T-bone. Agree more is merrier! It will help if I tell you where I'm coming from. I have a 41 A with mag ign and no battery. So I start with the hand crank. I refused to believe this at first, but the generator actually works with no battery input: lights work and everything. So the answer to your first question is yes! This voltage comes from the armature, and is induced by the magnetic field that comes from the current flowing in the (grounded) field windings. On the second question, by "limit" I mean design limitation. There is no "topping" governor or overload relay on these generators, like larger (multi KW or MW) ones do. As you reduce the resistance in the output current circuit, more current will flow. Using my example, if you replaced the output wire with ga. 2 copper and grounded it, the total circuit resistance would be only 0.1 ohm (plus a little bit for the ga. 2 copper, which is close to zero, maybe 0.0001, but we'll round off to 0.1 to keep the math simple). The armature produces 12 volts as always, but since the circuit resistance is only 0.1 ohm, that means 120 amps is flowing! 120 amps through the gen windings (since the gen is part of the circuit) will heat them up and melt them. Whatever amperage is the winding "melting" amperage is the limit of the generator's output. Big generators have water cooling in the windings to delay this as much as possible and also improve efficiency (work better cold). By "open circuit" or "no load" I mean no load applied to the generator: lights off, no battery to charge, etc. The generator is doing no work. Even with this no-load condition, a voltmeter across the terminals shows 8.4 volts for me. This is because there is current flowing in the field windings, which are grounded. Taking ALL the wires off the generator "should" result in nothing, just like you said. I think this is where the misunderstanding is. You're supposed to disconnect the field ground from the generator when running without the battrey for this reason: running an excited generator (one that's producing voltage) for a long time with no load will hurt it. Taking the field wire off disables the field current, and kills (most of) the magnetic field. I say "most of" and "should" because even with the field circuit open (not grounded) the iron core retains magnetism, and even this little bit produces some voltage in the armature! I took the resistor in my four-positon switch out, so my "low charge" is now equivalent to a open field circuit. This way, I just leave the switch in the off position, and don't have to worry about re-connecting wires when I do get a battery (and a box to put it in). I'll just flip to "high charge" and go. I gather off this board that that is where this switch was usually kept anyway, unless you're running for a really long time, which I don't do. It's called a voltage regulator because it regulates output voltage. A simple one like mine accomplishes this by a relay which opens if voltage is too low (cutout). A better one will keep output voltage from going too high by inserting a resistor into the field circuit, reducing the field current when rpms go up. I think the big confusion was caused by my use of the term "open circuit," which I meant to apply to the output circuit only. Hope that all makes sense. Dave
|