Motor Back EMF

[Grouch1980]

When current is applied to a motor load, you have the current inrush in the beginning due to the lack of back EMF from the motor load. Once back EMF is induced, then that 'resistance' lowers the current draw.

My question is, what exactly is generating this back EMF? is it the coils in the stator (since those are inductors), or is it the rotor spinning in the magnetic field, or is it both?

 

[drktmplr12]

the concept of back EMF is almost exclusively used to describe the behavior of DC motors. In DC motor, the back EMF is a result of the resistance of the motor. there was an experiment we did in college lab where you just placed a voltage across a DC motor, nothing happened but current was flowing. there was no back emf. the trick was to put a resistor in series with the motor and then it started turning.

try looking at the equivalent circuit of a AC induction motor and compare it to that of a DC motor. you will see that in AC induction motors the equivalent resistance varies with slip. the stator magnetic field will match the 3-phase source. As the machine begins to rotate slip, and therefore resistance changes until it reaches steady state. (usually around 0.97-0.98 slip)

 

[ptonsparky]

IIRC, a magnetic field expands as current flows in a wire. With AC it rises and falls. For US 60 times per second. The falling field meets the rising creating the impedance that limits the current in inductors, motors.

 

[LarryFine]

With AC it rises and falls. For US 60 times per second.

120, but who's counting?

 

[ptonsparky]

120, but who's counting?

Glad you're paying attention. I figured someone would correct me.

 

[gar]

230802-2000 EDT

In a DC motor with a fixed field, a constant current flowing in a field coil, or a permanent magnet, one has a fixed magnetic field of some relatively constant value. In this constant magnetic field we have a coil of wire (coils) on a rotor connected to a rotating switch ( commutator ). This rotating switch keeps a coil of wire at a maximum force as the rotor rotates.

Two major things are taking place. Current in the rotor produces a magnetic field force on the wire causing the rotor to rotate, and simultaneously because the rotor wire is moving in the magnet field a voltage is induced in the wire. The induced voltage in the rotor coil is in opposition to the externally applied power supply voltage. Thus. the voltage drop across the rotor internal to the motor is lowered. Thus we have two voltage drops in series in the motor. One voltage in combination with the motor current is power converted into the mechanical load on the motor. The other voltage is the induced voltage in the rotor.

This system automatically balances so the required mechanical power is obtained from the input power to the motor.

As mechanical load is increased the speed of the motor drops. This is a relatively a linear function.

A useful equation for a shunt wound DC motor is:

RPM = V - R * I / K

where
RPM = motor shaft RPM
V = voltage applied to armature
R = rotor resistance
I = armature current
K = a scaling constant

.

 

[Grouch1980]

try looking at the equivalent circuit of a AC induction motor and compare it to that of a DC motor. you will see that in AC induction motors the equivalent resistance varies with slip. the stator magnetic field will match the 3-phase source. As the machine begins to rotate slip, and therefore resistance changes until it reaches steady state. (usually around 0.97-0.98 slip)

So a few follow up questions:
1. When you say “equivalent resistance”… are you talking about rotor resistance?

2. Is rotor resistance equal to slip times rotor reactance?

3. How does back emf relate to rotor resistance? Is there an equation?

I tried googling, but as usual you fall into a rabbit hole and only find a few answers.

 

[ptonsparky]

Search : impedance

 

[Carultch]

When current is applied to a motor load, you have the current inrush in the beginning due to the lack of back EMF from the motor load. Once back EMF is induced, then that 'resistance' lowers the current draw.

My question is, what exactly is generating this back EMF? is it the coils in the stator (since those are inductors), or is it the rotor spinning in the magnetic field, or is it both?

This is what is necessary to satisfy conservation of energy, as energy departs the electrical domain, and converts to the mechanical domain. This generates an apparent voltage difference that opposes the current through the motor winding. It is a consequence of Faraday's law of induction, as the winding experiences a changing projection of the magnetic field on its winding area.

Motor start-up is governed by a system of differential equations, that establishes the relationship between current and rotation rate, such that it spirals from its starting rest state, to its steady state. The apparent voltage across the winding (the back EMF) is proportional to the rotation rate, and the torque is proportional to the current.

 

[Joethemechanic]

In my world of dealing almost exclusively in AC motors "Back EMF" is a term that is used to try to BS me when people do something stupid and let the magic smoke out

 

[drktmplr12]

So a few follow up questions:
1. When you say “equivalent resistance”… are you talking about rotor resistance?

2. Is rotor resistance equal to slip times rotor reactance?

3. How does back emf relate to rotor resistance? Is there an equation?

I tried googling, but as usual you fall into a rabbit hole and only find a few answers.

1. equivalent resistance refers to the equivalent circuit model. the model derivation is something you can research


2 & 3. i'm not sure, you would need to refer to some literature. Electromechanical Energy Devices and Power Systems by Yamayee and Bala, Jr. is the book used in my course work year ago.

 

[Jraef]

When current is applied to a motor load, you have the current inrush in the beginning due to the lack of back EMF from the motor load. Once back EMF is induced, then that 'resistance' lowers the current draw.

My question is, what exactly is generating this back EMF? is it the coils in the stator (since those are inductors), or is it the rotor spinning in the magnetic field, or is it both?

It’s not exactly “back EMF” per se. It is impedance caused by mutual inductance. The magnetic fields in the stator induce current to flow in the rotor bars. The rotor bars then create magnetic fields that oppose those of the stator, which is why the rotor spins. But the magnetic fields of the rotor bards also induce counter EMF in the stator fields, which results in impeding the current.

The rate at which the rotor bar fields cut the lines of force in the stator fields determines the impedance. That’s why, when you add a load and the rotor slows down, the stator draws more current. The slower rotor cuts fewer lines of force, so less mutual induction and less impedance. Less impedance means more current flow, which creates more torque and brings the motor speed back to equilibrium, but the stator is drawing more current in the process.

 

[Grouch1980]

It is impedance caused by mutual inductance.

Great, thanks! Is there an equation for this? I checked on Google, but couldn't find an equation relating the two.

 

[Carultch]

Great, thanks! Is there an equation for this? I checked on Google, but couldn't find an equation relating the two.

Consider this simplified model of a motor, with the winding simplified to a rectangle:

In this motor, there are n windings in the wire loop with an area of A. A uniform magnetic field (B) points from the north pole to the south pole.

The motor rotates CCW as viewed from the top of the screen. Assign angle θ= 0 at the position shown, positive in the direction it rotates.

The magnetic flux at any given instant, is A*B*sin(θ). It's zero at the position shown, and its full value, 90 degrees from the position shown. The rate of change in magnetic flux which we'll need for Faraday's law, dΦ/dt is therefore: dΦ/dt = -A*B*cos(θ)*dθ/dt. With n windings, and rotation speed ω=dθ/dt, the back EMF that the winding produces, is n*A*B*ω*cos(θ). The commutator will put absolute value signs on the cosine term, when the brushes switch to the opposite stationary contacts, after the rotor spins past its stall point, 90 degrees from the position shown.

There will be a time-averaged value of this back-EMF that you will measure as the effective back-EMF of the motor. For this simplified model, the average value of abs(cos(θ)) = 2/π. Thus, the time-average of the back-EMF will be 2*n*A*B*ω/π. Real motors have more winding groups, to smooth out the operation, and eliminate the stall points.

This is the circuit diagram that models the electrical behavior of a motor winding. It is treated as a self-inductance (L) and resistance (R), in series with the motor's back EMF. The self-inductance (L) is what the winding would do, if it weren't part of a motor, and were just a stationary coil, reacting to changes in current. On net, the motor produces an ohmic voltage drop from the parasitic resistance, a back EMF from the coil's self-inductance, plus a back-EMF from the coil's motion in the motor's magnetic field.

 

[Grouch1980]

This is the circuit diagram that models the electrical behavior of a motor winding. It is treated as a self-inductance (L) and resistance (R), in series with the motor's back EMF. The self-inductance (L) is what the winding would do, if it weren't part of a motor, and were just a stationary coil, reacting to changes in current. On net, the motor produces an ohmic voltage drop from the parasitic resistance, a back EMF from the coil's self-inductance, plus a back-EMF from the coil's motion in the motor's magnetic field.
View attachment 2566694

Thanks for this. So 2 follow up questions:
1. The back-emf is from 2 sources? the coil's self-inductance, plus the coil's motion in the magnetic field?
2. What about the stator windings? Do those windings produce back emf as well or no?

 

[Grouch1980]

But the magnetic fields of the rotor bards also induce counter EMF in the stator fields, which results in impeding the current.

So what is actually impeding the current? The impedance produced by mutual induction, or the back emf?

 

[Carultch]

Thanks for this. So 2 follow up questions:
1. The back-emf is from 2 sources? the coil's self-inductance, plus the coil's motion in the magnetic field?
2. What about the stator windings? Do those windings produce back emf as well or no?

Yes. The back-emf is from two sources. Initially, the self-inductance dominates, but this is quickly overtaken by the back-EMF due to the coil's motion.

My example shows a rotor winding, and a stator permanent magnet. Some motors are built the other way around, but the same principles still apply, and stator windings will have a back-EMF. Rather than being from their motion in a magnetic field, a stator's back-EMF comes from the magnetic fields' transient behavior across its area. Some motors have no permanent magnets, and both the stator and rotor windings will have back-EMF as a consequence of Faraday's law.

 

[Grouch1980]

Yes. The back-emf is from two sources. Initially, the self-inductance dominates, but this is quickly overtaken by the back-EMF due to the coil's motion.

My example shows a rotor winding, and a stator permanent magnet. Some motors are built the other way around, but the same principles still apply, and stator windings will have a back-EMF. Rather than being from their motion in a magnetic field, a stator's back-EMF comes from the magnetic fields' transient behavior across its area. Some motors have no permanent magnets, and both the stator and rotor windings will have back-EMF as a consequence of Faraday's law.

Sorry... just trying to put this together... so in an AC 3-phase motor, the stator and rotor windings will initially have a back-EMF, due to the coils. But will be quickly overtaken by the back EMF generated by the coil's motion in the magnetic field, once the armature starts rotating?

 

[Carultch]

Sorry... just trying to put this together... so in an AC 3-phase motor, the stator and rotor windings will initially have a back-EMF, due to the coils. But will be quickly overtaken by the back EMF generated by the coil's motion in the magnetic field, once the armature starts rotating?

I have a simulation where I've worked this out. I'll share some of the results and show you how it all fits together.

 

[ggunn]

It's also why an 8 ohm speaker doesn't show 8 ohms to a DVM.