Page 185 - Electric Drives and Electromechanical Systems

Chapter 6 Brushless motors 179

6.2.1 Torque characteristics

The force on a group of stator conductors of length L, within the angle dq of the stator, is
determined by the product of the flux and the stator current, i;
N sin pq cosðpq aÞdq
F ¼ Bil (6.17)
b
2
The resultant torque on a rotor of radius r, including the contribution of the opposite
winding, is given by,
T ¼ 2Fr (6.18)
The total output torque of the motor can therefore be determined by integration over
the whole air gap; giving for a p -pole pair motor,
pr Bli Nsin a
Z p=2
b
T ¼ 2Frdq ¼ (6.19)
0 2
The peak torque will be generated when the rotor’s magnetic north axis lags the axes
of the stator’s ampere-conductor distribution by 90 degrees.
Thecasewhere therotor is stationary relative to the stator was considered in the
analysis above. In order to produce a constant torque with the rotor rotating at a
constant speed, the stator’s ampere-conductor distribution must rotate in synchronism
with the rotor. This is achieved by using a three-phase winding supplied with a three-
phase balance current. If the r.m.s. phase current is I, then for a motor where the
winding are 120 degrees apart, the rotating ampere conductor distribution can be
shown to be,
N 2p N 2p 2p N 2p
icos ut sin pq þ icos ut sin pq þ icos ut þ sin pq þ
b
b
b
2 3 2 3 3 2 3
p ffiffiffi
3 2I
¼ N sinðpq aÞ (6.20)
4
The rotating magnetic-flux distribution is given by,
(6.21)
BðqÞ¼ BcosðpqeuteaÞ
b
If this is combined with Eq. (6.19), the output torque can be calculated to be,
3 p ffiffiffi prl BN2
b
T ¼ 2I sin b (6.22)
2 2
The angle b (which equals a), is termed the torque angle, and it is held positive for
motoring; any variation in b will require adjustment of the phase current to hold a
constant torque. This equation shows that the peak torque, and hence a motor’s
efficiency, is optimal when b ¼ p/2. To ensure that the ampere-conductor distribution
remains in synchronism with the rotor’s magnetic field, the stator’s supply frequency, f,
is made equal to the rotor’s rotational frequency, u s , hence,

u s ¼ 2pf (6.23)