Minimum voltage to keep an unloaded motor spinning.

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philly

Senior Member
With an uncoupled motor I know that as you reduce the supply voltage the current will decrease due to the fact that most of the current is magnetizing current and very little is current across the rotor producing torque. So since the torque current is so small, when you reduce the voltage the magnetizing current drops (less flux density?) and therefore the overall motor current drops.

How much or what percent can you decrease the supply voltage on an uncoupled motor to where it will still turn? Is there a minimum value for which below it will not turn?

I have seen that for determining values of Rm paramater an unloaded motor is run at a reduced voltage that usually provides full load current. How come at a very reduced value of voltage with motor unloaded the current eventually increases instead of decreasing as we said above? Is this because the torque required to overcome the friction and winding losses decreses with reduced voltage, and therefore current must increase to keep motor spinning?
 

kwired

Electron manager
Location
NE Nebraska
With motors if you change voltage the motor is still going to try to so the work. An uncoupled motor still requires a certain amount of input to turn the rotor. If you reduce the voltage it is still going to try to turn the rotor at the same speed as it does at normal voltage, this will result in higher current but power (disregarding losses) will still be about the same VA as at normal voltage. Increasing voltage will have opposite effect.

By changing frequency with change in voltage you are no longer trying to do the same amount of work, and motor will now have a different power output.

Add: reducing voltage to an unloaded motor will probably reduce current until it can no longer produce required torque at that current level then the current will go up to try to provide the required torque. I am not an expert at this but I believe it has to do with saturation level in the windings.
 
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hurk27

Senior Member
In a synchronous motor the motor poles trys to stay in sync with the 60hz. as a load is placed on the motor a slip is created and the current increases if voltage is decreased there will be even more slip, this causes more current.

Now with an unloaded motor it doesn't require as much energy to maintain synchronous sync so as voltage is reduced at first the current will reduce, but at the point you stop providing enough energy to maintain sync the pole will start to slip out of sync again causing current to rise, if you had a RPM meter on the motor this would be at the point the motor just starts to slow down from its rated speed.

to understand this, it is just like a simple solenoid, if the armature is puled out of the coil while it is energized the current on the coil will rise put the armature back in and it lowers.
 

broadgage

Senior Member
Location
London, England
I have just tried a small fan motor, rated at 240 volt, 50 cycles, or 277 volt, 60 cycles. it just rotated on 25 volts, but would not turn on 12 volts.

As well as strictly electrical concerns, it probably depends on the type of bearings.
 

PowerQualityDoctor

Senior Member
Location
Israel
When you reduce the voltage you reduce the maximum torque the motor can generate. It can be as low as the no-load torque on the motor is lower than the new torque the motor can generate.

When the voltage is reduced on partially loaded motor three things happen on the same time:
1. Since the voltage goes down, the current goes high, as the motor maintains same power.
2. The Power Factor of the motor is improved (as the magnetization has low power factor) -> current is reduced.
3. The motor efficiency is improved -> current is reduced.

As 1 is current increase and 2 and 3 are decrease, the result can be either increase or decrease of current. The point where you can still reduce the voltage is where the current in the motor is still going down. If it starts to increase - the slip gets too high.

You can read more details at http://www1.cetim.fr/eemods09/pages/programme/077-Broshi-final.pdf
 

Jraef

Moderator, OTD
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Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
It's going to be different for every motor. If you look up the formula for accelerating a load, it will tell you that the torque required has everything to do with the Moment of Interia (WK^2) of the load and motor rotor together. If the load is removed (uncoupled), then you still have the rotor WK^2 to contend with. The more mass in the rotor, the more torque it will require to accelerate it, to any speed. At the same time, torque is proportional to the square of the applied voltage. So for example at 50% voltage you get 25% torque, at 25% voltage you get only 6-1/4% torque. So the point at which the minimum accelerating torque requirements of the rotor mass intersects with the developed torque in the motor as voltage is reduced, you stop moving. At some point in that process, as others have already pointed out, your slip will increase and CURRENT will increase in spite of the voltage reduction, but the first part of your question was about rotation.

I helped design Motor Winding Heaters once upon a time, we had to ensure that the output voltage was never enough to support rotation in a motor (if it were already spinning when energized). The value we felt comfortable with not knowing the WK^2 of any given rotor was 15% voltage. Some that we tested kept spinning at 20%, some stopped at 50%.

As usual, the answer is "It depends".
 

Besoeker

Senior Member
Location
UK
The more mass in the rotor, the more torque it will require to accelerate it, to any speed.
With respect, that's not quite right. The applied torque and rotor inertia will determine acceleration rate and thus the time to reach any given speed.
Greater torque will get it to that speed sooner.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
With respect, that's not quite right. The applied torque and rotor inertia will determine acceleration rate and thus the time to reach any given speed.
Greater torque will get it to that speed sooner.
Yes, I'll concede that point. After reading it with your comment, I realized I was applying the acceleration principals, which isn't what he asked.
 
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