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High starting current; 3-point starter

Short answer

At start, back emf is zero, so I_a = V/R_a, drawing 15–20 times full-load current, causing damage, voltage dip, and excessive torque. A starter inserts external resistance in series with the armature to limit current, cutting it out as speed rises. A 3-point starter provides no-volt and overload release protection.

Why does a dc motor draw a large current at starting? Explain the necessity of a starter and describe the working of a 3-point dc motor starter with neat diagram.

Why the starting current is large

Reason: the armature current of a DC motor is

Ia=V−EbRaI_a = \frac{V - E_b}{R_a}

At the instant of starting the armature is at rest, so the back emf Eb=ΦZNP60A=0E_b = \dfrac{\Phi ZNP}{60A} = 0. The current is then limited only by the very small armature resistance:

Ia(start)=VRaI_{a(start)} = \frac{V}{R_a}

For example, a 220 V motor with Ra=0.5 ΩR_a = 0.5\ \Omega would draw 220/0.5=440220/0.5 = 440 A, about 15–20 times its full-load current (say 25 A).

Starter

Necessity of a starter: this large current would

  • burn the armature winding and damage the commutator and brushes (heavy sparking),
  • cause a large voltage dip on the supply and affect other consumers,
  • produce a very high starting torque that can shock and damage the shaft, gears and load.

A starter inserts an external resistance in series with the armature at starting and removes it step by step as the motor gains speed and EbE_b builds up. It limits the starting current to about 1.25–2 times full-load current:

Rstart=VIstart−RaR_{start} = \frac{V}{I_{start}} - R_a

It also gives protection (no-volt and overload release).

3-point starter

Parts of a 3-point starter: three terminals L (line), A (armature) and F (field); a starting resistance divided into sections with studs; a spring-loaded handle (starting arm) with a brass arc; a no-volt release coil (NVC), an overload release coil (OLR) and a soft-iron piece on the handle.

   L (+)
    |      OLR (overload coil)
    +------[~~~]---+
                   |   handle (spring loaded)
   OFF  1   2   3  4  ON   o----- soft iron
        |   |   |  |   |     piece, held by NVC
        +[R]+[R]+[R]+---+---------> A (armature)
                        |
             +--- brass arc ---[NVC]----[ Field ]--> F
   L (-) ---------------------------- supply -

Working:

  1. With the handle at OFF, the circuit is open. The handle is moved slowly to stud 1. The full starting resistance is now in series with the armature, and the shunt field gets full supply voltage through the brass arc and NVC. This gives full flux, so the motor starts with high torque and low current.
  2. As the motor speeds up, EbE_b rises. The handle is moved stud by stud, cutting out resistance sections so the armature current stays within limits.
  3. At the last stud (ON), the whole resistance is cut out and the motor is directly across the supply. The soft-iron piece on the handle touches the NVC electromagnet, which holds the handle there against the spring.
  4. No-volt protection: if the supply fails, the NVC loses its magnetism, the spring returns the handle to OFF, and the motor does not restart with the resistance out when the supply returns.
  5. Overload protection: if the armature current exceeds a set value, the OLR electromagnet lifts its armature and short-circuits the NVC. The NVC releases the handle, and the spring returns it to OFF.

Drawback: the NVC is in series with the field. If the field current is reduced for speed control, the NVC may be too weak to hold the handle. A 4-point starter places the NVC in a separate branch across the supply to avoid this.

✓ Answer checked by Claude Opus 5.5 (high effort)

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