1. DC Machines

2. DC Machines
A DC Machine
Armature along with
the commutator

3. Significant Features of DC Machines
• Conventional DC generators are being replaced by the
solid state rectifiers where ac supply is available.
• The same is not true for dc motors because of
– Constant mechanical power output or constant torque
– Rapid acceleration or deceleration
– Responsiveness to feedback signals
• 1W to 10,000 hp
• Applications – in electric vehicles to extend their range
and reduce vehicle weight, in steel and aluminum rolling
mills, traction motors, electric trains, overhead cranes,
control devices, etc.

4. Introduction
Electromagnetic Energy Conversion:
1. When armature conductors move in a magnetic field produced
by the current in stator field winding, voltage is induced in the
armature conductors.
2. When current carrying armature conductors are placed in a
magnetic field produced by the current in stator field winding,
the armature conductors experience a mechanical force.
These two effects occur simultaneously in a DC machine
whenever energy conversion takes place from electrical to
mechanical or vice versa.

5. Electromagnetic Force, f
f=Bli, where B, f and i are mutually perpendicular. Turn
the current vector i towards the flux vector B. If a right
hand screw is turned in the same way, the direction in
which the screw will move represents the direction of the
force f.

6. Motional Voltage, e
e=Blv, where B, v and e are mutually perpendicular. The
polarity of the induced voltage can be determined from
the right hand screw rule. Turn the vector v towards the
vector B. If a right hand screw is turned in the same way
the motion of the screw will indicate the direction of
positive polarity of the induced voltage e.

7. Constructional Features of DC Machines
• Commutator along with the armature
on the rotor
• Salient-pole on the stator and, except
for a few smaller machines,
commutating poles between the main
poles.
• Field windings (as many as 4):
– Two fields that act in a corrective
capacity to combact the detrimental
effects of armature reaction, called
the commutating (compole or
interpole) and compensating
windings, which are connected in
series with the armature.
– Two normal exciting field
windings, the shunt and series
windings

8. Schematic Connection Diagram of a DC Machine

9. Equivalent Circuit of a DC Machine
aaat
fff
RIEV
RIV
±=
=
Ia_gen
If
Vf Vt
Rf
+
- Ea
+
-
Ia_mot
Ra
+
Ia
If
Vt
Rf
Ea
-
IL
Ra
+
+
-

10. Generated emf and Electromagnetic Torque
aaat
fff
RIEV
RIV
±=
=
mdaa KE ωφ=
adae IKT φ=
meaaem TIEP ω==
Voltage generated in the armature circuit due the flux of the stator field current
Electromagnetic torque
Ka: design constant
Motor: Vt > Ea
Generator: Vt > Ea

11. Comparison between the Shunt and Series Connected DC Machines

12. Armature Reaction
If a load is connected to the terminals of the dc
machine, a current will flow in its armature windings.
This current flow will produce a magnetic field of its
own, which will distort the original magnetic field from
the machine’s field poles. This distortion of the magnetic
flux in a machine as the load is increased is called the
armature reaction.

13. Types of DC Machines
Both the armature and field circuits carry direct current in the case
of a DC machine.
Types:
Self-excited DC machine: when a machine supplies its own
excitation of the field windings. In this machine, residual
magnetism must be present in the ferromagnetic circuit of the
machine in order to start the self-excitation process.
Separately-excited DC machine: The field windings may be
separately excited from an eternal DC source.
Shunt Machine: armature and field circuits are connected in parallel.
Shunt generator can be separately-excited or self-excited.
Series Machine: armature and field circuits are connected in series.

14. Separately-Excited and Self-Excited DC Generators
If
IL
If
DC Supply Vt
Rf
+
- Ea
+
-
Ra
+
Ia
Vt
Rf
Ea
-
IL
Ra
+
Separately-Excited Self-Excited

15. Example 1
A 100-kW, 250-V DC shunt generator has an
armature resistance of 0.05 Ω and field circuit
resistance of 60 Ω. With the generator operating at
rated voltage, determine the induced voltage at (a) full
load, and (b) half-full load.

16. Solution to Example 1
(a) At full load,
Vt=Ea-IaRa
If=250/60=4.17 A
IL_FL=100,000/250=400 A
Ia=IL_FL+If=400+4.17=404.17 A
Ea=Vt+IaRa=250+404.17*0.05=270.2 V
(b) At half load,
If=250/60=4.17 A
IL_HL=50,000/250=200 A
Ia=IL_HL+If=200+4.17=204.17 A
Ea=Vt+IaRa=250+204.17*0.05=260.2 V

17. DC Generator Characteristics
In general, three characteristics specify the steady-state
performance of a DC generators:
1. Open-circuit characteristics: generated voltage versus field
current at constant speed.
2. External characteristic: terminal voltage versus load current
at constant speed.
3. Load characteristic: terminal voltage versus field current at
constant armature current and speed.

18. DC Generator Characteristics
Open-circuit and load characteristics
The terminal voltage of a dc
generator is given by
( )[ ]
aa
mf
aaat
RI
dropreactionArmatureIf
RIEV
−
−=
−=
ω,

19. DC Generator Characteristics
It can be seen from the external
characteristics that the terminal
voltage falls slightly as the load
current increases. Voltage regulation
is defined as the percentage change
in terminal voltage when full load is
removed, so that from the external
characteristics,
External characteristics
100
V
VE
regulationVoltage
t
ta
×
−
=

20. Self-Excited DC Shunt Generator
Schematic diagram of connection
Open-circuit characteristic
Maximum permissible value of the field
resistance if the terminal voltage has to build
up.

21. Speed Control in DC Motors
© N. Chowdhury of U of Saskatchewan

22. Speed Control in DC Motors
Shunt motor:
Electromagnetic torque is Te=Ka φd Ia, and the conductor emf is Ea=Vt - RaIa.
For armature voltage control: Ra and If are constant
For field control: Ra and Vt are constant
For armature resistance control: Vt and If are constant
( )221 etm TKVK −=ω
( )
( )32 e
ff
a
ff
t
m T
IK
R
IK
V
−=ω
( )
( )12
da
ae
da
t
m
a
da
e
tmda
K
RT
K
V
R
K
T
VK
φ
−
φ
=ω






φ
−=ωφ
( )
( )42 e
da
adja
da
t
m T
K
RR
K
V
φ
+
−
φ
=ω

23. Speed Control in Shunt DC Motors
Armature Voltage Control:
Ra and If are kept constant and the armature
terminal voltage is varied to change the motor
speed.
For constant load torque, such as applied by an
elevator or hoist crane load, the speed will
change linearly with Vt. In an actual
application, when the speed is changed by
varying the terminal voltage, the armature
current is kept constant. This method can also
be applied to series motor.
( )
.constis;
K
K;
K
K
TKVK
d
dada
etm
φ
φ
=
φ
=
−=ω
221
21
11

24. Field Control:
Ra and Vt are kept constant, field rheostat is varied to
change the field current.
For no-load condition, Te=0. So, no-load speed varies
inversely with the field current.
Speed control from zero to base speed is usually
obtained by armature voltage control. Speed control
beyond the base speed is obtained by decreasing the
field current. If armature current is not to exceed its
rated value (heating limit), speed control beyond the
base speed is restricted to constant power, known as
constant power application.
Speed Control in Shunt DC Motors
mm
aa
e
meaaat
.constIE
T
TIEconstIVP
ω
=
ω
=
ω====
( ) e
ff
a
ff
t
m T
IK
R
IK
V
2
−=ω

25. Armature Resistance Control:
Vt and If are kept constant at their rated value,
armature resistance is varied.
The value of Radj can be adjusted to obtain
various speed such that the armature current Ia
(hence torque, Te=KaφdIa) remains constant.
Armature resistance control is simple to
implement. However, this method is less
efficient because of loss in Radj. This resistance
should also been designed to carry armature
current. It is therefore more expensive than the
rheostat used in the field control method.
Speed Control in Shunt DC Motors
( )
ee
da
adja
da
t
m TKKT
K
RR
K
V
652
−=
φ
+
−
φ
=ω

26. Armature Voltage Control:
A variable dc voltage can be applied to a series motor
to control its speed. A variable dc voltage can be
obtained from a power electronic converter.
Torque in a series motor can be expressed as
Speed Control in Series DC Motors
( )[ ]
sae
t
sa
sa
sae
t
m
samsa
tsa
asaadae
KKT
V
KK
RR
KKT
V
,or
RRKK
VKK
IKKIKT
≈
+
−=ω
++ω
=
=φ=
2
2
2
( )
( )
( ) ( )
samsa
t
a
saamasa
saamda
saaat
asd
RRKK
V
I
RRIIKK
RRIK
RRIEV
IK
++ω
=
++ω=
++ωφ=
++=
=φ

27. Field Control:
Control of field flux in a sries motor is achieved by
using a diverter resistance.
The developed torque can be expressed as.
Speed Control in Series DC Motors
ds
d
sa
aa
ds
d
saadae
RR
R
andKKK,where
IKI
RR
R
KKIKT
+
=ρ=
ρ=





+
=φ= 22
( ) ( )
( )
asm
t
a
aasm
aasmasa
aasamda
aaa
ds
ds
at
RRK
V
I,or
IRRK
IRRIKK
RIRIK
RII
RR
RR
EV
+ρ+ρω
=
+ρ+ρω=
+ρ+ωρ=
+ρ+ωφ=
+





+
+=

28. Speed Control in Series DC Motors
2






+ρ+ρω
ρ=
asm
t
e
RRK
V
KT

29. Armature Resistance Control:
Torque in this case can be expressed as
Rae is an external resistance connected in series with
the armature.
For a given supply voltage and a constant developed
torque, the term (Ra+Rae+Rs+Kωm) should remain
constant. Therefore, an increase in Rae must be
accompanied by a corresponding decrease in ωm.
Speed Control in Series DC Motors
( )
K
RRR
KT
V
,or
V
T
K
KRRR,or
T
KV
KRRR
sadja
e
t
m
t
e
msadja
e
t
msadja
++
−=ω
=ω+++
=ω+++
2
2
( )2
2
msadja
t
e
KRRR
KV
T
ω+++
=

30. Power Division in DC Machines
Input from
prime-mover
Elec-magnetic
Power =EaIa
Arm. terminal
power = Vta Ia
Output power
= Vt IL
No-load rotational loss (friction
+windage+core)+stray load loss
Arm. copper loss
Ia
2
Ra+brush contact loss
Series field loss IL
2
Rs
+shunt field loss If
2
Rf
Input power from
mains =Vt IL
Elec-magnetic
Power =EaIa
Arm. terminal
power = Vta Ia
Output available
at the shaft
No-load rotational loss (friction
+windage+core)+stray load loss
Arm. copper loss
Ia
2
Ra+brush contact loss
Series field loss IL
2
Rs
+shunt field loss If
2
Rf
DC Motor
DC Generator

31. Efficiency
InputPower
Losses
InputPower
LossesInputPower
InputPower
OutputPower
−=
−
=
=η
1
The losses are made up of rotational losses (3-15%), armature
circuit copper losses (3-6%), and shunt field copper loss (1-5%).
The voltage drop between the brush and commutator is 2V and
the brush contact loss is therefore calculated as 2Ia.

32. DC Machines Formulas

33. Problem 9-1 to 9-7 (Page 621)

34. Solution to Problem 9-1 (Page 621)

35. Solution to Problem 9-2 (Page 621)

36. Solution to Problem 9-5 (Page 621)

37. Problem 9-13 (Page 623)

38. Solution to Problem 9-13 (Page 623)

39. Solution to Problem 9-13 (Page 623)

40. The End