16.1 POWER SEMICONDUCTOR-CONTROLLED DRIVES 767

(^0) π/3 ωt
α
T 4 T 6
va va
ia ia
T 5 T 1 T 5
T 2
T 3
T 4
E
π 2 π
AB AC BC BA CA CB
Figure 16.1.15Continuous conduc-
tion mode for the motoring operation.
Va=
3
π
∫α+ 2 π/ 3
α+π/ 3
Vmsinωt d(ωt)=
3
π
Vmcosα=Vaocosα (16.1.42)
where the line voltagevAB=Vmsinωtis taken as the reference. From Equations (16.1.32),
(16.1.37), and (16.1.42), we get
ωm=
3 Vm
πKm
cosα−
Ra
K^2 m
Ta (16.1.43)
For normalization, taking the base voltageVBas the maximum average converter output voltage
Vao= 3 Vm/π, and the base current as the average motor current (that will flow whenωm= 0
andVa=VB)IB=VB/Ra= 3 Vm/π Ra, the normalized speed and torque are given by
Speedωmn=
E
VB

πE
3 Vm
(16.1.44)
TorqueTan=Ian=
πRa
3 Vm
(Ia) (16.1.45)
EXAMPLE 16.1.3
Consider the 220-V, 1800-r/min dc motor of Example 16.1.2, controlled by a three-phase fully
controlled rectifier from a 60-Hz ac source. The armature-circuit resistance and inductance are
1.5and 30 mH, respectively.
(a) When the motor is operating in continuous conduction, find the ac source voltage required
to get rated voltage across the motor terminals.
(b) With the ac source voltage obtained in part (a), compute the motor speed corresponding
toα=60° andTa=25 N·m assuming continuous conduction.
(c) Let the motor drive a load whose torque is constant and independent of speed. The
minimum value of the load torque is 1.2 N·m. Calculate the inductance that must be
added to the armature circuit to get continuous conduction for all operating points,
given thatψ = tan−^1 (ωLa/Ra) = 1 .5 rad,Tan > 0 .006, and all points on the
Tan–ωmnplane lie to the right of the boundary between continuous and discontinuous
conductions.