Basis of Motor Action - University of Southern Maine
Basis of Motor Action
Oersted and Ampere showed that an electric current produces a magnetic field in the space around it and the field is circular around the wire carrying the current.
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The following description and link describe motor action at a more fundamental level than we consider in ELE 100. We have chosen to describe motor action in terms of electromagnets, as was described in class. You can use either explanation.
Ampere and others showed that electric currents attract bits of iron and that parallel currents attract each other. From later work by others it was found that when a charged particle moved in a magnetic field a force is exerted on the particle that changes the direction of motion of the particle. When this finding is applied to a current flowing in a straight piece of wire of length L that is placed in a uniform field of density B, the magnitude of the force can be shown to be the product of the length of the wire and the magnitudes of the current and the field density.
F = BIL
The direction of the force is perpendicular to both the direction of the wire and the direction of the field.
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This process of switching the polarity of the applied voltage and therefore the direction of the current is called commutation, and the physical component of the motor that accomplishes that is called the commutator.
In order to understand the mechanical operation of a motor we need to introduce the concept of torque, the “force” that causes rotation.
Suppose we have a large person on one end of a see-saw and a small person on the other end.
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Weight W1 is a force that causes a rotation counterclockwise around the fulcrum. W2 is a force that causes a rotation clockwise around the fulcrum. The see-saw balances when
W1L1=W2L2. This product of force times the distance between the force and the axis around which rotation occurs as the result of the force is called torque. Torque is defined as the product of the length of the radius and the force perpendicular to the radius.
You might know from experience that it is easier to unscrew a nut from a bolt if the wrench has a longer handle than if it has a short handle; or that a screwdriver with a larger diameter handle makes it easier to turn a screw than one with a small diameter handle. This results directly from the phenomenon of torque, or twisting ability, being the product of the force and the distance from the center of the circle that the force is applied.
In the case of the electric motor, the force causing the rotation is the result of the interaction of the magnetic field created by the motor “electromagnet” and the fixed magnetic field. Once the torque has been so generated in the motor axle or shaft it can be transferred to do useful work by connecting the motor shaft to a mechanical system. The motor is an electromechanical device that transforms electrical energy into mechanical work. If we take into account the rate at which it does this we can say it transforms electrical power into mechanical power.
Now we need to ask a very important question: If the torque is making the motor shaft rotate why doesn’t it just keep going faster and faster? To answer that question we need to learn two important concepts: Newton’s laws of motion and counter-electromotive force.
Newton’s Laws of Motion
Tippler, 78
First Law: An object at rest of moving at a constant velocity will remain at rest of at a constant velocity unless acted upon by an external force.
Second Law: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
This law can be expressed mathematically as F=MA where it is seen that the constant of proportionality between acceleration and force is the mass, M. If you think about this relationship, it makes sense that the larger the mass of an object, the more force will be required to accelerate it. In units, one Newton is the magnitude of the force required to accelerate one kilogram at the rate of 1 meter per second per second.
Third Law: For every action, there is an equal and opposite reaction.
Notice that the second law states that there will be acceleration so long as there is an unbalanced net force. If equal and oppositely directed forces are applied, there will be no net force and no acceleration. Under such conditions, the first law applies and either the object will not move or, if it is already moving, it will continue to move in the same direction at a constant velocity.
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With respect to our motor, Newton’s law says that there will be a rotational acceleration until there are forces that balance the torque producing capabilities of the motor. These two oppositional forces are of two types: the counter-electromotive force and the load.
Counter-electromotive Force
Following the work of Oersted and Ampere, in 1831 Faraday did experiments that revealed that if the magnetic field enclosed by a loop of wire changes, a voltage appears at the ends of the wire. This voltage is usually called an electromotive force or emf. His experiment is shown in the following animation.
We note in the animation that the field must be changing in order for a voltage to be produced, and that the more rapidly the field changes, the greater the voltage that is produced. This phenomenon is the basis for an electric generator. If we rotate a loop of wire in a magnetic field, the amount of magnetic field within the loop changes and a voltage is produced. This is shown in the following animation. (You might need to copy and paste the url into Netscape in order to get the animation to show up.)
When the coil of a dc motor passes through a magnetic field, Faraday’s law holds and a voltage is produced that is of a polarity that opposes the direction of the current produced by the power supply external to the motor. Thus, it is called the counter-electromotive force, or counter-emf, because it is counter, or in opposition to, the direction of flow of the current that produced the motion in the first place.
Applying Kirchhoff’s voltage law, the sum of the voltage rises around a loop equals the sum of the voltage drops. We can represent the electrical portion of the motor as a resistance (of wires, etc.) and a generator of a voltage, Em.
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Writing the loop equation: VB = IR + Em or VB – Em = IR.
Now we need to determine Em. It turns out that Em is proportional to the speed that the wire passes through the magnetic field:
Em = kv
where k is a motor constant that includes things like the field strength and the radius between the center of rotation and the wire, and lower case v is the velocity of the wire passing through the magnetic field.
The net result is that when a voltage is applied to the terminals of the motor a current flows that creates a torque that accelerates the motor (makes it rotate faster). As it rotates faster the motor generates a counter-emf that increases with the speed of rotation. As the counter-emf increases, the current decreases thus decreasing the torque and consequently reducing the acceleration. When the counter-emf balances the applied voltage the current is zero so the torque is zero so the acceleration is zero so the velocity is constant.
Mechanical Load
Actually, the load is never zero because there is always internal friction, so torque is generated that balances those losses. If there is a mechanical load, that is if we are asking the motor to do useful mechanical work, then sufficient current will be delivered to match internal losses plus the external load, and the motor will rotate at a velocity slower than it would with no internal losses or mechanical load.
Quantitative Motor Relationships
So far we have been describing motor action in qualitative terms. Now we can derive some quantitative relationships among voltage and speed, torque and current.
First we need to introduce angular velocity. Linear velocity is the change of distance per unit time; for a constant velocity, v = Δd/Δt where Δd is the change in distance and Δt is the change in time. Angular velocity is the change of angle per unit time; for a constant angular velocity, ω ’ Δθ/Δt, where ω is measured in radians, and Δθ is the change in radians and Δt is the change in time. One radian is 180/π degrees.
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As a point moves along the circumference from its position at t1 to its position at t2, it moves along a length of arc, s. The relationship between the angle (in radians), the arc length and the radius is s = rθ. Since velocity is the rate of change of linear distance with time and angular velocity is the rate of change of angle with time,
Δs/Δt = rΔθ/Δt
v = rω
We saw earlier that the counter-emf produced by the motor is Em=kv = krω. Since r is a constant radius, we can rewrite the equation, using a new constant K=kr, as Em=Kmω. Τhat is, the counter-emf is proportional to the angular velocity of the armature.
Relationships among applied voltage, current, torque, and motor speed
In the following analysis, we assume that the field is produced by permanent magnets, so it is constant.
We saw that, according to Kirchhoff’s voltage law, the equation for the motor circuit is:
VB = IR + Em
We have just seen that Em=Kmω so VB = IR + Kmω.
Now suppose that a dc motor is running at some constant angular velocity ω at some external load. What happens if we increase the external load? We expect that the increase in load will cause the motor to slow down. We have experienced this driving a car and coming to a hill. If we don’t press on the accelerator, the automobile slows down as it ascends the hill. If we try to push a lawn mower at a constant speed and come to a thick patch of grass, we need to use more force to maintain the constant speed. In the case of the dc motor, if the applied voltage VB stays the same, and the angular velocity, ω, decreases as the result of an increase in the load, then current must increase to maintain the equality between the left and right sides of the equation VB = IR + Kmω.
We learned earlier that the force on a current-carrying wire is F = BIL, i.e. force is proportional to current. Since torque is force times the radius around which rotary motion takes place, torque is proportional to current. Therefore, an increase in current increases the torque produced by the motor. An increase in torque accelerates the motor until the balance between applied voltage VB and IR+Kmω is re-established. The angular velocity is not restored to its original value because when current increases, the IR term on the right side of the equation increases, so the motor accelerates to an angular velocity less than it was before.
We can express the proportionality of torque and current as:
T = KTI
where KT includes such things as the field strength, the length of wire in the field, and radius of rotation. (We are assuming here that field strength is constant.) If we substitute for I in the loop equation, we get:
VB = (T/KT )R+ Kmω
This equation expresses the relationships among the variables of interest in motor operation: applied voltage, torque, and angular velocity. It turns out that KT and Km are equal to one another but different motors have different K. So we finally get:
VB = (R/K)T + Kω
Suppose that the load and therefore the torque is constant. How does angular velocity change as applied voltage changes? We can rewrite the equation with voltage as the independent variable and velocity as the dependent variable.
ω = (1/K)VB– (R/K2 )T
For a constant load, the torque is constant, so this is a linear equation with ω axis intercept –(R/K2 )T and slope (1/K). Below is a chart showing the relationship between angular velocity and applied voltage for K=1, R=1 and different values T.
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We see that velocity increases linearly with speed for a given load. For a given voltage, the velocity is less as the load (torque) increases. It appears that the velocity is negative at low voltages. Actually the velocity is zero until the voltage is high enough to produce enough current to produce enough torque to make the motor begin to move.
Now let’s look at how the angular velocity changes as the load changes but applied voltage stays the same.
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We see that for a set applied voltage, as the load (torque) increases the angular velocity decreases. These are performance curves for the motor. Different motors have different performance curves.
Since angular velocity is a function of both applied voltage and load, we can create a surface plot showing angular velocity for any combination of voltage and load.
ω = (1/K)VB– (R/K2 )T
For K= 1 and R=1 ω = VB –T
The curves are shown in the figure below.
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For linear, non-rotational systems, mechanical work is force times distance if the force is in the same direction as the direction of motion. Power is the rate of change of work, so power is a constant force times the rate of change of distance; i.e. Power = Force x Velocity.
For rotational systems, mechanical work is torque times the change in angle, and power is torque times the rate of change of angle; i.e. Power = Torque x Angular Velocity.
Let’s see how mechanical power varies with torque. Below is the performance curve which has been extended to the point where the velocity becomes zero.
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From the curve, we can calculate power at each load (torque). For example, at a torque of 0.5, the angular velocity is 1.5, so Power = 0.5 x 1.5 = 0.75 watts. If we do this for all points on the curve, we get the power versus torque curve below.
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It is seen that the output power varies with the load, and has a maximum midway between zero load and the load which causes the motor to stop. Therefore, in order to get the maximum power output from the motor it is necessary to identify this point on the performance curve. This can be done experimentally by measuring velocity at several loads, creating the performance curve, extending it to the “stopping load” and finding the torque at one-half that “stopping load.” Alternatively, we can find the best fit curve to the data points of power versus torque, provided that the data points are in the vicinity of maximum power output.
Mechanical power is measured in watts. Efficiency of an electromechanical system can be calculated as output power divided by input power, with the ratio multiplied by 100 to convert to percent.
Efficiency = (output mechanical watts/input electrical watts) x 100
Maximum efficiency usually occurs close to maximum power output.
-----------------------
L2
L1
W2
W1
F1
F2
F = F1-F2 Unbalanced force produces acceleration to the right
F1
F2
F=F1-F2 = 0 Balanced forces; no acceleration
+Em
R
Motor
+VB
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