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For further detail, I plan to extract an alternating current and convert it to direct current. Suppose that the AC flowed through the wires; how can I make it a direct current using the full bridge rectifier and use it to power electronic devices?

Thank you for your help.

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    \$\begingroup\$ you may wish to consider accepting one of the answers. It's an important part of the process. \$\endgroup\$ – Stephen Collings Jan 8 '14 at 16:27
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A few specs are important: input voltage, output voltage, output current, whether the input is single-phase or three-phase, acceptable output voltage ripple, and output voltage tolerance.

For typical single-phase full-wave rectification, you'll need four diodes and a capacitor. Three-phase would require an additional two diodes.

The DC voltage output will be roughly the peak of the AC line. You'll need a transformer to step it up or down from your source. Even if you just want to rectify the AC line, a transformer is probably a good idea, just to help limit the available energy on the output side in case something goes wrong. The transformer needs to be spec'd to supply at least as much power as you'll be pulling with your load.

Make sure the diodes can handle all the current you want to pull, and the voltage of the DC bus. You can use individual diodes, or a package containing multiple diodes already arranged as a rectifier.

The capacitor should be selected to handle the DC bus voltage. The capacitance and the load current set the ripple voltage. For a 60 Hz line, the cap will be charged once every 8.3 mS. How far the cap drops in between line peaks is determined by the equation. I = C dv/dt, or rearranged, dv = dt I/C. How much ripple is acceptable depends on your application.

Even with relatively low ripple, you'll need a regulator downstream of the capacitor if you want any sort of precise voltage, as the AC line will fluctuate over time.

Also, appropriate use of fuses is important. Put a fuse with a rating somewhat higher than your expected load current in the DC output link. Having a switch in that link may also be nice, just for ease of use, depending on your application.

Power supply topology https://secure.wikimedia.org/wikipedia/en/wiki/Rectifier#Full-wave_rectification

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The basic full wave bridge circuit looks like this:

This would normally be followed by a capacitor to smooth out the DC. Otherwise the DC level will ripple between 0 and the maximum of the AC input wave minus two diode drops.

However, your question is broader than this in that you are really asking about a AC to DC "power supply" in general. That's too broad for a post here, but hopefully this will get you started.

Added:

Someone commented on noise pushed back onto the AC line due to the non-linear nature of the diodes and especially due to short term high currents during the reverse recovery time of the diodes. Here is a comment I tried to add to that post but it got deleted before I submitted it:

Good point, +1. This is also another argument to use Schottky diodes if your AC voltage is low enough, which is usually a few 10s of volts before the problems with Schottkys outweighs their advantages of fast recovery time and lower forward drop compared to regular silicon diodes. If your input is 20 VAC, for example, then Schottkys make a lot of sense. If it's "wall power", then you will want full silicon diodes.

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