Generator AC to DC

Discussion in 'The Projects Forum' started by Macgyverasb, Nov 14, 2009.

  1. Macgyverasb

    Thread Starter New Member

    Nov 14, 2009
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    Hello all! I would appreciate any help that you can offer.

    I have a crude AC generator that I would like to convert its output to a stable DC voltage. I purchased a full wave bridge rectifier, but I think I need a capacitor as well. What else would I need to make it as equal as I can to a DC battery supply?

    I will be calling it a night shortly, but I will return in the morning.

    Thanks.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    What do you mean by "crude AC generator"? Is it a home-built, or manufactured by a company?

    Is it a single split phase output? Three phase? Polyphase? (more than three)

    Does it have a field winding in the rotor, or does it use permanent magnets?

    Do you know what the output voltage is? What is the current, VA or wattage rating?

    How much of a load do you expect to put on it?

    Do you want the output regulated, or is ripple on the DC voltage OK?
     
  3. Macgyverasb

    Thread Starter New Member

    Nov 14, 2009
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    It is home built, using permanent magnets and copper wire.

    I'm not sure how to identify that.

    Permanent magnets.

    The AC output voltage peaks at 13.1

    I would like to get 5v of DC out of it.

    I would like the output regulated, between 3v and 5v DC.
     
  4. Macgyverasb

    Thread Starter New Member

    Nov 14, 2009
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    0
    Thanks for the help on this!

    Alberto, I'll try what you suggested and get back to you about my progress.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    You will need one bridge rectifier per coil in your home-built motor.

    The AC inputs go to either end of a single coil.

    All of the + outputs get wired together, and connected to the + side of an electrolytic capacitor.
    All of the - outputs get wired together, and connected to the - side of an electrolytic capacitor.

    If you were reading 13.1VAC peak RMS with no load, you should read about 16V to 17V DC peak across the capacitor with no load.

    You could then use a 7805 regulator with the INPUT terminal connected to the + side of the capacitor, and the GND terminal connected to the - side of the capacitor.

    You should use a small 1uF capacitor between the OUT and GND terminal. You should then get a fairly constant 5v on the output. Current will be a maximum of 1A.

    You should use a large heat sink on the 7805 regulator.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Here is another way to create a poly-phase bridge rectifier and regulated supply. Only 3 phases are shown. The boxes on the left represent your coils with voltage being output. Note that one end of each coil has been connected together, and that point serves as a common ground.

    The rectifier diodes could be 1N400x series for currents up to 1A, or 1N540x series for up to 3A.

    The 1N540x rectifier diodes will have less of a voltage drop across them than the 1N400x series diodes.

    The points labeled (A), (B), etc have their waveforms displayed on the simulated O-scope at the bottom in their respective colors.
     
  7. Macgyverasb

    Thread Starter New Member

    Nov 14, 2009
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    Thanks to both of you for your suggestions and help in my project. I purchased the recommended components and got the predicted results.

    As an aside, is there any usefulness to having a generator that runs off of a 5v battery and can produce over 10v without the use of a transformer? I should know better than to think that I'm getting more out than I'm putting in, but it would seem to be the case. I think I just need someone to tell me that it's not really the case.
     
  8. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    The ratio of power in to power out will tell that tale. Power is the product of current and voltage. If you have a meter, you can make measurements.

    For a fixed load, the product of input current times that voltage will always be larger than the voltage output and current through the load.
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    You will likely discover that a switching "boost" type or flyback supply will be more efficient than a motor driving a generator/alternator. There are losses in power incurred by the rotational friction of the bearings, air resistance, etc.

    In this particular circuit, there are also substantial losses of power in the voltage drop across the diodes and the very inefficient linear regulator.

    The suggested components have the advantage of being relatively easy to obtain, understand and use.

    More recent developments in power design use power MOSFETs as "ideal diodes" for rectification. When a MOSFET is turned on, it has very low resistance, thus practically no voltage drop occurs across the source and the drain terminals. The control circuitry for such "ideal diodes" is somewhat complex and difficult to understand in comparison to a rectifier diode.

    A typical rectifier diode will have about a 1v drop from anode to cathode at it's rated current. If you are rectifying a sine wave that is 10v peak to peak, you lose 10% of the voltage across the rectifier alone.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You will likely discover that a switching "boost" type or flyback supply will be more efficient than a motor driving a generator/alternator. There are losses in power incurred by the rotational friction of the bearings, air resistance, etc.

    In this particular circuit, there are also substantial losses of power in the voltage drop across the diodes and the very inefficient linear regulator.

    The suggested components have the advantage of being relatively easy to obtain, understand and use.

    More recent developments in power design use power MOSFETs as "ideal diodes" for rectification. When a MOSFET is turned on, it has very low resistance, thus practically no voltage drop occurs across the source and the drain terminals. The control circuitry for such "ideal diodes" is somewhat complex and difficult to understand in comparison to a rectifier diode.

    A typical rectifier diode will have about a 1v drop from anode to cathode at it's rated current. If you are rectifying a sine wave that is 10v peak to peak, you lose 10% of the voltage across the rectifier alone.
     
  11. rvh002@gmail.com

    Active Member

    May 15, 2009
    118
    2
    With reference and acceptance of all the above, you will loose at least 17% of the planned output
     
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