Good explanation.
As @directive0 said, the diodes (triangles in the above circuit) only conduct in one direction.
On the AC wave (top left) in the positive half, two diodes are conducting. In the negative half, the other two diodes are conducting.
The two sets of diodes are connected so that the positive half and the negative half of the sine wave come out on the same line. Since the waveform on the right of the image is all positive (ignore the bumps), it is considered DC.
It will not work… if you look at a simple AC to DC converter (half-wave rectifier) either output = 0 volts or output = input - conduction losses, depending on how you polarized the DC source.
In another circuit (full wave rectifier), the diodes will be polarized such at two will allow current flow. output = input - conduction losses
If the rectifier is something more complex (12 pulse rectifier)… this circuit relies on transformers. If you put DC on the primary of the transformer then the output will be 0 volts and thus the rectifier output will also be 0 volts. In this example I’ve ignored the fact that the transformers have 3 terminals and the DC source would only have 2.
I’m less knowledgeable of more elegant rectification methods (such as active rectification)…
Interestingly… the fundamental DC to AC converter (inverter) is the same circuit as the full wave rectifier, with the diodes replaced with controllable switches (SCRs, IGBTs, IGCTs, MOSFETs, ect.)
not seeing a cog :/
Edit: I see the cog. :D
Edit 2: I’m not sure what to change to enable infinite scrolling on this page
You could ask lemmy! :D
You tease