Motor torque above rated current

My turn to post a non-plc related question. Believe it or not, it's drive related.

Most drive manufacturers specify the overload capacity of their drives in terms of percent of rated current for some time (150% for 1 minute, for example). However, what does that mean in terms of torque when using field oriented control?

For the purposes of this discussion assume a perfectly matched drive/motor pair operating in field oriented control (or some other 'vector' equivalent). With a typical induction motor in the 5HP - 30HP range roughly half the current goes to develop the rotor field and the other half generates torque. If you assume the rotor field is controlled at it's optimum point (near saturation) this MIGHT infer that all the additional current goes to producing torque. Since rated torque is developed using 50% of the drives rated current this would mean that a drive developing 150% rated current will develop 200% rated torque in the motor connected to it.

Am I all wet with this or is this correct? The question comes from calculating how quickly I can decelerate a load with a motor given the drives overcurrent rating.

Thanks,
Keith

Just saw this post.

The voltage felt on the stator windings is a direct result of the amount of rotor flux and speed. Please refer back to the "How and why does motor current go up?" Here

In that thread, I sorta touched on the fact that the stator sees voltage from the back EMF produced by the motor's own rotation. From that, you get the idea that stator voltage is directly related to rotor flux.

So, since motor torque is derived from rotor flux times stator flux, if you increase rotor flux, then you increase torque without increasing stator current above a given maximum. You may have heard of controlling motors using "Volts per Hertz." Another way of saying this is that if the Voltage goes up at a given frequency/speed, then it is necessary that the rotor flux has increased.

So, assuming that the motor is at nameplate speed and voltage, then the rotor flux is at its nominal value, and only stator current increases will affect the torque produced. If it is at half speed and half voltage, the rotor flux is exactly the same. Or any ratio where the Volts/Hz is constant.

I will probably have to edit this post to make it understood, but the information is in here somewhere....

Did my post answer your question? If not, I'll try to rephrase it:

The Voltage on the motor terminals is actually a representation of the amount of Counter EMF produced by the motor. If the Voltage is proportional to the speed at any given speed, then the rotor flux is constant. This is easily seen by taking a resistance reading of the motor, then dividing that into the name plate voltage. Ohm's law tells you that something else must be keeping the current below FLA. The "something else" is the CEMF.

The torque felt on the shaft is a function of rotor flux times stator flux. By controlling the "Volts per Hertz" (V/Hz) at a constant level, your rotor flux becomes constant. Therefore, shaft torque becomes proportional to stator flux. If you have a 460V motor at 60 Hz, and you run it at 30 Hz with 230V, then the rotor flux is the same in both instances.

It then follows that by controlling the stator flux (stator current), you can get any torque value you desire, above, below, or at the motor's nameplate.

For a more technical explanation:

The motor's stator current has two effects. One effect is to magnetize the rotor. We call this current Id or magnetizing current. The second effect is to produce motor torque. We call this current Iq or torque producing current. With older drives, computers weren't fast enough to compute Iq and Id, and therefore this was mostly trivia useful only to drive and motor designers. You can think of the two vectors as rotor current (Id) and stator current (Iq), if you like.

In modern drives, knowing the characteristics of a motor allows us to formulate an equation that gives us precise values of Id and Iq. The Id gives us the rotor flux/magnetization. If you discount IR drop, etc., from your calculations then Id becomes constant if the CEMF (Voltage) is kept proportional to the speed (Hz), in other words, the Volts/Hz on the motor is kept constant. Of course, in the real world, you can't discount anything, so modern drives contain motor models (mathematical equations the describe what's really happening inside the motor) that can be adjusted by changing multipliers or constants within its drive parameters.

For example, one check almost all drives do is to have you lock the rotor and then supply various currents at various frequencies. Since the rotor isn't turning, all you're getting is IR drop at the various frequencies. From this, you can calculate the resistance and reactance of the stator. You now know how much of the voltage you measure in your "CEMF" is really from stator drop, and not from actual counter EMF. Knowing the true CEMF allows you to calculate rotor flux. You now know how much of your current is Id and how much is Iq, and how much is IR (copper loss), etc.

These two currents are at 90 degree angles, and therefore, you know all of the characteristics of your supplied current, in detail. The "q" in Iq actually means "Quadrature" which means "perpendicular to" (at right angles).

So, if you know how much flux you have in your rotor (coming from Id) and how much flux you have in your stator (coming from Iq), then you know how much torque you have (phi q * phi d; that is flux in the stator times flux in the rotor).

In a modern flux vector drive, the Id is kept constant throughout its speed range. Therefore the Iq, or more precisely, the flux developed by Iq becomes the only factor that varies when determining motor torque.

Now, it is possible for some motor models to fudge a bit and allow the Id to increase at lower speeds. I've seen this done, and it is typically called a "boost" because you can get more torque since you're increasing both the rotor and stator flux. I've only seen this type of boost done on very high inertia, cubic loads. These need very, very high starting torque to overcome the inertia, and every little bit helps.

A good paper on this is at this site but I don't know if you really want that much detail.

Hope this helps.

Originally posted by DonsDaMan (presumably):

In a modern flux vector drive, the Id is kept constant throughout its speed range. Therefore the Iq, or more precisely, the flux developed by Iq becomes the only factor that varies when determining motor torque.

Click to expand...

So taking that back to my original question, if you assume the 50% of the full load current value goes to produce rotor flux and ANYTHING ELSE produces stator flux AND I send 150% FLA through the motor I will get 200% rated motor torque.

Keith

I think I misunderstood your original question.

Let's give it some (made up) numbers to make this easier for me to understand your original question.

You're saying that a 10 Amp FLA motor will need 5 Amps to magnetize the rotor. Therefore, if you give it 15 amps, 5 go to the rotor, and 10 (5 times 2) go to the stator. You'll get 5 amps times 5 amps worth of torque under normal conditions, but twice that under the specified conditions: 5 rotor times (5 * 2) stator.

Well, on the surface this is correct. I had never thought of this situation in this way. It's questions like this that got me wondering about what's REALLY happening inside a motor, and thereby led to the conclusions (some would say fantasies) I've tried to explain in my motor-related posts.

I could quibble about current to flux not being linear (which your situation presupposes), but your main point IS correct with the assumptions given.

Let's explore this further, then. I'm not really quibbling in what follows. Think of it as an externally presented internal monologue.

Assume a motor at full nameplate speed, at full load, and at FLA:

As load increases the motor slows slightly, slip increases, and therefore the amount of current in the stator that becomes rotor flux decreases, due to the faster relative motion between the rotor's position and the stator's magnetic field's position. In other words, you get more rotor flux due to the faster relative motion, not due to the stator increasing its current.

As load increases, slip tends to increase. The vector drive, though, compensates for this in its model, and keeps the rotor current at a constant as I outlined in my other posts. In a standard motor across the line, I don't doubt that rotor flux is changing along with the slip and stator excitation; it may follow your 50% rule for all I know, and pull more stator amps, overheating the motor.

So, with a heavier load (larger slip) at any given excitation frequency, more torque is produced per unit of stator current than with a smaller slip at lesser loads with the same excitation frequency. So, your assumption that 50% of the current always goes to the rotor is too simplistic. As the load changes, so does the amount of stator current that becomes Id or rotor magnetization current.

Running at 150% current will give you much more torque. Will it be 200%? Probably at some load and speed level, it will be, but on either side of that point, it will be more or less than 200% depending on the slip/load/motor characteristics.

I'm going to have to give your point more consideration to come to a conclusion on this. What I said above is my first response to your actual question... conjecture, if you will. I've also not considered what happens with larger stator currents at slower excitation frequencies/speed.

I'd like to hear what others think about this.