7. How to identify the direction of rotation and how it can be reversed?
Rotating Magnetic Field in a Three-Phase AC Motor 2 / 10
8. Pulsating Magnetic Field
Oriented in space along the magnetic axis
Rotating Magnetic Field in a Three-Phase AC Motor 3 / 10
Axis of U
U
U’
Fig. 2. MMF vector of single-phase winding
ωt
π 2π
Im
0
-Im
A
Fig. 1. Winding current
9. Balanced three-phase winding of an AC motor
Axis of U
Axis of V
Axis of W
U
U’
W’
W
V
V’
Fig. 1. Relative location of magnetic axes
Fig. 2. Schematic diagram
of stator windings
U
V
W
Equal number of turns in each
winding Z
In space, the adjacent phases should
be displaced by 120◦
electrical
The mechanical angle displacement
θm =
2θe
p
The windings can be represented by
UU’= Z cos(θ)
VV’= Z cos(θ + 120◦
)
WW’= Z cos(θ − 120◦
)
Rotating Magnetic Field in a Three-Phase AC Motor 4 / 10
10. Balanced three-phase supply
Equal in magnitude
Time phase difference of 120◦
electrical
IA
IB
IC
iA = Im sin(ωt)
iB = Im sin(ωt − 120◦
)
iC = Im sin(ωt + 120◦
)
Rotating Magnetic Field in a Three-Phase AC Motor 5 / 10
ωt
π 2π
Im
0
-Im
A B C
Winding currents
11. Balanced three-phase supply
Equal in magnitude
Time phase difference of 120◦
electrical
IA
IB
IC
iA = Im sin(ωt)
iB = Im sin(ωt − 120◦
)
iC = Im sin(ωt + 120◦
)
Phase sequence - 1 Phase sequence - 2
ABC CBA
BCA ACB
CAB BAC
Rotating Magnetic Field in a Three-Phase AC Motor 5 / 10
ωt
π 2π
Im
0
-Im
A B C
Winding currents
12. Balanced three-phase supply
Equal in magnitude
Time phase difference of 120◦
electrical
IA
IB
IC
iA = Im sin(ωt)
iB = Im sin(ωt − 120◦
)
iC = Im sin(ωt + 120◦
)
Phase sequence - 1 Phase sequence - 2
ABC CBA
BCA ACB
CAB BAC
IA
IC
IB
iA = Im sin(ωt)
iB = Im sin(ωt + 120◦
)
iC = Im sin(ωt − 120◦
)
Rotating Magnetic Field in a Three-Phase AC Motor 5 / 10
ωt
π 2π
Im
0
-Im
A B C
Winding currents
13. Rotating Magnetic Field
Rotating Magnetic Field in a Three-Phase AC Motor 6 / 10
Fm
Axis of U
Axis of V
Axis of W
U
U’
W’
W
V
V’
Fig. 2. Individual MMF vector of three-phase motor
1.5Fm
Uaxis
Vaxis
Waxis
U
U’
W’
W
V
V’
Fig. 3. Resultant MMF vector
ωt
π 2π
Im
0
-Im
A B C
Fig. 1. Balanced three-phase
winding currents
14. Rotating Magnetic Field
Rotating Magnetic Field in a Three-Phase AC Motor 7 / 10
Fm
Axis of U
Axis of V
Axis of W
U
U’
W’
W
V
V’
Fig. 2. Individual MMF vector of three-phase motor
1.5Fm
Uaxis
Vaxis
Waxis
U
U’
W’
W
V
V’
Fig. 3. Resultant MMF vector
ωt
π 2π
Im
0
-Im
A C B
Fig. 1. Balanced three-phase
winding currents
15. Speed of the Rotating Magnetic Field
ωe = 2πf
ωs =
2ωe
p
=
4πf
p
(rad/sec)
Multiplying above expression with
60
2π
gives angular speed in revolution per minute
Ns =
120f
p
(rpm)
Rotating Magnetic Field in a Three-Phase AC Motor 8 / 10
16. Derivation - Rotating Magnetic Field
Three-phase balanced windings can be
represented by
UU’= Z cos(θ)
VV’= Z cos(θ + 120◦
)
WW’= Z cos(θ − 120◦
)
Rotating Magnetic Field in a Three-Phase AC Motor 9 / 10
17. Derivation - Rotating Magnetic Field
Three-phase balanced windings can be
represented by
UU’= Z cos(θ)
VV’= Z cos(θ + 120◦
)
WW’= Z cos(θ − 120◦
)
Three-phase balanced supply is
iA = Im sin(ωt)
iB = Im sin(ωt − 120◦
)
iC = Im sin(ωt + 120◦
)
Rotating Magnetic Field in a Three-Phase AC Motor 9 / 10
18. Derivation - Rotating Magnetic Field
Three-phase balanced windings can be
represented by
UU’= Z cos(θ)
VV’= Z cos(θ + 120◦
)
WW’= Z cos(θ − 120◦
)
Three-phase balanced supply is
iA = Im sin(ωt)
iB = Im sin(ωt − 120◦
)
iC = Im sin(ωt + 120◦
)
MMF produced by individual winding are
FUA = ImZ sin(ωt)cos(θ)
FV B = ImZ sin(ωt − 120◦
)cos(θ + 120◦
)
FW C = ImZ sin(ωt+120◦
)cos(θ −120◦
)
Rotating Magnetic Field in a Three-Phase AC Motor 9 / 10
19. Derivation - Rotating Magnetic Field
Three-phase balanced windings can be
represented by
UU’= Z cos(θ)
VV’= Z cos(θ + 120◦
)
WW’= Z cos(θ − 120◦
)
Three-phase balanced supply is
iA = Im sin(ωt)
iB = Im sin(ωt − 120◦
)
iC = Im sin(ωt + 120◦
)
MMF produced by individual winding are
FUA = ImZ sin(ωt)cos(θ)
FV B = ImZ sin(ωt − 120◦
)cos(θ + 120◦
)
FW C = ImZ sin(ωt+120◦
)cos(θ −120◦
)
sin(A) cos(B) = 0.5[sin(A + B) + sin(A − B)]
Hence,
FUA = 0.5Fm [sin(ωt + θ) + sin(ωt − θ)]
FV B = 0.5Fm [sin(ωt + θ) + sin(ωt − θ + 120◦
)]
FW C = 0.5Fm [sin(ωt + θ) + sin(ωt − θ − 120◦
)]
Rotating Magnetic Field in a Three-Phase AC Motor 9 / 10
20. Derivation - Rotating Magnetic Field
Three-phase balanced windings can be
represented by
UU’= Z cos(θ)
VV’= Z cos(θ + 120◦
)
WW’= Z cos(θ − 120◦
)
Three-phase balanced supply is
iA = Im sin(ωt)
iB = Im sin(ωt − 120◦
)
iC = Im sin(ωt + 120◦
)
MMF produced by individual winding are
FUA = ImZ sin(ωt)cos(θ)
FV B = ImZ sin(ωt − 120◦
)cos(θ + 120◦
)
FW C = ImZ sin(ωt+120◦
)cos(θ −120◦
)
sin(A) cos(B) = 0.5[sin(A + B) + sin(A − B)]
Hence,
FUA = 0.5Fm [sin(ωt + θ) + sin(ωt − θ)]
FV B = 0.5Fm [sin(ωt + θ) + sin(ωt − θ + 120◦
)]
FW C = 0.5Fm [sin(ωt + θ) + sin(ωt − θ − 120◦
)]
The resultant MMF is sum of above three MMF’s
FR = FUA + FV B + FW C
FR = 1.5Fm sin(ωt + θ)
Resultant MMF is distributed in space and time. It can be
termed as a rotating magnetic field with sinusoidal space
distribution, whose space phase angle changes linearly with
time as ωt.
Rotating Magnetic Field in a Three-Phase AC Motor 9 / 10
21. Summary
Balanced winding and balanced supply are needed to create rotating magnetic field
in a three-phase motor
This concept is valid for m-phase system
Rotating MMF moves from the axis of the leading phase to that of the lagging
phase at synchronous speed Ns =
120f
p
The direction of the rotating MMF can be reversed by simply changing phase
sequence of the supply
Rotating Magnetic Field in a Three-Phase AC Motor 10 / 10