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Motor starting voltage dip calculator
Estimate how far the voltage drops at the motor and at the supply bus when an induction motor starts, for direct-on-line, star-delta, autotransformer and soft starters. Free, no signup.
- DOL, star-delta, autotransformer, soft starter
- Results update as you type
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Voltage dip at the motor terminals
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Voltage during the start: —
Check the inputs: positive values, power factors between 0 and 1, efficiency up to 100 %.
- Voltage dip at the supply bus
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- Starting current, with the dip
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- Starting current at full voltage
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- Rated current
- —
- Starting power at full voltage
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- Starting torque, share of full-voltage locked-rotor torque
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- Supply impedance
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- Cable impedance
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Steady-state estimate at the moment of starting, with the rest of the network unloaded. With star-delta, autotransformer and soft starters the voltage is on the supply side of the starter.
How the calculation works
While it starts, an induction motor behaves like a fixed impedance at a low power factor. The supply and the cable are in series with it, so the voltage divides between them: the larger the starting current, the deeper the dip.
- Rated and starting currentIn = P / (√3 · U · η · cos φ), Ist = k · (ILR/In) · Ink = 1 direct on line, 1/3 star-delta, tap² for an autotransformer and Ilimit/ILR for a soft starter.
- Motor impedance while startingZM = U / (√3 · Ist) at cos φst
- Supply and cableZS = (uk/100) · U² / SrT or U² / S″k, ZC = (R + jX) · L / nSplit into R and X with the X/R ratio.
- Voltage at the motor and dipUM / U = |ZM| / |ZS + ZC + ZM|, dip = 1 − UM / U
- Starting torqueTst / TLR = kT · (UM / U)²kT = 1 direct on line, 1/3 star-delta, tap² autotransformer, (Ilimit/ILR)² soft starter.
Worked example
A 110 kW, 400 V motor (η = 95 %, cos φ = 0.86, ILR/In = 7, cos φst = 0.3) started direct on line from a 1000 kVA, 6 % transformer through 100 m of 95 mm² copper cable:
| Rated current | 194.3 A |
|---|---|
| Starting current at full voltage | 1,360 A |
| Voltage dip at the motor | 12.8 % |
| Voltage dip at the transformer bus | 4.8 % |
| Same motor with a star-delta starter | 4.6 % at the motor |
When one calculation is not enough
- Other loads on the same transformer lower the voltage before the start and add to the dip.
- How long the dip lasts depends on the load torque and inertia: that needs a dynamic run-up simulation.
- Several motors, generators or a weak grid connection change the picture.
Electrisim’s motor starting study handles these on your whole network: steady-state with pandapower or a dynamic start with ANDES. See the motor starting guide.
Get these formulas and more in a free two-page PDF cheat sheet
Frequently asked questions
What voltage dip is acceptable when a motor starts?
There is no single limit. Common practice keeps the dip at the motor terminals within about 10–15 %, so the motor develops enough torque, and the dip at other loads lower, for example 5–10 %, so contactors hold in and lights do not flicker noticeably. Check the motor and load requirements and the network operator’s rules.
Why does a star-delta starter reduce the dip?
In star the windings see U/√3, so the line current and the starting torque both drop to a third of the direct-on-line values. The supply sees a third of the starting kVA and the dip is much smaller, but the motor must still accelerate the load with a third of the torque.
Where do I find the locked-rotor current and starting power factor?
The locked-rotor (starting) current is on the motor datasheet as IA/IN or ILR/In, typically 5–8 for standard induction motors. The starting power factor is rarely listed; 0.2–0.4 is typical, lower for large motors.
Does the calculator include the run-up time?
No. It gives the dip at the moment of starting, when the current is highest. How long the dip lasts depends on the load torque and inertia; a dynamic study, such as Electrisim’s motor starting study with ANDES, shows the run-up.
What if the motor is fed from a generator?
A generator’s transient reactance is usually much higher than a transformer’s impedance, so the dip is larger. Enter SrG / x′d as the fault level at the bus. The voltage regulator then brings the voltage back, which this estimate does not include.