Transcript 02_DC Machines - UniMAP Portal

SHAIFUL NIZAM MOHYAR
UNIVERSITI MALAYSIA PERLIS
SCHOOL OF MICROELECTRONIC
2007/2008
2.0 INTRODUCTION
Motor
Electrical
Energy
Mechanical
Energy
Generator
Electromechanical Energy Conversion
Pelec = vi = Teωm = Pmech
The machine’s magnetic field is the medium of
(energy) conversion.
GLOSSARY
 Torque, T – A force that produces rotation on a axis and also
defined as a linear force multiplied by a radius. In an electric
motor, this is the force from the interaction of the magnetic
fields produced by the flow of current through the armature
and field windings/coils.
 Flux, Φ - The magnetic field which is established around an
energized conductor or permanent magnet. The field is
represented by flux lines creating a flux pattern between
opposite poles. The density of the flux lines is a measure of
the strength of the magnetic field.
 Magnetic field, B - A region of space that surrounds a moving
electrical charge or a magnetic pole, in which the electrical
charge or magnetic pole experiences a force that is above the
electrostatic ones associated with particles at rest.
 The voltage in any real machine will
depend on three factors:
1. The flux, Φ in the machine.
2. The speed of rotation, ω.
3. A constant representing the
construction of the machine.
2.0 INTRODUCTION…contd
Flemming’s Left Hand Rule (Motor Rule)
 Use: To determine the direction of a force on a current carrying conductor
in a magnetic field.
2.0 INTRODUCTION…contd
 The carbon rod is NOT magnetic.
 When no current flows, the rod is stationary
 When we turn on the current, the rod experiences a
force that makes it move.
 The direction of the force is determined by Fleming' Left
Hand Rule
2.1 CONSTRUCTION
Cutaway view of a dc motor
2.1 CONSTRUCTION
….contnd
Stator with visible poles
2.1 CONSTRUCTION ….contnd
Rotor of a dc motor.
2.2 CONSTRUCTION….contnd
General arrangement of a dc machine
2.1 CONSTRUCTION….contnd
 The stator of the dc motor
has poles, which are excited
by dc current to produce
magnetic fields.
 In the neutral zone, in the
middle between the poles,
commutating
poles
are
placed to reduce sparking of
the
commutator.
The
commutating
poles
are
supplied by dc current.
 Compensating windings are
mounted on the main poles.
These
short-circuited
windings
damp
rotor
oscillations. .
2.1 CONSTRUCTION….contnd
 The poles are mounted on
an iron core that provides a
closed magnetic circuit.
 The motor housing supports
the iron core, the brushes
and the bearings.
 The rotor has a ring-shaped
laminated iron core with
slots.
 Coils with several turns are
placed in the slots. The
distance between the two
legs of the coil is about 180
electric degrees.
2.1 CONSTRUCTION….contnd
 The coils are connected in series
through
the
commutator
segments.
 The ends of each coil are
connected to a commutator
segment.
 The commutator consists of
insulated
copper
segments
mounted on an insulated tube.
 Two brushes are pressed to the
commutator to permit current
flow.
 The brushes are placed in the
neutral zone, where the magnetic
field is close to zero, to reduce
arcing.
2.1 CONSTRUCTION….contnd
Commutator of a dc motor
2.1 CONSTRUCTION….contnd
 The rotor has a ring-shaped
laminated iron core with slots.
 The commutator consists of
insulated copper segments
mounted on an insulated tube.
 Two brushes are pressed to
the commutator to permit
current flow.
 The brushes are placed in the neutral zone, where the
magnetic field is close to zero, to reduce arcing.
2.1 CONSTRUCTION
 The commutator switches the
current from one rotor coil to
the adjacent coil.
 The switching requires the
interruption of the coil current.
 The sudden interruption of an
inductive current generates
high voltages .
 The high voltage produces
flashover and arcing between
the commutator segment and
the brush.
2.2 DC MACHINE OPERATION
BASIC THEORY
B
I
I
w
h
shaft
Next slide looks
down the shaft
2.2 DC MACHINE OPERATION
BASIC THEORY….contnd
B
Current
coming
toward you
Current leaving
away from you
Shaft
2.2 DC MACHINE OPERATION
BASIC THEORY….contnd
B
direction by left-hand rule
Thumb = current
Forefinger = B
Rest = Force
F
F
T = 2hFcos
= 2hIwBNcos
2.2 DC MACHINE OPERATION
BASIC THEORY….contnd
B
2.2 DC MACHINE OPERATION
BASIC THEORY….contnd
B
2.2 DC MACHINE OPERATION
BASIC THEORY….contnd
B
2.2 DC MACHINE OPERATION
BASIC THEORY….contnd
B
2.2 DC MACHINE OPERATION
BASIC THEORY….contnd
B
2.2 DC MACHINE OPERATION
BASIC THEORY….contnd
B
2.2 DC MACHINE OPERATION
BASIC THEORY….contnd
B
2.2 DC MACHINE OPERATION
BASIC THEORY….contnd
B
2.2 DC MACHINE OPERATION
BASIC THEORY….contnd
B
2.3 DC MACHINE OPERATION
 In a dc motor, the stator
poles are supplied by dc
excitation
current,
which
produces a dc magnetic field.
 The rotor is supplied by dc
current through the brushes,
commutator and coils.
 The
interaction
of
the
magnetic field and rotor
current generates a force that
drives the motor.
2.3 DC MACHINE OPERATION
v
S
B
a
N
1
30
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
 The magnetic field lines enter
into the rotor from the north
pole (N) and exit toward the
south pole (S).
 The
poles
generate
a
magnetic
field
that
is
perpendicular to the current
carrying conductors.
 The interaction between the
field
and
the
current
produces a Lorentz force.
 The force is perpendicular to
both the magnetic field and
conductor.
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
2.3 DC MACHINE OPERATION
v
S
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
 To avoid the reversal of the force
direction,
the
commutator
changes the current direction,
which
maintains
the
counterclockwise rotation.
N
30
 At this point, the magnetic field
becomes practically zero together
with the force.
 However, inertia drives the motor
beyond the neutral zone where
the direction of the magnetic field
reverses.
B
a
1
 The generated force turns the
rotor until the coil reaches the
neutral point between the poles.
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
2.3 DC MACHINE OPERATION
v
S
B
a
N
1
30
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
 Before reaching the neutral zone,
the current enters in segment 1
and exits from segment 2,
 Therefore, current enters the coil
end at slot a and exits from slot b
during this stage.
 After passing the neutral zone, the
current enters segment 2 and
exits from segment 1,
 This reverses the current direction
through the rotor coil, when the
coil passes the neutral zone.
 The result of this current reversal
is the maintenance of the rotation.
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
2.4 DC Generator Operation
2.4 DC Generator Operation
Fleming’s Right hand rule (Generator Rule)
 Use: To determine the direction of the induced
emf/current of a conductor moving in a magnetic field.
2.4 DC Generator Operation..contnd.
v
B
a
S
N
1
30
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
 The N-S poles produce a dc
magnetic field and the rotor
coil turns in this field.
 A turbine or other machine
drives the rotor.
 The conductors in the slots
cut the magnetic flux lines,
which induce voltage in the
rotor coils.
 The coil has two sides: one is
placed in slot a, the other in
slot b.
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
2.4 DC Generator Operation..contnd.
v
B
a
S
N
1
30
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
 In Figure (a), the conductors
in slot a are cutting the field
lines entering into the rotor
from the north pole,
 The conductors in slot b are
cutting the field lines exiting
from the rotor to the south
pole.
 The cutting of the field lines
generates voltage in the
conductors.
 The voltages generated in
the two sides of the coil are
added.
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
2.4 DC Generator Operation..contnd.
B
a
S
N
1
30
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2
(slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
 The induced voltage is connected
to the generator terminals
through the commutator and
brushes.
 In Figure (a), the induced
voltage in b is positive, and in a
is negative.
 The
positive
terminal
is
connected
to
commutator
segment 2 and to the conductors
in slot b.
 The
negative
terminal
is
connected to segment 1 and to
the conductors in slot a.
v
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
2.4 DC Generator Operation..contnd.
B
a
S
N
1
30
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
 When the coil passes the neutral
zone:
 Conductors in slot a are then
moving toward the south pole
and cut flux lines exiting from
the rotor
 Conductors in slot b cut the
flux lines entering the in slot
b.
 This changes the polarity of the
induced voltage in the coil.
 The voltage induced in a is now
positive, and in b is negative.
v
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
2.4 DC Generator Operation..contnd.
v
B
a
S
N
1
30
Vdc
2
 The
simultaneously
the
commutator reverses its terminals,
which assures that the output
voltage (Vdc) polarity is unchanged.
b
v
Ir_dc
B
S
2
a
30
v
v
N
Vdc
1
 In Figure (b)
 the
positive
terminal
is
connected
to
commutator
segment
1
and
to
the
conductors in slot a.
 The
negative
terminal
is
connected to segment 2 and to
the conductors in slot b.
(a) Rotor current flow from segment 1 to 2
(slot a to b)
b
Ir_dc
(b) Rotor current flow from segment 2 to 1
(slot b to a)
2.5 DC Machine Type
There are generally five major types of
DC motors:
 The separately excited dc motor
 The shunt dc motor
 The permanent magnet dc motor
 The series dc motor
 The compounded dc motor
2.6 DC Machine Equivalent Circuit
The magnetic field produced by the stator poles
induces a voltage in the rotor (or armature) coils
when the generator is rotated.
This induced voltage is represented by a voltage
source.
The stator coil has resistance, which is connected
in series.
The pole flux is produced by the DC excitation/field
current, which is magnetically coupled to the rotor
The field circuit has resistance and a source
The voltage drop on the brushes represented by a
battery
2.6 DC Machine Equ. Circuit..contnd.
1. Permanent magnet
•
The poles are made of permanent
magnets.
•
No field winding required.
•
Small size.
•
Disadvantage is low flux density, so low
torque.
2.6 DC Machine Equ. Circuit..contnd.
2.
Separately excited
The field flux is derived from a separate power source
independent of the generator itself.
B
Field
winding
Armature
winding
2.6 DC Machine Equ. Circuit..contnd.
3.
Self-excited
shunt machine
•
–
The field flux is derives
by connecting the field
directly
across
the
terminals
of
the
generator.
B
2.6 DC Machine Equ. Circuit..contnd.
3.
Self-excited series machine
•
field are connected in
series with armature
B
2.6 DC Machine Equ. Circuit..contnd.
3. Self-excited –
Compounded dc
motor
-
both a shunt and a series field
are present
2.6 DC Machine Equ. Circuit..contnd.
3.
Self-excited
•
Cumulatively compounded
B
•
B
Differentially compounded
B
B
2.6 DC Machine Equ. Circuit..contnd.
 The armature is represented by an ideal voltage source
EA and a resistor RA.
 The brush voltage drop is represented by a small
battery Vbrush opposing the direction of the current flow
in the machine.
 The field coils, which produce the magnetic flux, are
represented by inductor LF and RF.
 The separate resistor Radj represents an external
variable resistor used to control the amount of current
in the field circuit.
Equivalent Circuit of a DC Motor.
 The brush drop voltage is often only a very tiny fraction of the
generated voltage in the motor.
 Therefore, in cases where it is not critical, the brush drop
voltage may be left out or approximately included in the value
of RA.
 Also, the internal resistance of the filed coils is sometimes
lumped together with the variable resistor, and the total is
called RF , Figure below.
A Simplified Equivalent Circuit eliminating the Brush Voltage
Drop and Combining Radj with the Field Resistance .
2.5 DC Machine Equ. Circuit..contnd.
The Equivalent Circuit of Separately Excited dc Motor.
From the above figure,
VF
IF 
RF
VT  EA  I A RA
IL  IA
Shunt DC motors
The Equivalent Circuit of a Shunt dc Motor
From the above figure,
VF VT  E A  I A RA
IF 
RF I L  I A  I F
Torque Equation
T  k AI A
T = torque of armature (N-m)
kA = geometry constant
= flux/pole (Wb)
IA = armature current (A)
Geometry Constant
pN
pN
'
kA 
(rad / s ), k A 
(rpm)
2M
60 M
p = number of field poles
N = number of active conductors on armature
M = number of parallel paths in armature winding
(=p for lap winding, =2 for wave winding)
Power Equation
P  EI A  T
P=power (W) – not counting losses
E = EMF induced in armature (back EMF)
IA = armature current (A)
T = torque of armature (N-m)
 = speed of rotation (rad/s)
Note that Pin = VLIL which will be higher than P
because of loss in the field and armature windings as
well as rotational (friction) losses.
EMF Equation
E  k A  k n
'
A
60
n
2
E = EMF induced in armature (V)
kA = geometry constant
= flux/pole (Wb)
 = speed of rotation (rad/s)
n = speed of rotation of armature
(rpm)
Terminal Voltage Equation
RA
+
+
E
VT
-
VT  E  I A RA
-
VT = voltage at motor terminals
E = EMF induced in armature (V)
IA = armature current (A)
RA = armature resistance
Speed Equation
VT  I A R A
n
'
k A
(applies to shunt connected motor only)
Note that  can also be written as kfIf where kf is
/If (normally a constant ratio)
Ratio Equation
n2 E 2

n1 E1
Speed-Torque
Speed
Differential Compound
Shunt
Cumulative Compound
Series
Torque
Example 1
A 6 pole, 3.0 hp 120V DC lap-wound shunt motor has 960 conductors
in the armature. It takes 25.0 A from the supply at full load. Armature
resistance is 0.75, flux/pole=10.0 mWb, field winding current is 1.20A.
Find the speed and torque.
 746W 
  2.24kW
P  3hp 
 hp 
E  K A

I A  I L  I F  25 A 1.2 A  23.8 A
E
102V

 66.9rad / s
3
K A 153 10 x10


E  VT  I A RA  120V  23.8 A0.75  102V

pN
6960
KA 

 153
2M 2 6
 60 
n  
  638rpm
2



T
P


2.24kW
 33.5 N  m
66.9rad / s
Example 2
A 10hp, 115V Dc series motor takes 40A at its full load speed of
1800rpm. What is the torque at 30A?

2n 2 1800 

 188rad / s
60
60
T  K AI A  K A K F I F I A
IF  I A
 746W 
  7.46kW
P  10hp 
 hp 
P  T
P
7.46kW
T 
 39.6 N  m
 188rad / s
T  K AKF I A
K AKF 
2
T
39.6 N  m

 0.025
2
2


IA
40 A
Tnew  K A K F I Anew  0.02530 A  22.2 N  m
2
2
Example 3 (a)
A 220V DC shunt motor draws 10A at 1800rpm. The armature
resistance is 0.2 and field winding resistance is 440. (a) What is the
torque?
IF 
VT
220V

 0.5 A
RF 440
I A  I L  I F  10 A  0.5 A  9.5 A
E  VT  I A RA  220V  9.5 A0.2  218V
P  EI A  218V 9.5 A  2.07kW

2n 2 1800 

 188rad / s
60
60
T
P


2.07kW
 11.0 N  m
188rad / s
Example 3 (b)
A 220V DC shunt motor draws 10A at 1800rpm. The armature
resistance is 0.2 and field winding resistance is 440. (b) What will
be the speed and line current at a torque of 20 N-m (if field current is
constant)?
I L  I A  I F  17.3A  0.5 A  17.8 A
E  K A
K A 
E


218V
 1.16
188rad / s
T  K AI A
IA 
T
20 N  m

 17.3 A
K A
1.16
E  VT  I A RA  220V  17.30.2  217V

E
217V

 187rad / s
K A 1.16
n
60
 1.79 x103 rpm
2
(shunt is constant speed)