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Most explanations of full wave bridge rectifiers start off with a diagram like the one below, and
then go on about what will happen on the positive and negative cycle of the AC input voltage. This is all fine and well, but I would like to understand what goes on at a deeper level.


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During the positive cycle, diodes D2 and D3 will conduct, but let's stop right here: the depletion region of D2 will only be forward-biased (start to conduct) if its anode is more positive than its cathode. But for this to work, terminal 'a' would have to be at a lower voltage than D2's anode initially. I don't see any defined potential for terminal 'a' in the beginning, but with 'a' floating, how can the bridge begin to rectify?

Clearly, some of my assumptions must be faulty, but I don't see it.

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'a' is not floating, it is connected through the load to 'b', which in turn is connected through D3 to the negative terminal of the supply.

When the upper AC terminal is positive, there will be a conducting path through D2, 'a', Rl, 'b', and D3 to the lower AC (negative) terminal.

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  • \$\begingroup\$ You made me remember that everything happens 'at the same time', so to speak: For the positive cycle, a voltage potential across D2 is also seen across RL, and at D3 simultaneously. Like tentatively going around to see which doors can be opened a bit more... \$\endgroup\$ Apr 1, 2021 at 2:10

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