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?