Lecture 32
DC Motors Part II
Learning Objectives

Analyze the circuit equivalent of a permanent magnetic
DC motor that accounts for armature resistance, induced
electromotive force (back EMF), developed
electromagnetic torque, and applied (input) voltage.

Define the power output of a permanent magnetic DC
motor in terms of developed electromagnetic torque and
angular velocity. Relate power output in terms of horse
power.

Determine the efficiency of a permanent magnetic DC
motor using the given or calculated power in and power
out.
Basic DC Motor Operations
Parts of a DC Motor
SOURCE: Gears Educational Systems.
Rotary DC Motor
 , TLOAD

Torque developed
Td=Kv Ia

Power Developed
Pd = Td ω = Kv Ia ω

Back EMF
Ea = Kv ω

Angular velocity
ω = 2  (RPM/60)
DC motor power flow
Pelec loss
Pmech loss
Pin
Pout
Pdev = Pout + Pmech loss
Electrical Power losses

Electrical loss occurs due to the armature
resistance and is expressed as
Pelec loss = Ia2 Ra.
Pelec loss
Torque losses (Tloss)

Pmech loss represents losses due to the friction of
mechanical parts, magnetic inefficiencies of
the material, and losses coupling brushes
and commutator and is expressed as
Pmech loss = Tloss m
Pmech loss
Output Power

Pd represents the power developed by the motor which
includes power out and mechanical losses (Ploss). It is
expressed
Pd = Pmech loss + Pout = Tdev  = Kv IA m

Power out is the power that ultimately gets to the load
and is expressed Pout = TLOADm
Pout
Pd = Pout + Pmech loss
Power Conversion Diagram
Electrical
Mechanical
Pd = Ea I a = Td 
Motor Efficiency

Developed power is:
Pd = Ea I a = Pout  Pmech loss

If we Ignore rotational losses, Pd=Pout, and machine
efficiency can be calculated as:
Pout
Ea I a
PD
=
100 = 100 =
100
Pin
Pin
VDC I a
Ea
=
(Tloss = 0)
VDC
Magnetic field



Instead of permanent magnet, we could raise
the field strength B with an electromagnet.
The wires wrapped around a ferromagnetic core
are known as field windings.
The field windings are stationary and are part of
the stator.
Magnetic field
2 poles
Magnetic poles

Increasing the number of poles will increase and
smooth the output torque.
four-pole dc machine
eight-pole dc machine
Example Problem 1
A 24 V DC motor is rated for 15 A.
RA = 0.20 Ω
Assumed no rotational losses.
Determine:
a) The input power
b) The power loss due to the resistance of the armature
c) The power developed
d) The back EMF (EA)
e) The efficiency assuming no mechanical power loss
f) Draw a power conversion diagram and fill-in the values
for power in, electrical power loss, power developed,
mechanical power loss and power out.
Example Problem 2
A permanent magnet DC motor is rated for 120V, 17A and
1200 rpm.
The machine is 90% efficient at rated conditions
Tloss = 0.0334 N·m
Find Ra and Kv and torque developed by the motor .
Example Problem 3
We wish to design a 1/4 hp, 28 V DC motor with an
efficiency of 96%. What current can we expect to draw? If
the machine constant is Kv= 0.2139 ν·s, determine Tout if we
ignore mechanical losses. Calculate rated speed in rpm.
Why is there so much field wiring on
the Practical Exercise?
FIELD
DC Machine
Excitation Supply
Field Rheostat
1
T
T
2
1
10
0 - 150 V. DC-1A max
+
Shunt Field
1
T
T
T
2
Armature
1
T
T
Voltage Adjust
DC Current
T
2
1
10
0 - 125 V. DC-5A max
Series Field
1
T
T
2
+
T
ARMATURE
T
+
-
T
T
PE 20 Diagram
FIELD
ARMATURE
 , TLOAD