Speed control of d.c. Motors:Thyristor Controller Starters

Thyristor Controller Starters

The moving parts and metal contacts etc., of the resistance starters discussed in Art. 30.21 can be eliminated by using thyristors which can short circuit the resistance sections one after another. A thyristor can be switched on to the conducting state by applying a suitable signal to its gate terminal. While conducting, it offers zero resistance in the forward (i.e., anode-to-cathode) direction and thus acts as a short-circuit for the starter resistance section across which it is connected. It can be switched off (i.e., brought back to the non-conducting state) by reversing the polarity of its anode-cathode voltage. A typical thyristor-controlled starter for d.c. motors is shown in Fig. 30.49.

After switching on the main supply, when switch S1 is pressed, positive signal is applied to gate G of thyristor T1 which is, therefore, turned ON. At the same time, shunt field gets established since it is directly connected across the d.c. supply. Consequently, motor armature current Ia flows via T1, R2, R3 and R4 because T2, T3 and T4 are, as yet in the non-conducting state. From now onwards, the starting procedure is automatic as detailed below :

1. As S1 is closed, capacitor C starts charging up with the polarity as shown when Ia starts flowing.

2. The armature current and field flux together produce torque which accelerates the motor and load.

3. As motor speeds up, voltage provided by tachogenerator (TG) is proportionately increased because it is coupled to the motor.

4. At some motor speed, the voltage provided by TG becomes large enough to breakdown Zener diode Z2 and hence trigger T2 into conduction. Consequently, R2 is shorted out and now Ia flows via motor armature, T1, T2, R3 and R4 and back to the negative supply terminal.

5. As R2 is cut out, Ia increases, armature torque increases, motor speed increases which further increases the voltage output of the tachogenerator. At some speed, Z3 breaks down, thereby triggering T3 into conduction which cuts out R3.

6. After sometime, R4 is cut out as Z4 breaks down and triggers T4 into conduction. In fact, Zener diodes Z2, Z3 and Z4 can be rated for 1/3, 1/2 and 3/4 full speed respectively.

For stopping the motor, switch S2 is closed which triggers T5 into conduction, thereby establishing current flow via R1. Consequently, capacitor C starts discharging thereby reverse-biasing T1 which stops conducting. Hence Ia ceases and, at the same time, T2, T3 and T4 also revert back to their non-conducting state.

Incidentally, it may be noted that the function of C is to switch T1, ON and OFF. Hence, it is usually called commutating capacitor.

The function of the diodes D1 and D2 is to allow the decay of inductive energy stored in the motor armature and field when supply is disconnected. Supply failure will cause the thyristors to block because of this current decay, thereby providing protection usually given by no-voltage release coil.

Recently, thyristor starting circuits have been introduced which use no starting resistance at all, thereby making the entire system quite efficient and optimized as regards starting time. These are based on the principle of ‘voltage chopping’ (Art. 30.12). By varying the chopping frequency, the ratio of the time the voltage is ON to the time it is OFF can be varied. By varying this ratio, the average voltage applied to the motor can be changed. A low average voltage is needed to limit the armature current while the motor is being started and gradually the ratio is increased to reach the maximum at the rated speed of the motor.

Tutorial Problems

1. A shunt-wound motor runs at 600 r.p.m. from a 230-V supply when taking a line current of 50 A. Its armature and field resistances are 0.4 W and 104.5 W respectively. Neglecting the effects of armature reaction and allowing 2 V brush drop, calculate (a) the no-load speed if the no-load line current is 5A (b) the resistance to be placed in armature circuit in order to reduce the speed to 500 r.p.m. when motor is taking a line current of 50 A (c) the percentage reduction in the flux per pole in order that the speed may be 750 r.p.m. when the armature current is 30 A with no added resistance in the armature circuit. [(a) 652 r.p.m. (b) 0.73 W (c) 1.73 %]

2. The resistance of the armature of a 250-V shunt motor is 0.3 W and its full-load speed is 1000 r.p.m.

Calculate the resistance to be inserted in series with the armature to reduce the speed with full-load torque to 800 r.p.m., the full-load armature current being 5A. If the load torque is then halved, at what speed will the motor run ? Neglect armature reaction. [0.94 W; 932 r.p.m.]

3. A 230-V d.c. shunt motor takes an armature current of 20 A on a certain load. Resistance of the armature is 0.5 W. Find the resistance required in series with the armature to half the speed if (a) the load torque is constant (b) the load torque is proportional to the square of the speed.

[(a) 5.5 W (b) 23.5 W]

4. A 230-V series motor runs at 1200 r.p.m. at a quarter full-load torque, taking a current of 16 A. Calculate its speed at half and full-load torques. The resistance of the armature brushes, and field coils is 0.25 W. Assume the flux per pole to be proportional to the current. Plot torque/speed graph between full and quarter-load. [842 r.p.m. ; 589 r.p.m.]

5. A d.c. series motor drives a load the torque of which is proportional to the square of the speed. The motor current is 20 A when speed is 500 r.p.m. Calculate the speed and current when the motor field winding is shunted by a resistance of the same value as the field winding. Neglect all motor losses and assume that the magnetic field is unsaturated. [595 r.p.m. ; 33.64 A]

(Electrical Machines-I, Aligarh Muslim Univ. 1979)

6. A d.c. series motor, with unsaturated magnetic circuit and with negligible resistance, when running at a certain speed on a given load takes 50 A at 500 V. If the load torque varies as the cube of the speed, find the resistance which should be connected in series with machine to reduce the speed by 25 per cent. [7.89 W]

(Electrical Engg-I, M.S. Univ. Baroda 1980)

7. A series motor runs at 500 r.p.m. on a certain load. Calculate the resistance of a divertor required to raise the speed to 650 r.p.m. with the same load current, given that the series field resistance is 0.05 W and the field is unsaturated. Assume the ohmic drop in the field and armature to be negligible.

[0.1665 W]

8. A 230-V d.c. series motor has armature and field resistances of 0.5 W and 0.3 W respectively. The motor draws a line current of 40 A while running at 400 r.p.m. If a divertor of resistance 0.15 W is used, find the new speed of the motor for the same armature current.

It may be assumed that flux per pole is directly proportional to the field current. [1204 r.p.m.]

(Electrical Engineering Grad. I.E.T.E. June 1986)

9. A 250-V, d.c. shunt motor runs at 700 r.p.m. on no-load with no extra resistance in the field and armature circuit. Determine :

(i) the resistance to be placed in series with the armature for a speed of 400 r.p.m. when taking a total current of 16 A.

(ii) the resistance to be placed in series with the field to produce a speed of 1,000 r.p.m. when taking an armature current of 18 A.

Assume that the useful flux is proportional to the field. Armature resistance = 0.35 W, field resis- tance = 125 W. [(i) 7.3 W (ii) 113 W] (Elect. Engg. Grad. I.E.T.E., June 1984)

10. A d.c. series motor is operating from a 220-V supply. It takes 50 A and runs at 1000 r.p.m. The resistance of the motor is 0.1 W. If a resistance of 2 W is placed in series with the motor, calculate the resultant speed if the load torque is constant. [534 r.p.m.]

11. A d.c. shunt motor takes 25 A when running at 1000 r.p.m. from a 220-V supply.

Calculate the current taken form the supply and the speed if the load torque is halved, a resistance of 5 W is placed in the armature circuit and a resistance of 50 W is placed in the field circuit. Armature resistance = 0.1 W ; field resistance = 100 W

Assume that the field flux per pole is directly proportional to the field current.[17.1 A; 915 r.p.m.]

(Elect. Technology, Gwalior Univ. Nov. 1977)

12. A 440-V shunt motor takes an armature current of 50 A and has a flux/pole of 50 mWb. If the flux is suddenly decreased to 45 mWb, calculate (a) instantaneous increase in armature current (b) per- centage increase in the motor torque due to increase in current (c) value of steady current which motor will take eventually (d) the final percentage increase in motor speed. Neglect brush contact drop and armature reaction and assume an armature resistance of 0.6 W.

[(a) 118 A (b) 112 % (c) 5.55 A (d) 10% ]

13. A 440-V shunt motor while running at 1500 r.p.m. takes an armature current of 30 A and delivers a mechanical output of 15 h.p. (11.19 kW). The load torque varies as the square of the speed. Calcu- late the value of resistance to be connected in series with the armature for reducing the motor speed to 1300 r.p.m. and the armature current at that speed. [2.97 W, 22.5 A]

14. A 460-V series motor has a resistance of 0.4 W and takes a current of 25 A when there is no addi- tional controller resistance in the armature circuit. Its speed is 1000 r.p.m. The control resistance is so adjusted as to reduce the field flux by 5%. Calculate the new current drawn by the motor and its speed. Assume that the load torque varies as the square of the speed and the same motor efficiency under the two conditions of operation.

[22.6 A; 926 r.p.m.] (Elect. Machines, South Gujarat Univ. Oct. 1977)

15. A 460-V, series motor runs at 500 r.p.m. taking a current of 40 A. Calculate the speed and percent- age change and torque if the load is reduced so that the motor is taking 30 A. Total resistance of armature and field circuit is 0.8 W. Assume flux proportional to the field current.

[680 r.p.m. 43.75%]

16. A 440-V, 25 h.p (18.65 kW) motor has an armature resistance of 1.2 W and full-load efficiency of 85%. Calculate the number and value of resistance elements of a starter for the motor if maximum permissible current is 1.5 times the full-load current. [1.92 W, 1.30 W, 0.86 W; 0.59 W]

(Similar example in JNTU, Hyderabad, 2000)

17. A 230-V, d.c. shunt motor has an armature resistance of 0.3 W. Calculate (a) the resistance to be connected in series with the armature to limit the armature current to 75 A at starting and (b) value of the generated e.m.f. when the armature current has fallen to 50 A with this value of resistance still in circuit. [(a) 2.767 W (b) 76.7 A]

18. A 200-V, d.c. shunt motor takes full-load current of 12 A. The armature circuit resistance is 0.3 W and the field circuit resistance is 100 W. Calculate the value of 5 steps in the 6-stud starter for the motor. The maximum starting current is not to exceed 1.5 times the full-load current.

[6.57 W, 3.12 W, 1.48 W, 0.7 W, 0.33 W]

19. The resistance of a starter for a 200-V, shunt motor is such that maximum starting current is 30 A. When the current has decreased to 24 A, the starter arm is moved from the first to the second stud.

Calculate the resistance between these two studs if the maximum current in the second stud is 34 A. The armature resistance of the motor is 0.4 W. [1.334 W]

20. A totally-enclosed motor has thermal time constant of 2 hr. and final temperature rise at no-load and 40° on full load.

Determine the limits between which the temperature fluctuates when the motor operates on a load cycle consisting of alternate period of 1 hr. on full-load and 1 hr. on no-load, steady state conditions having been established. [28.7° C, 21.3°C]

21. A motor with a thermal time constant of 45 min. has a final temperature rise of 75°C on continuous rating (a) What is the temperature rise after one hour at this load ? (b) If the temperature rise on one- hour rating is 75°C, find the maximum steady temperature at this rating (c) When working at its one- hour rating, how long does it take the temperature to increase from 60°C to 75°C ? [(a) 55 °C (b) 102 °C (c) 20 min]

(Electrical Technology, M.S. Univ. Baroda. 1976)