Motor control is core craftsmanship in industrial and automation installations. Here are the four start methods (DOL, star-delta, soft start, frequency converter), the correct setting of motor protection, the frequency converter's EMC cabling and leakage currents, the correct RCD choice — and machinery safety according to DS/EN 60204-1.
Introduction
An induction motor is the most common motor type in industry — robust, cheap and reliable. But it draws a strong starting current, and must be protected against overload in a different way than an ordinary circuit. Motor installation is therefore about three things: getting the motor running without unnecessarily loading the grid, protecting it against overheating, and being able to control its speed. The governing standards are DS/EN 60204-1 (electrical equipment of machinery) and DS/EN 60947-4-1 (contactors and motor starters).
The starting current problem
An induction motor draws typically 5–8 times its rated current in the brief starting moment until it builds up speed. On small motors it does not matter, but on larger ones the starting current can cause voltage dips in the network and require a larger supply. Therefore several starting methods exist that reduce the starting current at the expense of starting torque.
| Starting method | Starting current (approx.) | Starting torque | Typical use |
|---|---|---|---|
| Direct (DOL) | 5–8 · In | Fully | Small motors where starting current is unproblematic |
| Star-delta (Y/Δ) | ~1/3 of DOL | ~1/3 of DOL | Pumps and fans (low starting torque sufficient) |
| Soft starter | Adjustable (soft ramp) | Reduced, controlled | Conveyor belt where soft start is desired |
| Frequency converter (VFD) | Low, controlled | Full from 0 rpm | Where variable speed is desired |
Star-triangle — why a third?
With star-delta starting, the motor is started in star connection (Y) and switched to delta (Δ) when it has reached speed. In star, each motor winding only gets the phase voltage (line voltage divided by √3), approximately 58% of delta voltage. Since both current and torque depend on voltage in a squared relationship, both starting current and starting torque drop to approximately one-third of direct starting.
Example: A motor that draws 6 · In at direct start will draw only about 2 · In in the star stage at star-delta start. The price is that the starting torque is also only a third — therefore the method is only suitable for loads with low resistance torque at start, such as centrifugal pumps and fans, not for a fully loaded conveyor. (Web-verified: Y/Δ reduces starting current AND torque to about 1/3.)
Read the motor's nameplate
Before you connect a motor, you read the type plate. The voltage is shown as TWO numbers each with its own connection symbol, e.g. '230 V Δ / 400 V Y' or '400 V Δ / 690 V Y'. The low number is the voltage each individual winding may have across it. The choice of triangle (Δ) or star (Y) in the terminal box is therefore determined by the MAINS voltage: the winding must have its rated voltage — neither more nor less.
- 01Motor rated 230 V Δ / 400 V Y on a 400 V supply → connect in STAR (Y). Each winding then gets 400/√3 = 231 V ≈ 230 V, exactly what the winding can take. Delta here would put 400 V across a 230 V winding, and the motor burns out.
- 02Motor marked 400 V Δ / 690 V Y on a 400 V network → connect in DELTA (Δ). Each winding gets the full 400 V that it can handle. It is precisely this motor type that can be star-delta STARTED on 400 V (start in Y, switch to Δ), because the winding can handle 400 V during operation.
Worked example — a typical rating plate may look like this: »3~ 400 V Δ / 690 V Y · 4.0 kW · cos φ 0.83 · 1450 rpm · IE3 · In 8.3 A (400 V Δ)«. Procedure: (1) The supply is 400 V → connect in DELTA, because the 400 V winding must have 400 V. (2) Read the rated current In = 8.3 A directly from the plate — do not calculate it yourself when the figure is there. (3) The motor protection (thermal relay) is set to In, adjustable 100–115 %, i.e. 8.3–9.5 A. cos φ and the efficiency are used if you need to calculate the power INPUT: the plate output is the shaft power delivered, and power input = P/η, while P ≈ √3 · U · In · cos φ · η links the figures together.
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