When a three-phase induction motor is switched straight onto the supply, it draws a huge starting current, often six or seven times its normal running current. That surge can trip breakers and stress the motor windings. The four main starting methods are direct-on-line (DOL), star-delta, soft starter and variable frequency drive (VFD).
Why Starting Current Is So High
A three-phase induction motor works by induction: the stator creates a rotating magnetic field, and that field induces current in the rotor. At switch-on, the rotor is standing still, so the slip is 100 percent. The current is therefore very high, and it falls as the motor speeds up.
Direct-On-Line (DOL) Starting
DOL starting is the simplest method: a contactor connects the motor straight to the full supply voltage. The motor draws about 6 to 7 times its full-load current and produces its full starting torque. The starter is just a contactor, an overload relay and a start/stop station.
DOL suits small motors, usually up to about 5 to 10 horsepower, and loads that need full starting torque from the first moment. The disadvantage is the high current and the sudden mechanical jerk, so it is not suitable for large motors or fragile loads.
Star-Delta Starting
A star-delta starter reduces the starting current by first connecting the motor windings in star, then changing them to delta. In star, each winding gets only 1/√3, about 58 percent, of the line voltage, so the starting current falls to about one third of the DOL value, roughly 2 to 2.5 times full-load current. A timer then switches the windings to delta.
The starter uses three contactors — main, star and delta — plus a timer. The motor must be able to start with only one third of its normal starting torque, so star-delta is used for light-load starts, such as a fan or a pump. The motor must have all six winding ends brought out.
Diagram: Typical starting current as a multiple of full-load current: DOL about 6.5 times, star-delta 2.2, soft starter 3.5, VFD 1.5.
Soft Starters
A soft starter controls the voltage applied to the motor with thyristors, also called SCRs. At switch-on it applies a low voltage and ramps it up smoothly, so the current rises gradually instead of surging. The starting current is limited to about 3 to 4 times full-load current, and the mechanical jerk is almost gone.
Soft starters cost more than a star-delta starter but far less than a VFD. They are used for pumps, fans and conveyors where a smooth start matters but speed control is not needed. Once running, the motor runs at full speed on the mains supply, usually through a bypass contactor.
Variable Frequency Drives (VFD)
A VFD is the most capable method, and the most expensive. It converts the fixed mains supply to DC and then back to AC at a variable frequency and voltage, so the motor starts at a very low speed and ramps up smoothly. The starting current stays near full-load current, about 1.5 times.
The real advantage of a VFD is that it controls the motor speed for the whole running time, not just the start. That saves a large amount of energy on fans and pumps, where the power needed falls with the cube of the speed: a fan at 80 percent speed needs only about half the power. The trade-offs are cost, shielded cables, and harmonics put back into the supply.
Diagram: A star-delta starter. The main contactor K1 stays closed while running. The star contactor K2 joins the winding ends U2, V2 and W2 together at start. A timer then closes the delta contactor K3 for full-voltage running.
Key Takeaways
Every starting method is a trade-off between current, torque, cost and control. DOL is cheap and gives full torque, but it draws 6 to 7 times full-load current. Star-delta cuts the starting current to about one third, but only suits light-load starts. A soft starter ramps the voltage smoothly and limits the current to 3 to 4 times full-load, but it cannot control speed. A VFD starts at about 1.5 times full-load current, but it costs the most.
Objective Questions
1. Why does an induction motor draw a very high current at the moment of switch-on?
a) The windings are damaged
b) The rotor is standing still, so the slip is 100 percent and the motor behaves almost like a short circuit
c) The supply voltage is too low
d) The overload relay is set too high
Answer: (b) The rotor is standing still, so the slip is 100 percent and the motor behaves almost like a short circuit
Explanation: At switch-on the rotor is at standstill, so the slip is 100 percent and the motor draws a very high current, which falls as the motor speeds up.
2. About how much starting current does a DOL starter draw?
a) 1.5 times full-load current
b) 2.2 times full-load current
c) 3.5 times full-load current
d) 6 to 7 times full-load current
Answer: (d) 6 to 7 times full-load current
Explanation: A DOL starter connects the motor straight to full voltage, so it draws about 6 to 7 times its full-load current at start.
3. In a star-delta starter, how much voltage does each motor winding get while connected in star?
a) Full line voltage
b) About 58 percent of the line voltage
c) Half of the line voltage
d) Double the line voltage
Answer: (b) About 58 percent of the line voltage
Explanation: In star, each winding gets 1/√3, about 58 percent, of the line voltage, so the starting current falls to about one third of the DOL value.
4. What is the main advantage of a soft starter over a star-delta starter?
a) It is cheaper
b) It ramps the voltage smoothly, so the current rises gradually and there is almost no mechanical jerk
c) It can control the motor speed for the whole run
d) It needs no contactors
Answer: (b) It ramps the voltage smoothly, so the current rises gradually and there is almost no mechanical jerk
Explanation: A soft starter uses thyristors to ramp the voltage up over a few seconds, limiting the starting current to about 3 to 4 times full-load and removing the mechanical shock.
5. A fan running on a VFD is slowed from 100 percent speed to 80 percent speed. About how much power does it now need?
a) 80 percent
b) About half
c) 20 percent
d) The same as before
Answer: (b) About half
Explanation: For fans and pumps, power falls with the cube of the speed: 0.8 × 0.8 × 0.8 is about 0.51, so the fan needs only about half the power at 80 percent speed.
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