0
\$\begingroup\$

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?

enter image description here

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?

\$\endgroup\$
4
  • \$\begingroup\$ Yeah seems your calculation is correct ! \$\endgroup\$
    – user215805
    Jul 3, 2021 at 11:38
  • \$\begingroup\$ @user215805 think again. \$\endgroup\$
    – Andy aka
    Jul 3, 2021 at 11:44
  • \$\begingroup\$ @Andy aka isn't for instantaneous voltage OP's calculation is right (at least in assumed condition )? And if his/her intention was to find maximum reverse voltage which Op didn't mention then i was wrong \$\endgroup\$
    – user215805
    Jul 3, 2021 at 15:20
  • \$\begingroup\$ The peak instantaneous reverse voltage across a reverse biased diode is what I explained in my answer @user215805 \$\endgroup\$
    – Andy aka
    Jul 3, 2021 at 19:00

1 Answer 1

2
\$\begingroup\$

what's the reverse voltage applied to it?

Consider 3-phase voltage waveforms from here: -

enter image description 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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.