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I've had this puzzle for a couple of years but never been able to figure it out. Regulator manufacturers for typical 3-phase alternators provide the 14.4V by 'shorting' the phases. Obviously this comes with great losses but I cannot understand why this design is still in use. My electromagnetic theory is basic but I assume that somehow after a certain angular velocity back emf induces positive torque into the rotor which lessens the braking effect. I did intend to test that on a bench but I assume some of you at least have tried it or know the theory behind it. Is there a graph somewhere that will show braking torque vs RPMs? This could also apply to wind generators.

Edit: The stator is on the bike, contains several wound inductors connected into a 3-phase delta configuration. The flywheel connected to the engine/crankshaft has permanent magnets. The configurations I've seen had magnets inside the flywheel, which would then spin outside of the wound stator. Three wires from the stator go into the 'black box', another (usually) two wires come out.

The rectifier part is a standard 6-diode bridge. Regulator part is a zener-like part that triggers the shunt to all three (incoming) phases simultaneously to the ground after the bridge. Shorting is done via thyristor-like parts, so the zener triggers the gate for the SCRs. Typical generator power is about 400W@9000rpm, typical regulator/rectifier rating is about 20A. They get really hot, especially when no consumers are turned on. I suspect also consumers (headlights) also play a role in smoothing the resulting waveform.

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    \$\begingroup\$ so they regulate to 14.4 by just throwing extra load on the alternator, essentially just electrically bogging it down to get 14.4V? that does seem rather wasteful. \$\endgroup\$ – JustJeff Jun 21 '11 at 0:58
  • \$\begingroup\$ Yes, they do that exactly, it's USUALLY not a wound stator/ wound rotor like in cars but a permanent magnet stator and wound tri-phase rotor with P-P voltages of 80-200V. To bring the voltage down they shunt(?) the phases to common ground. Still looking for an answer to this one. \$\endgroup\$ – brainwash Jun 25 '11 at 19:48
  • \$\begingroup\$ do they really put the windings on the rotor? \$\endgroup\$ – JustJeff Jun 25 '11 at 20:54
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This is an old thread but I thought I would add something as there seems to be a widespread misunderstanding of this and there is not much information on the web. I think the designers and manufacturers of these devices want to keep the general population in the dark. I think what most people are missing is that these "fixed field" permanent magnet alternators have an inherently poor load regulation curve. With no load the output voltage can vary from 20V at idle to well over 100V at high revolutions per minute. However as the load current is increased a magnetic field is created in the stator which is in opposition to the field of the permanent magnets. Thus even short circuited it will not produce more than 40A or so. When the short is made by an SCR with a voltage drop of 1-2V this is only about 80W. The SCR is only turned on for part of the a.c cycle as it only turns on when the battery voltage exceeds 14.4V. Before it turns on the current is being delivered to the battery and all the other loads, the SCR only turns on when the other loads cannot absorb the current. When the SCR does short the windings the current is so high the the field in the stator almost cancels out the permanent magnet field so the torque load on the engine is reduced. This may seem inefficient but remember even if the field was produced by a wound rotor with slip-rings and brushes like in a car alternator that would also require power and there would also be power lost in the regulator so it isn't as bad as would first appear.

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The same constant that relates machine voltage to RPM also relates its current to the torque. If you can determine how much current is due to the 'shorting', then that current should reflect back mechanically as braking torque, using the aforementioned constant.

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  • \$\begingroup\$ Good call, actually since the heat dissipation on the regulator is likely less than 50W then that should be proportional to the lost power. Assuming blindly that in the worst case the same power is lost in the windings then it would amount to 100W, let's be generous and say 200W. That amounts to 0.26 HP. \$\endgroup\$ – brainwash Jun 10 '15 at 10:05

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