I wonder if the 'Neutral Point' as mentioned in the attached image is correctly labeled or not? Is there any mistake in this 3-phase circuit design?

enter image description here Edit 1:

Also it looks like the 'Neutral point' is electrically at the same potential as DC Bus (+) point. So can we connect the 'Neutral point' with either DC Bus (+) or (-) point? Please correct me if I am wrong.

Edit 2:

What condition will turn-off the opto-coupler 'U6'? It looks like it will always be in ON state.

Edit 3:

Do we really need diodes D1, D2, D3? If the three resistors R5, R7, R8 are used will they also not give the same result?

  • 1
    \$\begingroup\$ It is a high impedance neutral at best. R5, R7, and R8 are 470K each, so if the phases are an exact match and so are the resistors, in theory that 'point' should be at zero volts. This is a delta feed with no reference to ground. \$\endgroup\$ – Sparky256 Feb 18 at 1:04
  • \$\begingroup\$ Can i use this this point as neutral point to see the phase voltages on the oscilloscope? Also what will I get at the opto output? \$\endgroup\$ – scico111 Feb 18 at 1:16
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    \$\begingroup\$ The optocoupler output will be a DC voltage of about 4.3 volts with some ripple on it. \$\endgroup\$ – Sparky256 Feb 18 at 1:30
  • \$\begingroup\$ Will there be any situation when the opto will be off and I do not get this 4.3 V.. for example if any phase is missing or unbalanced mains or something else? \$\endgroup\$ – scico111 Feb 18 at 1:36
  • \$\begingroup\$ Also this 'Neutral point' will be at same voltage level as 'DC Bus (-)'.. please correct me if I am wrong. \$\endgroup\$ – scico111 Feb 18 at 1:59

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