CHAPTER 7

Additional Problems

Solved Problems

7.1 A dual converter three phase circuit supplies power to a 540 V, 40 A separately excited d.c. motor with an armature resistance of 1.2 :. The voltage drops on the bridge thyristors are 12 V at rated motor current. Power is supplied by an ideal three-phase source with an rms line voltage 400 V, 50 Hz. Obtain the necessary firing delay angle and motor back emf for:

(a) Motoring operation at rated load current with terminal voltage of 400 V.

(b) Regeneration operation at rated load current with terminal voltage of 400 V.

(c) Motor plugged at rated load current with a terminal voltage of 400 V and a current limiting resistor of 5:.

Sol. we have

Edco = 3 2 3 2 400rmsE

S S

� � = 540.12 V

= Maximum average voltage and motor terminals.

For steady-state operation,

Edc = Eb + Ia.Ra + 'Eb + armature reaction volt-drops.

? Edc = Edc D – VT = Eb + IaRa

(a) With Edc = 400 V, Idc = 40 A.

? Eb = Edc – Ia.Ra = 400 – 40 � 1.2 = 352 V.

EdcD = 400 + VT = 400 + 12 = 412 V.

= Edco . cosD = 540 � cosD.

? cosD = 412

540 = 0.7627 = cos 40.3°

? D = 40.

(b) Edc = – 400 V, Idc = 40 A.

Eb = – 400 – 48 = – 448 V

EdcD = – 400 + 12 = – 388.

? cosD = 388

504dc

dco

E

ED

�

= – 0.77

= cos 140.34, ? D = 140.34°

Solution Manual 2

(c) Eb = – 400 – 48 V = – 448 V.

EdcD = – 400 + 12 + 5 � 40 = – 188 V

? cosD = 188

504dc

dco

E

ED

�

= – 0.348

= cos 110.37° ? D = 110.37°

7.2 A Single-phase fully controlled doubled bridge converter is operated from a 120 V, 60 Hz supply and the load resistance is 10 ohms. The circulating inductance is 40 mH. Firing delay angle for converter I and converter II are 60° and 120° respectively. Determine the peak circulating current and current through converters.

Sol. W = 2 Sf = 377 rad/sec.

DI = 60°, DS = 120°

Em = 2 E� = 2 120� = 169.7 V.

Now, icp = 2 2

r

E

w L¹ (1 – cosDI)

= 3

2 2 120

377 40 10�

�

� �

. (1 – cos 60) = 11.25 A.

Peak value of load current, ILP = 2 169.7

10

E

R = 16.97 A

Current though converter, I = ICP + ILP

= 11.25 + 16.97 = 28.22 A.

Current through converter II = ICP = 11.25 A.