Download Basis of Motor Action

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.
http://micro.magnet.fsu.edu/electromag/electricity/generators/
If we form a wire into a loop and apply the right-hand-rule, we see that the magnetic field
inside the loop is generally in the same direction as shown in the figure.
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current
field
If we wrap several turns of the wire around a cylindrical object, the magnetic field of
each loop adds to the total field, which, in the cylinder is in the same direction (right to
left in the figure below). This field configuration is the same as that for a permanent bar
magnet. Therefore, we can create an electromagnet in this fashion.
An advantage of an electromagnet is that we can change the direction of the magnet by
changing the direction of the current.
In the following figure the far left and the far right permanent magnets are fixed. The
center magnet is free to rotate around a center point, which it does because of the
attraction of opposite magnetic poles.
N
N
N
S
N
S
S
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Because of momentum, the magnet will overshoot as shown in the figure below.
S
N
S
N
N
S
But then the magnetic attraction of opposite poles causes the rotation to reverse and the
magnet quickly settles down in alignment with the fixed magnets.
Now, suppose we can reverse the north and south poles of the rotating magnet just as it
passes the fixed magnets. We can do this by changing the direction of the current in the
wire of an electromagnet. Then, the poles near one another are the same and repel one
another so the electromagnet continues to rotate in the same direction.
N
N
S
N
S
S
Change direction of current to change poles of the electromagnet.
N
N
S
N
S
S
S
N
N
N
S
S
S
S
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. Notice in the applet found by clicking on the
following link that the direction (red arrows) of the current in the wire changes as the
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electrical contacts on the rotating shaft alternately connect to a positive and negative
voltage.
http://www.walter-fendt.de/ph11e/electricmotor.htm
Torque
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.
L1
L2
W1
W2
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, as illustrated above. 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
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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 produced by a net force on the object 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.
F1
F2
F = F1-F2 Unbalanced force produces acceleration to the right
F1
F=F1-F2 = 0
F2
Balanced forces; no acceleration
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.
http://micro.magnet.fsu.edu/electromag/java/faraday2/
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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.)
http://www.walter-fendt.de/ph14e/generator_e.htm
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.
Motor
+Em
+VB
R
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.
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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 = d/t where d is the change in distance and t is
the change in time. Angular velocity is the change of angle per unit time; for a constant
angular velocity, t, where  is measured in radians, and is the change in
radians and t is the change in time. One radian is degrees.
2, t2

r
1, t1
2-1)/(t2-t1)

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 = r Since velocity is the rate of change of linear distance with
time and angular velocity is the rate of change of angle with time,
s/t = rt
v = r
We saw earlier that the counter-emf produced by the motor is Em=kv = kr. Since r is a
constant radius, we can rewrite the equation, using a new constant K=kr, as
Em=Kmhat is, the counter-emf is proportional to the angular velocity of the armature.
Relationships among applied voltage, current, torque, and motor speed
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=Km so VB = IR + Km
Now suppose that a dc motor is running at some constant angular velocity  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, ,
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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 + Km

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+Km 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+ Km
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 + K
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.
 = (1/K)VB– (R/K2 )T
For a constant load, the torque is constant, so this is a linear equation with 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.
These are the curves you got in the laboratory exercise, although K and R may have been
different sot the slope and intercepts were different. The form or shape of the curves is
the same, however.
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Velocity for Different Loads, K=1
Angular Velocity, rad/s
5
4
3
2
1
0
-1 0
1
2
3
4
5
6
T=1
T=2
T=3
-2
-3
-4
Applied Voltage, Volts
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.
Angular Velocity, rad/s
Angular Velocity Versus Torque for Different
Voltages, K=1
3.5
3
2.5
V=1
V=2
V=3
2
1.5
1
0.5
0
0
0.2
0.4
0.6
0.8
1
1.2
Torque, n-m
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.
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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.
Angular Velocity, rad/s
Angular Velocity Versus Torque for an Applied
Voltage of 2 Volts, K=1
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
Torque, n-m
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.
Motor Mechanical Power, K=1
1.2
Power, w
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
Torque, n-m
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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 “stalling load” and finding the
torque at one-half that “stalling 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.
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