Why does the current in a rectifier circuit not go through the diodes marked with the question marks in the picture and create a short circuit when the current is returning from the negative side of the load? Is it because the two mentioned diodes are already reverse biased in the first half of the circuit?
Figure 1. The same circuit with a couple of voltage measurements.
- D1 doesn't conduct because its cathode is at 12 V and its anode is at only 0.7 V. The diode is reverse biased.
- D4 doesn't conduct because its cathode is at 11.23 V and its anode is at 0 V. The diode is reverse biased.
Figure 2. A non-return valve analogy. Image source: LEDnique.
Diodes are electrical non-return valves. If you look at the check-valve in the figure above, it should be clear that the spring normally keeps the ball in position and prevents back-flow. When “forward-biased” the ball shut-off can be moved against the spring but it will take some initial pressure to move the ball. This results in a pressure drop across the valve: the pressure downstream will be less than the inlet pressure.
In a similar manner the PN junction causes a voltage drop. For silicon it is about 0.7 V. And similar to the valve the diode PN junction prevents current flowing backwards through the valve. Only when the voltage at the anode is higher than the voltage at the cathode will current flow.
From the comments:
Why the current does not go through those same marked diodes which would be in forward bias relative to the returning current from the negative side of the load?
They're not forward biased. Forward biased means the anode voltage is higher than the cathode voltage - not that current near it appears to be going the right direction.