The speed of series DC motor is controlled by a single-phase full bridge diode rectfier and
boost converter. The diode rectfier is operated from a 380 V, 60 Hz supply.
The boost converter has f s = 1.5 kHz , L = 0.1 mH and C = 100uF . The armature and field
resistances are 0.08 and 0.15 Ω and Kφ = 1.2 V/rpm. The converter and ac supply are
considered to be ideal.
(I) If the motor current is 200A at 0.7 duty ratio of boost converter, find motor speed,
developed torque.
(II) For the same load current (200) and 1100 rpm speed, determine the required duty ratio of
boost converter.
Solution:
Vd =
2Vm
π
∴ I OB =
=
2 * 308 2
π
= 342.12 V
TS Vo
T V
342.12
D(1 − D) 2 = S d D(1 − D) =
* 0.7 * (1 − 0.7) = 239.5 A
2L
2L
1500 * 2 * 0.0001
Q I o < I OB , then, The boost work in discontinuous conduction mode
342.12
T V 


Q I o =  S d  D∆1 = 200 = 
 * .7 * ∆1
 1500 * 2 * .0001 
 2L 
Then ∆1 = 0.2505
Vo
Vo
∆ +D
0.2505 + 0.7
= 1
=
=
Vd
∆1
342.12
0.2505
Then Vo = 1298.2V
Then Ea = Vo − I a Ra + R f = 1298.2 − 200 * (0.08 + 0.15) = 1252.2V
(
)
Ea = Kφ Ν = 1252.2 = 1.2 * N
Then N = 1043.5 rpm
E I
1252.2 * 200
Tdev = a a =
= 2292 N .m
2 * π * 1043.5 / 60
ω
2
(II) In series dc motor Tdevα I a , due to the torque is still the same as in part (I) then the current also still the same.
Then I a = 200 A
Ea 2 = Kφ N Then Ea 2 = 1.2 * 1100 = 1320V
Then, Vo = Ea + I a Ra + R f = 1320 + 200 * (0.08 + 0.15) = 1366V
(
)
2
1366
2 TS Vo
I OB, max =
=
= 674.57 A
1/ 2
1/ 2
27
27
1500
*
0
L
.
0001
 4 Vo  Vo
 Io 
 4 1366  1366
200 


∴D = 
− 1
=
− 1
= 0.7245



 27 342.12  342.12  674.57 
 27 Vd  Vd
 I oB, max 