Transcript DC Motors - United States Naval Academy

Lecture 31
DC Motors
Learning Objectives

Identify and define the components of a two pole
permanent magnetic DC motor (stator, armature,
commutator and brushes).

Given the direction of a magnetic field in a two pole
permanent magnetic DC motor, determine the direction
of force applied to a single armature loop (Lorentz/Force
Law).

Understand the effect of multiple armature loops in a DC
motor.

Understand the induced effects of rotating a currentcarrying closed loop conductor in a magnetic field
(Faraday/Lenz/Electromotive Force)
Basic DC Motor Operations
Basic DC Motor Operations

The current passing
through the coil
creates an electromagnet with a
North/South pole
Basic DC Motor



This electro-magnet
interacts with the
permanent magnetic field
The opposite poles repel
each other, while the like
poles attract each other
This causes the motor to
spin
Torque on wire
Commutation


In order to provide continuous rotation, the armature
current Ia must change direction every 180 of rotation.
This process (commutation) is accomplished by brushes
and a segmented commutator bar.
Commutator
Brushes
Forces on DC motor rotor
SOURCE: FISIK.FREE.FR. Applied Physics.
Developed Force and Torque

The force between the rotating electromagnet and the
stationary magnetic field is given by the Lorentz Force Law
F d  I a LxB

Torque=Force x Distance:
r
T  FDr  (I a LxB)r (N  m)
where r = radius from central axis

If power is only applied to the armature wire in the optimum
position, the cross product becomes simple multiplication:
T  BLI a r

In our simple motor we have 2 wires being acted upon:
T  2BLI a r
Torque

We can significantly increase torque by increasing:




The magnetic field
The current
The number of wires being acted upon
The most practical is to increase the number of wires
being acted upon:
T  2NBLI a r
where N = number of turns of wire

This equation is greatly simplified by using the
MACHINE CONSTANT (KV) (V*sec)
T  Kv I a
where Kv  2NBLr
Torque

Multiple sets of windings are used on the rotor to ensure that
torque is applied smoothly throughout the rotation of the
motor.

Each winding is only briefly supplied current when it is at the
position where it would apply maximum torque to the rotor

This ensures maximum efficiency of
the motor, since power isn’t wasted
on forces which try to pull the rotor
apart
F
Torque Balance

Under steady-state conditions:

Td  Tload  Tloss
N m
Watt 
sec

N  m=Watt  sec=V  A  sec
Td
Tloss
Tload
Torque

Developed power is:
Pd  Td   Kv I a

Ignoring rotational losses, this developed power is
the mechanical output, and machine efficiency can
be calculated as:
Pout
PD

100  100 (Tloss  0)
Pin
Pin
N m
Watt 
sec

N  m=Watt  sec=V  A  sec
REVIEW
Faraday’s Law


Faraday’s Law states that a voltage is
induced in a circuit when a conductor is
moved through a magnetic field
In the case of DC linear motors we know
that
Einduced= B L u
u
= velocity
 B = magnetic field
 L = length of conductor
Faraday’s Law in motors

Induced voltage in motors is given by also
Einduced = Ea = Kv ω


Kv is the motor constant (V*sec)
ω is the angular velocity (rad/sec)

Also called Counter EMF (CEMF) or “back EMF”
because it opposes the applied voltage.

Angular velocity can be calculated by:
ω = 2  (RPM/60)
Parts of a DC Motor
SOURCE: Gears Educational Systems.
Actual DC Motor
POLES
(or FIELD WINDINGS)
COMMUTATOR
ARMATURE
WINDINGS
BRUSH
ROTOR
(entire rotating part)
STATOR
(stationary part)
Points to remember





Torque developed
Mechanical Power Developed
Back EMF
Angular velocity
KVL Vdc-IaRa-Ea=0
Td=Kv Ia
Pd = Td ω = Kv Ia ω
Ea = Kv ω
ω = 2  (RPM/60)
Example problem 1
A 120V permanent magnet DC motor is rated for
15A at 3600 rpm. The motor is 85% efficient at
rated conditions. Assume no rotational losses.
Find:
a. KV(Machine constant) .
b. Find the back EMF at 3600 rpm.
c. Find Ra
Example problem 2
A permanent magnet DC motor has a machine
constant, KV = 1.0 ν·s. The no-load torque is 2.0
N-m assuming rotational power loss is linear
with speed. (Pd=Pmech-loss)
a.
b.
c.
Determine the mechanical power developed by
the unloaded motor at 3600 rpm.
Find the back EMF at 3600 rpm
Determine the armature current.
Example problem 3
A 120 VDC permanent magnet DC motor with KV = 0.95 V·s
is measured under the following two conditions (same
voltage applied):
1. Motor is unloaded, IA = 2 A.
2. Motor is loaded until IA = 15 A and 1200 rpm.
Find:
 Torque due to rotational loss.
 Torque due to the load in condition 2 assuming
rotational power loss is linear with speed.
(Pd=Pmech-loss)
 Machine efficiency at condition 2?