Starting system: starter motor and drives

Why it matters

An IC engine cannot start itself: it must be spun fast enough to draw in charge, compress it and fire, and the starter motor does this for a few seconds while drawing a few hundred amperes from the battery. Knowing how the motor, solenoid and drive work lets you size a starter for an engine, diagnose "click but no crank" complaints, and understand why start-stop vehicles use heavier-duty starters.

Key ideas

What the starter has to do. It must turn the crankshaft at a minimum cranking speed against compression, friction and cold, thick oil. Typical minimum cranking speeds are about 70–150 rpm for petrol engines and 100–200 rpm for diesels, which need more speed and much more torque because of their higher compression ratio. The required torque rises steeply in cold weather.

The motor. Most starters are DC motors running on 12 V (24 V on heavy vehicles):

• Series-wound DC motor: field winding in series with the armature, so the same current flows through both. Torque is roughly proportional to I² (until the iron saturates), which gives a very high torque at stall — exactly what is needed to break the engine away. Speed rises as load falls; a series motor must never run unloaded for long or it over-speeds.
• Permanent-magnet motors: field from ferrite or rare-earth magnets, lighter and smaller, usually combined with an internal planetary reduction gear so the small, fast armature still gives high torque.
• Main parts: yoke and pole shoes with field coils (or magnets), armature with a laminated core and heavy copper conductors, commutator, carbon–copper brushes, end frames and bushes, the solenoid and the drive.

Gearing. The small pinion (about 9–11 teeth) meshes with the ring gear on the flywheel (about 100–150 teeth), giving a reduction of roughly 10:1 to 15:1. Gear-reduction starters add an extra stage (about 3:1 to 5:1) so the armature can spin faster and be smaller.

The solenoid. A heavy-duty electromagnetic switch mounted on the starter. When the key (through a starter relay and the neutral-safety or clutch switch) energises it:

• the pull-in winding (low resistance, in series with the motor) and hold-in winding (higher resistance, to earth) together pull the plunger in;
• the plunger moves the shift (fork) lever, pushing the pinion into mesh, then closes the main contacts that connect the battery cable to the motor;
• once the contacts close, the pull-in winding is shorted out (both its ends are at battery voltage) and only the hold-in winding keeps the plunger in, reducing current and heating.

Drives.

• Bendix (inertia) drive: the pinion sits on a helical sleeve on the armature shaft. When the armature suddenly spins, the pinion's inertia makes it run along the helix into mesh. When the engine fires and the ring gear drives the pinion faster than the armature, it is thrown back out. Simple, but noisy and harsh; now rare.
• Pre-engaged drive (most cars today): the solenoid lever pushes the pinion into mesh before full current flows, so engagement is gentle. A roller-type overrunning (one-way) clutch between the pinion and the shaft transmits torque from motor to engine but freewheels when the engine drives the pinion faster, so the engine cannot over-speed the armature while the key is still held.
• Sliding-armature and multi-plate clutch drives: on large diesel engines, for very high torque.

Circuit and protection. Thick, short battery and earth cables keep voltage drop low (a typical limit is about 0.5 V on each side during cranking). Starters are rated for short-time duty: crank for no more than about 10–15 s, then rest to let it cool.

Links. The battery (previous topic) sets how much current is available; the ignition and fuel systems (later topics) must fire within the first few revolutions.

Formulas

P_elec = V × I

• P_elec: electrical input power (W), V: terminal voltage at the motor while cranking (V), I: current (A).

P_mech = T × ω, ω = 2π n / 60

• P_mech: mechanical power (W), T: torque (N·m), ω: angular speed (rad/s), n: speed (rpm).

G = z_ring / z_pinion, n_motor = G × n_engine, T_motor = T_engine / (G × η_gear)

• G: gear ratio (dimensionless), z: number of teeth, η_gear: gear efficiency (dimensionless). Valid while the pinion is in mesh.

E_b = V − I × (R_a + R_f)

• E_b: back emf (V), R_a: armature resistance (Ω), R_f: series field resistance (Ω). Gross mechanical power developed = E_b × I.

T ∝ Φ × I; for a series motor below saturation T ∝ I²

• Φ: flux per pole (Wb). Use as a ratio between two operating points.

η_motor = P_mech / P_elec

Worked examples

Example 1 (standard). A petrol engine needs 120 N·m to crank at 150 rpm. The ring gear has 120 teeth and the pinion 10 teeth, gear efficiency 90 %. The motor efficiency is 55 % and the terminal voltage while cranking is 10 V. Find the motor torque, motor speed, motor output power and battery current.

1. G = 120 / 10 = 12.
2. T_motor = T_engine / (G × η_gear) = 120 / (12 × 0.9) = 11.11 N·m.
3. n_motor = G × n_engine = 12 × 150 = 1800 rpm, ω = 2π × 1800 / 60 = 188.5 rad/s.
4. P_mech = T × ω = 11.11 × 188.5 = 2094 W (check: engine needs 120 × 15.71 = 1885 W; 1885 / 0.9 = 2094 W).
5. I = P_mech / (η_motor × V) = 2094 / (0.55 × 10) = 381 A.

Answer: 11.1 N·m, 1800 rpm, about 2.09 kW, about 381 A.

Example 2 (GATE level). A series starter motor has R_a + R_f = 0.012 Ω and sees 10.5 V at its terminals. At 300 A it delivers 10 N·m (neglect rotational losses). Find (a) the back emf and speed at 300 A, (b) the torque at 400 A assuming no saturation, (c) the stall current.

1. E_b = V − I(R_a + R_f) = 10.5 − 300 × 0.012 = 6.9 V.
2. Gross power = E_b × I = 6.9 × 300 = 2070 W; ω = P / T = 2070 / 10 = 207 rad/s; n = 207 × 60 / (2π) = 1977 rpm.
3. T ∝ I²: T_400 = 10 × (400 / 300)² = 17.8 N·m.
4. At stall E_b = 0, so I_stall = V / (R_a + R_f) = 10.5 / 0.012 = 875 A (in practice the battery voltage sags at this current, so the real stall current is lower).

Answer: 6.9 V and about 1980 rpm; about 17.8 N·m; 875 A.

Common mistakes

• Calling V × I the "power output"; it is electrical input. Output is T × ω, smaller by the motor efficiency.
• Forgetting the pinion-to-ring-gear ratio: motor torque is the engine torque divided by the ratio, and motor speed is multiplied by it.
• Using the battery's open-circuit voltage instead of the much lower terminal voltage during cranking.
• Thinking the overrunning clutch pulls the pinion out of mesh. It only freewheels; the return spring and solenoid release withdraw the pinion.
• Assuming torque is proportional to current for a series motor; below saturation it is proportional to current squared.
• Diagnosing a "click, no crank" as a bad starter without checking battery state and cable voltage drop first.

For GATE ME

Starter questions reduce to standard machine and electrical numericals: gear ratio from teeth, torque and speed transfer through a gear pair with efficiency, P = Tω, motor efficiency from electrical input, and DC series-motor relations (back emf, T ∝ I², stall current). Practise converting rpm to rad/s and keeping track of where efficiency divides or multiplies.

Quick check

1. Why is a series-wound DC motor suited to engine starting?
2. What is the function of the overrunning clutch in a pre-engaged starter?
3. Why is the pull-in winding cut out once the main contacts close?
4. Pinion 9 teeth, ring gear 117 teeth, engine cranked at 200 rpm: what is the armature speed?

Answers: 1. It gives very high torque at low speed and at stall because torque rises roughly with current squared. 2. It transmits torque from the motor to the engine but freewheels when the engine overruns, protecting the armature. 3. Both its ends are then at battery voltage, so it carries no current; the hold-in winding alone keeps the plunger in with less heating. 4. 13 × 200 = 2600 rpm.