UNIT - 3 : RECTIFIER CONTROLLED FED DC DRIVES
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ExAMPLE 5.13
A 200 V, 875 rpm, 150 A separately excited de motor has an armature resistance of 0.06 2. It is
fed from a single phase fully-controlled rectifier with an ac source voltage of 220 V, 50 Hz.
Assuming continuous conduction, calculate
(i) firing angle for rated motor torque and 750 rpm.
i ) firing angle for rated motor torque and (-500) rpm.
(ii) motor speed for a = 160° and rated torque.
Solution
At rated operation E = 200 - 150 x 0.06 = 191 Vv
i) E at 750 rpm, E750
7x 191 = 163.7 V
V = E + , R = 163.7 + 150 x 0.06 172.7 V
Now 2 cos a = Va
or
2x 2202
cos a = 172.7
or cos & = 0.872 or = 29.3°
-S00
(ii) At -500 rpm E X 191 -109 V
R75 =
Since Va= E+ lRa
V= -
109 + 150x 0.06 = -
100 V
Now 2Vm cos a =
V
or 2x220N2 COs - 100
or cOs a = - 0.5 or a= 120
(ii) At a= 160°
V, = cos x-2x 220 2 cos 160° = -186 V
Since Va=E+,Ra
-
186 = E + 150 x 0.06
E = - 195 V
or
-195
Speed191 x 875 =-893.2 rpm
5.15 RECTIFIER CONTROL OF dc SERIES MOTOR
Single-phase controlled rectifier fed dc series motors are employed in traction. A single-phase
half-controlled rectifier-fed dc series motor is shown in Fig.
5.37(a). Equivalent circuit of motor
isalso shown. Since back emf decreases with armature
current, discontinuous conduction occurs
only in a narrow range of operation. Hence, it will be neglected here. The waveforms of va. ia and
instantaneous back emf e for continuous conduction are shown in Fig. 5.37(b). Although, in
steady state, fluctuations in speed are negligible, e is not constant but fluctuates with ia For a
given speed, e is related to i, through magnetization curve of motor, which is nonlinear owing
to saturation. Thus
e
=f(ia) m (5.102)
a Series
motors
D
a
27
T D e=fi,)on
Ti, D 7, D
D, D2 D, D2
(a) Drive circuit (b) Waveforms
Fig. 5.37 Single-phase half-controlled rectifier fed series motor
Motor operation is described by following equations for duty and freewheeling intervals respectively,
Vm sin ot= R,i, + L, dt +fi,)0m. for
a s ot ssr (5.103)
0 = Raia +L dia
+ f l i , )0m, for 7 wr s (r+ ) (5.104)
Because of the presence of termfli), Eqs. (5.103) and (5.104) are nonlinear differential equations
and can only be solved numerically. A simple method of analysis is obtained when e is replaced
by its average value E such that
E = K,m (5.105)
where K =f (5.106)
due to dc component of armature current l is zero
Since the drop across the inductance La
V= E + 1,R,
m (5.107)
or
Ka
and T Kala (5.108)
For continuous conduction, Va for half-controled and fully-controlled single-phase rectifiers
1S given by Eqs. (5.93) and (5.83), respectively.
used to
sequence of steps
are
Following
calculate speed-torque characteristic for a given
ataking into account non-linearly of the
magnetic circuit: A value is chosen for I.
Increasing
a
from
Corresponding value of K, is obtained
the magnetization characteristic of the motor.
For the known value of a, calculate V, from
Eq. (5.93) or (5.83), depending on the rectilier
Circuit used. Now and T are obtained from
T
Eqs. (5.107) and (5.108), respectively. Nature
of speed-torque characteristics for the drive of Fig. 5.38 Speed torque curves of series motor
fed from a controlled rectifier
Fig. 5.37(a) is shown in Fig. 5.38.
