Motor Back EMF

[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?

Here is the setup I've used for my simulation:
DC motor
3 winding groups on the rotor

Electrical data:
Loop Area: 0.02 m^2
Magnetic Field: 0.1 Tesla
QTY of turns per winding group: 50 turns
QTY of winding groups: 3
Supply Voltage: 24 Vdc
Parasitic resistance: 250 milliohms
Self-inductance: 100 milli-Henries

Mechanical data:
Drag constant: 0.015 N-m/(rad/s), linear drag assumed
Rotational inertia: 0.2 kg-m^2
Load torque: 0.5 Newton-meters

Here are the graphs that I've simulated:

The voltage distribution shows how the 24V supply voltage distributes among the three elementary behaviors of the windings. In reality, these elementary behaviors are distributed throughout the windings, rather than lumped as idealized elements. The reason for the roughness is due to abs(trig) terms, that relate the mechanical back-EMF to speed, and the torque to current. If you had lots of winding groups, equally distributed around the rotor, the roughness would smooth out, and these plots would be sine waves enveloped by exponential decay, settling on the steady-state condition in the long run.

As you can see, the back EMF of the self-inductance initially dominates, and the back-EMF of motion overtakes it pretty quickly. There is a brief time when the self-inductance's back-EMF is negative, where it is assisting the voltage source and pushing its stored energy back in to the rest of the circuit, as the inrush current settles to its full load current. In the long run, the back-EMF of motion dominates the voltage that the motor consumes.

The current grows from zero first, and the rotation speed follows shortly thereafter. The current starts growing right away, since slow-moving motor behaves like an LR circuit. The rotation speed curve lags behind the current curve, because the current has to first grow from zero, to provide the starting torque to speed it up from rest.

 

[Grouch1980]

Great stuff. The charts are great. They especially clarify the beginning (negative self-inductance, how current builds up, etc.). what'd you use for the simulation if you don't mind my asking? or is it a program like a python script?

Two final questions, one of which I believe you answered, but just making sure:
1. the back-EMF self inductance, this happens in both the stator and rotor windings? assuming no permanent magnets.
2. post #12 from Jraef: is it impedance or back-EMF that limits the current? impedance makes sense, since that is what always limits current, but any literature on the topic always says back-EMF impedes the current.

 

[Carultch]

Great stuff. The charts are great. They especially clarify the beginning (negative self-inductance, how current builds up, etc.). what'd you use for the simulation if you don't mind my asking? or is it a program like a python script?

Great stuff indeed. For this simulation, I used an Excel spreadsheet and Euler's method of numerically solving differential equations. If you message me with your email address, I can share a copy. Imagine doing this in Euler's time, when all he had was quill and paper.

To summarize how it works;
Set up your constants (resistance, inertia, etc) at the top of the spreadsheet, and assign names to the cells, to make absolute references. You set your timestep (dt) as a constant, which I set as 0.01 sec. Then create columns for time and each time-varying quantity (e.g. current, dI/dt, position, speed, acceleration). Use governing equations (like acceleration = net torque / inertia) to establish relationships among values in the same row. We we don't have a governing equation, use an approximation of constant acceleration during each time interval, to establish the relationship to the same variable in the previous row. Then you pull down the formulas, until you have hundreds of rows, enough to produce the graph.

 

[Carultch]

Great stuff. The charts are great. They especially clarify the beginning (negative self-inductance, how current builds up, etc.). what'd you use for the simulation if you don't mind my asking? or is it a program like a python script?

Two final questions, one of which I believe you answered, but just making sure:
1. the back-EMF self inductance, this happens in both the stator and rotor windings? assuming no permanent magnets.
2. post #12 from Jraef: is it impedance or back-EMF that limits the current? impedance makes sense, since that is what always limits current, but any literature on the topic always says back-EMF impedes the current.

To clarify one thing, it's not really negative self-inductance, since self-inductance is a constant. It's really more like a "negative voltage drop", or a "voltage boost", as a consequence of the current decreasing.

To answer your two remaining questions:
1. yes, both the back-EMF of self-inductance and the back-EMF of motion, happen in both rotor and stator windings. Since they are windings, they have a self-inductance by virtue of their geometry and presence of magnetic material. The back-EMF of self-inductance models what these components would do if they were just inductors, i.e. stationary windings that aren't part of a motor. Think of it like "electrical inertia", such that a water analogy would be like a free-spinning paddlewheel, that applies a back-pressure, any time you attempt to change its speed. Connect this paddlewheel to a load, and you make a turbine as an analogy for a motor.

2. Lots of moving parts to this question. If you expand your definition of impedance, the short answer is yes. Most of us think of impedance in the Fourier transform. This is a special case of the Laplace transform, which also models the start-up behavior. In a general case, think of generalized impedance as the ratio between an input and output signal, in the Laplace domain. The reciprocal of the transfer function between the input and output.

I'll start with some simplifying assumptions:
1. Frictional drag is proportional to speed. Not consistent with experimental results, but any other model makes it much harder to solve.
2. An "infinite" number of winding groups, so that the torque-current relationship is independent of position. When doing so, we can equate Torque = K*I, and motional EMF = K*ω, where K is a constant of the winding geometry, core material and magnetic field.

The system of differential equations modelling the motor's behavior:
L*I'(t) = -R*I(t) - K*ω(t) + V(t)
J*ω'(t) = K*I(t) - D*ω(t) - τ(t)

In the Laplace transform domain:
L*s*H(s) = -R*H(s) - K*Ω(s) + U(s)
J*s*Ω(s) = K*H(s) - D*Ω(s) - T(s)

The short answer for what s means is complex frequency. For the Laplace transform, the short explanation: a method for converting time-domain calculus into s-domain algebra. Integrating with t becomes dividing by s, and derivatives with t become multiplication by s. Then we invert the transform, to get the time-domain result we want. I'm using the neighbors (H & U) of I & V, for their Laplace-domain functions.

A visual representation (block diagram) for the interaction of all these functions is shown below. Each block it multiplies its input in the Laplace domain to produce an output, and each summing junction will add & subtract to combine values. With block diagram algebra, you can show that it produces the above equations. This shows how many moving parts there are, and the feedback loops they produce.

When solving for H(s) and Ω(s), we get the following:
H(s) = [(J*s + D)*U(s) + K*T(s)]/[(L*s + R)*(J*s + D) + K^2]
Ω(s) = [-(L*s + R)*T(s) + K*U(s)]/[(L*s + R)*(J*s + D) + K^2]
For the case of a constant DC source and load torque, U(s) and T(s) are both the Laplace of a constant, which is U(s) = V/s, and T(t) = τ/s.

For a motor with no load other than its own friction, we can calculate Z(s) = U(s)/H(s), to find the effective impedance of the unloaded motor:
Z(s) = U(s)/H(s) = [(L*s + R)*(J*s + D) + K^2]/(J*s + D)

In the long run, with the final value theorem, the no-load current approaches the following:
I(infinity) = V*D/(K^2 + D*R)

So its no-load steady state impedance would be Z_infinity = (K^2 + D*R)/D.

 

[Grouch1980]

To answer your two remaining questions:
1. yes, both the back-EMF of self-inductance and the back-EMF of motion, happen in both rotor and stator windings. Since they are windings, they have a self-inductance by virtue of their geometry and presence of magnetic material. The back-EMF of self-inductance models what these components would do if they were just inductors, i.e. stationary windings that aren't part of a motor. Think of it like "electrical inertia", such that a water analogy would be like a free-spinning paddlewheel, that applies a back-pressure, any time you attempt to change its speed. Connect this paddlewheel to a load, and you make a turbine as an analogy for a motor.

Got it! Thank you so much. It's extremely difficult finding answers on some of these topics... browsing the internet is sometimes futile, and searching just throws you down a rabbit hole. Or you keep finding articles that repeat the same things over and over with no new information.

I'll need some time to go through your answer to my 2nd question... lot of good stuff to go through.

 

[Grouch1980]

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.

Making sure I understand motor torque.

If you increase the supply voltage, or increase the supply current, torque increases.

If you decrease the supply voltage, or decrease the supply current, torque decreases.

Is this correct?

 

[Carultch]

Making sure I understand motor torque.

If you increase the supply voltage, or increase the supply current, torque increases.

If you decrease the supply voltage, or decrease the supply current, torque decreases.

Is this correct?

Approximately, yes, that is correct. More voltage means more current, which also means more torque. But there are a lot of interdependencies and calculus operations that govern the relationships among the various quantities involved, so it isn't a simple proportionality.

The torque on the rotor is directly determined by the current, with a simple proportionality (tau = K*I). This torque has 3 destinations: frictional drag, acceleration of the rotor, and output torque. It's usually output torque we want to know, while the inertia and drag are mechanical impedances that get in the way.

A similar thing happens with the voltage applied. It has three destinations as well: storing energy in the inductance, dissipating energy in the resistance, and delivering mechanical energy to the rotor, through the back-EMF of motion.

It's common that you see graphs, such as this, to illustrate the steady-state behavior of a motor. All curves and all conditions on this chart, are based on the same source voltage. We also need to know the mechanical constraint between torque and speed, to determine where on this chart, the motor will operate.

Given a particular voltage, the highest current will be drawn at stall conditions, when the rotor is locked. The coil acts like an RL circuit, with a current that follows an exponential approach curve, from initially zero to its maximum. This is a dangerous way to run a motor, because it will heat up the coil the most, and it could damage the voltage source as well. This is what an overload protection will prevent: both too much current, and too much load torque.

The lowest current is drawn, when the motor can freely spin without any load torque. If there were no drag, the motor would draw zero current, and would spin freely, just as it would if you shut off the power and spun it manually. The back-EMF would equal the source voltage, and effectively create an open circuit. There would be no ohmic heating or inductive ramping of current, with all the source voltage opposed by the motional back-EMF. This is unrealistic, since all real motors have some internal friction, in some form or another. So it will take a little power and current to sustain its maximum speed.

Two maxima are of interest: maximizing power output, and maximizing efficiency. Maximum power occurs at about half its maximum speed, and maximum efficiency occurs at about 90% its max speed, depending on the details of the drag. This is why motors usually are built to run at high speed, and use reduction gearing, when used with a machine that needs to provide high torque.

 

[ggunn]

Making sure I understand motor torque.

If you increase the supply voltage, or increase the supply current, torque increases.

If you decrease the supply voltage, or decrease the supply current, torque decreases.

Is this correct?

How would one increase or decrease the supply current?

 

[Grouch1980]

How would one increase or decrease the supply current?

if the supply voltage is decreased for whatever reason, the current will then increase. So the increased current will increase the torque.

 

[ggunn]

if the supply voltage is decreased for whatever reason, the current will then increase. So the increased current will increase the torque.

Well, OK, but the question was "...If you increase the supply voltage, or increase the supply current...", implying that you could change one independent of the other. You can change the voltage directly but not the current.

 

[Grouch1980]

Well, OK, but the question was "...If you increase the supply voltage, or increase the supply current...", implying that you could change one independent of the other. You can change the voltage directly but not the current.

correct, I could've worded it better.