DC Motors
How does a DC Motor work?
26.1
CrossCross-section of DC motor
Soft Iron Core (Rotor)
Magnetic field vector, B
Wire length vector, dL
N
S
N
X
Force vector, df
Permanent Magnet (Stator)
Current, i
CrossCross-section of DC motor
Rotor supported on bearings (free to rotate)
N
X
S
Generating ___________ torque
CrossCross-section of DC motor
Switch direction of current flow after 90 degrees
(the current switching process is called _______________)
X
N
N
Rotation past 90 degrees
N
S
N
X
Now generating _____ torque!
2 Commutator Bars
DC
Two segment
commutation
on rotor
X
N
S
N
S
Now generating ______ torque!
%Torque vs.
Angular
Position
Torque
N
CrossCross-section of DC motor
100
50
0
0
90
180
270
Angular Position
360
DC Motors
26.2
4 Commutator Bars
24 Commutator Bars
DC
DC
Torque
%Torque vs.
Angular
Position
N
S
N
Sixteen segment
commutation on
rotor
N
50
0
0
90
180
270
360
%Torque vs.
Angular
Position
100
50
0
0
Angular Position
Wire length vector, dL
Velocity vector, v
Induced EMF, dV (voltage)
Note that an induced EMF always gives rise to a current whose magnetic field
opposes the original change in magnetic flux - Lenz’s Law
ME 372 Circuit Model for
Permanent Magnet DC Motor
Va
ia
La
360
that a DC motor always begins to act
like a generator once the rotor wires start to
move through the magnetic field
the induced “back EMF” is proportional to
____________________________
“back EMF” generates a ___________________
which opposes the applied current,
_________________ the force (torque) output
of the motor
ME 360 Circuit Model for
Permanent Magnet DC Motor
ia
+
Vb
-
-
+
Ra
Va
-
Va= applied armature voltage
Vb= back EMF
Ra= armature resistance
ia= armature current
La= armature inductance
270
DC Motors & Generator
+
Ra
180
► Note
Magnetic field vector, B
Assuming B, v, and
dL are mutually
perpendicular,
90
Angular Position
How does a DC Generator work?
+
S
N
100
Torque
Four segment
commutation on
rotor
Va= applied armature voltage
Ra= armature resistance
Vb
-
Vb= back EMF
ia= armature current
DC Motors
26.3
PMDC Motor SteadySteady-State
Equations
ia =
1
Ra
(
)
from circuit
PMDC Motor SteadySteady-State
Equations
► For
a given motor, Ra, Ka, and Kb are
_________________
► Armature
Vb =
from dV= B v dL and v = rω
τ=
from df = ia dL X B and τ = rf
voltage Va, speed ω, and output
torque τ are related by the 3 equations
► If
InIn-Class Exercise #1
=
Ra
=
Plot for InIn-Class Exercise #1
10
48 volts
=
0.17 N-m/amp
=
0.17 volt/rad/s
0.9 ohms
Determine the output torque, τa, for the speed
assigned to your group
1) find back-EMF, Vb for your speed
2) find current, ia for your speed
3) find torque, τa for your speed
9
8
Torque, N-m
A small DC motor V a
has these
Ka
parameter
constants
Kb
we know 2 of these, can solve for 3rd one
7
6
5
4
3
2
1
0
0
500
1000
1500
2000
2500
3000
Speed, RPM
Manufacturer’s Data
2nd InIn-Class Exercise
A small DC motor V a
has these
Ka
parameter
constants
Kb
=
Ra
=
?
volts
=
3.60 oz-in/amp
=
2.67 volt/KRPM
50 ohms
On a single graph, we will plot the torque vs.
speed relationship for different input voltages 24, 18, 12, 6 VDC
DC Motors
InIn-Class Exercise #2
26.4
DC Tachometer Equations
2.00
Mechanical construction of a DC tachometer is
essentially identical to DC motor
1.80
Torque, oz-in
1.60
1.40
ia =
1.20
1.00
1
(Va − Vb ) = 0
Ra
High impedance load,
No current
0.80
Vb = k bω = Va
0.60
0.40
0.20
Output voltage proportional to
angular velocity, ω
τ = k aia = 0
0.00
0
No current
1000 2000 3000 4000 5000 6000 7000 8000 9000
Speed, RPM
PMDC Motor Equation Part #2
Unit Conversions
1V = 1
N ⋅m
→1 =
A⋅ s
Substitute into units for the back-EMF constant, kb,
kb :
Vb = k bω
Va = R a i a + Vb
V
N ⋅m
=
⋅
rad / s rad V ⋅ A ⋅ s
Va = R a i a + k bω
τ = k a ia
k b = k a →τ = k bia
Multiply both sides by ia
Conclusion: __________________________
Electrical Power = Power Dissipated + Mechanical Power
(Input)
(as heat)
(useful output)
PMDC Motor Equation Part #2
Va = R a i a + Vb
Vb = k bω
Va = R a i a + k bω
τ = k a ia
k b = k a →τ = k bia
PMDC Motor Equation Part #2b
Va ia = R a ia2 + τω
Electrical Power = Power Dissipated + Mechanical Power
(Input)
(as heat)
(useful output)
Multiply both sides by ia
Efficiency =
Electrical Power = Power Dissipated + Mechanical Power
(Input)
(as heat)
(useful output)
Power Out
Power In
DC Motors
Circuit Model for Permanent
Magnet DC Generator
PMDC Generator Equations
Vgen + R a ia = Vb
DC Motor/Generator
ia
+
Vgen
Rexternal
-
k b = k a →τ = k bia
-
PMDC Motor Equation Part #3
Torque, τ
τ = k a ia
Multiply both sides by ia
Vb
Vgen= generated armature voltage Vb= back EMF
Ra= armature resistance
ia= armature current
τ stall
Vb = k bω
R externalia + R a ia = k bω
+
Ra
26.5
ω =0
Va = R a i a,stall → τ stall = k a
constant Va
Speed, ω
At any point on load curve,
V − k bω
Ra = a
ia
ωNL=“no-load” speed (ia=0)
Va = k bω NL
DC Motor Commutation
► DC
motors require periodic switching of
currents to maintain rotation
(“commutation”)
conventional DC motors use ______________
to provide commutation, but
"brushless" DC motors which use
________________________ commutation
have been developed.
Va
Ra
Electrical Power + Power Dissipated = Mechanical Power
(useful output)
(as heat)
(Input)
DC Motors
DC Brushed Motor Advantages
► _________________________________,
_________________________________,
requiring only a voltage source, power opopamp, and analog control input for variable
speed operation.
► __________________________________
through design variations
► __________________________________
without additional power input
26.6
DC Brushed Motor Disadvantages
► ____________________________,
____________________________,
the wear
producing small particles which can affect the
cleanliness of surrounding operations.
► High current through the brushes can cause them
to burn out rapidly
► _________________________________________
which is primarily conducted away through the
rotor shaft
► _______________________ are generated at the
brush/rotor interface
DC Brushless Motor
DC Brushless Motor
► The
magnetic field in the rotor is provided
by permanent magnets on the ___________
Permanent Magnet Rotor
► Hall
effect sensors (or resolver output) are
used to signal a motor driver when to switch
the current in the ___________ windings
► Motor
driver depends on the controller to
set desired torque output
Wound Wire Stator
DC Brushless Motor
Disadvantages
DC Brushless Motor Advantages
► _________________________________________
and hence the heat conducted to the shaft is
minimized.
► Due to the lack of brushes, motors can be
operated at high torque and zero rpm indefinitely
as long as the winding temperature does not
exceed the limit.
► ________________________________________
or contaminate the surroundings
► Torque
ripple ____________________________
by design
► Motor operation requires the purchase of an
_______________________________________
► Rotor magnets can become demagnetized in high
current or temperature environments
► Most motor drivers brake DC brushless motors by
applying reverse current, in which almost as much
power is expended to stop the motor as was
required to start it moving