Why is there a current spike when voltage is first applied to a motor?

[jrive]

Is the current when voltage is first applied to a motor the same as the stall current?

Also, shouldn't this current show an exponential rise governed by the L/R of the motor (analogous to the voltage rise in a cap)? Why would there be a current spike when voltage is first applied to a motor, if the motor is an inductor?

Thanks,
Jorge

 

[vk6kro]

Is the current when voltage is first applied to a motor the same as the stall current?

Yes, it is.

Also, shouldn't this current show an exponential rise governed by the L/R of the motor (analogous to the voltage rise in a cap)? Why would there be a current spike when voltage is first applied to a motor, if the motor is an inductor?

The current spike exists only compared to what happens when the motor starts turning.

If you imagine a motor that is already turning, you have a coil rotating in a magnetic field, just like a generator. There is a voltage generated that opposes the current flowing into the coil from the supply.
This causes a reduction in supply current as the motor turns.
If you stall the motor, this generated voltage vanishes and the motor then draws a much larger current.

This effect is called Back EMF or Counter EMF and is very important in motor behaviour.
You can read more about it here:
http://en.wikipedia.org/wiki/Counter_emf

 

[jrive]

thanks for the response,vk6kro.

I understand the concept of back emf. My question was more about the initial current rise in a motor, and whether this current should be an exponential rise due to the motor as an inductor. Although the back emf is 0 at this point, the motor should still be an inductor which should resist the instantaneous change in current. Thus, I expect it to be an inherent in-rush current limiter, but I'm not sure it is. Should it be? What is the current profile when voltage is first applied to a motor?

Thanks again!

 

[hisham.i]

When we apply voltage in the stator of 3-phase motor for example, a rotating magnetic field is created which causes, a variation on the flux on the rotor, thus creating e.m.f=-d(flux)/dt...
When we start the motor we have maximum variation of the flux, thus we have maximum current passing through the rotor...this is the high current because of..
then when the rotor starts to rotate the variation will be decreased so the current on the rotor is decreased also.

 

[jrive]

Thanks hisham.i.

I'm not sure I agree with that...when voltage is first applied, the rotor hasn't begun rotating, so there shouldn't be any big change in flux. That's why the emf is low when we first start and why maximum current flows through the motor.

However, this fails to answer the crux of my inquiry. If the motor is a coil, and current cannot change instantaneously in a coil, why isn't the motor an inherent in-rush current limiter? Ie, why is there is large initial current, and why isn't it an exponential rise with time constant L/R, at least until the back emf is generated? Or, it is an exponential rise, just a "big" one? ;-)

 

[vk6kro]

The delay in a series L/R circuit depends on the inductance and the series resistance.

There would be a small delay in the rise of current, but these coils have very little resistance, so the delay would only be a few milliseconds while the motor may take a second or more to start rotating and generating back EMF.

 

[bejoynp]

Hi. Kindly indicate if the below could be a reason for the large inrush current when a motor is connected to supply.

Can we say that the instantaneous short circuit of the motor coil responsible for the large inrush current. Any coil (whether AC motor or DC motor) has negligible resistance. Once the motor picks up speed, the back emf stabilizes the current through the coil.

 

[jrive]

thanks again for staying with me on this thread.

I think vk6kro gave me the answer I was looking for. Yes, the motor coil is an inductor,but since the R is small, the time constant of the series LR circuit is much shorter than the time for the motor to begin to rotate. Thus, the current reaches it's maximum prior well in advance of the motor rotation generating the back emf.

I'm not sure about about the "instantaneous short circuit current" bejoynp suggests, though. Again, since the motor windings are an inductor, it will initially resist the flow of current. However, to bejoynp's point, like any coil it will become a short at DC, so if it takes a long time for the motor to start rotating, the response will approach DC performance (which is similar to what vk6kro said above as well).

Excellent. Thanks for the input!

Jorge

 

[cabraham]

Actually, a small R results in a large time constant since tau = L/R. The inductance does limit the current upon startup, but the L value times the radian frequency results in the reactance X. This reactance is still small enough to result in a large stall current. The winding resistance is smaller than X, so that X is what is chiefly responsible for limiting current. If it is a dc motor, then X is zero, but upon start up, the L value still limits the stall current, more so than R. Does this explain it?

Claude

 

[sophiecentaur]

Is this question about an AC or DC motor?

 

[jrive]

DC motor...

 

[sophiecentaur]

. . . . . only I am reading comments about 3 phase motors here . . .

In answer to the op, there are two effects. First the short-lived 'back emf' caused by the fact that the windings are inductive when stationary, which means the current builds up to a peak - limited by the resistance of the windings. Then there is a decrease in current due to further back emf caused by the build up of rotation.
The two time constants involved are very different. There will be a build up of current, initially, in a matter of a few tens of ms, then a decrease over a matter of a second or so.

 

[m.s.j]

thanks for the response,vk6kro.

My question was more about the initial current rise in a motor, and whether this current should be an exponential rise due to the motor as an inductor

I think your concerned subjects is transient motor inrush current in motor starting situation, it is true that in additional of big RMS starting current which can be 6 times of motor rated current approximately, we face to exponential rise of current in few cycle of starting current due to dc offset phenomenon. For instantaneous maximum starting current we can write:

Imax = √3 Irms (starting current)

For applicable technical discaussion regarding this subject you can refer to http://electrical-riddles.com/topic.php?lang=en&cat=6&topic=614"

 

[sophiecentaur]

This is all very well but the OP was about a DC motor and the comments still come in about induction motors.

 

[jrive]

Yes, I am asking about DC motors specifically...

All good responses, and I appreciate all the input.
Fundamentally, I expected the DC motor to exhibit an initial rise in current (before the motor starts rotating) modeled by :
I(t)= V/R*(1-exp(-tR/L)), where R is the winding resistance and L the inductance of the winding. Thus I expected the inductance to serve as an inrush current limiter. I now realize that it is indeed doing that, but the current still rises to its max value (V/R) way in advance of the start of the motor rotation (which would generate the back emf which would lower my "rotation" current). If I need to limit this motor start current, is adding a small series resistor a viable/practical option?

 

[sophiecentaur]

See my post no.12 about the inductance and the time constant associated with it. In traction DC motors they used to use a switched 'ballast' resistor on startup, to limit the current.

 

[Airbag79]

When the motor is started, the slip speed is equal to the synchronous speed, as the rotor speed is zero (slip equal to 1), so the induced EMF in the rotor is large. As a result, a very high current flows through the rotor. This is similar to a transformer with the secondary coil short circuited, which causes the primary coil to draw a high current from the mains.

 

[sophiecentaur]

Yep. That makes sense - a shorted turn - several shorted turns, even. The induced emf is not balanced by a back emf.