Reverse voltage for diode in three phase circuit?

Question

I'm trying to understand what happens when a diode is not conducting in this circuit, more specifically, what's the reverse voltage applied to it?

I know only one diode will be working at a time, but what about the other two diodes?

If D1 is ON, and I assume 0.7v drop on it, that leaves

VR = VRN - 0.7v

Then if the voltage on the cathode of the diodes is equal to VR, that means in the case of D2 the reverse voltage is:

VD2 = - (VRN - 0.7 - VSN)

Is that right?

If that's true, then for D3, the reverse voltage is:

VD3 = - (VRN - 0.7 - VTN)

Is that the reverse voltage for the diodes in this circuit?

Answer 1

what's the reverse voltage applied to it?

Consider 3-phase voltage waveforms from here: -

When Phase_1 is maximum, the other two phases voltages are not at negative maximum but at 50% of their peak negative voltage so, the maximum reverse voltage that any single diode can see is not twice the peak voltage minus one diode drop but somewhat less than that.

However, it's not 1.5 times either; it's somewhere in between 1.5 times and twice the individual peak voltages.

If you do 3-phase theory, you'd know that \$\sqrt3\$ is involved somewhere. In fact, if you plotted each instance of the black and red phase voltages in the picture above you'd find that the peak difference is \$\sqrt3\$ higher than the peak of either.

This is because if one diode is conducting (assume 0 volts dropped) the other diode is subject the the full "line-voltage" and not the phase voltage.

So, the peak verse voltage rating for each diode has to be at least \$\sqrt3\cdot V_{PHASE} - D_{DIODE}\$.

However, anyone designing a three phase rectifier will choose a diode reverse voltage that is far in excess of this value in order to account for surges, drop-outs, imbalances and back-emfs from loads.