About Motor Starting & Voltage Dip Calculator
The motor starting calculator estimates how much current a three-phase induction motor draws when it starts and how far the supply voltage sags as a result. A motor accelerating from standstill draws its locked-rotor current, typically five to eight times the running full-load amps, and that brief surge can dim lights, trip relays, or stall the motor if the bus voltage collapses.
This tool computes the full-load current, the locked-rotor kVA and current from either the NEMA MG-1 code letter or a locked-rotor-current multiple, and then the bus voltage dip caused by the inrush flowing through the supply transformer impedance. It also compares reduced-voltage starting methods, star-delta, autotransformer, and soft starter, so you can see how each cuts both the starting current and, unavoidably, the starting torque.
How It Works
- Enter the motor HP, line-to-line voltage, full-load power factor and efficiency; the tool computes FLA = HP*746 / (sqrt(3)*V*PF*eff).
- Choose how the starting kVA is found: a NEMA code letter (locked-rotor kVA = HP * kVA/hp factor) or a locked-rotor-current multiple of FLA. The locked-rotor current follows from I = kVA*1000 / (sqrt(3)*V).
- Pick a starting method. Direct-on-line draws the full locked-rotor current; star-delta cuts line current and torque to one third; an autotransformer scales both by the tap squared; a soft starter scales current by the voltage ratio and torque by its square.
- Enter the supply transformer kVA and %Z. The bus voltage dip is approximated as %Z * (starting kVA / transformer kVA), and the residual voltage during start is 100 minus the dip.
Worked Example
A 50 HP motor at 460 V, PF 0.85 and 90% efficiency draws FLA = 50*746 / (1.732*460*0.85*0.90) = 61.2 A. With NEMA code letter G at 6.0 kVA/hp, the locked-rotor kVA is 50*6.0 = 300 kVA and the locked-rotor current is 300000 / (1.732*460) = 376.5 A, about 6.2 times FLA. Started direct-on-line off a 1000 kVA transformer of 5.75% impedance, the bus dip is 5.75 * (300/1000) = 1.73%. Switching to star-delta cuts the starting kVA to 100 kVA and the dip to 0.58%, but the available starting torque also drops to one third.
Formulas
- Full-load current (three-phase)
FLA = HP*746 / (sqrt(3) * V * PF * eff)- Locked-rotor (starting) kVA & current
kVA = HP * codeFactor; I_lr = kVA*1000 / (sqrt(3) * V)- Reduced-voltage method factors
star-delta: I&T x1/3; autotransformer: I&T x tap^2; soft starter: I x ratio, T x ratio^2- Bus voltage dip (transformer %Z)
Vdip% = %Z * (Sstart / Stransformer)
Standards & References
- NEMA MG-1 (Motors and Generators, locked-rotor code letters)
- NEC Article 430 (Motors, Motor Circuits and Controllers)
- IEEE 399 (Brown Book, power system analysis)
Frequently Asked Questions
Why does star-delta starting reduce the current to one third?
In wye (star) the phase windings see the line voltage divided by the square root of three, so the phase current falls to 1/sqrt(3) of the delta value; the line current, which is the phase current in wye, ends up at one third of the direct-on-line delta inrush. The starting torque, proportional to voltage squared, also drops to one third.
How much voltage dip is acceptable on motor start?
A common guideline is to keep the bus dip below about 10% to 15% at the motor terminals so contactors hold in and the motor develops enough torque to accelerate. Sensitive equipment on the same bus may require tighter limits; check NEMA and utility requirements.
What is a NEMA code letter?
The NEMA MG-1 locked-rotor indicating code letter, stamped on the motor nameplate, gives the locked-rotor apparent power per horsepower. Code A is the lowest inrush at about 1.6 kVA/hp and the letters increase to roughly 22 kVA/hp, letting you estimate starting kVA without testing.
Does reducing starting current also reduce starting torque?
Yes. Motor torque is roughly proportional to the square of the applied voltage, so any reduced-voltage method that cuts the inrush also cuts the available accelerating torque. The motor must still develop enough torque to break away and run up the load, which limits how far you can reduce the voltage.