Losing field current in a DC machine
Losing field current in a DC machine
(OP)
Hello!
I'm simulating a DC Motor with matlab. I have this problem with simulating the situation where the field current is lost when the machine is running in steady state. My results show that the armature current becomes very high and the rotation speed drops to zero. But what should happen with the voltages and torque? I have very little experience with these kinds of machines, and I get some results for these also, but they don't seem to be reasonable. I would be very happy for all tips and information anyone can give.
Thank you.
I'm simulating a DC Motor with matlab. I have this problem with simulating the situation where the field current is lost when the machine is running in steady state. My results show that the armature current becomes very high and the rotation speed drops to zero. But what should happen with the voltages and torque? I have very little experience with these kinds of machines, and I get some results for these also, but they don't seem to be reasonable. I would be very happy for all tips and information anyone can give.
Thank you.





RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
The three main types of DC motors are: Permanent Magnet, Shunt wound, and Series wound.
Permanent Magnet
These motors have a permanent magnet which generates a magnetic field. Rotating in this field is the armature through which current is passed. The armature will be wound from only a few turns of quite thick copper, giving it a low resistance. When a voltage is applied this will draw a very large current from the battery. When the motor starts to rotate the current will drop due to an effect called "back EMF". This is the voltage generated by the armature as if the motor was a dynamo. For example for a motor running at 12V with an armature resistance of 0.1 ohm, at start up the current drawn by the motor would be 120A (known as the stall current). The action of the armature wires rotating in the field may generate a back EMF voltage of 11V giving a total drive voltage across the motor of 1V. This would give a current of 10A. If load is applied to the motor the speed of rotation would drop, this would also cause the back EMF voltage to drop, and hence the drive voltage to the armature to increase. This increase in armature current would then cause the motor to try and speed up. i.e. the motor will try and maintain a constant speed.
Shunt wound
These are very similar to a permanent magnet except the magnetic field is supplied by passing a current through a fixed coil positioned around the armature. Operation is exactly the same as the permanent magnet motor, except these motors also have the possibility of field current control. This is where it gets interesting (or very boring). If the field current is reduced the motor speeds up. This can be explained by considering the example above. If the field current is reduced the magnetic field will drop causing the back EMF voltage to drop. This increase in the drive voltage to the armature will cause the motor to speed up until the back EMF is once again at its previous level. The penalty for this speed increase is loss of torque - with such a low field magnetic field the motor will be quite weak. However, by using a high field current the motor will turn quite slowly but with very high torque - the penalty here being potentially wasting current in the field when the torque requirement is low.
Series Wound
In a series wound motor the field coil is in series with the armature coil. The field coil is wound from thick wire to cope with the large armature current and has a resistance of only a few ohms. The advantage of this type of motor is its high torque output. Consider a series motor running with a given load. If the load is now increased the motor will slow causing the back EMF to drop. As with the previous type of motor the current drawn will be the difference between the voltage applied and the back EMF produced. This large increase in current will cause both the armature and the field magnetic strength to increase, resulting in a very high output torque. Hence this motor is used in high torque situation i.e. starter motors. The disadvantages are the large current drawn will soon flatten batteries, and if run for prolonged periods the motor will soon overheat. The other point that should be noted is the motor without substantial modification will only run in one direction as reversing the armature current will also reverse the field current.
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
VA = E + IA*RA
VA= line voltage applied to the armature through the brushes.
IA = armature current
RA = armature resistance
E = induced voltage in the armature
E= k*Phi*n
k= winding constant
Phi = Field flux
n = rotational speed
Then; VA = k*Phi*n + IA*RA
Simplifying for n;
n = (VA-IA*RA)/ (k*Phi)
From the last equation, when Phi becomes 0 (lost of field), the denominator is zero and then:
n becomes infinity
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
You are theoretically correct and possibly are correct for a brand new motor that has never been energised, but the motors we have to live with in the real world almost always maintain some small residual magnetism in the core even after the field current has decayed to zero. The result is the terrifying scenario described by eemotor where the armature current increases to a very high value and the speed increases very rapidly. An accurate model would take in to account the residual magnetism of the armature, but I think this would be a challenging thing to model accurately (for me at least!). Best of luck to blunt80fi if he chooses to extend his model to account for this factor.
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If we learn from our mistakes,
I'm getting a great education!
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
I've seen the results of this on a medium sized motor (~200HP) and, believe me, trying it is among the last things I intend to do.
I'm often wrong, although on this occasion I don't think I am. Assuming I am wrong, please explain how you account for the real-world experiences of several people above who have all witnessed the destruction of series-wound machines caused by loss of field current? What is the real mechanism behind the motor accelerating to several times rated speed prior to the armature disintegrating?
------------------------------
If we learn from our mistakes,
I'm getting a great education!
RE: Losing field current in a DC machine
When the machine is running, the magnetic circuits handle high flux densities leaving a residual magnetic flux (B). This residual flux is small but combined with the very high armature current ( i ) still produces a high torque (T).
Certainly the load speed-torque demand could restrain acceleration, but in some applications destructive speeds are still reached. If for some reason the motor looses the load restrain , surely the speed reached will be very high.
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
Any motor with this configuration cannot fully loose its field since the armature current will always provide some even if the shunt field is totally de-energized.
While I have never experienced a runaway based on residual magnetism alone and certainly don't want to, it seems to me that the torque would be so small that almost any connected load would limit speed. In the case of motors with shaft fans (ODP and TEFC enclosures), I would think the fan load alone would limit speed.
Of course, with compound wound motors, torque would likely be much higher than with residual magnetism alone and destruction would occur quickly with rapid acceleration.
If I am overlooking something with regard to residual magnetism and torque, I would welcome someone explaining a more correct understanding of it.
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
T = K*(flux)*Ia
where
T = electromagnetic torque
K = motor constant
flux= flux produced by field current or remanent flux
Ia = armature current
Now, the decay in the term flux is compensated by an increase in armature current the overall effect being a higher torque.
Any mistakes in this reasoning?
RE: Losing field current in a DC machine
There you go cbarn, there's one of the mistakes I was talking about making!!
------------------------------
If we learn from our mistakes,
I'm getting a great education!
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
As far as i'm concerned the increase in speed occurs due to an accelerating torque produced by an increase in electromagnetic torque (unless you intentionally reduce load torque at the same moment the field gets lost)
If not, How can you explain shaft accelaration? I really would like to know.
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
RE: Losing field current in a DC machine
The armature actually blew through the end casing shooting straight up to the ceiling (40 ft) breaking the roof struts and falling to the floor on the opposite side of the extruder. There was copper shrapnel all over the place and luckily no one was hurt.
We immediately put a procedure in place for maintenance to follow when starting up these motors.
RE: Losing field current in a DC machine
Even with the shunt field disconnected, it had a weak field from its series field. This would be just enough to cause the runaway you describe.
RE: Losing field current in a DC machine
No doubt you have had plenty to digest here, and everybody has raised some very good points. Technically speaking you are right about the torque going to zero with Zero field current, and, if your motor is a pure Shunt connection (which I have never seen in the sizes I deal with) it should stop.
You need to keep in mind the dynamics of the model. In a real motor, the shunt field will have a time Constant of about 4 seconds. If you remove the field excitation and let the field decay naturally, the Field current will decay exponentially to 39% of it's rated value in the first 4 seconds, The remaing drop to 2% of it's rated value will take a further 3 time constants ( 12 seconds) during this time, you need to do intermediate calculations of speed, current and torque.
I think you will find, that if you have a machine running at operating speed, with no restraining load torque, and you remove the motor field excitation, that the motor does in fact run away in speed up until the time the field is so weak that it no longer provides any motor torque.
Unfortunately it takes far less time to accelerate to destruction then it does for the field to decay to a level where the motor stalls. As others have said, IT DOES HAPPEN.
If the motor does have a load torque then the behaviour during the time the field takes to decay, will be dictated by speed torque characteristic of the load.
Most motors above a couple of hundred Kw will have a series component of armature current in the field. This is used to help with commutation and to limit Motor speed droop under load.
I remember, as a young engineer, doing a current regulator CFB ploatiry test on a 300 HP Motor with the field disconnected. We had no shaft brake but I had the wooden handle of a sledge hammer wedged in the shaft mounted fan. The CFB was the incorrect polarity and we had uncontrolled armature current run away. The torque developed was enough to snap the wooden handle, but Luckily it did not let the motor get away. So Much for the "No field, No Torque" camp.
I think it would be remiss of any DC Drive manufacturer, to suggest doing a stall check with no field, without suggesting the armature be restrained in some way.
I know the company I work for ( A well known DC Motor Manufacturer) does not suggest you try that.
Tom