1 - principles of electrical machines and power electronics p_c_sen

(from Eq. 10.3) 
For a semiconverter (from Eq. 10.5) 
V2v Vt = __ P ( 1 + COS a) 
1T 
Three-phase input. For a full converter (from Eq. 10.10) 
3\16V vt = p cos a 
1T 
For a semiconverter (from Eq. 10.10a) 
3\16V 
vt = 2rr p (1 + cos a) 
(4.58) 
(4.59) 
(4.60) 
(4.61) 
where Vr is the rms value of the ac supply phase voltage. The variation 
of the motor terminal voltage Vt as a function of the firing angle a is 
shown in Fig. 4.61 for both semiconverter and full-converter systems. If 
the laRa drop is neglected (Vt =E.) the curves in Fig. 4.61 also show the 
variation of E. (hence speed) with the firing angle. 
Although instantaneous values of voltage v t and current i. are not constant 
but change with time, in terms of average values the basic de machine 
equations still hold good. 
T= K/PI. 
(4.61a) 
(4.61b) 
(4.61c) 
FIGURE 4.61 Controlled-rectifier char-
acteristics. 
'+' 
I 
Speed Control 191 
EXAMPLE 4.11 
The speed of a 10 hp, 220 V, 1200 rpm separately excited de motor is 
controlled by a single-phase full converter as shown in Fig. 4.60 (or Fig. 
10.21a). The rated armature current is 40 A. The armature resistance is 
R. = 0.25 fl and armature inductance is La= 10 mH. The ac supply voltage 
is 265 V. The motor voltage constant is Ka<l> = 0.18 V/rpm. Assume that 
motor current is continuous and ripple-free. 
For a firing angle a = 30° and rated motor current, determine the 
(a) Speed of the motor. 
(b) Motor torque. 
(c) Power to the motor. 
Solution 
(a) From Eq. 4.58 the average terminal voltage is 
(b) 
The back emf is 
vt = 2Vl x 265 cos 300 
1T 
= 206.6 v 
= 206.6 - 40 X 0.25 
= 196.6 v 
Hence the speed in rpm is 
196.6 
N = 0. 18 = 1092.2 rpm 
Ka<l> = 0.18 V/rpm 
0.18X60V I d 
= 21T ·sec ra 
= 1. 72 V · sec/rad 
T = 1.72 X 40 
=68.75N·m 
(c) The power to the motor is 
Since ia is ripple-free (i.e., constant), 
(ia)nns = (iJaverage = fa 
'" I 
" 
'• 
.,. 
192 chapter 4 DC Machines 
EXAMPLE 4.12 
P = fiR. + E.I. 
= 206.6 X 40 
= 8264W • 
The speed of a 125 hp, 600 V, 1800 rpm, separately excited de motor is 
controlled by a 34> (three-phase) full converter as shown in Fig. 4.60 (or Fig. 
10.27a). The converter is operated from a 34>, 480 V, 60 Hz supply. The 
rated armature current of the motor is 165 A. The motor parameters are 
R. = 0.0874 !1, L. = 6.5 mH, and K.<l> = 0.33 V/rpm. The converter and ac 
supply are considered to be ideal. 
(a) Find no-load speeds at firing angles a = 0° and a = 30°. Assume 
that, at no load, the armature current is 10% of the rated current and 
is continuous. 
(b) Find the firing angle to obtain the rated speed of 1800 rpm at rated 
motor current. 
(c) Compute the speed regulation for the firing angle obtained in part (b). 
Solution 
(a) No-load condition. The supply phase voltage is 
V: = 480 = 277V pv'3 
From Eq. 4.60 the motor terminal voltage is 
3v6 X 277 vt = cos a = 648 cos a 
For a= 0° 
1T 
Vt = 648V 
= 648- (16.5 X 0.0874) 
= 646.6 v 
No-load speed is 
For a= 30° 
E. 646.6 N 0 =-=--= 1959rpm K.<l> 0.33 
vt = 648 cos 30° = 561.2 v 
E.= 561.2- (16.5 X 0.0874) = 559.8 V 
T 
f 
T Speed Control 193 
The no-load speed is 
559.8 N0 = 0_33 = 1696 rpm 
(b) Full-load condition. The motor back emf E. at 1800 rpm is 
Ea = 0.33 X 1800 = 594 V 
The motor terminal voltage at rated current is 
Therefore, 
vt = 594 + (165 x 0.0874) 
= 608.4 v 
648 cos a = 608 V 
608.4 
cos a= 648 = 0.94 
a=20.1° 
(c) Speed regulation. At full load the motor current is 165 A and the speed 
is 1800 rpm. If the load is thrown off, keeping the firing angle the same 
at a = 20.1 °, the motor current decreases to 16.5 A. Therefore 
E.= 608.4- (16.5 X 0.0874) 
= 606.96 v 
and the no-load speed is 
606.96 N0 =~= 1839.3rpm 
The speed regulation is 
1839.3- 1800 X 100o/t = 2 180/ 1800 ° . 0 • 
Choppers 
A solid-state chopper converts a fixed-voltage de supply into a variable-
voltage de supply. A schematic diagram of a chopper is shown in Fig. 4.62a. 
The chopper is a high-speed on-off switch as illustrated in Fig. 4.62b. The 
switch S can be a conventional thyristor (i.e., SCR), a gate turn-off (GTO) 
thyristor, or a power transistor. The operation of choppers is described in 
detail in Chapter 10. 
When the switching device in the chopper (Fig. 4.62b) is on, v1 = V (supply 
voltage) and motor current i. increases. When it is off, motor current i. 
decays through the diode (D), making v1 = 0. The waveforms of voltage v1 
and current i. are shown in Fig. 4.62c. The output voltage v1 is a chopped 
voltage derived from the input voltage V. The average output voltage V1 , 
1
, •. , 
··' 
1 
••. 
... ·· 
~..., .. 
l 
r' 
'• . 
''I'' I 
t t t t 
on off on off 
~7--
(c) 
FIGURE 4.62 Chopper circuit and its operation. 
which determines the speed of the de motor, is 
=aV 
where ton is the on time of the chopper 
T is the chopping period 
a is the duty ratio of the chopper 
(4.62) 
(4.63) 
Speed Control 195 
From Eq. 4.63 it is obvious that the motor terminal voltage varies linearly 
with the duty ratio of the chopper. 
EXAMPLE 4.13 
The speed of a separately excited de motor is controlled by a chopper as 
shown in Fig. 4.62a. The de supply voltage is 120 V, the armature circuit 
resistance Ra = 0.5 fl, the armature circuit inductance La = 20 mH, and the 
motor constant is Ka<P = 0.05 V/rpm. The motor drives a constant-torque 
load requiring an average armature current of 20 A. Assume that motor 
current is continuous. 
Determine the 
(a) Range of speed control. 
(b) Range of the duty cycle a. 
Solution 
Minimum speed is zero, at which Ea = 0. Therefore from Eq. 4.61a 
Vt = laRa = 20 X 0.5 = 10 V 
From Eq. 4.63 
10 = 120a 
1 
a= 12 
Maximum speed corresponds to a = 1, at which V1 = V = 120 V. Therefore 
Ea = Vt- laRa 
From Eq. 4.61b 
= 120- (20 X 0.5) 
= llOV 
Ea 110 
N = Ka<P = 0.05 = 2200 rpm 
(a) The range of speed is 0 < N < 2200 rpm. 
(b) The range of the duty cycle is -b. < a < 1. • 
4.5.3 CLOSED-LOOP OPERATION 
DC motors are extensively used in many drives where speed control is de-
sired. In many applications where a constant speed is required, open-loop 
operation of de motors may not be satisfactory. In open-loop operation, if 
load torque changes, the speed will change too. In a closed-loop system, the 
speed can be maintained constant by adjusting the motor terminal voltage 
as the load torque changes. The basic block diagram of a closed-loop speed 
.• 
;.:'11 
196 chapter 4 DC Machines 
N Speed controller 
Power 
supply 
FIGURE 4.63 Closed-loop speed control system. 
control system is shown in Fig. 4.63. If an additional load torque is applied, 
the motor speed momentarily decreases and the speed error EN increases, 
which increases the control signal Vc. The control signal increases the con-
verter output voltage (the control signal decreases the firing angle if the 
converter is a phase-controlled rectifier or increases the duty ratio if the 
converter is a chopper). An increase in the motor armature voltage develops 
more torque to restore the speed of the motor. The system thus passes 
through a transient period until the developed torque matches the applied 
load torque. 
There are other advantages of closed-loop operation, such as greater accu-
racy, improved dynamic response, and stability of operation. In a closed-
loop system the drive characteristics can also be made to operate at constant 
torque or constant horsepower over a certain speed range, a requirement 
in traction systems. Circuit protection can also be provided in a closed-
loop system. In fact, most industrial drive systems operate as closed-loop 
feedback systems. 
In a de motor, the armature resistance (R.) and inductance (L.) are small. 
The time constant ( T1 = L.fRa) of the armature circuit is also small. Conse-
quently, a small change in the armature terminal voltage may result in a 
quick and large change in the armature current, which may damage the 
solid-state devices used in