ECE 441 1

Quadrature-Field Theoryand 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 2

Single-Phase Squirrel-Cage Induction Motor

There are two “Main Poles”

Squirrel-Cage Rotor

Single-Phase Mains Supply

ECE 441 3

Excite the Main WindingStator 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.

Magnetic axis of the rotor is in line with the magnetic axis of the stator – no rotation!

ECE 441 4

“Main” pole flux (Φ) increasing in the downward direction

Rotor mmf develops in the upward direction

Current “into” the page

Current “out of” the page

ECE 441 5

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 6

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

ECE 441 7

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 8

Resultant Flux

• Determine from

ECE 441 9

Resultant Flux Rotates CW

ECE 441 10

Phase-SplittingSplit-Phase Induction Motor

ECE 441 11

Provides “direct” flux

Provides quadrature flux

Ensures phase difference between winding currents

Start winding

ECE 441 12

Equivalent Circuit

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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

sin

mw aw

lr sp mw aw

i i

T k I I

ECE 441 14

Example 6-1

• The main and auxiliary windings of a hypothetical 120 V, 60 Hz, split-phase motor have the following locked-rotor parameters:– Rmw=2.00 Ω Xmw=3.50 Ω

– Raw=9.15 Ω Xaw=8.40 Ω

• The motor is connected to a 120 V system. Determine

ECE 441 15

Example 6-1 continued

• The locked-rotor current in each winding

2.00 3.50 4.0311 60.2511

9.15 8.40 12.4211 42.553mw mw mw

aw aw aw

Z R jX j

Z R jX j

ECE 441 16

120 029.8 60.3

4.0311 60.2511120 0

9.66 42.612.4211 42.5530

T

mw

mw

T

aw

aw

VI A

ZV

I AZ

Example 6-1 continued

ECE 441 17

Example 6-1 continued

• The phase displacement angle between the main and auxiliary currents

60.3 42.6 17.7mw awi i

ECE 441 18

Example 6-1 continued

• The locked-rotor torque in terms of the machine constant

sin

(29.8)(9.66)sin17.7 87.52

lr sp mw aw

lr sp sp

T k I I

T k k

ECE 441 19

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 20

Example 6-1 continued

• Phasor diagram for the new conditions

' 60.3 30 30.3awi

ECE 441 21

Example 6-1 continued

' '

' '

'

'

030.3

'

30.3aw

aw

T T

aw aw

aw aw Z

Z

aw x aw aw

V VI I

Z Z

Z R R jX

ECE 441 22

Example 6-1 continued

'

'

tan

tan

8.409.15 14.38 9.15 5.23

tan30.3

aw

aw

aw

Z

aw x

aw

x aw

Z

XR R

XR R

R

ECE 441 23

Example 6-1 continued

• Locked-rotor torque for the condition in d

'

'

'

sin

120 09.15 5.23 8.40

7.2 30.29

(29.8)(7.2)sin30 107.1

107.1

lr sp mw aw

T

aw aw

aw

aw

lr sp sp

lr sp

T k I I

VI I

Z jI

T k k

T k

ECE 441 24

Example 6-1 continued

• % increase in locked-rotor torque due to the adding of additional resistance

107.1 87.52100% 22.37%

87.52sp sp

sp

k kX

k