Star (Y) and delta (Δ) connection of three-phase windings

Two ways to connect three windings

A three-phase motor, transformer or generator has three identical windings, one per phase. Each winding has two ends, a start and a finish, and they can be joined in two ways.

• Star (Y, wye): one end of each winding is tied to a common point, the neutral point. The other three ends go to the supply lines L1, L2 and L3.
• Delta (Δ): the three windings are joined head to tail in a closed triangle. Each corner of the triangle goes to one supply line.

The supply line values are called line quantities (V_L, I_L). The values across and through one winding are called phase quantities (V_ph, I_ph). The two connections differ in how these relate.

Star connection: V_L = √3 × V_ph

In star, each line is attached to one winding, so the line current is the winding current:

I_L = I_ph

The line voltage is measured between two lines, i.e. across two windings in series, whose voltages are 120° apart. Their phasor difference is √3 times one phase voltage:

V_L = √3 × V_ph

On a 400 V network each winding therefore sees 400 / √3 ≈ 231 V. A star connection also provides a neutral point, which a generator or transformer can bring out as a fourth wire.

Delta connection: I_L = √3 × I_ph

In delta, each winding is connected directly between two lines, so the winding sees the full line voltage:

V_ph = V_L

Each line current splits into two winding currents, 120° apart, and the line current is their phasor difference:

I_L = √3 × I_ph

On a 400 V network each winding of a delta-connected machine sees 400 V, which is √3 times more than in star. The same winding at the same line voltage therefore draws √3 times the phase current and three times the line current and power of the star connection.

Power is the same formula for both

Total power is the sum of three phase powers. In star, 3 × V_ph × I_ph × cos φ with V_ph = V_L / √3 gives

P = √3 × V_L × I_L × cos φ

In delta, 3 × V_ph × I_ph × cos φ with I_ph = I_L / √3 gives exactly the same expression. This is why the nameplate and the power formula always use line values, whatever the connection.

The angle φ is the phase angle between a phase voltage and the current in the same phase, and cos φ is the power factor.

The terminal box and the 230/400 V nameplate

A standard motor brings both ends of each winding to the terminal box: U1, V1, W1 are the starts and U2, V2, W2 are the finishes. They are arranged in two rows, with U1, V1, W1 under W2, U2, V2.

• Star: three links join U2, V2 and W2 together. The supply goes to U1, V1 and W1.
• Delta: three links join the terminals one above the other (U1 with W2, V1 with U2, W1 with V2). The supply goes to the three link points.

The nameplate says which connection suits which supply, for example Δ/Y 230/400 V. Each winding is designed for 230 V. On a 230 V line-to-line supply you use delta, and on a 400 V line-to-line supply you use star. In both cases the winding sees 230 V, so the motor produces the same torque, speed and power. A motor plate marked 400/690 V Δ/Y follows the same rule for 400 V and 690 V networks.

Connecting a 230/400 V motor in delta on 400 V puts 400 V across windings designed for 230 V. The current rises sharply and the winding overheats.

Worked example

A 4 kW motor is marked Δ/Y 230/400 V, 8.2 A at 400 V, cos φ = 0.82, and runs on a 400 V network, so it is wired in star.

1. Winding voltage: V_ph = 400 / √3 ≈ 231 V. 2. Winding current: I_ph = I_L = 8.2 A. 3. Input power: P = √3 × 400 × 8.2 × 0.82 ≈ 4.66 kW. With 4 kW at the shaft, the efficiency is about 86%.

Now the same motor on a 230 V network, wired in delta. Each winding still sees 230 V and carries 8.2 A, so the line current is

I_L = √3 × I_ph = 1.732 × 8.2 ≈ 14.2 A

and P = √3 × 230 × 14.2 × 0.82 ≈ 4.64 kW, the same within rounding. The lower voltage requires more line current for the same power.

Finally, star-delta starting. Take a motor marked Δ/Y 400/690 V on a 400 V network. It runs in delta, with 400 V across each winding. For starting, the windings are first connected in star, so each gets 400 / √3 ≈ 231 V, which is 1/√3 of the running voltage. The phase current falls by 1/√3, and because the star line current equals the phase current, the line current is 1/√3 × 1/√3 = 1/3 of the direct-on-line value. Torque follows the square of the winding voltage, so it also falls to 1/3.

Try it in the three-phase induction motor simulator

The simulator does not have a switch for the terminal box connection; the motor is described by its line values. It does, however, model the star-delta connection in the starting method. Open the three-phase induction motor simulator and keep the default 4 kW, 400 V, 8.2 A, four-pole motor.

• Under Starting method, choose Direct on line (DOL) and note the Peak starting current (× In) and Start time readouts.
• Switch to Star-delta. The simulator applies 1/√3 of the voltage until the switch-over time (default 3 s, set by Switch-over time to full voltage) and then full voltage. The chart marks the change with a Y→Δ label.
• Compare the Peak starting current (× In) of the two runs. During the star stage the line current is about one third of the direct start; the peak after the switch-over to delta can be higher than that, so compare the readout with the shape of the current curve.
• Raise Mechanical load while in star-delta. With only a third of the torque available during the star stage, expect a longer Start time under a heavy load.

The wound-rotor induction motor simulator has the same Starting method control. The 3D motor model includes a terminal box cover, but the star and delta links themselves are not modelled.

Try it in the simulator