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Why in a rectifier circuit the current does 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?

rectifier circuit image from https://www.allaboutcircuits.com/textbook/semiconductors/chpt-3/rectifier-circuits/

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  • \$\begingroup\$ Pick a point in the A/C cycle and draw what you think the current flow through the circuit is at that point. Try a few other points. I think you will see why right away. \$\endgroup\$ Aug 4 at 18:57
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    \$\begingroup\$ Because reverse biased diodes block current. \$\endgroup\$ Aug 4 at 19:01
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    \$\begingroup\$ as @user_1818839: because they are diodes. That's what they do. So, if this is surprising, maybe your question is "what does a diode do?" (which is complex enough without the embedding in the rectifier circuit) \$\endgroup\$ Aug 4 at 19:06
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The two diodes are in reverse bias, so they don't conduct. Only the two diodes that are in forward bias will conduct.

The point of a bridge rectifier is to make the current go through load in one direction only, and not to short circuit anything.

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  • \$\begingroup\$ In the first half of circuit before the current reaches the load (the upper part) the two marked diodes bock the current, but when the current is returning in the second half of circuit after passing through the load (the lower part) 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? \$\endgroup\$
    – mhmdghfr
    Aug 4 at 19:14
  • \$\begingroup\$ Sorry I don't even understand what you mean. Only two diodes are forward biased, because the power supply biases them so, and current only flows via forward biased diodes, and not via diodes that are in reverse. \$\endgroup\$
    – Justme
    Aug 4 at 19:27
  • \$\begingroup\$ Actually Transistor (electronics.stackexchange.com/a/580253/290573) has explained it quite good. \$\endgroup\$
    – mhmdghfr
    Aug 4 at 20:04
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schematic

simulate this circuit – Schematic created using CircuitLab

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.

enter image description here

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.

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