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It occurred to me that since single-phase AC induction motors can act as generators, it might be possible to brake an induction motor with a resistive load. The basic idea is that when you want to stop the motor, you would disconnect the motor from the AC power source, and connect a resistive load to the motor. Well, I just bought a radial arm saw which takes a long time to coast down, so I thought I would try it. Here is the plan. If anyone knows whether this will work at all, or can help suggest what the load should be, it would be much appreciated.enter image description here

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This won't work but you are close. In your situation, the induction motor doesn't have any stator current, therefore the motor doesn't act like a generator.

However, all you need to do is introduce a DC current into the stator and the motor will come to a stop very quickly. The rotor acts as a shorted turn in the presence of the DC magnetic field and converts all the kinetic energy to heat.

Do note that you must NOT leave the DC current applied to the stator or you will burn it out.

There are commercial units that work this way and they all have a timer that shuts the DC current off after sufficient time has elapsed for the motor to come to a complete stop.

How quickly the motor stops is a function of how much DC current you feed into the stator. Several Amps is normal but I can't give you a specific value - it depends on your motor.

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  • \$\begingroup\$ This is a great answer. Thanks! I will try to think of a good and fail safe way to do this. But I just want to double-check something. You say that the motor won't act as a generator. However, induction motors DO act like generators. There are many videos on youtube showing it. Due to residual magnetism in the rotor core if I remember correctly. \$\endgroup\$ – mkeith Feb 25 '15 at 5:54
  • \$\begingroup\$ You may be correct in saying that residual rotor magnetism may produce some energy. But it's quite small compared to what you need. I think that when you see an induction motor acting as a large generator, the stator is energised with the power supply that the motor is feeding. The motor acts like a generator because you are over-driving it. That is: trying to make the motor spin faster than the incoming power would normally make it spin. When you do that, the motor then converts that extra rotational energy into output current that feeds the incoming supply line. \$\endgroup\$ – Dwayne Reid Feb 25 '15 at 6:02
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    \$\begingroup\$ Of course, the easy way to confirm this is to just try it. Don't modify your saw. Just start it up, then pull the plug from the wall socket and put a short across the terminals. If there is sufficient magnetism to allow the rotor to act as a brake, you will see it very quickly. However, I don't think that you will see any significant braking occur. \$\endgroup\$ – Dwayne Reid Feb 25 '15 at 6:04
  • \$\begingroup\$ I don't want to re-invent the wheel. I will figure out how to do the DC injection. But if I experiment with the resistive load idea, I will follow-up with the result. \$\endgroup\$ – mkeith Feb 25 '15 at 6:46
  • \$\begingroup\$ I did eventually try the resistive load. I connected two 300 Watt incandescent bulbs in parallel with the saw motor (these are two series 120V bulbs, and the motor is 240 V single phase). When I cut power to the system, the bulbs glow distinctly for around one or two seconds, and you can hear the deceleration of the saw blade. But then the bulbs go out, and from that point there seems to be no electric braking effect. Even though there is one or two seconds of braking, the net gain compared to no braking is not much. It still takes 20 or 30 seconds for the rotor to coast to a stop. \$\endgroup\$ – mkeith Oct 1 '15 at 5:11
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The reason induction motors can act as efficient generators is that they are connected to an already powered grid, which maintains their field. Just rotate them faster than synchronous speed instead of slower and they will put power back into it according to the (now negative) slip speed. (This is also why the mains frequency increases when lightly loaded : reducing the power taken out reduces the slip speed, bringing the grid frequency closer to the rotation speed)

It follows from this that you could in principle achieve regenerative braking by monitoring the motor speed, and adjusting the applied AC at a frequency lower than synchronous speed. The motor will supply power, absorbing power from its own inertia and the drive shaft.

As its speed decreases you would have to continually reduce the AC frequency to maintain a negative slip frequency appropriate to the required braking torque.

This is obviously quite a complex problem, in practice the simplicity of DC braking usually makes it the better choice.

NOTE : DC injection current shouldn't be greater than the motor's short term rated operating current. That means much lower voltages - maybe 12V for a 110V or 230V AC motor as a starting point. Measure the winding resistance, decide on a current, and that'll give you a suitable voltage.

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