Quadrature-Field Theory and Induction-Motor Action
• Single-phase induction motor cannot develop a rotating magnetic field
• Needs an “auxiliary” method
– That method is another (auxiliary) winding
ECE 441 1

Single-Phase Squirrel-Cage
Induction Motor
There are two
“Main Poles”
Squirrel-Cage Rotor
Single-Phase Mains Supply
ECE 441 2

Excite the Main Winding
Stator flux is produced across the air gap – as shown, it is increasing in the downward direction.
The squirrel-cage rotor responds with a mmf in the opposite (upward) direction.
ECE 441
Magnetic axis of the rotor is in line with the magnetic axis of the stator – no rotation!
3

Current “out of” the page
Rotor mmf develops in the upward direction
ECE 441
“Main” pole flux (Φ) increasing in the downward direction
Current “into” the page
4

Cause the rotor to turn clockwise
Rotor conductors cut through the main pole flux.
Current is induced in the rotor bars as shown, producing a magnetic flux perpendicular to the main pole flux. This is known as “Quadrature” flux.
ECE 441 5

The quadrature flux is sustained as the rotor conductors shift their positions – other conductors replace them.
ECE 441 6

Phase Relationship Between the
Direct and Quadrature Flux
The “speed” voltage is in phase with the flux that created it, and the flux due to current is in phase with the current that caused it. The instantaneous amplitudes of the direct and quadrature flux are shown above.
ECE 441 7

Resultant Flux
• Determine from
ECE 441 8

Resultant Flux Rotates CW
ECE 441 9

Phase-Splitting
Split-Phase Induction Motor
ECE 441 10

Provides “direct” flux
Start winding
Provides quadrature flux
Ensures phase difference between winding currents
ECE 441 11

Equivalent Circuit
ECE 441 12

Purpose of the “Phase-Splitter”
• Make the current in the Auxiliary Winding out of phase with the current in the Main Winding.
• This results in the quadrature field and the main field being out of phase.
• The locked-rotor torque will be given by
T lr
 k I I sp mw aw sin

   i mw
  i aw
ECE 441 13

Example 6-1
• The main and auxiliary windings of a hypothetical 120 V, 60 Hz, split-phase motor have the following locked-rotor parameters:
– R mw
=2.00 Ω
– R aw
=9.15 Ω
X mw
=3.50 Ω
X aw
=8.40 Ω
• The motor is connected to a 120 V system. Determine
ECE 441 14

Example 6-1 continued
• The locked-rotor current in each winding
Z mw

R mw
 jX mw
Z aw

R aw
 jX aw

2.00
 j 3.50

4.0311 60.2511

9.15
 j 8.40

12.4211 42.553

ECE 441 15


Example 6-1 continued
I
I m w aw

V
T

Z m w

V
Z
T aw

120 0

4.0311 60.2511

120 0

12.4211 42.5530



29.8

60.3

A
9.66

42.6

A
ECE 441 16

Example 6-1 continued
• The phase displacement angle between the main and auxiliary currents
   i mw
  i aw
 
60.3
  
42.6
 
17.7

ECE 441 17

Example 6-1 continued
• The locked-rotor torque in terms of the machine constant
T lr
 k I I sp mw aw sin

T lr
 k sp
(29.8)(9.66) sin 17.7
 
87.52
k sp
ECE 441 18

Example 6-1 continued
• External resistance required in series with the auxiliary winding in order to obtain a 30
 phase displacement between the currents in the two windings.
ECE 441 19

Example 6-1 continued
• Phasor diagram for the new conditions

'
 
60.3
 
30
  
30.3
 i aw
ECE 441 20

Example 6-1 continued
I aw
'

Z aw
'

V
T
'

Z aw

30.3

I aw
'

30.3
 
V
Z aw
'
T
Z ' aw

R x

R aw
 jX aw

0

'
Z aw
ECE 441 21

Example 6-1 continued t an
R x
R

Z
' aw

X aw
R aw

R x

X t an aw

Z aw
'

R aw

8.40
t an 30.3


9.15

14.38

9.15

5.23

ECE 441 22

Example 6-1 continued
• Locked-rotor torque for the condition in d
T lr
 k I I sp m w aw sin

I aw
I aw
'

I ' aw

V
T
'

Z aw

9.15
7.2

30.29


120 0
5.23

 j 8.40
T lr
 k sp
(29.8)(7.2) sin 30
 
107.1
k sp
T lr

107.1
k sp
ECE 441 23

Example 6-1 continued
• % increase in locked-rotor torque due to the adding of additional resistance
107.1
k sp

87.52
k sp X 100%

22.37%
87.52
k sp
ECE 441 24