Direct – Current Motor
Characteristics and Applications
• Straight Shunt Motor
– Essentially a constant speed motor
• Compound or Stabilized – Shunt Motors
– Has both shunt and series field windings
– Series field generates mmf in the same direction as the shunt field mmf.
ECE 441 1

Circuit Diagram of a Compound Motor
ECE 441 2

Differential Connection of Fields
• Both the series and shunt fields must provide fluxes that are additive.
• If the series field is reversed with respect to the shunt field, the net flux decreases, and the speed increases.
• The time constant of the series field is such that the current increases faster than the shunt field current.
ECE 441 3

Differential Connection of Fields
• If the series field is reversed
,
– The motor will start in the wrong direction
– Depending upon the load and the structure of the series field, the motor could
• slow down and stop, tripping the breaker
• slow down, stop, reverse direction, and accelerate
• slow down, stop, reverse direction, slow down, stop, reverse direction, etc. until a breaker trips
ECE 441 4

Reversing the Direction of
Compound Motors
• Reverse either the armature current or reverse both the series and shunt fields.
– If only one field is reversed, a “differential” connection results!
– The field mmfs will be reduced, resulting in excessive speed!
ECE 441 5

Reversing the Armature Current
ECE 441 6

Using NEMA standard terminal markings
ECE 441 7

Series Motor
• Series field
– Heavy windings
– Must conduct the armature current
• Potentially dangerous problem if the shaft load is removed!
ECE 441 8

ECE 441
Field winding is in series with the armature
9

More Details
• When shaft load is removed, T
D
>T load
– Motor speed increases
– cemf increases
– armature current decreases
– series field flux decreases
ECE 441 10

Reversing the Direction of a Series Motor
• Reverse the current in the armatureinterpole-compensating branch
• Reverse the current in the series field windings
ECE 441 11

Reversing the Armature Current
ECE 441 12

Using NMEA standard
Terminal Markings
ECE 441 13

Using NEMA standard terminal markings
Reversing the series field
ECE 441 14

Effect of Magnetic Saturation on
DC Motor Performance
• Pole flux is not directly proportional to the applied mmf due to magnetic saturation
• Net mmf is made up of the following components, as applicable
– F net
– F net
= F f
+ F s
- F d
= net mmf (A-t/pole)
– F f
– F s
= shunt field mmf (N f
I f
)(A-t/pole)
= series field mmf (N s
I a
)(A-t/pole)
– F d
= equivalent demagnetizing mmf due to armature reaction (A-t)/pole
ECE 441 15

Effect of Magnetic Saturation on
DC Motor Performance
• Note that F d is not exactly proportional to the armature current, but is assumed to be.
• If a compensating winding is used, F d
= 0.
ECE 441 16

Developed Torque and Speed
T
D

B I k p a M n

V
T

I R a acir
 k p G
0
R acir

R a

R
IP

R
CW

R s
ECE 441 17

Defining Parameters
• R acir
• R a
= resistance of armature circuit ( Ω)
= resistance of armature windings (Ω )
• R
IP
• R
CW
= resistance of interpole windings (Ω)
= resistance of compensating windings (Ω)
• R s
• B p
• Φ p
= resistance of series field winding (Ω)
= air-gap flux density (T)
= pole flux (Wb)
ECE 441 18

Solve Problems with Proportions
T
D 1
T
D 2

[ B I p a
]
[ B I p a
]
2
1




V
T


I R a acir n n
1
2

V
T

 k p G
I R a acir k p G




 
1
2
  p
B p

A
,
 
0 n n
1
2



V
T

I R a acir
B p
B
 

1 p
V
T

I R a acir


2
ECE 441 19

Example 11.1
• A 240-V, 40-hp, 1150 r/min stabilizedshunt motor, operating at rated conditions, has an efficiency at rated load of 90.2%.
The motor parameters are
• R
R a s
= 0.0680 Ω R
IP
= 0.0198 Ω
= 0.00911 Ω R shunt
= 99.5 Ω
• Turns/pole series - ½ shunt - 1231
ECE 441 20

Example 11.1 (continued)
• The circuit diagram and magnetization curve are shown on the next slide.
Determine (a) the armature current when operating at rated conditions; (b) the resistance and power rating of an external resistance required in series with the shunt field in order to operate at 125% rated speed. Assume the shaft load is adjusted to a value that limits armature current to
115% of rated current.
ECE 441 21

ECE 441 22

Solution for Armature Current
P

V I
T T

I
T

P

V
T
I
T

137.84
A

I f

V
R
T f

240
99.5
I a

I
T

I f


2.4121
A
ECE 441
A
23

Solution for External Resistance
• The series field of a compound motor is designed to be approximately equal and opposite to the equivalent demagnetizing mmf of armature reaction. Therefore, the net flux is due to the shunt field alone.
F net
 F f

N I f f
  
2969.2

/
ECE 441 24

net mmf = 0.70 T
ECE 441 25

R acir

R a

R
IP

R s
R acir
  
0.0969
 n n
1
2



V
T

I R a acir
B p
B
 

1 p
V
T

I R a acir


2
B p 2

B p 1
 n
1
 n
2
[
[
V
V
T
T


I R a acir
I R a acir
]
2
]
1
B p 2

0.70

1150


B p 2

0.56
T
ECE 441 26

ECE 441
F f
= 2.3 X 1000 = 2300 A-t/pole
27

F
I f f

N I f f

I f

F
N f f


1.87
A
2300
1231
I f

V
T
R f

R x

R x

V
R
T f

R f
R x

240

187
99.5

28.8

R x
 2
P I R f x

(1.87)
2  
W
ECE 441 28

Linear Approximations
• If the magnetization curve is not available
– rough approximation obtained by assuming magnetization effects are negligible
– Do not use approximations if the motor is operating under heavy overload or locked rotor conditions.
• If the net mmf is to be reduced below its rated value, approximation using the linear assumption is OK.
ECE 441 29

Approximate Equations for
Torque and Speed
T
D 1
T
D 2 n n n n
1
2
1
2

[
[ B I p a
B I p a
]
]
1
2

[
[ F
F
I net a
I net a
]
]
1
2



V
T

I R a acir
B p
B
 

1 p
V
T

I R a acir


2




V
T

I R a acir
F net
 
F net
V
T

I R a acir


2
,
 
0
ECE 441 30

For the Series Motor
If the range of operation is in the unsaturated region, and armature reaction effects are either negligible or compensated for,
T
D 1
T
D 2
T

[ F I net a
]
1
[ F I net a
]
2

I a
2

T
 F I net a

N I s a
The developed torque is proportional to the square of the armature current.
ECE 441 31

Example 11.2
• Example 11.1 is re-solved using the linear approximation, and the solution is compared to the results obtained in
Example 11.1.
ECE 441 32

I f 1

V
R
T f

240
99.5
I a 1

I
T

I f

F net
 F f

N I f f

2.412
A
  
A
2969.2

/ n n
1
2



V
T

I R a acir
F net
 
V
T

F net
I R a acir
F net 2
 F net 1
 n n
1
2

[
[
V
V
T
T


I R a acir
I R a acir
]
2
]
1


2
F net 2
F net 2



1150
2969.2
1

2354.8

/

ECE 441 33

I f

F
N net
 f
2354.8

1.91
A
1231
I f

V
T
R f

R x

R x

V
I
T f

R f
R x

240
1.91

99.5

26.15

From Example 11.1, the value of resistance was determined to be 28.8 Ω
ECE 441 34

Calculate the Percent Error
% error

R actual

R approx 
100%
R actual
% error
% error

28.8

9.2%

100%
This lower value of resistance would cause a slightly higher field current, and therefore, a speed slightly lower than 1437.5 r/min.
ECE 441 35

Comparison of Steady – State
Operating Characteristics of DC Motors
• The steady-state operating characteristics of typical shunt, compound, and series motors of the same torque and speed ratings are shown on the next slide.
ECE 441 36

ECE 441 37

Comparisons (continued)
• Shunt Motor
– relatively constant speed from no-load to full-load
– does not have high starting torque
– essentially constant flux
– torque varies linearly with armature current
– speed regulation around 5%
ECE 441 38

Relatively
Constant Speed
ECE 441
Linear Torque
39

Comparisons (continued)
• Compound Motor
– Higher torque, lower speed than shunt motor
– speed regulation between 15 and 25%
– used with loads requiring high starting torques or have pulsating loads
• smoothes out the energy required by the pulsating load, lowering the demand on the electrical supply
ECE 441 40

Lower Speed at
Higher Torque
ECE 441
Higher Torque above base speed than Shunt motor
41

Comparisons (continued)
• Series Motor
– high starting torque
– wide speed range
– REMOVING THE LOAD CAUSES IT TO RUN
AWAY!
• CONNECT LOAD BY GEARS OR SOLID
COUPLING – NO BELT DRIVES!
ECE 441 42

Wide Speed Range
ECE 441
High Starting Torque
43

Dynamic Braking, Plugging, and Jogging
• Dynamic Braking is the deceleration of the motor by converting the energy stored in the moving masses into electrical energy and dissipating it as heat via resistors.
Also called resistive braking .
ECE 441 44

Dynamic Braking (continued)
• Disconnect the armature from the electrical supply lines and connect across a suitable resistor while maintaining the field at full strength.
• The motor behaves as a generator, feeding current to the resistor, dissipating heat.
ECE 441 45

Dynamic Braking (continued)
• Choose the resistance for current between
150 and 300% of rated current.
• The armature current is in a direction to oppose the armature motion, producing a negative, or, counter-torque, slowing down the load.
ECE 441 46

Compound Motor Example
Normal Operation
Dynamic Braking
ECE 441 47

Normal Operation
Closed
ECE 441
Open
48

Dynamic - Braking
Open
ECE 441
Closed
49

Regenerative Braking
• Convert energy of overhauling loads into electrical energy and pumps it back into the electrical system.
• The overhauling load drives a DC motor faster than normal, causing the cemf to become greater than the supply voltage and results in generator action.
• Trains, elevators, hybrid automobiles
ECE 441 50

Plugging
• The electrical reversal of a motor before it stops
• Reverse the voltage applied to the armature
• Current in the series and shunt fields is not reversed
• Insert resistance in series with the armature to limit the current
ECE 441 51

Normal Operation
ECE 441 52

Plugging
ECE 441 53

Jogging
• Very brief application of power to a motor
• Fraction of a revolution
• Used for positioning the load
• Place resistance in series with the armature to limit the current
ECE 441 54

Example 11.7
• A 240-V, compensated shunt motor driving a 910 lb-ft torque load is running at 1150 r/min. The efficiency of the motor at this load is 94.0%. The combined armature, compensating winding, and interpole resistance is 0.00707
Ω, and the resistance of the shunt field is 52.6Ω. Determine the resistance of a dynamic-braking resistor that will be capable of developing 500 lb-ft of braking torque at a speed of 1000 r/min. Assume windage and friction at 1000r/min are essentially the same as at 1150 r/min.
ECE 441 55

Circuit for Dynamic Braking
ECE 441 56

P shaft

T
 n
5252

P in

P shaft


5252
0.940

199.257
hp

158134 W
P in

V I
T T

I
T

158134
240

658.89
A
I f

V
R
T f

240
52.6

4.56
A
I a

I
T

I f

658.89 4.56

654.33
A
V
T

E a 1

I R a 1 acir
E a 1

235.37
V

E a 1
 
ECE 441 57

T
T
1
2

[
[ B I p a
B I p a
]
]
1
2

I

I a 2

I a 1 a 2
I a 1

T
2
T
1
I a 2
E a 1
E a 2


[
910 n
 k p G
]
1
[ n
 k p G
]
2

359.52
A
 n
1

E a 2
 n
2 n
2 n
1
E a 1
E
E a a
2
2

1000
1150

235.37

204.67
V

I a 2
( R acir

R
DB
)

R
DB

E a 2

I R a 2 acir
I a 2
R
DB

R
DB

0.562

359.52

0.562

ECE 441 58