Field weakening in DC motors 2

[gsjhand]

How does field weakening affects the incoming AC power to the DC drive? Is there any benefit of having field weakening speed below the nominal speed of the motor? As an example I have a DC motor with 268 RPM at 750 volts and 2500 Amps. When I checked the programming of the drive I found that motor has been forced to go into field weakening at 500 volts. AC supply to the drive is 500volts 60Hz.

 

[waross]

Weakened field will result in higher current and motor speed.
If you have a situation where low incoming voltage is dropping the speed but the motor is not fully loaded and not drawing full rated current, field weakening may be used carefully to maintain the speed. If the motor is loaded to maximum rated current, field weakening may result in over current in the armature circuit overheating of the motor.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Marke]

You normally use field weakening to increase the maximum speed of a DC motor.
If you are able to operate at your required speed without field weakening, that is the best way to go. If your motor is rated at 268RPM and you need to operate at 280RPM, then field weakening will enable this to happen provided that nothing is overloaded.

Best regards,


Mark Empson
L M Photonics Ltd

 

[waross]

Let me state my post a little more clearly. The speed of a DC motor is closely related to the armature voltage. If low supply voltage is limiting your speed, the speed may be increased by field weakening, but the allowable horsepower will be reduced.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[DickDV]

A typical DC drive with SCR rectification will output a maximum of about 550 volts DC on a three phase input of 500 volts AC.

The motor nameplate data indicates a motor base voltage of 750VDC and 268rpm. Assuming a shunt-wound field, with a full field, it will take 750V to get to 268rpm. Since your drive cannot supply that much voltage, you could not hope to get to full speed with a fully excited field.

As has been mentioned, you can weaken the field somewhat. When you do that, you essentially trade speed for torque since the torque will decrease in the inverse proportion of speed increase.

It sounds like that is what has been done in your case. Your application probably needs the 268rpm so the field was weakened to get that speed at 550V instead of 750V. The maximum available torque would be reduced but probably your application didn't need all of it anyway.

If your drive has a simple fixed field supply, then the field has been weakened over the whole speed range. However, if your drive has an active field regulator, the field will remain fully excited up to the 500V point and then weakened above that point. With that arrangement, the motor will supply full torque up to the 500V point and reduced torque above that speed.

 

[waross]

You said it better than I did, DickDV.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[DickDV]

waross, thanks, but I also used three times more words. I guess that would be less efficient than your way of saying it!