Starting and speed control of induction motors
Why induction motors need starters, DOL, star–delta, autotransformer, soft and rotor-resistance starting, and speed control by poles, voltage, V/f and rotor resistance.
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Why it matters
An induction motor switched straight onto the supply draws five to eight times its full-load current, which dips the plant voltage and stresses the machine and coupling. Choosing a starter (DOL, star–delta, autotransformer, soft starter, rotor resistance) and a speed-control method (pole changing, voltage, V/f drive, rotor resistance) is a routine design and maintenance decision, and variable-frequency drives on pumps and fans are one of the biggest energy savers in industry.
Key ideas
Why starting current is high. At standstill s = 1, so the rotor branch resistance R2'/s is small and the motor looks like a transformer with a short-circuited secondary. Starting current Isc is typically 5–8 times full-load current, at a poor power factor. Starting torque, however, is only about 1–2 times full-load torque.
Starting torque relation. Since T ∝ I2²·R2/s, at s = 1 and at full-load slip sfl: Tst/Tfl = (Ist/Ifl)²·sfl. Any method that reduces the phase voltage by a factor x reduces starting current per phase by x and starting torque by x².
Starting methods for cage motors
- Direct-on-line (DOL): full voltage; simplest; used for small motors (a few kW) where the supply can tolerate the current.
- Stator resistance or reactor: series impedance drops the voltage; wasteful, rarely used now.
- Autotransformer starter: tapping x (e.g. 0.5–0.8) on the motor side. Motor current falls by x, but the line current falls by x² (autotransformer action), as does the torque. Best torque per ampere drawn from the line.
- Star–delta starter: for motors designed to run in delta. Started in star, each phase gets V/√3, so phase current and torque both fall to 1/3 of the DOL value, and so does the line current. Equivalent to an autotransformer with x = 1/√3. Cheap; the torque reduction is fixed.
- Soft starter: back-to-back thyristors ramp the voltage; smooth, adjustable current limit, but torque still falls as V².
Slip-ring (wound-rotor) motors are started with external rotor resistance. This lowers the starting current and raises the starting torque (up to Tmax when R2' = X), then the resistance is cut out in steps. Used for cranes, hoists and mill drives with heavy starting loads.
Speed control. N = Ns·(1 − s) = (120·f/P)·(1 − s), so speed can be changed by P, f or s.
- Pole changing: separate windings or a consequent-pole (Dahlander) winding give two or more fixed speeds (e.g. 1500/750 rpm). Cage rotor only, because a cage automatically adapts to any pole number.
- Stator voltage control: reducing V lowers torque (∝ V²) so slip rises. Narrow range, poor efficiency (high slip means high rotor loss); suitable only for fan-type loads.
- Frequency control (V/f): a variable-frequency drive (rectifier + PWM inverter) changes Ns smoothly. V is varied in proportion to f to keep the air-gap flux (∝ V/f) constant: lowering f at full V would saturate the core. Below base frequency this gives constant maximum torque; above base frequency V stays at rated and flux weakens — constant-power region. With constant V/f and low slip, the slip speed at a given torque is roughly constant, so the speed–torque lines shift parallel to each other.
- Rotor resistance (slip-ring motors): at a given torque in the low-slip region s ∝ R2, so speed falls as resistance is added. Simple but the slip power s·Pag is wasted in the resistors.
- Slip-power recovery (static Kramer / Scherbius drives) returns the rotor slip power to the supply through converters instead of burning it.
- Cascade connection of two motors gives additional speeds 120·f/(P1 ± P2).
Formulas
Tst / Tfl = (Ist / Ifl)² · sfl— starting-to-full-load torque ratio (DOL values for Ist).Reduced voltage x·V: I_motor = x·Isc ; Tst = x²·Tst,DOL.Autotransformer starter: I_line = x²·Isc.Star–delta: I_line = Isc,DOL / 3 ; Tst = Tst,DOL / 3.Ns = 120·f / P ; N = Ns·(1 − s)— speed (rpm).V / f = constant— keeps air-gap flux constant below base speed.Low-slip region at constant torque: s ∝ R2 (rotor resistance control); s ∝ 1/V² (voltage control).N_cascade = 120·f / (P1 ± P2)— cascaded motor speeds (rpm).
Worked examples
Example 1 (standard): comparing starters. Given: a delta-connected cage motor has DOL starting current 6 times full-load current and full-load slip 4 %.
- DOL:
Tst/Tfl = (Ist/Ifl)²·sfl= 6² × 0.04 = 1.44; line current 6·Ifl. - Star–delta: phase voltage 1/√3, so Tst = 1.44/3 = 0.48 Tfl; line current = 6/3 = 2 Ifl.
- Autotransformer with 70 % tapping (x = 0.7): motor current 0.7 × 6 = 4.2 Ifl; line current = 0.7² × 6 = 2.94 Ifl; Tst = 0.49 × 1.44 = 0.71 Tfl.
- Star–delta draws less line current but gives too little torque if the load needs more than about half its full-load torque at standstill; the autotransformer gives more torque per line ampere.
Example 2 (GATE level): V/f control and rotor resistance. Given: a 4-pole, 50 Hz, 400 V motor runs at 1440 rpm at rated load torque. (a) A V/f drive reduces the supply to 30 Hz and 240 V; load torque unchanged.
- Slip speed at rated torque: 1500 − 1440 = 60 rpm. With constant V/f and low slip, slip speed at a given torque stays the same.
- New synchronous speed: Ns = 120 × 30/4 = 900 rpm.
- New speed = 900 − 60 = 840 rpm. (b) Instead, the motor is a slip-ring motor whose rotor-circuit resistance is doubled (50 Hz, 400 V, same torque).
- At constant torque in the low-slip region s ∝ R2, so s = 2 × 0.04 = 0.08.
- Speed = 1500 × (1 − 0.08) = 1380 rpm; rotor copper loss doubles to 8 % of air-gap power.
Common mistakes
- Saying star–delta reduces line current to 1/√3; it reduces it to 1/3 (phase current falls by √3 and the delta-to-star change removes another √3).
- Forgetting that line current with an autotransformer starter falls as x², not x.
- Lowering frequency at constant voltage — flux rises and the core saturates.
- Using pole changing on a wound rotor; its winding has a fixed pole number.
- Believing rotor resistance control is efficient; the slip power is lost as heat.
- Assuming starting torque equals full-load torque times the current ratio; torque scales with current squared times slip.
For GATE IN
Expect starting-torque and line-current comparisons (DOL, star–delta, autotransformer), speed after a frequency or rotor-resistance change, and MCQs on V/f control, pole changing and which method suits which rotor. VFD and soft-starter operation links to power electronics and to process-control actuators.
Quick check
- DOL Ist = 5 Ifl, sfl = 0.05. Find Tst/Tfl.
- What fraction of DOL line current does a star–delta starter draw?
- Why must voltage be reduced along with frequency?
- Which speed-control method suits only cage rotors?
Answers: 1. 25 × 0.05 = 1.25. 2. One third. 3. To keep V/f, and hence flux, constant so the core does not saturate. 4. Pole changing.
Interview questions
All Electrical Machines interview questionsTry answering each one aloud before you open it.
1.What is the purpose of a starter in an induction motor?Concept
At standstill the motor behaves like a short-circuited transformer and would draw 5–8 times full-load current, causing a supply voltage dip, heating and mechanical shock. A starter limits this current, either by reducing the applied voltage (star–delta, autotransformer, soft starter) or, for slip-ring motors, by adding rotor resistance, which also raises starting torque. It also provides overload and no-volt protection. Small motors can be started direct-on-line when the supply permits.
2.Explain the working principle of a star-delta starter.Concept
A motor designed to run in delta is first connected in star, so each phase sees V/√3. Phase current falls by √3 and, because of the star connection, the line current drawn is one third of the DOL value; starting torque, proportional to phase voltage squared, also falls to one third. Near running speed the starter switches the windings to delta for full voltage. It is cheap but the torque reduction is fixed, so it suits loads with low starting torque such as pumps and fans.
3.Why is rotor resistance control used for speed control in slip ring induction motors?Application
Rotor resistance control is used in slip ring induction motors to vary the speed by inserting external resistors into the rotor circuit. By increasing the rotor resistance, the slip increases, which reduces the speed of the motor. This method is effective for speed control but results in power loss due to the resistors, making it suitable for applications where speed control is required for short durations.
4.What happens if an induction motor is started directly on line without a starter?Application
If an induction motor is started directly on line without a starter, it will draw a very high inrush current, typically 5 to 7 times its full-load current. This can cause a significant voltage drop in the power supply network, potentially affecting other equipment. Additionally, the high current can cause excessive heating and mechanical stress on the motor windings, leading to reduced lifespan or damage.
5.How does a variable frequency drive (VFD) control the speed of an induction motor?Concept
A VFD rectifies the supply to a DC link and uses a PWM inverter to produce a variable-frequency, variable-voltage output, changing the synchronous speed 120·f/P. Below base speed it keeps V/f constant so the air-gap flux and the available torque stay at rated values; above base frequency the voltage is held at rated and the flux weakens (constant-power region). Because the motor always runs at low slip, efficiency stays high, and the drive also gives soft starting and controlled braking.
6.Why is a soft starter preferred over a direct-on-line starter for large motors?Application
A soft starter is preferred over a direct-on-line starter for large motors because it provides a gradual increase in voltage, reducing the inrush current and mechanical stress on the motor. This results in smoother acceleration, less wear and tear on the motor components, and reduced electrical disturbances in the power supply network. Soft starters are particularly beneficial in applications where frequent starting and stopping are required.
7.Calculate the starting current of a 3-phase induction motor with a full-load current of 50 A if started directly on line.Numerical
If the motor is started directly on line, the starting current can be approximately 5 to 7 times the full-load current. Assuming a factor of 6, the starting current would be: Starting Current = 6 × Full-load Current = 6 × 50 A = 300 A.
8.A 4-pole induction motor is connected to a 50 Hz supply. Calculate its synchronous speed.Numerical
The synchronous speed (Ns) of an induction motor is calculated using the formula: Ns = 120 × Frequency / Number of Poles. For a 4-pole motor connected to a 50 Hz supply, Ns = 120 × 50 / 4 = 1500 RPM.
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