How does voltage distributes in full wave rectifier with resistive load?

Question

I am a bit confused with how voltages distribute themselves in the circuit, for the positive cycle the diodes D1, D2 are turned on.

We have positive input voltage terminal connected to D1's anode and there's a negative input voltage terminal connected to D2's cathode.

The load resistor is in between these two diodes for positive half of cycle.

If the negative terminal of input voltage is taken as 0 volt, which is connected to diode D2's cathode, and as per figure the anode of D2 is also connected to 0 volt of the negative load resistance terminal, then there would not be any voltage drop in diode D2 and it wouldn't conduct?

Answer 1

Of course if AC source were defined have 0V on bottom pin, it would mean D2 is bypassed with short circuit as resistance has 0V too.

But in this circuit, the AC circuit has no other reference to 0V, except via the diode, so don't make an assumption the AC source is directly connected to 0V.

Answer 2

If they are typical silicon diodes, then they drop about 0.6V when conducting. Other types of diode are available, but silicon ones are commonly used for rectifiers, so that's what I am assuming here.

You have declared that the negative terminal to the load is at 0V. Nothing in the circuit diagram indicates that the bottom terminal of the AC supply is also at 0V, so don't make that assumption.

The anode of D2 is at 0V, because you have said it is. If the AC supply is in its positive half cycle, then the cathode of D2 must be at -0.6V.

Similarly, the cathode of D1 is at +V, so the anode of D1 must be at (+V + 0.6V).