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Module PE5
Problem 1
Recall that the speed of a DC motor is given by w  Vt  R a T . A particular separately excited DC motor
2
k
k 
has k  1.44 , Ra  0.86 . When it is operating at 150 rad/sec, the armature current is I a  40 A . The terminal
voltage Vt is held constant under all conditions.
(a) Compute Vt
(b) Compute the no-load speed in rad/sec
Solution to problem 1
(a) Ea  kwm  1.44   150   216V
 Vt  E a  I a ra  216  40   0.86 
 250 .4V
(b) "No load" speed occurs when T=0 (No load !), therefore, w 
w
Vt
RT
V
 a 2  t when T=0
k k 
k
250 .4
 173 .9rad / sec
1.44
Problem 2
A separately excited DC motor runs at 100 rpm supplying rated torque when the applied voltage is 50 volts
and the armature current is 10A. The armature resistance is Ra = 1 ohm. A thyristor- controlled chopper
circuit is used to decrease the applied voltage to 40 volts, at constant field flux, to drive a load at 75% of
the rated torque. The conducting time T on of this circuit is fixed at 0.2 seconds. Compute the speed of the
motor at 40 volts and the blocking time T off of the chopper circuit.
Solution to problem 2
T1  k  I a1
T2  0.75T1  k  I a 2  0.75(k  I a1 )  I a 2  0.75I a1  (0.75)(10 A)  7.5 Amps
E a1  Vt1  I a1 Ra  50  (10)(1)  40Volts
E a 2  Vt 2  I a 2 Ra  40  (7.5)(1)  32.5Volts
E a1  k  1 , E a 2  k   2 
E a1
1

Ea 2
2
 2 
Ea 2
32.5
1 
(100)  81.25rpm
E a1
40
VLoad , DC
Ton
0.2
40



 Toff  0.05 sec onds
Ton  Toff
V
0.2  Toff 50
Problem 3
A separately excited D.C motor is rated for a full load torque of 100Nm. For an applied voltage of 240V
and for flux ( ) per pole of 30 mWb, the no-load and full load speeds of the motor are 1595 r.p.m and
1195 r.p.m respectively.
Find
a)
b)
i)
The constant k
ii)
The armature resistance Ra
If the applied voltage and the flux per pole are reduced to 200V and 20mWb respectively,
Find
i) The corresponding no load speed
ii) The corresponding full load speed
Solution to problem 3
Vt
RT
 a 2
. From the problem, we have:
k (k )
 100 Nm Vt  240V   30mWb nnoload  1595rpm n fullload  1195rpm
a) From the text, we know that
T fullload

At no load,
1595rev 1 min 2rad
 167.03rad / sec
min 60 sec rev
V
Vt
240
 167.03  t  k 

 47.9
k
(167.03) (167.03)(0.03)
T  0and noload 
At full load,
TFL  100Nm
2
 125.1rad / sec
60
V
RT
(k ) 2
 125.1  t  a FL2  Ra 
k (k )
TFL
 FL  (1195)
 Vt


 125.1
 k


(( 47.9)(0.03)) 2 
240

 Ra 
 125.1  0.865
100
 (47.9)(0.03)

b) Vt = 200 V,  = 0.02 Webers.
V
200
 60 
 noload  t 
 204.08rad / sec  n NL  204.08   1949.8rpm
k (49)(0.02)
 2 
V
RT
0.865(100)
 fullload  t  a 2  204.08 
 88.265rad / sec
k (k )
[( 49)(0.02)] 2
 60 
 n FL  88.265   843.3rpm
 2 
Problem 5
An engineer wants to construct a half-wave rectifier for speed control of a DC motor using a thyristor, but
there is no gate-pulse circuit available. The implication is that he/she can supply no gate pulse. Without a
gate pulse circuit, can the engineer just use the thyristor like a diode, so that at least the circuit can supply
one DC voltage level to make the motor run at a single speed? Indicate yes or no, and identify the
feature(s) of a thyristor that support your answer.
Solution to problem 5
Answer 1: No, it cannot be done, because without the gate pulse, the thyristor will not be able to turn on
and it will appear as open-circuit at all times.
Answer 2: Yes, it can work if the applied AC voltage has a peak that is significantly higher than the
forward breakover voltage of the thyristor.
Problem 6
A square wave voltage waveform having a period of 2 seconds is applied through a half wave thyristorcontrolled rectifier circuit to the armature terminals of a separately excited DC motor driving It is known
that the DC motor has k=1.44 (a constant under all conditions), and Ra=0.86 ohms. When Ia=40 amperes,
the speed is 150 rad/sec. The source voltage and the thyristor firing time is shown in the figure below.
(a) Draw this circuit, including the rectifier and the motor.
(b) What is the effective DC voltage at the motor armature terminals under this condition ?
(c) What is the peak value of the square wave under this condition ?
Vsource
Time  (sec)
0.5
1
2
Fire thyristor
Solution to problem 6
a)
Ia
V(t)
_
Ra
Ea
b) Ea = k = (1.44)(150) = 216, Vt = 216 + IaRa = 250.4 Volts
c)
2
Vdceff 
1
1
1
1
V (t )dt   V (t )dt  V  250.4  V  1001.6Volts

T0
2 0.5
4