Is this the correct way to simulate a three-phase full bridge (6 pulses) rectifier?

I ask because I have tried this first without the two capacitors, just one, and the capacitor seems to have discharged (or charged in other direction) in some parts of the wave.

I think this is because the line-to-line without ground reference, then I added two capacitors with a ground (that is in common to neutral) like show in the image, and it seems to work fine, but it should difficult some simulations I think.

(The line-to-neutral RMS voltage is 220V and line-to-line RMS voltage is 380V).

I'm correct about the assumption?

three-phase full-bridge rectifier simulation

Also, is it acceptable (yes that depends on regulations and so on, but lets take as acceptable) to just wire a plug with line to line wiring (plus ground for safety), without the neutral wire as the equipment will not be connected to neutral?


@ConnorWolf suggested doing the differential measurement across the two rails. Although I have that in mind too, I didn't have doing it (but needed sometimes) and as it says its just implementing it by math using the difference between the two measured points. It worked, I just don't know how other components that my refer ground would get to work.

The V-BUS(+) in the graph is the difference between V-BUS(+) and V-BUS(-) so, V-BUS(+) - V-BUS(-).

three-phase full-bridge rectifier simulation

Also, from the comments of the non-reader @Naib I put the power supply floating removing the ground reference from its neutral (that I think is not what @Naib is suggesting, I will edit if it is), and putting the ground in the DC-Link without any high-impedance resistor.

three-phase full-bridge rectifier simulation

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    \$\begingroup\$ Why do you have a large resistor to EARTH from the supply? This point should be EARTH. A high resistor is needed from DC- (and maybe DC+) to keep the simulator package happy - they generally do not like floating nodes and like to reference to their EARTH \$\endgroup\$
    – user16222
    Feb 9, 2014 at 23:55
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    \$\begingroup\$ @Naib - It looks like it's because it was in Tinchito's answer. \$\endgroup\$ Feb 10, 2014 at 6:29
  • \$\begingroup\$ @Naib Read the article better, it clearly says its because of a user answer, be more clever and delete that comment as my design is not that, I just wanted to show the user it didn't worked. Also I comment just about the floating nodes and reference to the earth (what's wrong calling it ground?). Anyway I didn't understand exactly where you want the ground to be. You mean I should remove the ground from the supply? \$\endgroup\$ Feb 10, 2014 at 10:48
  • \$\begingroup\$ @Naib removing the ground reference from the supply and putting the reference only on the rectifier section worked. Although you say things before reading, I would mark as an answer if you post it. \$\endgroup\$ Feb 10, 2014 at 11:00
  • \$\begingroup\$ @Naib ops, what you said is to put ground directly to the supply, and use a high-impedance resistor (high resistor), just from the rectifier to the ground. If this is the point, no it gives wrong results. \$\endgroup\$ Feb 10, 2014 at 11:12

2 Answers 2


Maybe you can add a high impedance path to the neutral, like this:


simulate this circuit – Schematic created using CircuitLab

This works for me when i want to simulate a 3 phase rectifier.

  • \$\begingroup\$ I tried it and it don't worked, I posted an image on the original question. Thanks. \$\endgroup\$ Feb 9, 2014 at 23:37
  • \$\begingroup\$ @DiegoCNascimento Can you tell me the name of the software you're using? \$\endgroup\$ Feb 10, 2014 at 3:21

Basically, the problem you are having is that simulations aren't perfect, and they get really cranky when all the nodes don't have a path to ground.

The results you're getting from the modifications as suggested in @Tinchito's answer is exactly what I would expect.

Basically, since one of the three phases is always more negative then the earth, the negative power node is basically just that voltage referenced to ground. Remember, there is zero capacitance between the earth and the negative node of your capacitors. Therefore, the time-constant of even a 100 MΩ resistor is zero, and you directly see the mains frequency applied as an AC signal superimposed on the rectified DC.

At this point, I kind of have to ask what specifically are you trying to do? Really, the second simulation would work. If you take the differential voltage from the HVDC+ and HVDC- rails, it will be what you would expect.

  • \$\begingroup\$ thanks for at least reading that the uptade has an answer. I don't think if you mean by the second solution is the first with the two capacitors with reference to ground (that's my solution and yes it worked), I know what is the problem, I just want a better solution, as I don't want to place a split capacitor every time I need a capacitor, and this probably would cause problems when I extend the design. Its a theoretical question, but at first I used it to rough simulate the PF of the rectifier. \$\endgroup\$ Feb 10, 2014 at 10:53
  • \$\begingroup\$ The issue is basically you're basically designing a system where both power rails are live. What you are seeing is what you'd actually get. There is no reason you need the split capacitor. If you set up your simulator to display the voltage across C2 rather then from each terminal of C2 to ground, it will still work fine, and you will get the results you expect. \$\endgroup\$ Feb 10, 2014 at 10:55
  • \$\begingroup\$ sure, I have tried that, but I didn't found any way to measure differential potential not referenced to ground. Anyway removing the ground reference from the supply and putting it just on the DC-Link worked. Thanks. \$\endgroup\$ Feb 10, 2014 at 10:58
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    \$\begingroup\$ If you can float your three-phase source (which is what removing it's ground-linkage does) will make the circuit behave like I think you expect it to. For what it's worth, if your simulation package can't do differential measurements, you should get a better simulation package. Anything spice-based should be able to do measurement math with almost zero effort. \$\endgroup\$ Feb 10, 2014 at 11:00
  • \$\begingroup\$ Well I think I get what you are trying to say, its the math between the two points, so DC-BUS(+) - DC-BUS(-). This worked too, but I don't know how it would affect other components. \$\endgroup\$ Feb 10, 2014 at 11:22

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