Large starting current; 3-point starter

Why the starting current is large

The armature current is I_a = (V − E_b) / R_a. At the instant of starting the armature is at rest, so the back emf E_b = KΦN = 0. Hence

I_a(start) = V / R_a

The armature resistance R_a is very small (about 0.1 to 1 Ω). For example, with V = 220 V and R_a = 0.5 Ω, I_a = 440 A, while the rated current may be only 20 A, i.e. more than 20 times. As the motor speeds up, E_b grows and the current falls to the normal value.

This large current would damage the commutator and brushes by sparking, overheat the winding, produce a heavy torque shock on the shaft and cause a voltage dip in the supply. So a starter is used to limit the current to about 1.5 to 2 times full-load current.

Working of the 3-point starter

It has three terminals L (line), A (armature) and F (field).

 L o--[OLR]--+--handle--> o--o--o--o  (studs with R1..R4)
             |                          |
             +--[NVC]--- F (field)      A (armature)

1. Start: the handle is moved from OFF to the first stud. The supply is connected to the armature through the full starting resistance, and to the shunt field through the no-volt coil. The current is limited.
2. Acceleration: as the speed rises, the back emf increases. The handle is moved to the next studs slowly, cutting the resistance out step by step.
3. Run: at the last stud the whole resistance is out and the soft-iron piece on the handle is held by the electromagnet of the no-volt coil (NVC). The motor runs with full voltage.
4. No-volt protection: if the supply is interrupted, the NVC loses its magnetism and the spring returns the handle to OFF, which avoids a dangerous restart.
5. Overload protection: the overload release (OLR) electromagnet short-circuits the NVC on excess current, releasing the handle.

The shunt field is always in circuit from the first stud, so that the full field is available at starting and high starting torque is obtained.