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I've made a mini generator for a mini wind turbine. An oscilloscope voltage measurement shows a +0.4/-0.4 V sinusoidal AC current. I'd like to convert it into DC, in order to put a charge and detect amperes.

I have tried to convert it with a classical "4 diodes bridge", with the smallest diodes I found (1N4148), albeit unsuccessfully: the diodes seem to reduce the voltage too much, and at the end, no current at all is delivered (the oscilloscope does not detect any voltage).

How can I accomplish the AC->DC conversion then?

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    \$\begingroup\$ google for active rectification \$\endgroup\$
    – PlasmaHH
    Commented Jul 31, 2015 at 13:36
  • \$\begingroup\$ "measuring the current we see +-400mV" Do you mean you are measuring the voltage or are you actually measuring the current? \$\endgroup\$
    – user16222
    Commented Jul 31, 2015 at 13:40
  • \$\begingroup\$ we are measuring the voltage. \$\endgroup\$
    – Anc
    Commented Jul 31, 2015 at 13:43
  • \$\begingroup\$ You could have a look at the Greinacher voltage doubler circuit, with only two diodes instead of four. \$\endgroup\$
    – DamienWontContributeToAITheft
    Commented Jul 31, 2015 at 16:10
  • \$\begingroup\$ How come there is such a low voltage? Did you design the alternator? How much current is available? At what voltage is the maximum power available? \$\endgroup\$
    – Kevin White
    Commented Jul 31, 2015 at 22:50

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Half-wave with a Schottky diode (eg. 1N5817) would give you some voltage at very low efficiency.

A better way would be to use a step-up transformer to provide more like >10V RMS where an ordinary rectifier or Schottky bridge would be reasonably efficient.

You have 285mV RMS, so >30:1 would be good. If you can find a 240V:5V transformer that should give you about 14V RMS which can be easily rectified (at 1/50th the current, of course). That could provide about 18VDC which could be converted with a SMPS converter to something useful.

There may be some chips designed for harvesting energy that would be effective directly at the 285mV level with an inductor rather than a transformer.

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    \$\begingroup\$ 30:1 would probably not be good, it implies scaling impedances by 900:1, at which point the transformer's own losses tend to absorb most of the available power. Diminishing returns probably start between 5 or 10:1. It might also be useful to know the frequency range, and source resistance of the generator. \$\endgroup\$
    – user16324
    Commented Jul 31, 2015 at 13:57
  • \$\begingroup\$ @BrianDrummond What losses are you thinking of? If the source is really high-Z then magnetizing current might be an issue, but transformer internal losses (core and winding I^2R) will be very low at 6% of design flux density (60Hz) and from a few Hz to a few hundred Hz. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 31, 2015 at 14:19
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    \$\begingroup\$ Transformer directly first, transformers don't like DC. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 31, 2015 at 14:58
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    \$\begingroup\$ Done, and work fine ! : after the what we call the "CI" turbine, we put a transformer 5vac/230Vac for printed circuits, then we put a diodes bridge made with 4 1N5817 Schottky diodes, then we put a 50V/220µF capacitor, and at the end a ruban of LEDs. We turn the turbine at 60 RPM (1 turn per second). The current just after the turbine gives an 1volt AC. The current just after the transformer gives a 30 volts AC. The current just after the diodes bridge gives a not smoothed DC 15 volt. The current after the capacitot gives 15 volt DC totaly planed (a great line). \$\endgroup\$
    – Anc
    Commented Sep 19, 2015 at 9:53
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    \$\begingroup\$ If you want to see: wikilabomedia \$\endgroup\$
    – Anc
    Commented Sep 19, 2015 at 9:55

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