AC Motor Braking
AC Motor Braking
(OP)
I need to select a braking unit for my 415V 37KW 3phase induction motor. I have been told that a general rule of thumb for calculating the maximum braking current is 4*KW. This gives me ~150A. So i should select the next size unit. Is this correct, and does anybody have a more technical solution rather than a 'rule of thumb'





RE: AC Motor Braking
RE: AC Motor Braking
http://www.rdmengineering.co.uk/drivloc.htm
http://www.deltaww.com/products/motordrives/vfdb.htm
(for specification, resistors, diagram, etc.)
http://www.northstar-mn.com/ambitech1.html
etc. for more info
RE: AC Motor Braking
In the case of pulsed DC, the maximum brakeing torque is quite restricted. The line frequency component can keep the rotor spinning at full speed with no resultant braking occuring.
DC injection results in a high current flow in both the stator and the rotor. This translates into heat and care must be taken not to exceed the motor ratings. The kintetic energy in the driven load is dissipated in the rotor during stop.
Best regards,
Mark Empson
http://www.lmphotonics.com
RE: AC Motor Braking
It provides an interesting parallel to the discussion of rotor heating during an unloaded (inertia-only) start, which was also equal to the [final] kinetic energy of the motor.
In the case of dynamic braking it seems somewhat easier to understand the result from physical considerations. The stator provides a dc stationary field, and the rotor acts like a generator feeding the rotor resistance (load). Neglecting other motor losses (stator I^2*R), it is easy to see there is no work done by the stator current or stator field.
Stator Field Power = Torque x Speed = Torque x 0 = 0.
If there is no work done by the stator field then conservation of energy tells us the rotor kinetic energy has to go somewhere and the only place it can go is to the "generator load" (rotor resistance heat dissipation).
As I said the intuition is a little tougher for the unloaded start. (Using similar assumptions of no stator losses or other losses). In that case there is energy input by the stator field and the amount of that energy by conservation of energy must be TWICE the final kinetic energy of the rotor. (1 times kinetic energy goes to kinetic energy of rotor, and 1 times kinetic energy goes to rotor heating). It doesn't follow intuitively, but maybe it follows somewhat intuitively (?) from analogy with the braking case.
RE: AC Motor Braking