ECE 441 1

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 2

Circuit Diagram of a Compound Motor

ECE 441 3

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 4

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 5

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 6

Reversing the Armature Current

ECE 441 7

Using NEMA standard terminal markings

ECE 441 8

Series Motor

• Series field– Heavy windings – Must conduct the armature current

• Potentially dangerous problem if the shaft load is removed!

ECE 441 9

Field winding is in series with the armature

ECE 441 10

More Details

• When shaft load is removed, TD>Tload

– Motor speed increases– cemf increases– armature current decreases– series field flux decreases

ECE 441 11

Reversing the Direction of a Series Motor

• Reverse the current in the armature-interpole-compensating branch

• Reverse the current in the series field windings

ECE 441 12

Reversing the Armature Current

ECE 441 13

Using NMEA standard Terminal Markings

ECE 441 14

Using NEMA standard terminal markings

Reversing the series field

ECE 441 15

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– Fnet = Ff + Fs - Fd

– Fnet = net mmf (A-t/pole)

– Ff = shunt field mmf (NfIf)(A-t/pole)

– Fs = series field mmf (NsIa)(A-t/pole)

– Fd = equivalent demagnetizing mmf due to armature reaction (A-t)/pole

ECE 441 16

Effect of Magnetic Saturation on DC Motor Performance

• Note that Fd is not exactly proportional to the armature current, but is assumed to be.

• If a compensating winding is used, Fd = 0.

ECE 441 17

Developed Torque and Speed

0p

D p a M

T a acir

p G

acir a IP CW s

T B I k

V I Rn

k

R R R R R

ECE 441 18

Defining Parameters

• Racir = resistance of armature circuit (Ω)• Ra = resistance of armature windings (Ω )• RIP = resistance of interpole windings (Ω)• RCW = resistance of compensating

windings (Ω)• Rs = resistance of series field winding (Ω)• Bp = air-gap flux density (T)• Φp = pole flux (Wb)

ECE 441 19

Solve Problems with Proportions

11

2 2

1 1

2

2

1

2 21

[ ]

[ ]

, 0

p aD

D p a

T a acir

p G

T a acir

p G

p p

pT a acir

p T a acir

B IT

T B I

V I R

kn

n V I R

k

B A

BV I Rn

n B V I R

ECE 441 20

Example 11.1

• A 240-V, 40-hp, 1150 r/min stabilized-shunt motor, operating at rated conditions, has an efficiency at rated load of 90.2%. The motor parameters are

• Ra = 0.0680 Ω RIP = 0.0198 Ω Rs = 0.00911 Ω Rshunt = 99.5 Ω

• Turns/pole series - ½ shunt - 1231

ECE 441 21

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 22

ECE 441 23

40 746

0.902 240

137.84

2402.4121

99.5

137.84 2.41 135.43

T T TT

T

Tf

f

a T f

PP V I I

V

I A

VI A

R

I I I A

Solution for Armature Current

ECE 441 24

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.

1231 2.412 2969.2 /net f f fN I A t pole F F

ECE 441 25

net mmf = 0.70 T

ECE 441 26

1

2 21

212 1

2 1

2

2

0.0680 0.0198 0.0091 0.0969

[ ]

[ ]

1150 240 1.15 135.43 0.09690.70

1.25 1150 240 135.43 0.0969

acir a IP s

acir

pT a acir

p T a acir

T a acirp p

T a acir

p

p

R R R R

R

BV I Rn

n B V I R

V I RnB B

n V I R

B

B

0.56T

ECE 441 27

Ff = 2.3 X 1000 = 2300 A-t/pole

ECE 441 28

2 2

2300

1231

1.87

24099.5 28.8

187

(1.87) 28.8 100.7x

ff f f f

f

f

T Tf x f

f x f

x

R f x

N I IN

I A

V VI R R

R R R

R

P I R W

FF

ECE 441 29

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 30

Approximate Equations for Torque and Speed

1 11

2 2 2

1

2 21

1

2 1 2

[ ] [ ]

[ ] [ ]

, 0

p a net aD

D p a net a

pT a acir

p T a acir

T a acir net

net T a acir

B I IT

T B I I

BV I Rn

n B V I R

V I Rn

n V I R

FF

FF

ECE 441 31

For the Series Motor

11,

2 2

2,

[ ]

[ ]net aD

D series net a s aD net a

D series a

ITT I N I

T I

T I

FF

F

If the range of operation is in the unsaturated region, and armature reaction effects are either negligible or compensated for,

The developed torque is proportional to the square of the armature current.

ECE 441 32

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 33

1

1

1

2 1 2

212 1

2 1

2

2402.412

99.5

137.84 2.412 135.43

1231 2.412 2969.2 /

[ ]

[ ]

11502969.2

1

Tf

f

a T f

net f f f

T a acir net

net T a acir

T a acirnet net

T a acir

net

VI A

R

I I I A

N I A t pole

V I Rn

n V I R

V I Rn

n V I R

F F

FF

F F

F

2

240 1.15 135.43 0.0969

.25 1150 240 135.43 0.09692354.8 /net A t pole

F

ECE 441 34

2354.81.91

1231

24099.5 26.15

1.91

netf

f

T Tf x f

f x f

x

I AN

V VI R R

R R I

R

F

From Example 11.1, the value of resistance was determined to be 28.8 Ω

ECE 441 35

% 100%

28.8 26.15% 100%

28.8% 9.2%

actual approx

actual

R Rerror

R

error

error

Calculate the Percent Error

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 36

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 37

ECE 441 38

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 39

Relatively Constant Speed

Linear Torque

ECE 441 40

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 41

Higher Torque above base speed than Shunt motor

Lower Speed at Higher Torque

ECE 441 42

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 43

Wide Speed Range

High Starting Torque

ECE 441 44

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 45

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 46

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 47

Compound Motor Example

Normal Operation

Dynamic Braking

ECE 441 48

Normal Operation

Closed

Open

ECE 441 49

Dynamic - Braking

Open

Closed

ECE 441 50

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 51

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 52

Normal Operation

ECE 441 53

Plugging

ECE 441 54

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 55

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 0.00707Ω, 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 56

Circuit for Dynamic Braking

ECE 441 57

1 1 1

1

910 1150199.257

5252 5252

199.257 746158134

0.940

158134658.89

240240

4.5652.6

658.89 4.56 654.33

240 654.33 0.00707

235.37

shaft

shaftin

in T T T

Tf

f

a T f

T a a acir a

a

T nP hp

PP W

P V I I A

VI A

R

I I I A

V E I R E

E V

ECE 441 58

1 1 1 212

2 2 2 1

2

11 1 22 1

2 2 2 1

2

2 22 2

2

[ ]

[ ]

654.33 500359.52

910[ ]

[ ]

1000235.37 204.67

1150

( )

204.67 359.52 0.00707

359.52

p a a aa

p a a

a

p Gaa a

a p G

a

a a acira a acir DB DB

a

DB

B I I I TTI

T B I I T

I A

n kE n nE E

E n k n n

E V

E I RE I R R R

I

R

0.562

0.562DBR