# Field Control Method

• Field control is the most common method and forms one of the outstanding advantages of shunt motors. The method is, of course, also applicable to compound motors. Adjustment of field current and hence the flux and speed by adjustment of the shunt field circuit resistance or with a solid-state control when the field is separately excited is accomplished simply, inexpensively, and without much change in motor losses.

The speed is inversely proportional to the field current

• The lowest speed obtainable is that corresponding to maximum field current; the highest speed is limited electrically by the effects of armature reaction under weak- field conditions ill causing motor instability and poor commutation.
• Since, voltage across the motor remains constant, it continues to deliver constant output.

This characteristic makes this method suitable for fixed output loads. The performance curve of a D.C. motor with voltage and field control is shown in Fig. 78.

Merits. The merits of this method are:

1.Good working efficiency.

2.Compact controlling equipment.

3.Capability of minute speed control.

4.The speed is not effected by load, and speed control can be performed effectively even at light loads.

5.Relatively inexpensive and simple to accomplish, both manually and automatically.

6. Within limits, field control does not affect speed regulation in the cases of shunt, compound, and series motors.

7. Provides relatively smooth and stepless control of speed.

Demerits. The demerits of field control as a method of speed control are:

1.Inability to obtain speeds below the basic speed.

2.Instability at high speeds because of armature reaction.

3.Commutation difficulties and possible commutator damage at high speeds.

Shunt Motors:

• The flux of a D.C. shunt motor can be changed by changing shunt field current (Ish) with the help of a shunt field rheostat as shown in Fig. 79. Since the field current is very small, the power wasted in the controlling resistance is very small.

• In non-interpolar machines the speed can be increased by this method in the ratio 2 : 1. In machines fitted with interpoles a ratio of maximum to minimum speeds of 6: 1 is fairly _ common.

Series Motors. In a series motor, variations afflux can be brought about in anyone of the following ways:

(i)                           Field divertors

(ii)                        Armature divertor

(iii)                      Tapped field control

(iv)                      Paralleling field coils.

(i) Field divertors. A variable resistance, known as field divertor (Fig. 80) shunts the series windings. Any desired amount of current can be passed through the divertor by adjusting its resistance. Hence, the flux can be decreased and consequently the speed of the motor increased.

Example 36. A 220 V D.C. shunt motor draws a no-load armature current of 2.5 A when running at 1400 r.p.m. Determine its speed when taking an armature current of 60 A, if armature reaction weakens the flux by 3 per cent.

Take armature resistance = 0.2 Ω

Solution. Supply voltage,             V = 220 Volts

No-load current, Ia0 = 2.5 A

No-load speed, N0 = 1400 r.p.m.

Armature resistance, Ra = 0.2 Ω

Armature current, Ia = 60 A

Full-load flux, ɸ = 0.97 ɸ0

Back e.m.f. at no-load, Eb0 = V –Ia0Ra = 200 – 2.5 × 0.2 = 219.5 V

Back e.m.f on load, Eb = V -IaRa = 220-60 × 0.2 = 208 V.

Hence, load speed = 1367.7 r.p.m. (Ans.)

Example 37. The armature and shunt field resistances of a 500 V shunt motor arc 0.2 Ω and 100 n respectively. Find the resistance of the shunt field regulator to increase the speed from 800 r.p.m. to 1000 r.p.m., if the current taken by the motor is 450 A. The magnetisation characteristic may be assumed as a straight line.

Solution. Supply voltage,             V = 500 Volts

Armature resistance, Ra = 0.2 Ω

Shunt field resistance, Rsh = 100 Ω

Initial speed, N1 = 800 r.p.m.

Final speed, N2 = 1000 r.p.m.

Current drawn by the motor, I = 450 A

Resistance of shunt field regulator, R:

Since the magnetization characteristic is a straight line,

Example 38. A 500 V series motor has an armature resistance of 0.4 ohm and series field resistance of 0.3 ohm. It takes a current of 100 A at a speed of 600 r.p.m. Find the speed of the motor if a divertor of resistance 0.6 ohm is connected across the field, the load torque being kept constant.

Neglect armature reaction and assume that flux is proportional to the current.

Solution. Supply voltage, V = 500 Volts

Armature resistance, Ra = 0.4 ohm

Series field resistance, Rse = 0.3 ohm

Divertor resistance, Rdiv. = 0.6 ohm

Ia1 = 100 A, N1 = 600 r.p.m.

Speed.N2 :

Back e.m.f., Eb1 = V – Ia1(Ra + Rse)
= 500 – 100(0.4 + 0.3) = 430 V.

Let Ia2 be the current taken and ɸ2 be the flux produced when a divertor is connected across the series field (Fig. 84)

Since the torque remains constant,

ɸ1Ia1 = ɸ2Ia2 … (i)

But ɸ current through series field

Example 39. A D.C. series motor drives a load the torque of which varies are square of the speed. The motor takes a current of20 A when the speed is 800 r.p.m. Calculate the speed and current when the motor field winding is shunted by a divertor of the same resistance as that of the field winding.

Neglect all motor losses and assume that the magnetic circuit is unsaturated.

Solution. Ia1 = 20 A, N1 = 800 r.p.m.

Ia2=?, N2=?

When the field winding is shunted by a divertor of equal resistance, then current through either is half the armature current, If Ia2 is the new armature current, then Ia2/2 passes through the winding (Fig. 85).