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Old 03-27-2012, 10:05 PM   #51 (permalink)
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Beautiful explanation, mort. The flywheel analogy was perfect!

Unfortunately I do not have a clamp ammeter at home. I may steal the one from work tonight. But I did put 12V to the field windings through my $3.68 eBay Chinese multimeter today and it showed 7.25A. Mind you, I think of that $3.68, $0.03 went into the leads because they quickly got scarily hot and stinky, so there was some serious resistance there. I don't know what the amps will be when there is no $3.68 multimeter in the way. Maybe 8? 9? And eventually when I power it with [maybe] 36V on startup, that would make it 24-27A as a guess. Who knows. I will measure the resistance with an ohmmeter tonight and edit it into this post.

EDIT: The field is 1.1 ohms. But that's with a $3.68 multimeter, so who knows. Can I assume 12.6V/1.1 ohms = 11.4A, or does an inductive resistance change when there is current?

Since all these cheaper diodes come in packs of 10, maybe I will hook them all up together to make one strong diode. Add some redundancy to it at the same time!

Last edited by mechman600; 03-28-2012 at 05:20 AM..
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