Motors Drawing Higher Current At Low Speeds 2

[rockman7892]

I am trying to determine why induction motors tend to draw higher current at lower speeds?

For example with a VFD I have heard that below 10% that you may need to have a V/HZ boost in order to boost the torque and reduce the amount of current.

Also I have noticed with several drives that througout the range of the drive the current stayed pretty much the same, however when the motor was run below 10% the current seemed to increase dramatically. For example when stopping a motor from 60hz to 0hz through a ramp time, I notice that the current stays the same all the way to about 10hz when it increases for the range of 10HZ down to 0Hz before it is cut off to 0

Does the drive have a hard time producing enough flux at this lower range, and therefore must use current to compensate for producing required torque?

 

[CJCPE]

The V/Hz boost is needed because the resistance of the windings becomes more significant in comparison to the inductive impedance at lower frequencies. The motor needs the magnetizing V/Hz to remain constant. The magnetizing voltage is what is left over after you subtract the IR drop.

Even if the load is a fan, there is a certain amount of static friction to overcome to get the load moving. Once the load starts moving, the initial friction is likely to drop quite a bit. Sometimes, it takes so much boost voltage to get the load moving that the current doesn’t drop to a normal level once the load breaks away. That isn’t usually too much of a problem if the drive accelerates to a higher frequency fairly quickly. When the drive decelerates to a stop, the friction at low speed is not what it was when starting and the inertia may be reducing the driving load. As a result, the V/Hz may be higher than it should be as the motor comes to a stop.

In a good sensorless vector drive, the microprocessor should be able to calculate what is really needed. Whatever the case, the motor should be OK if the current doesn’t exceed rated current on a continuous basis and doesn’t exceed the intermittent rating of the drive or make the drive trip off.


CJC

http://www.vfdriveinfo.com

 

[ScottyUK]

That last sentence is a little risky: running the drive at rated current at low speed tends to let out the magic smoke out of the motor windings unless there's an auxiliary blower to move the cooling air that the shaft-mounted fan isn't moving.


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If we learn from our mistakes I'm getting a great education!

 

[rockman7892]

It sounds like that at low speeds there is more of a voltage drop in the windings due to resistance and it is this voltage drop that causes a reduced voltage and thus more current, and a need for a voltage boost.

I cant seem to figure out why the resistance of the windings becomes more signifigant at lower speeds.

 

[electricpete]

Let’s compare normal speed and low speed at same current. We have the same resistive voltage drop in the stator resistance R1 in both situations, but it is a higher fraction of the applied voltage when at low speed (low voltage).

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[CJCPE]

About that last sentence.

I should have said that the adjustments are probably OK if the current doesn't exceed the rated current on a continuous basis. The motor can still overheat if it operates at too low a speed for too long without appropriate cooling provisions.

CJC

http://www.vfdriveinfo.com

 

[rockman7892]

Lets then assume that the load torque stays the same throughout the speed range. If we are running at base speed at a given V/Hz then then we will have a current related to the required torque. As we reduce the speed of the motor the V/Hz output will adjust proportionally and thus the current will stay the same to produce the required torque. This current will cause a voltage drop in the stator windings however this drop will be somewhat negligable in relation to the ouputed Voltage.

As we continue to reduce speed and draw the same current the same voltage drop is seen across the stator windings, however as the reduced voltage from the drive gets lower and lower this voltage drop has more of an effect on the voltage seen at the rotor.

So finally when we get down in to the 10% speed region the voltage drop becomes more signifigant and leads to a signifigantly reduced voltage for the V/Hz ration and thefefore has trouble maintaining the flux to produced the required torque with the given current. The current must then rise to compensate for this reduced V/Hz ratio?

Is this somewhat on the right path?

 

[CJCPE]

The current does not need to rise to compensate for the reduced V/Hz across the magetizing part of the equivalent circuit, only the V/Hz needs to rise. The problem is that it is difficult to get the V/Hz high enough for the worst case starting condition without it being too high under some other condition. That is why sensorless vector drives are needed for applications with difficult and/or varying starting conditions and for running at low speeds with high torque.

CJC

http://www.vfdriveinfo.com

 

[rockman7892]

CJCPE

You mention that the current does not need to increase but rather the V/Hz needs to increase to maintain flux and torque.

However what if the drive is in V/Hz control mode and therfore cannot adjust the V/Hz ratio for these low speeds. Is it then that the current will increase to compensate for the reduced V/Hz as a result of reduced voltage from voltage drop in stator?

I understand that in sensorless vector mode the drive will be able to adjust this V/Hz value and raise it in order to correct reduced V/Hz as a result of voltage drop.

 

[CJCPE]

In V/Hz mode, the drive will have a boost adjustment to manually boost the V/Hz below some some frequency. That usually involves a breakpoint in the voltage vs frequency line at some frequency and an adjustable voltage at zero frequency intercept point. The breakpoint frequency may or may not be adjustable. As the frequency approaches zero, the V/Hz increases. If the adjustment is set to allow the motor to develop a high torque to break the load away on start, but the motor is lightly loaded after the load breaks away, the result can be a current that increases out of proportion to the load as the speed is reduced.

CJC

http://www.vfdriveinfo.com