Synchronous motor: load angle, pull-out torque and V curves

Constant speed: the rotor locks to the field

The stator of a three-phase synchronous motor is wound like that of an induction motor and produces a rotating field at

N_s = 120 × f / p

The rotor carries a DC-excited field winding (fed through two slip rings and brushes) or permanent magnets, so it is a magnet of fixed polarity. A rotating magnet is pulled around by the rotating stator field and locks to it, like two hands held together while one turns.

There is no slip. A 4-pole motor on 50 Hz turns at 1500 rpm at no load and at full load alike. Mechanical load does not slow the rotor; it makes the rotor lag the field by a larger angle. Speed can only be changed by changing the supply frequency.

Load angle and the power equation

In the per-phase model the motor is a supply voltage V opposed by an internal (excitation) voltage E_f produced by the rotor field, with a synchronous reactance X_s between them. The angle by which E_f lags V is the load angle δ. For a round-rotor machine with negligible stator resistance, the power drawn is

P = 3 × V × E_f × sin δ / X_s

with V and E_f as phase voltages. Torque is P divided by the synchronous angular speed.

At no load δ is near zero. As the shaft load grows, δ grows and the motor takes more power from the supply. The maximum power occurs at δ = 90°:

P_max = 3 × V × E_f / X_s

This is the pull-out (breakdown) power. Beyond it the electrical torque falls as δ increases further, the rotor cannot keep up and the motor loses synchronism. Stable operation is therefore only on the rising part of the curve, below 90°, and in practice engineers keep δ well below it, typically well below 40° at full load, to leave a margin for transients.

Because P_max is proportional to E_f, raising the field current raises the pull-out torque. It is also proportional to V, so a voltage dip lowers the margin.

V curves, excitation and power-factor correction

At constant shaft power, the in-phase component of the armature current is fixed, because it carries the real power. What the field current changes is the reactive component:

• Under-excited (E_f small): the motor takes lagging current from the grid, like an induction motor or an inductor.
• Normal excitation: the current is purely in phase with V, the power factor is unity and the current is at its minimum.
• Over-excited (E_f large): the motor takes leading current, so it behaves like a capacitor and delivers reactive power to the grid.

If you plot armature current against field current at constant load, you get a V-shaped curve with its minimum at unity power factor. Each load has its own V, and heavier loads give higher curves whose minimum sits at a higher field current. Joining the minima gives the unity-power-factor line. The left side of each V is lagging, the right side is leading.

At heavy load the left side of the V ends where the field is too weak to hold synchronism, because a weak field means a low P_max.

Most industrial loads (induction motors, transformers) draw lagging current. An over-excited synchronous motor draws leading current, so it can drive a load and, at the same time, supply the reactive power the other machines need. This raises the plant power factor, reduces line current and losses, and can reduce utility penalties.

A synchronous motor running with no mechanical load, only to supply or absorb reactive power, is called a synchronous condenser (or synchronous compensator). Because its excitation is adjustable, it can be set to generate or absorb reactive power continuously, which large grids use for voltage support.

The cost is the exciter, the slip rings and brushes (or a brushless exciter), and the extra copper loss in the field winding.

Worked example

Take a four-pole, 50 Hz, three-phase motor on a 400 V line (V = 230 V per phase) with X_s = 2 Ω per phase, carrying 50 kW of electrical power. Ignore resistance.

1. Speed: N_s = 120 × 50 / 4 = 1500 rpm, independent of load. 2. In-phase current: I_p = 50 000 / (3 × 230) ≈ 72.5 A. This is fixed at constant power. 3. The quantity E_f × sin δ must equal P × X_s / (3 × V) = 50 000 × 2 / 690 ≈ 145 V.

Now vary the excitation voltage E_f:

• E_f = 200 V: sin δ = 145 / 200, δ ≈ 46°. The reactive current is (E_f cos δ − V) / X_s ≈ −46 A, so the motor is under-excited. Armature current ≈ 86 A, power factor ≈ 0.84 lagging.
• E_f ≈ 272 V: sin δ = 0.53, δ ≈ 32°, reactive current ≈ 0. Armature current ≈ 72.5 A, unity power factor. This is the bottom of the V.
• E_f = 300 V: δ ≈ 29°, reactive current ≈ +16 A. Armature current ≈ 74 A, power factor ≈ 0.98 leading.

Pull-out power at E_f = 300 V is 3 × 230 × 300 / 2 ≈ 103.5 kW, about twice the load, so the motor has a comfortable margin. At E_f = 200 V it is only 69 kW.

Starting: why a synchronous motor cannot start itself

At standstill the stator field sweeps past the rotor at synchronous speed. The rotor, with its inertia, cannot accelerate fast enough to lock on: the torque alternates between forward and backward and averages zero. A synchronous motor therefore has no net starting torque on its own.

The usual fix is a damper (amortisseur) winding, a cage of bars in the pole faces. The motor starts as an induction motor with the field winding unexcited (often closed through a discharge resistor), accelerates to about 95% of synchronous speed, then the DC field is applied and the rotor pulls into step. Other methods are a small pony motor, or a variable-frequency drive that raises the stator frequency from zero so that the rotor can follow.

How the simulator models the synchronous motor

The simulator treats the motor as a round-rotor machine per phase, V = E_f + j × X_s × I, with X_s set to 1.0 per unit of the rated base impedance. The excitation voltage follows the field current with a mild saturation curve, calibrated so that rated field current gives rated voltage, current and power factor.

The power-angle relation sin δ = P × X_s / (3 × V × E_f) fixes the load angle, and the armature current splits into the in-phase and reactive parts described above. If the load exceeds P_max the model reports loss of synchronism: the rotor drops out of step, speed falls to zero and no stable operating point is shown.

Simplifications to know about: although the 3D rotor has salient poles, the reluctance torque of saliency is neglected; the model does not solve the swing equation, so there are no hunting oscillations; and saturation applies only to the open-circuit curve. The start-up is a simplified sequence on the damper winding, with the field applied near full speed. It is a teaching model, not a tool for selecting or protecting real machines.

Try it in the synchronous motor simulator

Open the synchronous motor simulator and start from one of the three templates (100, 250 or 500 kW).

• Move the Field (excitation) current slider. Its hint shows the value that gives unity power factor at the present load. Below it the motor is under-excited (lagging), above it over-excited (leading), and the Reactive power readout changes sign (negative means delivered to the grid).
• Read the V curves chart (armature current vs. field current) at 0%, 50% and 100% load. The operating point slides along the curve for the present load, and the dashed unity power factor line joins the minima.
• Raise the mechanical load and watch the Load angle δ and Stability margin readouts, and the point climbing the Power–angle curve.
• Push the load past 100% and reduce the field current until the loss-of-synchronism warning appears (the chart then shows "Out of step at this field current"). Raise the field current again to restore the margin.
• Watch the 3D rotor: when locked it follows the field a load angle behind, and the model shows the slip rings, brush gear and a DC excitation supply.

The lesson is that load sets δ, while the field current sets the power factor.

Try it in the simulator