Actuators: injectors, solenoids and stepper motors

Why it matters

Sensors tell the ECU what is happening; actuators are how it acts. Every gram of fuel, every shift in an automatic gearbox and every movement of an idle-air valve, EGR valve or air-conditioning flap is made by an electrically driven actuator. Understanding how injectors, solenoids and stepper motors produce force and position — and their delays and limits — explains how fuel quantity is controlled and why some faults appear only hot, cold or at low battery voltage.

Key ideas

Actuators in general. An actuator converts an electrical command into mechanical motion: linear (solenoids, injectors), rotary in steps (stepper motors) or continuous (DC and brushless motors, e.g. the electronic throttle body, which uses a DC motor with a return spring and a position sensor for closed-loop control). The ECU usually switches them with a transistor on the earth side (low-side driver) and controls the average effect by pulse-width modulation (PWM). Inductive loads need a flyback (freewheel) diode or clamp to absorb the voltage spike at switch-off.

Solenoids. A coil around a movable iron plunger. Current creates flux; the magnetic circuit tries to close its air gap, pulling the plunger in against a spring. The pull is a reluctance force: roughly proportional to (N·I)² and inversely proportional to the gap squared, so it is weak at the start of the stroke and strong near the end. Uses: starter solenoid, door locks, canister purge and EGR valves, shift and pressure-control solenoids in automatic transmissions (on/off or PWM-proportional), ABS valves. Current builds up with the coil's L/R time constant, so response is not instantaneous.

Fuel injectors.

• Port fuel injection (PFI): a solenoid lifts a needle or ball off its seat against a spring; fuel at a regulated pressure (about 3–4 bar above manifold pressure) sprays onto the back of the intake valve. With constant pressure difference, the quantity injected is set by the pulse width (opening time).
• Dead time (opening delay): the injector does not open the instant it is energised — current has to build up and the needle has to accelerate. This delay, typically about 0.5–1.5 ms, grows as battery voltage falls, so the ECU adds a voltage-dependent correction.
• Drivers: high-impedance injectors (about 12–16 Ω) are switched directly ("saturated"); low-impedance ones (about 2–3 Ω) use a "peak-and-hold" current profile for fast opening without overheating.
• Gasoline direct injection (GDI): injects into the cylinder at about 100–350 bar, needing high-voltage, high-current drivers.
• Diesel common rail: solenoid or piezoelectric injectors at very high rail pressure (well above 1000 bar) with several injections per cycle; piezo stacks switch faster.
• Spray quality (atomisation, cone angle, targeting) and flow matching between injectors matter for emissions.

Stepper motors. Rotate in fixed angular steps when their windings are energised in sequence; position is known by counting pulses, so no position sensor is needed (open loop) as long as the motor never misses a step (overload, too fast acceleration).

• Types: permanent-magnet (larger steps, e.g. 7.5°–15°), variable-reluctance, and hybrid (fine steps, typically 1.8°).
• Drive: full-step, half-step (doubles resolution) or micro-stepping.
• Torque: holding torque (energised, stationary) and detent torque (unenergised, PM types); available torque falls as step rate rises.
• Uses: idle-air control valves, instrument-cluster needles, headlamp levelling, HVAC blend flaps, EGR valves, and electronic expansion valves in EV heat pumps. A lead screw converts rotation to precise linear motion.

Links. Pulse widths come from the air-mass calculation in the sensors topic and the maps in the ECU topic.

Formulas

F ≈ (N × I)² × μ₀ × A / (2 × g²)

• F: solenoid pull (N), N: turns, I: current (A), μ₀ = 4π × 10⁻⁷ H/m, A: pole-face area (m²), g: air gap (m). Assumes the iron's reluctance and fringing are negligible; real forces are lower, especially at small gaps where the iron saturates.

I(t) = (V / R) × (1 − e^(−t/τ)), τ = L / R

• Current build-up in a coil after switch-on; V (V), R (Ω), L (H), t (s).

m_inj = q_s × (t_pw − t_d)

• m_inj: fuel mass per injection (kg), q_s: static (fully open) mass flow (kg/s), t_pw: pulse width (s), t_d: dead time (s).

q = q_rated × √(Δp / Δp_rated)

• Injector flow scales with the square root of the pressure difference across it (orifice flow).

Duty = t_pw / t_cycle, t_cycle = 120 / n (four-stroke, one injection per cycle)

• t_cycle: time per engine cycle (s), n: engine speed (rpm).

θ_s = 360° / S, n_step = 60 × f_p / S

• θ_s: step angle (degrees), S: steps per revolution, f_p: pulse rate (steps/s), n_step: shaft speed (rpm).

Worked examples

Example 1 (standard). A port injector flows 200 cm³/min of petrol (density 740 kg/m³) fully open and has a dead time of 0.9 ms. Each cylinder needs 15.6 mg of fuel per cycle at 3000 rpm. Find the pulse width and injector duty cycle.

1. q_s = 200 / 60 cm³/s × 0.74 g/cm³ = 2.467 g/s.
2. Effective opening time = m_inj / q_s = 15.6 × 10⁻³ g / 2.467 g/s = 6.32 ms.
3. t_pw = 6.32 + 0.9 = 7.22 ms.
4. t_cycle = 120 / 3000 = 0.040 s = 40 ms; Duty = 7.22 / 40 = 0.18.

Answer: pulse width about 7.2 ms, duty about 18 %.

Example 2 (GATE level). (a) The fuel pressure difference in Example 1 is raised from 3 bar to 4 bar. Find the new static flow. (b) An idle-air stepper (1.8° per step) drives a lead screw of 1 mm lead at 400 steps/s. Find the shaft speed and the time to move the pintle 3 mm. (c) A solenoid has 500 turns, 1 A, pole area 1 cm² and an initial gap of 1 mm. Estimate the pull at 1 mm and at 0.5 mm.

1. q = 2.467 × √(4/3) = 2.467 × 1.155 = 2.85 g/s.
2. S = 360 / 1.8 = 200; n = 60 × 400 / 200 = 120 rpm.
3. 3 mm needs 3 rev = 3 × 200 = 600 steps; time = 600 / 400 = 1.5 s.
4. F = (500 × 1)² × 4π × 10⁻⁷ × 1 × 10⁻⁴ / (2 × (1 × 10⁻³)²) = 15.7 N.
5. Halving the gap multiplies force by 4: F = 62.8 N at 0.5 mm.

Answer: 2.85 g/s; 120 rpm and 1.5 s; about 15.7 N rising to about 62.8 N (idealised).

Common mistakes

• Using F = B·I·l for a solenoid. That is the Lorentz force on a current-carrying conductor (motors, voice coils); a solenoid plunger is pulled by reluctance force.
• Assuming injected fuel is proportional to the whole pulse width — the dead time delivers no fuel.
• Forgetting that injector flow depends on the square root of pressure difference, not linearly.
• Using 60/n instead of 120/n for the four-stroke cycle time.
• Thinking a stepper knows its position. It only counts pulses; if it stalls or misses steps, the ECU's count is wrong until it re-homes.
• Leaving out the flyback diode in a solenoid driver; the turn-off spike can destroy the transistor.

For GATE ME

Questions draw on basic electromagnetics and fluid mechanics: RL current rise, magnetic force on a plunger, orifice flow ∝ √Δp, fuel quantity and duty cycle from engine speed, and step angle, pulse rate and speed of stepper motors. Practise keeping time units consistent (ms vs s) and the 120/n cycle time.

Quick check

1. Why does an injector's dead time increase at low battery voltage?
2. A stepper has 200 steps per revolution and is pulsed at 1000 steps/s. What is its speed?
3. Why is solenoid pull weakest at the start of its stroke?
4. What actuator usually drives an electronic throttle body?

Answers: 1. Coil current builds up more slowly, so the needle takes longer to open. 2. 60 × 1000 / 200 = 300 rpm. 3. The air gap is largest, and force varies roughly as 1/g². 4. A DC motor with a return spring and position feedback.