the min nor the max voltage, so it should be obvious why it is not participating in the current at that instant in time. Again, I can't read you mind to say why you think it shouldn't be so.

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Let me start with the basics,,,

1)There is no "Negative Voltage" ,it is interpreted as negative since the current changes its direction in the circuit,it just flowing in reverse direction.

2)Diodes block reverse flow of current,i.e ideal reverse biased diodes wont conduct

3)

At the time of the vertical green line, The simulator shows me that the current flows from the supply of phase 1 because it is the most positive. And then all the current goes to the supply of Phase 3 because it is the most negative. But Phase 2 is also negative and I expect that some of the current should return back to it. I know that Phase 2 is not the most negative but I think a small amount of current should return back to it.

You have a point there but potential difference comes into play here consider two voltage source connected in parallel say 10v and 20v if you probe across the source you'll will read 20V in a multimeter ,,,current chooses the path with greater potential deference hence you cannot expect any current from phase 2 as difference if > between phase 1 and phase 3

hope this helps

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