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B & S Dual Circuit alternator question


Horvik

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Hello;

 

If the diode is removed from the output of this alternator may it then be connected to a 10 amp regulator/rectifier and work correctly, producing upwards of 10 amps?

 

Thanks

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Interesting idea.  We need an EE to answer your question.  

The Briggs service write up says there are 2 separate alternator coils.  Which leads me to believe that the "other end" of the alternator wire is connected to ground since only one wire for each circuit comes out from the tin vs 2 in the normal setup:

 220px-Alternator_1.svg.png

I would think to have a balanced load, each coil would have to be the same and I think the lighting circuit has a higher output.  Since the individual circuits provide power, I guess it would be OK that the "center" of the coil setup would be grounded.  

If you removed the diode and had an oscilloscope, you could see the wave pattern and see how well they matched.   

 

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Phandad...I think you are onto something with the idea of "balanced load" I also think it may, going along with your idea, that if these two (currently running) coils are different, then they are in a sense " out of phase" with the plug in rectifier, or something like that! I say that as i cannot get it to output any measurable DC output out of the rectifier. Also, because of this idea, it then affects the regulator "not sensing" a load , though it has been thoroughly checked as being hooked up correctly. 

 

If this basic idea is right, then to make it work with the external regulator, you have to remove this existing stator and replace it with 592830 (not dual circuit). Just the way it's looking to me.

Thanks

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A rectifier is just a bank of diodes to change A/C to D/C and combine the charge circuits so you would just be swapping out one diode for another. Second, a regulator just changes the magnetics in an altenator to increase charge rate. Being these use permanent magnets you cannot regulate the rate except by rpm. The light circuit in most cases is an A/C circuit at 12 volts if you go through a diode you would cut that in half to 6 volts D/C. the charge circuit is 25-27 volts A/c before the diode. As Always if I am incorrect in my thoughts and experience let me know, I am not an expert.  That being said If both circuits show the same A/C volts by all means you could run it through a rectifier to combine them. 

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5 hours ago, firefoxz1 said:

The light circuit in most cases is an A/C circuit at 12 volts if you go through a diode you would cut that in half to 6 volts D/C. the charge circuit is 25-27 volts A/c before the diode.

Thanks for the explanation. 

Per an early post, per Briggs Repair Manual, the AC circuit is 12V:

image.png.f50e955c690ea9eadd8344270a6cc423.png

 

The diode is chopping off "half" of the wave, so it halves the DC voltage.  Also from the Briggs Repair Manual:

image.png.a644605419cf012b52d654f8090692bc.png

So if you connected one AC terminal of a rectifier to ground and the other AC terminal to the "charging wire" with the diode removed, would you get a 6 amp DC supply?  

 

 

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I did a voltage check today of the rascal in question: At about 3/4 throttle I got 10 VAC on the one leg, and 25 VAC. Got to look into that thing you mention.

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The factory headlights on this 3416 H were wired into the factory (Starter/Generator) DC harness.

That being said, may these same headlights be run on the AC off of the  light circuit?

(Better to ask now than to possibly blow out lights!)
 

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