Transformer Motor Starting Voltage Drop: Direct-On-Line, Star-Delta, Soft-Starter & VFD Starting, Starting Current Multiples & IEC 60364 Voltage Drop Limits

Abstract

When a large induction motor starts from a transformer, the starting current — typically 5-8× the motor's full-load current for direct-on-line (DOL) starting — flows through the transformer's series impedance, causing a voltage dip on the transformer secondary bus. This voltage dip affects not only the starting motor itself (reduced starting torque, ∝ V²) but also all other loads connected to the same bus — lighting flicker, contactor drop-out, VFD undervoltage trip, and electronic equipment malfunction. The production engineer specifying a transformer must verify that the transformer's impedance and rating are adequate to maintain the voltage at the motor terminals within acceptable limits during starting. This article provides the quantitative voltage-drop calculation for each of the four dominant motor starting methods — DOL, star-delta, soft-starter, and variable-frequency drive (VFD) — and references the relevant standards (IEC 60364-5-52 for steady-state voltage drop, IEC 61000-3-3 and IEEE 141 for motor starting voltage dip).

1. The Physics of Transformer Voltage Drop During Motor Starting

1.1 The Transformer as a Series Impedance

From the perspective of the motor, the transformer is a series impedance in the supply path. The transformer voltage drop under load is:

ΔV (%) = (I_load / I_rated) × (R_pu × cos φ + X_pu × sin φ) × 100

Where:

• I_load / I_rated = the transformer loading in per-unit (for motor starting, this is the starting kVA divided by the transformer rated kVA)
• R_pu = transformer per-unit resistance (typically 0.002-0.01 pu for power transformers)
• X_pu = transformer per-unit reactance (equal to the per-unit impedance u_k, since X_pu >> R_pu for transformers >500 kVA — typically u_k = 0.04-0.125 pu)
• cos φ = power factor of the starting current (typically 0.2-0.4 lagging for an induction motor at starting — the motor draws a large reactive magnetizing current)

For an induction motor at starting: The power factor is low (0.2-0.4), the current is high (5-8× FLC), and the reactive component (I × sin φ × X_pu) dominates the voltage drop. The voltage drop is predominantly a reactive drop across the transformer leakage reactance.

1.2 Motor Starting Current Multiples

Motor Starting Method	Starting Current (× FLC)	Starting Torque (× FLT)	Starting kVA (× Rated kVA)
Direct-On-Line (DOL)	5-8× (typically 6-7×)	1.5-2.5× (standard cage)	5-8×
Star-Delta (Y-Δ)	2.0-2.7× (1/3 of DOL)	0.5-0.8× (1/3 of DOL)	2.0-2.7×
Soft-Starter (current limit)	2.0-5.0× (adjustable)	Adjustable (0.1-1.0× FLT)	2.0-5.0×
Variable-Frequency Drive (VFD)	1.0-1.5× (with ramp)	1.0-1.5× FLT (constant V/f)	1.0-1.5×
Auto-transformer (65% tap)	2.8-4.5×	0.6-1.0×	2.8-4.5×
Primary resistance/reactance	Adjustable (typically 3-5×)	Reduced (∝ V²)	3-5×

DOL starting is the worst case for voltage drop — and it is the reference case that the transformer and the supply system must be designed for unless an alternative starting method is specified and permanently installed.

2. Voltage Drop Calculation for Each Starting Method

2.1 Direct-On-Line (DOL) Starting

Step 1 — Calculate the motor starting kVA:

S_start = √3 × V_rated × I_start

Where:

• V_rated = motor rated line voltage (kV)
• I_start = motor starting current = (Starting Current Multiple) × FLC

For a 500 kW, 6.6 kV motor with FLC = 52 A, starting current multiple = 6×:

I_start = 6 × 52 = 312 A S_start = √3 × 6.6 × 312 = 3,566 kVA

Step 2 — Calculate the motor starting load as a fraction of the transformer rating:

kVA_start_pu = S_start / S_transformer

For a 2 MVA transformer: kVA_start_pu = 3,566 / 2,000 = 1.78 pu (the starting kVA exceeds the transformer rating).

Step 3 — Calculate the voltage drop on the transformer secondary:

ΔV_transformer (%) ≈ kVA_start_pu × X_pu × (sin φ_start) × 100

For a transformer with u_k = 0.06 (6%), X_pu ≈ 0.06, sin φ_start ≈ 0.9 (cos φ_start ≈ 0.35 → sin φ_start ≈ 0.94):

ΔV_transformer = 1.78 × 0.06 × 0.94 × 100 = 10.0%

The voltage at the transformer secondary terminals during DOL starting of this motor is 90% of the nominal voltage.

Step 4 — Add the voltage drop in the motor feeder cable:

The cable voltage drop adds to the transformer drop. For a cable with impedance Z_cable = R_cable + jX_cable, the voltage drop is:

ΔV_cable (%) = (√3 × I_start × (R_cable × cos φ_start + X_cable × sin φ_start) / V_rated) × 100

For a 200 m, 95 mm² copper cable (R = 0.193 Ω/km, X = 0.080 Ω/km at 50 Hz):

• R_cable = 0.193 × 0.2 = 0.0386 Ω
• X_cable = 0.080 × 0.2 = 0.016 Ω

ΔV_cable = (√3 × 312 × (0.0386 × 0.35 + 0.016 × 0.94) / 6,600) × 100 = (540.4 × 0.0284 / 6,600) × 100 = 0.23%

Total voltage drop at motor terminals: 10.0% + 0.23% = 10.2% → motor terminal voltage = 89.8% of nominal.

2.2 Star-Delta (Y-Δ) Starting

Star-delta starting reduces the phase voltage applied to the motor windings to 1/√3 (58%) of the line voltage during the starting period (motor windings connected in star). The starting current is reduced to 1/3 of the DOL starting current:

I_start(Y) = I_start(DOL) / 3 S_start(Y) = S_start(DOL) / 3

For the same 500 kW motor: S_start(Y) = 3,566 / 3 = 1,189 kVA kVA_start_pu(Y) = 1,189 / 2,000 = 0.59 pu

ΔV_transformer = 0.59 × 0.06 × 0.94 × 100 = 3.3%

The voltage drop is reduced from 10.0% to 3.3%. The trade-off: the starting torque is also reduced to 1/3 of the DOL starting torque — the motor must be capable of accelerating the load with this reduced torque. Star-delta starting is unsuitable for high-inertia loads where the full DOL starting torque is required.

Important limitation: The motor must have both ends of each phase winding brought out to the terminal box (6 terminals) for star-delta connection. Motors with only 3 terminals (internally connected in star or delta) cannot be star-delta started.

Transition transient: When the star-delta contactor transitions from star to delta, there is a momentary interruption of the motor supply (open-transition star-delta). The motor's back-EMF at the transition instant generates a current transient that can exceed the DOL starting current — the "transition spike." Closed-transition star-delta starters (with a resistor in the transition) eliminate the spike but are more expensive.

2.3 Soft-Starter

A soft-starter uses back-to-back thyristors (SCRs) in each phase to control the RMS voltage applied to the motor by phase-angle control. The voltage is ramped up from an initial value (typically 30-50% of rated) to 100% over a user-adjustable ramp time (typically 5-30 seconds).

The starting current is limited to a user-set multiple of FLC — typically 2.5-4.0× FLC. The soft-starter's current-limit function actively regulates the thyristor firing angle to prevent the current from exceeding the set limit.

For a current limit of 3.0× FLC: I_start = 3.0 × 52 = 156 A S_start = √3 × 6.6 × 156 = 1,783 kVA kVA_start_pu = 1,783 / 2,000 = 0.89 pu

ΔV_transformer = 0.89 × 0.06 × 0.94 × 100 = 5.0%

Advantage of soft-starter over star-delta:

• Smooth ramp (no transition spike)
• Adjustable current limit and ramp time (can be optimized for the specific load)
• No requirement for 6 motor terminals (works with standard 3-terminal motors)
• Built-in motor protection features (overload, phase loss, phase unbalance, stall protection)

Disadvantage:

• Thyristor losses (≈1-2 W per ampere per phase) — continuous losses even after starting (unless a bypass contactor is provided)
• Harmonics during the starting ramp (thyristor phase-angle control generates harmonics)
• Higher cost than star-delta (2-4×)

2.4 Variable-Frequency Drive (VFD)

A VFD rectifies the incoming AC to DC, then inverts it to variable-voltage, variable-frequency AC. By ramping both voltage and frequency simultaneously (constant V/f ratio), the motor produces full rated torque at any speed up to rated speed, with starting current limited to approximately 1.0-1.5× FLC.

Starting current: 1.2× FLC (typical with ramp)

S_start = √3 × 6.6 × (1.2 × 52) = 713 kVA kVA_start_pu = 713 / 2,000 = 0.36 pu

ΔV_transformer = 0.36 × 0.06 × 0.94 × 100 = 2.0%

The voltage drop is essentially negligible.

Additional benefit: The VFD's DC-link capacitors and input rectifier reduce the reactive power drawn during starting (the VFD presents a near-unity power factor to the supply, regardless of the motor's power factor). The reactive component of the starting current — which dominates the voltage drop — is minimized.

Disadvantage:

• Highest cost (5-10× the cost of a DOL starter)
• VFD generates harmonics on the supply (IEEE 519 compliance required)
• VFD requires controlled environment (clean, dry, temperature-controlled) — not suitable for all industrial locations
• VFD is a single-point failure — if the VFD fails, the motor cannot be started (unless a bypass contactor is provided for DOL starting as backup)

3. Voltage Drop Limits per Standards

3.1 Steady-State (Running) Voltage Drop — IEC 60364-5-52

Installation Type	Maximum Voltage Drop (% of Nominal)	Reference
LV installation supplied from a public distribution system	3% for lighting, 5% for other loads	IEC 60364-5-52, Clause 525
LV installation supplied from a private supply (transformer)	8% total from source to load	IEC 60364-5-52
Motor running (steady state)	≤5% at motor terminals	NEMA MG-1, IEC 60034-1
Motor starting (transient)	≤15% at motor terminals, ≤3-5% flicker at the point of common coupling (PCC)	IEC 61000-3-3, IEEE 141 (Red Book)

3.2 Motor Starting Voltage Dip — Flicker Limits (IEC 61000-3-3)

The voltage dip at the point of common coupling (PCC — the point where other customers are connected) during motor starting is limited by flicker standards:

Parameter	Limit	Measurement
Short-term flicker severity (P_st)	≤1.0 (over 10 minutes)	Per IEC 61000-4-15 (flickermeter)
Long-term flicker severity (P_lt)	≤0.65 (over 2 hours)	Per IEC 61000-4-15
Maximum voltage change (ΔV_max)	≤3% (for infrequent starts, ≤1 per hour)	Per IEC 61000-3-3
	≤4% (for very infrequent starts, manual operation only)	Per IEC 61000-3-3
Relative steady-state voltage change (d_c)	≤3%	Per IEC 61000-3-3

If the calculated voltage dip at the PCC exceeds the flicker limit, the motor starting method must be changed (DOL → soft-starter → VFD) to reduce the starting kVA.

3.3 Motor Terminal Voltage — NEMA MG-1 / IEC 60034-1

Condition	Minimum Motor Terminal Voltage
Motor running (continuous)	≥95% of rated (±5% tolerance)
Motor starting (transient)	≥85% of rated (the motor must start and accelerate with 85% voltage)

During starting, the motor torque is proportional to V². At 85% voltage, the starting torque is 0.85² = 72% of the rated starting torque. The load torque must be less than 72% of the motor's rated starting torque for the motor to accelerate from standstill. If the voltage dip reduces the starting torque below the load's breakaway torque, the motor will stall.

FAQ

Q: How do I estimate the starting kVA of a motor if I only know the rated kW?

For an induction motor: Starting kVA = (Rated kW / Efficiency / Power Factor) × Starting Current Multiple. Typical values for a 4-pole, 400 V motor: efficiency = 0.92-0.95, power factor = 0.85-0.90, starting current multiple = 6-7×. For a 200 kW motor: Starting kVA = (200 / 0.93 / 0.87) × 6.5 = 247 kVA_rated × 6.5 = 1,606 kVA. Rule of thumb: Starting kVA ≈ Rated kW × 8-10 (for DOL starting of typical 4-pole cage induction motors). For the 200 kW example: 200 × 8 ≈ 1,600 kVA — close to the calculated 1,606 kVA. This rule of thumb is valid only for DOL starting of standard-efficiency induction motors; for high-efficiency (IE3/IE4) motors, use kW × 9-12 because the higher locked-rotor current of high-efficiency designs.

Q: Does the motor starting voltage drop affect other motors on the same bus?

Yes. When the voltage at the common bus sags during starting of a large motor, other running motors experience a voltage reduction. This causes: (1) a momentary reduction in motor torque (∝ V²) — a lightly loaded motor will continue running; a heavily loaded motor may decelerate and draw higher current as the slip increases, (2) if the voltage sag is deep (>20%) and prolonged (>1 second), the contactor coil voltage may drop below the drop-out threshold (typically 70-80% of rated), causing the contactor to open and the motor to trip on undervoltage — this is the most disruptive consequence of motor starting on a shared bus, and (3) VFDs on the same bus are particularly sensitive — most VFDs will trip on "DC bus undervoltage" if the input voltage sags below 65-75% for more than 1-2 cycles. For industrial installations, critical motors and VFDs should be supplied from a separate bus (or separate transformer) if a large DOL-started motor causes unacceptable voltage dips.

Q: Is the voltage drop calculation the same for a generator supply as for a transformer supply?

No. A generator has a significantly higher subtransient reactance (X''_d = 0.15-0.25 pu) than a transformer (X = 0.04-0.125 pu). The generator's automatic voltage regulator (AVR) will attempt to compensate for the voltage dip, but the AVR response is slow (0.5-2 seconds) compared to the motor starting transient (the dip occurs in the first 100-200 ms). The initial voltage dip during motor starting on a generator is governed by the generator's subtransient reactance, not the steady-state synchronous reactance — the generator's voltage will dip by:

ΔV_generator ≈ (S_start / S_generator) × X''_d × 100

For a 500 kVA generator with X''_d = 0.18 pu, supplying a DOL-started motor with S_start = 400 kVA:

ΔV_generator = (400/500) × 0.18 × 100 = 14.4%

The generator AVR will recover the voltage to the setpoint within 1-2 seconds, but the initial dip of 14.4% may cause contactor drop-out and VFD tripping on the same bus. For generator-fed motor starting, a soft-starter or VFD is strongly recommended over DOL starting.

Q: Can a motor be started DOL from a transformer if the starting kVA exceeds the transformer rating?

The starting kVA (5-8× the motor's rated kVA) routinely exceeds the transformer rating for medium and large motors — this is normal and acceptable, provided: (1) the voltage dip at the motor terminals remains ≥85% of rated (to ensure the motor starts), (2) the voltage dip at the PCC remains within flicker limits (to avoid affecting other customers), (3) the transformer's thermal time constant is long enough that the starting current (lasting 5-30 seconds) does not overheat the windings — a 2 MVA transformer with 1,600 kVA starting load (1.78 pu) for 10 seconds experiences an I²t that is a tiny fraction of the transformer's thermal rating (the transformer is designed for continuous loading at 1.0 pu), and (4) the transformer protection (overcurrent relays, fuses) is coordinated to allow the starting current without tripping. A standard inverse-time overcurrent relay with pickup at 1.2 × I_rated will NOT trip for a 6× starting current lasting 10 seconds because the I²t is within the relay's time-current characteristic.

Q: What is the effect of a long motor feeder cable on the starting voltage?

A long cable has significant impedance — primarily resistance at low voltage (where X/R < 1). During starting, the motor draws high current through this impedance, causing an additional voltage drop between the bus and the motor terminals. For a 100 kW motor (FLC = 180 A at 400 V, DOL starting current = 1,080 A), a 200 m length of 95 mm² cable (R = 0.193 Ω/km → 0.0386 Ω/200 m) produces:

ΔV = √3 × 1,080 × 0.0386 = 72 V → 72/400 = 18% voltage drop

The motor starting voltage at the terminals is only 82% of the bus voltage — borderline for successful starting. The solution: increase the cable cross-section (reduce R), or move the transformer closer to the motor (shorter cable). This is why large motors in industrial plants are typically located near the substation.

Q: Should I specify DOL, soft-starter, or VFD for a large motor on a limited-capacity transformer?

Decision matrix: (1) If the motor is ≤5% of the transformer rating (S_start ≤ 0.4 × S_transformer): DOL is acceptable. (2) If the motor is 5-15% of the transformer rating: Soft-starter or star-delta starting required to limit the voltage dip. (3) If the motor is >15% of the transformer rating: VFD recommended — the soft-starter's current limit of 2.5-3.5× FLC may still cause an unacceptable voltage dip, while the VFD limits the starting current to 1.0-1.5× FLC. (4) If the motor requires speed control during operation (not just soft-starting): VFD is the only choice — the VFD serves both the starting and the speed-control function, making it the economically optimal solution despite the higher capital cost.

References / Standards

Reference	Title
IEC 60364-5-52:2009	Low-voltage electrical installations — Selection and erection of electrical equipment — Wiring systems
IEC 61000-3-3:2013	Electromagnetic compatibility (EMC) — Limits — Limitation of voltage changes, voltage fluctuations and flicker in public low-voltage supply systems
IEC 61000-4-15:2010	Electromagnetic compatibility (EMC) — Testing and measurement techniques — Flickermeter
IEC 60034-1:2022	Rotating electrical machines — Part 1: Rating and performance
NEMA MG-1:2021	Motors and Generators
IEEE 141-1993 (Red Book)	IEEE Recommended Practice for Electric Power Distribution for Industrial Plants
IEEE 399-1997 (Brown Book)	IEEE Recommended Practice for Industrial and Commercial Power Systems Analysis

*Authored by Du Fu, Production Engineer at ZY POWER. Motor starting voltage drop is a system-level problem — the transformer, the cable, the starting method, and the other loads on the bus must be considered simultaneously. The cheapest solution (DOL starting) may cause the most expensive problems (production downtime from contactor drop-out and VFD tripping). Always calculate the starting voltage dip before the transformer is ordered, not after the motor fails to start.*